STUDENT SOLUTIONS MANUAL FOR
ELEMENTARY DIFFERENTIAL EQUATIONS AND
ELEMENTARY DIFFERENTIAL EQUATIONS WITH BOUNDARY VALUE PROBLEMS
William F. Trench Andrew G. Cowles Distinguished Professor Emeritus Department of Mathematics Trinity University San Antonio, Texas, USA
[email protected] This book has been judged to meet the evaluation criteria set by the Editorial Board of the American Institute of Mathematics in connection with the Institute’s Open Textbook Initiative. It may be copied, modified, redistributed, translated, and built upon subject to the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
This book was published previously by Brooks/Cole Thomson Learning Reproduction is permitted for any valid noncommercial educational, mathematical, or scientific purpose. However, charges for profit beyond reasonable printing costs are prohibited.
TO BEVERLY
Contents Chapter 1 Introduction
1
1.2 First Order Equations
1
Chapter 2 First Order Equations
5
2.1 2.2 2.3 2.4 2.5 2.6
Linear First Order Equations Separable Equations Existence and Uniqueness of Solutions of Nonlinear Equations Transformation of Nonlinear Equations into Separable Equations Exact Equations Integrating Factors
5 8 11 13 17 21
Chapter 3 Numerical Methods
25
3.1 Euler’s Method 3.2 The Improved Euler Method and Related Methods
25 29
ii Contents 3.3 The Runge-Kutta Method
34
Chapter 4 Applications of First Order Equations
39
4.1 Growth and Decay 4.2 Cooling and Mixing 4.3 Elementary Mechanics 4.4 Autonomous Second Order Equations 4.5 Applications to Curves Chapter 5 Linear Second Order Equations
39 40 43 45 46 51
5.1 5.2 5.3 5.4 5.5 5.6 5.7
51 55 58 60 64 75 79
Homogeneous Linear Equations Constant Coefficient Homogeneous Equations Nonhomgeneous Linear Equations The Method of Undetermined Coefficients I The Method of Undetermined Coefficients II Reduction of Order Variation of Parameters
Chapter 6 Applcations of Linear Second Order Equations
85
6.1 6.2 6.3 6.4
85 87 89 90
Spring Problems I Spring Problems II The RLC Circuit Motion Under a Central Force
Chapter 7 Series Solutions of Linear Second Order Equations
108
7.1 7.2 7.3 7.4 7.5 7.6 7.7
91 93 96 102 103 108 118
Review of Power Series Series Solutions Near an Ordinary Point I Series Solutions Near an Ordinary Point II Regular Singular Points; Euler Equations The Method of Frobenius I The Method of Frobenius II The Method of Frobenius III
Chapter 8 Laplace Transforms
125
8.1 Introduction to the Laplace Transform 8.2 The Inverse Laplace Transform 8.3 Solution of Initial Value Problems 8.4 The Unit Step Function 8.5 Constant Coefficient Equations with Piecewise Continuous Forcing Functions 8.6 Convolution
125 127 134 140 143 152
Contents iii 8.7 Constant Cofficient Equations with Impulses
55
Chapter 9 Linear Higher Order Equations
159
9.1 9.2 9.3 9.4
159 171 175 181
Introduction to Linear Higher Order Equations Higher Order Constant Coefficient Homogeneous Equations Undetermined Coefficients for Higher Order Equations Variation of Parameters for Higher Order Equations
Chapter 10 Linear Systems of Differential Equations
221
10.1 10.2 10.3 10.4 10.5 10.6 10.7
191 192 193 194 201 245 218
Introduction to Systems of Differential Equations Linear Systems of Differential Equations Basic Theory of Homogeneous Linear Systems Constant Coefficient Homogeneous Systems I Constant Coefficient Homogeneous Systems II Constant Coefficient Homogeneous Systems II Variation of Parameters for Nonhomogeneous Linear Systems
Chapter
221
11.1 Eigenvalue Problems for y00 C y D 0 11.2 Fourier Expansions I 11.3 Fourier Expansions II
221 223 229
Chapter 12 Fourier Solutions of Partial Differential Equations
239
12.1 12.2 12.3 12.4
239 247 260 270
The Heat Equation The Wave Equation Laplace’s Equation in Rectangular Coordinates Laplace’s Equation in Polar Coordinates
Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations 13.1 Two-Point Boundary Value Problems 13.2 Sturm-Liouville Problems
273 273 279
CHAPTER 1 Introduction
1.2 BASIC CONCEPTS 1.2.2. (a) If y D ce 2x , then y 0 D 2ce 2x D 2y. x2 c 2x c 2x 2 c x2 c 0 (b) If y D C , then y 0 D , so xy C y D C C D x2. 2 3 x 3 x 3 x 3 x (c) If 1 2 2 y D C ce x ; then y 0 D 2xce x 2 and 1 2 2 2 2 C ce x D 2xce x C x C 2cxe x D x: y 0 C 2xy D 2xce x C 2x 2 (d) If yD
1 C ce 1 ce
x 2 =2 x 2 =2
then y0
.1
D
ce
x 2 =2
/. cxe .1
x 2 =2
/ cxe
.1 C ce x 2 =2 /2
x 2 =2
/cxe
2
2cxe x =2 .1 ce x2 =2 /2
D and y
2
1
D D
1 C ce 1
ce
.1 C ce
x 2 =2 x 2 =2
x 2 =2 2
1
/ .1 ce x .1 ce 2 =2/2 2
D
!2
4ce x =2 ; .1 ce x2 =2/2 1
x 2 =2 2
/
x 2 =2
2
Chapter 1 Basic Concepts so
4cx C 4cx D 0: .1 ce x2 =2 /2 3 3 3 x x x C c , then y 0 D x 2 sec2 C c D x 2 1 C tan2 D x 2 .1 C y 2 /. (e) If y D tan 3 3 x 3 C c y D .c1 C c2x/e C sin x C x 2 ; then (f) If y 0 D .c1 C 2c2x/e x C cos x C 2x; 2y 0 C x.y 2
y0 2y 0 C y
1/ D
D .c1 C 3c2x/e x sin x C 2; and y 00 D c1 e x .1 2 C 1/ C c2 xe x .3 4 C 1/ sin x 2 cos x C sin x C 2 4x C x 2 D 2 cos x C x 2 4x C 2: 2 2 4 and y 00 D c1 e x C 3 , so .1 x/y 00 C xy 0 y D (g) If y D c1e x C c2x C , then y 0 D c1 e x C c2 2 x x x 4.1 x/ 2 2 4.1 x x 2 / c1 .1 x C x 1/ C c2 .x x/ C D x3 x x x3 c1 sin x C c2 cos x c c2 sin x c1 sin x C c2 cos x 1 cos x 0 (h) If y D C 4x C 8 then y D C 4 and 1=2 1=2 x x 2x 3=2 c1 sin x C c2 cos x c1 sin x c2 cos x 3 c1 sin x C c2 cos x 1 2 00 0 2 00 y D C , so x y Cxy C x yD 4 4 x 1=2 x 3=2 x 5=2 3 c1 x 3=2 sin x x 1=2 cos x C x 1=2 sin x C x 1=2 cos x 4 1 1=2 3 1 1=2 3=2 x sin x C x sin x x sin x C c2 x 3=2 cos x C x 1=2 sin x C x 1=2 cos x 2 4 4 1 1=2 1 1=2 1 1=2 3=2 2 x sin x x cos x C x cos x x cos x C 4x C x .4x C 8/ D 4x 3 C 8x 2 C 2 4 4 3x 2. R 1.2.4. (a) If y 0 D xe x , then y D xe x C e x dx C c D .1 x/e x C c, and y.0/ D 1 ) 1 D 1 C c, so c D 0 and y D .1 x/e x . r 1 2 0 2 (b) If y D x sin x , then y D cos x C c; y D 1 ) 1 D 0 C c, so c D 1 and 2 2 1 yD1 cos x 2 . 2 1 d sin x (c) Write y 0 D tan x D D .cos x/. Integrating this yields y D ln j cos xj C c; cos x cos x dx p p y.=4/pD 3 ) 3 Dp ln .cos.=4// C c, or 3 D ln 2 C c, so c D 3 ln 2, so y D ln.j cos xj/ C 3 ln 2 D 3 ln. 2j cos xj/. x5 32 37 (d) If y 00 D x 4 , then y 0 D C c1 ; y 0 .2/ D 1 ) C c1 D 1 ) c1 D , so y 0 D 5 5 15 x5 37 x6 37 64 47 . Therefore, y D .x 2/ C c2; y.2/ D 1 ) C c2 D 1 ) c2 D , so 5 15 30 15 30 15 6 47 37 x yD .x 2/ C . 15 5 30 Z R xe 2x 1 xe 2x e 2x xe 2x e 2x (e) (A) xe 2x dx D e 2x dx D . Therefore, y 0 D C c1 ; 2 2 2 4 2 4 1 5 5 xe 2x e 2x 5 xe 2x y 0 .0/ D 1 ) C c1 D ) c1 D , so y 0 D C ; Using (A) again, y D 4 4 4 2 4 4 4 e 2x e 2x 5 xe 2x e 2x 5 1 29 C x C c2 D C x C c2 ; y.0/ D 7 ) C c2 D 7 ) c2 D , so 8 8 4 4 4 4 4 4 2x 2x xe e 5 29 yD C xC . 4 R 4 4 4 R R (B) x cos x dx D x sin x R (f) (A) x sin x dx D x cos x00 C cos x dx D x cos x C sin x and sin x dx D x sin x C cos x. If y D x sin x, then (A) implies that y 0 D x cos x sin x C c1 ; y 0 .0/ D 3 ) c D 3, so y 0 D x cos x sin x 3. Now (B) implies that y D x sin x Ccos x Ccos x 3x Cc2 D x sin x C 2 cos x 3x C c2 ; y.0/ D 1 ) 2 C c2 D 1 ) c2 D 1, so y D x sin x C 2 cos x 3x 1.
Section 1.2 Basic Concepts
3
R R (g) If y 000 D x 2 e x , then y 00 D x 2 e x dx D x 2 e x 2 xe x dx D x 2 eRx 2xe x C 2e x C c1 ; y 00 .0/ D 3 ) 2 C c1RD 3 ) c1 D 1, so (A) y 00 D .x 2 2x C 2/e x C 1. Since .x 2 2x C 2/e x dx D .x 2 2x C 2/e x .2x 2/e x dx D .x 2 2x C 2/e x .2x 2/e x C 2e x D .x 2 4x C 6/e x , 0 x 0 (A) implies that y D .x 2 4x C 6/e R 2C x C c2 ; yx .0/ D 22 ) 6 C c2xD R 2 ) c2 xD 8, so (B) 0 2 x y D .x 4x C 6/e C x 8; Since .x 4x C 6/e dx D .x 4x C 6/e .2x 4/e dx D .x 2 x2 8x Cc3; 4x C6/e x .2x 4/e x C2e x D .x 2 6x C12/e x , (B) implies that y D .x 2 6x C12/e x C 2 2 x y.0/ D 1 ) 12 C c3 D 1 ) c3 D 11, so y D .x 2 6x C 12/e x C 8x 11. 2 cos 2x 1 7 C c1 ; y 00 .0/ D 3 ) C c1 D 3 ) c1 D , (h) If y 000 D 2 C sin 2x, then y 00 D 2x 2 2 2 cos 2x 7 sin 2x 7 00 0 2 0 so y D 2x C . Then y D x C x C c2 ; y .0/ D 6 ) c2 D 6, so 2 2 4 2 3 7 x cos 2x 7 2 1 7 0 2 sin 2x C x 6. Then y D C C x 6xCc3 ; y.0/ D 1 ) Cc3 D 1 ) c3 D , y Dx 4 2 3 8 4 8 8 x3 cos 2x 7 2 7 so y D C C x 6x C . 3 8 4 8 (i) If y 000 D 2x C 1, then y 00 D x 2 C x C c1; y 00 .2/ D 7 ) 6 C c1 D 7 ) c1 D 1; so y 00 D x 2 C x C 1. x3 x2 14 26 x3 x2 Then y 0 D C C .x 2/ C c2; y 0 .2/ D 4 ) C c2 D 4 ) c2 D , so y 0 D C C 3 2 3 3 3 2 4 3 x x 1 26 8 5 26 .x 2/ . Then y D C C .x 2/2 .x 2/ C c3 ; y.2/ D 1 ) C c3 D 1 ) c3 D , 3 12 6 2 3 3 3 4 3 x 1 26 5 x so y D C C .x 2/2 .x 2/ . 12 6 2 3 3 1.2.6. (a) If y D x 2 .1 C ln x/, then y.e/ D e 2 .1 C ln e/ D 2e 2; y 0 D 2x.1 C ln x/ C x D 3x C 2x ln x, so y 0 .e/ D 3e C 2e ln e D 5e; (A) y 00 D 3 C 2 C 2 ln x D 5 C 2 ln x. Now, 3xy 0 4y D 3x.3x C 2x ln x/ 4x 2.1 C ln x/ D 5x 2 C 2x 2 ln x D x 2 y 00 , from (A). x2 1 1 2 2 5 (b) If y D C x 1, then y.1/ D C 1 1 D ; y 0 D x C 1, so y 0 .1/ D C 1 D ; (A) 3 3 3 3 3 3 2 x2 2 2 00 2 0 2 y D . Now x xy C y C 1 D x x xC1 C C x 1 C 1 D x 2 D x 2 y 00 , from (A). 3 3 3 3 (c) If y D .1 C x 2 / 1=2 , then y.0/ D .1 C 02 / 1=2 D 1; y 0 D x.1 C x 2 / 3=2 , so y 0 .0/ D 0; (A) 00 y D .2x 2 1/.1Cx 2 / 5=2 . Now, .x 2 1/y x.x 2 C1/y 0 D .x 2 1/.1Cx 2 / 1=2 x.x 2 C1/. x/.1C .x 2 1/y x.x 2 C 1/y 0 x 2 / 3=2 D .2x 2 1/.1 C x 2 / 1=2 D y 00 .1 C x 2 /2 from (A), so y 00 D . .x 2 C 1/2 x2 1=4 1 x.x 2/ . 1=2/. 3=2/ (d) If y D , then y.1=2/ D D ; y0 D , so y 0 .1=2/ D D 3; 1 x 1 1=2 2 .1 x/2 .1 1=2/2 2 x2 x x 2 .x 2/ x2 0 (A) y 00 D . Now, (B) x C y D x C D and (C) xy y D D .1 x/3 1 x 1 x .1 x/2 1 x 2 3 3 0 x x x 00 2.x C y/.xy y/ . From (B) and (C), .x C y/.xy 0 y/ D D y , so y 00 D . 2 3 3 .1 x/ .1 x/ 2 x 1.2.8. (a) y D .x c/a is defined and x c D y 1=a on .c; 1/; moreover, y 0 D a.x a 1 a y 1=a D ay .a 1/=a . (b) if a > 1 or a < 0, then y 0 is a solution of (B) on . 1; 1/.
c/a
1
D
1.2.10. (a) Since y 0 D c we must show that the right side of (B) reduces to c for all values of x in some
4
Chapter 1 Basic Concepts
interval. If y D c 2 C cx C 2c C 1, x 2 C 4x C 4y
p
D x 2 C 4x C 4c 2 C 4cx C 8c C 4 D x 2 C 4.1 C c/x C 4.c 2 C 2c C 1/ D x 2 C 4.1 C c/ C 2.c C 1/2 D .x C 2c C 2/2 :
x 2 C 4x C 4y D x C 2c C 2 and the right side of (B) reduces to c if x > 2c 2. x.x C 4/ xC2 (b) If y1 D , then y10 D and x 2 C 4x C 4y D 0 for all x. Therefore, y1 satisfies 4 2 (A) on . 1; 1/. Therefore,
CHAPTER 2 First Order Equations
2.1 LINEAR FIRST ORDER EQUATIONS y0 2.1.2. D 3x 2; j ln jyj D x 3 C k; y D ce y 2.1.4.
y0 D y
2.1.6.
y0 D y
yD
e
3 ; x
ln jyj D 3 ln jxj C k D
1Cx D x
1 x
ln jxj3 C k;
1; j ln jyj D
x
ln jxj
.ln x/2 =2
yD
y0 D y y D 3x k . 2.1.10.
y10 D y1
k ; j ln jyj D x
3; ln jy1 j D
e 3x 1 C c; y D C ce 3 3 y10 D y1
3x
ln jxj
c . x3
x C k; y D
ln j sin xj C k D
k ln jxj C k1 D ln jx
3x; y1 D e
3x
ce x ; y.1/ D 1 ) c D e; x
k
ln jx sin xj C k; y D
j C k1 ; y D cjxj
; y D ue
3x
; u0 e
3x
k
;
c ; x sin x
y.1/ D 3 ) c D 3;
D 1; u0 D e 3x ; u D
.
2x; ln jy1 j D x 2 ; y1 D e 2 x x2 2 uD C c; y D e x Cc . 2 2
x2
y10 D y1
; y D ue
1 1 u ; ln jy1 j D ln jxj; y1 D ; y D ; x x x 3x 2 7 ln jxj 3x c u D 7 ln jxj C C c; y D C C . 2 x 2 x 2.1.16.
.
.
1 y0 D cot x; j ln jyj D y x y.=2/ D 2 ) c D ; y D . x sin x
2.1.14.
. y D ce
.x 1/
2.1.8.
2.1.12.
x3
5
x2
; u0 e
x2
D xe
x2
; u0 D x;
u0 7 7 D 2 C 3; u0 D C 3x; x x x
6
Chapter 2 First Order Equations 2
y10 D y1
2
1 e x ue x 2x; ln jy1 j D ln jxj x 2 ; y1 D ; yD ; x x x x4 x3 c 2 C c; y D e x C . u0 D x 3 ; u D 4 4 x
2.1.18.
2
u0 e x D x2e x
x2
;
y10 D tan x; ln jy1 j D ln j cos xj; y1 D cos x; y D u cos x; u0 cos x D cos x; u0 D 1; y1 u D x C c; y D .x C c/ cos x. ˇ ˇ ˇ .x 2/5 ˇ y10 4x 3 5 1 ˇ ˇ; D D ; ln jy1 j D 5 ln jx 2j ln jx 1j D ln ˇ 2.1.22. y1 .x 2/.x 1/ x 2 x 1 x 1 ˇ .x 2/5 u.x 2/5 u0 .x 2/5 .x 2/2 1 1 1 ; yD ; D ; u0 D ; uD C y1 D 3 x 1 x 1 x 1 x 1 .x 2/ 2 .x 2/2 1 .x 2/3 .x 2/5 c; y D Cc . 2 .x 1/ .x 1/ 2.1.20.
2.1.24.
y10 D y1
u D xe x
3 ; ln jy1 j D 3 ln jxj D ln jxj 3; x ex ex c e x C c; y D 2 C 3. 3 x x x
y1 D
1 u ; y D 3; 3 x x
u0 ex D ; u0 D xe x ; x3 x2
y10 D y1
4x 1 ; ln jy1 j D 2 ln.1 C x 2 / D ln.1 C x 2 / 2 ; y1 D ; y D 1 C x2 .1 C x 2 /2 0 u 2 2x C c u ; D ; u0 D 2; u D 2x C c; y D ; y.0/ D 1 ) 2 2 2 2 2 2 .1 C x / .1 C x / .1 C x / .1 C x 2 /2 2x C 1 c D 1; y D . .1 C x 2 /2 2.1.26.
2.1.28.
y10 D y1
2.1.30.
y10 D y1
u0 D cos x; u0 D sin x sin x sin2 x 1 sin x cos x; u D C c; y D C c csc x; y.=2/ D 1 ) c D 21 ; y D .sin x C csc x/. 2 2 2 cot x; ln jy1 j D
3
; ln jy1 j D
ln j sin xj; y1 D
3 ln jx
1 u ; y D ; sin x sin x
1j D ln jx
1j
3
; y1 D
1 1/3
x 1 .x u0 1 sin x 1 0 D C ; u D C sin x; u D ln jx 1j .x 1/3 .x 1/4 .x 1/3 x 1 ln jx 1j cos x C c ln jx 1j cos x ; y.0/ D 1 ) c D 0; y D . .x 1/3 .x 1/3
2.1.32. uD
; ln jy1 j D
3 ln jx
1j D ln jx
1j
3
; y1 D
1
u 1/3
.x
;
cos x C c; y D
y10 2 D ; ln jy1 j D 2 ln jxj D ln.x 2 /; y1 D x 2 ; y D ux 2 ; u0 x 2 D y1 x ln jxj C c; y D x 2 .c ln jxj/; y.1/ D 1 ) c D 1; y D x 2 .1 ln x/.
y10 3 D y1 x u0 1 C .x D .x 1/3 .x
2.1.34.
; y D
x; u0 D
; y D
u
1 ; x
; 1 .x .x 1/3 1/ sec2 x 1 ln jx 1j C tan x C c ; u0 D Csec2 x; u D ln jx 1jCtan xCc; y D ; 1/4 x 1 .x 1/3 ln jx 1j C tan x C 1 y.0/ D 1 ) c D 1; y D . .x 1/3 1/3
Section 2.1 Linear First Order Equations
7
y10 2x D 2 ; ln jy1 j D ln jx 2 1j; y1 D x 2 1; y D u.x 2 1/; u0 .x 2 1/ D x; y1 x 1 x 1 1 2 2 0 2 ; u D ln jx 1j C c; y D .x 1/ ln jx 1j C c ; y.0/ D 4 ) c D 4; u D 2 x 1 2 2 1 y D .x 2 1/ ln jx 2 1j 4 . 2 2.1.36.
y0 2 2 2 2 2.1.38. 1 D 2x; ln jy1 j D x 2 ; y1 D e x ; y D ue x ; u0 e x D x 2 ; u0 D x 2 e x ; u D y 1 Z x Z x Z x 2 2 2 2 2 cC t 2 e t dt; y D e x c C t 2 e t dt ; y.0/ D 3 ) c D 3; y D e x 3 C t 2 e t dt . 0
0
0
y10 e x tan x tan x D 1; ln jy1 j D x; y1 D e x ; y D ue x ; u0 e x D ; u0 D ; yZ1 x x Z x Z x x tan t tan t tan t uDcC dt ; y D e x c C dt ; y.1/ D 0 ) c D 0; y D e x dt. t t t 1 1 1
2.1.40.
y10 D y1
2
1 e x ue x u0 e x ex ; ln jy1 j D x ln jxj; y1 D ; y D ; D ; x Z x x x Z xx x x e 2 2 2 u0 D e x e x ; u D c C e t e t dt; y D cC e t e t dt ; y.1/ D 2 ) c D 2e; x 1 Z x1 1 .x 1/ x t t2 yD 2e Ce e e dt . x 1 2.1.42.
1
2.1.44. (b) Eqn. (A) is equivalent to 2 D x
y0
1 x
.B/
y10 2 1 D ; ln jy1 j D 2 ln jxj; y1 D x 2 ; y D ux 2 ; u0 x 2 D ; y1 x x 1 1 1 u0 D ; uD C c, so y D C cx 2 is the general solution of (A) on . 1; 0/ and .0; 1/. 3 2 x 2x 2 (c) From the proof of (b), any solution of (A) must be of the form 8 1 ˆ < C c1x 2 ; x 0; 2 yD .C/ ˆ : 1 C c2x 2 ; x < 0; 2
on . 1; 0/ and .0; 1/. Here
for x ¤ 0, and any function of the form (C) satisfies (A) for x ¤ 0. To complete the proof we must show that any function of the form (C) is differentiable and satisfies (A) at x D 0. By definition, y 0 .0/ D lim
x!0
if the limit exists. But
y.x/
y.x/ x 1=2
x
y.0/ y.x/ 1=2 D lim x!0 0 x D
c1 x; x > 0 c2 x; x < 0;
so y 0 .0/ D 0. Since 0y 0 .0/ 2y.0/ D 0 0 2.1=2/ D 1, any function of the form (C) satisfies (A) at x D 0. (d) From (b) any solution y of (A) on . 1; 1/ is of the form (C), so y.0/ D 1=2.
8
Chapter 2 First Order Equations
1=2 and c2 arbitrary is a solution x02 of the initial value problem on . 1; 1/. Since these functions are all identical on .0; 1/, this does not contradict Theorem 2.1.1, which implies that (B) (so (A)) has exactly one solution on .0; 1/ such that y.x0 / D y0 . A similar argument applies if x0 < 0. (e) If x0 > 0, then every function of the form (C) with c1 D
y0
2.1.46. (a) Let y D c1 y1 C c2 y2 . Then y 0 C p.x/y
D
.c1 y1 C c2 y2 /0 C p.x/.c1 y1 C c2 y2 /
D
c1y10 C c2y20 C c1p.x/y1 C c2 p.x/y2
D
c1.y10 C p.x/y1 / C c2 .y2 C p.x/y2 / D c1f1 .x/ C c2f2 .x/:
(b) Let f1 D f2 D f and c1 D c2 D 1. (c) Let f1 D f , f2 D 0, and c1 D c2 D 1. 3x 0 3x 2.1.48. (a) If ´ D tan y, then ´0 D .sec2 y/y 0 , so ´0 3´ D 1; ´1 D e 3x ; ´ D ue ; u e D 1; 3x e 1 1 u0 D e 3x ; u D C c; ´ D C ce 3x D tan y; y D tan 1 C ce 3x . 3 3 3 1 1 u u0 1 2 2 2 (b) If ´ D e y , then ´0 D 2yy 0 e y , so ´0 C ´ D 2 ; ´1 D 2 ; ´ D 2 ; 2 D 2 ; u0 D 1; x x x x x x 1=2 c 1 c 1 y2 u D x C c; ´ D C 2 D e ; y D ˙ ln C 2 . x x x x 0 y 2 y0 2 1 (c) Rewrite the equation as C ln y D 4x. If ´ D ln y, then ´0 D , so ´0 C ´ D 4x; ´1 D 2 ; y x y x x c c u u0 2 0 3 4 2 ´ D 2 ; 2 D 4x; u D 4x ; u D x C c; ´ D x C 2 D ln y; y D exp x C 2 . x x x x 1 y0 1 3 1 u u0 3 0 0 (d) If ´ D , then ´ D , so ´ C ´ D ; ´ D ; ´ D ; D ; 1 1Cy .1 C y/2 x x2 x x x x2 3 3 ln jxj C c 1 x u0 D ; u D 3 ln jxj c; ´ D D ; y D 1C . x x 1Cy 3 ln jxj C c
2.2 SEPARABLE EQUATIONS By 2.2.2. inspection, y k (k Dinteger) is a constant solution. Separate variables to find others: cos y 0 y D sin x; ln.j sin yj/ D cos x C c. sin y ln y .ln y/2 2.2.4. y 0 is a constant solution. Separate variables to find others: y0 D x2; D y 2 x3 C c. 3 1 2.2.6. y 1 and y 1 are constant solutions. For others, separate variables: .y 2 1/ 3=2 yy 0 D 2 ; x 2 1 1 C cx x x 2 1=2 2 1=2 2 .y 1/ D c D ; .y 1/ D ; .y 1/ D ; x x 1 C cx 1 C cx 2 2 !1=2 x x 2 y D1C ; y D˙ 1C . 1 C cx 1 C cx
Section 2.2 Separable Equations 2.2.8. By inspection, y 0 is a constant solution. Separate variables to find others: ln jyj D
y0 D y
c 1 ln.1 C x 2 / C k; y D p , which includes the constant solution y 0. 2 1 C x2
2.2.10. .y 1/2 y 0 D 2xC3;
1/3
.y 3
9
x ; 1 C x2
D x 2 C3xCc; .y 1/3 D 3x 2 C9xCc; y D 1 C 3x 2 C 9x C c/1=3 .
ˇ ˇ 2 ˇ y ˇ y0 1 1 0 ˇ ˇ D x C k; y D ce x2=2 ; y.2/ D D x; y D x; ln ˇ 2.2.12. y.y C 1/ y yC1 y C 1ˇ 2 yC1 x 2 =2 2 ce e e2 2 2 2 ; y D .y C 1/ce x =2 ; y.1 ce x =2/ D ce x =2 ; y D 1)cD ; setting c D 2 2 2 1 ce x =2 .x 2 4/=2 e yields y D . 2 2 e .x 4/=2 1 1 1 1 1 1 1 1 1 3 2 y0 2.2.14. D ; C y0 D ; C y0 D .y C 1/.y 1/.y 2/ xC1 6yC1 2y 1 3y 2 xC1 yC1 y 1 y 2 6 .y C 1/.y 2/2 c ; ln jy C 1j 3 ln jy 1j C 2 ln jy 2j D 6 ln jx C 1j C k; D ; xC1 .y 1/3 .x C 1/6 .y C 1/.y 2/2 256 y.1/ D 0 ) c D 256; D . 3 .y 1/ .x C 1/6 ! y0 1 y jyj y 2 0 2.2.16. D 2x; y D 2x; ln p D x 2 C k; p D ce x ; 2 2 2 2 y.1 C y / y y C1 y C1 y C1 2
ex 1 y 1 2 2 2 y.0/ D 1 ) c D p ; p D p ; 2y 2 D .y 2 C1/e x ; y 2 .2 e x / D e 2x ; y D p . 2 2x 2 2 y C1 2e 2 1 ˇ ˇ ˇy 2ˇ y0 1 1 ˇ D x 2 C k; y 2 D ce x2 ; 2.2.18. D 2x; y 0 D 2x; ln ˇˇ .y 1/.y 2/ y 2 y 1 y 1 ˇ! y 1 2 x2 x2 x2 x2 e e e e 4 e x 1 y 2 . y.0/ D 3 ) c D ; D ;y 2 D .y 1/; y 1 D2 ; yD 2 y 1 2 2 2 2 2 e x2 The interval of validity is . 1; 1/. ˇ ˇ ˇy 2ˇ y0 1 1 1 0 1 1 0 ˇ D 2x C k; 2.2.20. D 1; y D 1; y D 2; ln ˇˇ y.y 2/ 2 y 2 y y 2 y y ˇ y 2 y 2 D ce 2x ; y.0/ D 1 ) c D 1; D e 2x ; y 2 D ye 2x ; y.1 C e 2x / D 2; y y 2 yD . The interval of validity is . 1; 1/. 1 C e 2x
2.2.22. y 2 is a constant solution of the differential equation, and it satisfies the initial condition. Therefore, y 2 is a solution of the initial value problem. The interval of validity is . 1; 1/. y0 1 D ; tan 1 y D tan 2 1Cy 1 C x2 tan A C tan B tan.A C B/ D with A D tan 1 tan A tan B c D tan k. 2.2.24.
1
x C k; y D tan.tan
1
x and B D tan
1
1
x C k/. Now use the identity
c to rewrite y as y D
xCc , where 1 cx
10 Chapter 2 First Order Equations ) c D 0, so (A) cos y D sin x. To obtain 2 y explicity we note that sin x D cos.x C =2/, so (A) can be rewritten as cos y D cos.x C =2/. This equation holds if an only if one of the following conditions holds for some integer k: 2.2.26. .sin y/y 0 D cos x;
cos y D sin x C c; y. / D
(B) y D x C
C 2k I mbox.C / y D 2
x
C 2k: 2
Among these choices the only way to satisfy the initial condition is to let k D 1 in (C), so y D 2.2.28. Rewrite the equation as P 0 D
xC
3 : 2
1=˛/. By inspection, P 0 and P 1=˛ are P0 1 1 constant solutions. Separate variables to find others: D a˛; P 0 D a; P .P 1=˛/ P 1=˛ P ˇ ˇ ˇ P 1=˛ ˇ 1 ˇ D at C k; (A) P 1=˛ D ce ˛t ; P .1 ce ˛t / D 1=˛; (B) P D . ln ˇˇ ˇ P P ˛.1 ce ˛t / P0 1=˛ P0 From (A), P .0/ D P0 ) c D . Substituting this into (B) yields P D . P0 ˛P0 C .1 ˛P0/e at From this limt !1 P .t/ D 1=˛. a˛P .P
2.2.30. If q D rS the equation for I reduces to I 0 D
rI 2 , so
I0 D I2
r;
1 D I
rt
1 ; so I0
I0 and limt !1 I.t/ D 0. If q ¤ rS , then rewrite the equation for I as I 0 D rI.I ˛/ 1 C rI0 t ˇ ˇ ˇI ˛ˇ q I0 1 1 0 ˇ ˇD with ˛ D S . Separating variables yields D r; I D r ˛; ln ˇ r I.I ˛/ I ˛ I I ˇ I ˛ ˛ r ˛t C k; (A) D ce r ˛t ; I.1 ce r ˛t / D ˛; (B) I D . From (A), I.0/ D I0 ) I 1 ce r ˛t I0 ˛ ˛I0 c D . Substituting this into (B) yields I D . If q < rS , then ˛ > 0 and I0 I0 C .˛ I0 /e r ˛t q limt !1 I.t/ D ˛ D S . If q > rS , then ˛ < 0 and limt !1 I.t/ D 0. r I D
2.2.34. The given equation is separable if f D ap, where a is a constant. In this case the equation is y 0 C p.x/y D ap.x/:
.A/
Let P be an antiderivative of p; that is, P 0 D p. S OLUTION
BY
S EPARATION
OF
VARIABLES . y 0 D
p.x/.y
a/;
y0 y
a
D
p.x/; ln jy
aj D
P .x/ C k; y a D ce P.x/ ; y D a C ce P.x/ . S OLUTION BY VARIATION OF PARAMETERS . y1 D e P.x/ is a solution of the complementary equation, so solutions of (A) are of the form y D ue P.x/ where u0 e P.x/ D ap.x/. Hence, u0 D ap.x/e P.x/ ; u D ae P.x/ C c; y D a C ce P.x/ . 2 x5 2 yD . y1 D x 2 is a solution of y 0 y D 0. 2 x yCx x x5 x3 x Look for solutions of (A) of the form y D ux 2 . Then u0 x 2 D D ; u0 D ; 2 .u C 1/x uC1 uC1 p p .1 C u/2 x2 c .u C 1/u0 D x; D C ; u D 1 ˙ x 2 C c; y D x 2 1 ˙ x 2 C c . 2 2 2 2.2.36. Rewrite the given equation as (A) y 0
Section 2.3 Existence and Uniqueness of Solutions of Nonlinear Equations
11
2.2.38. y1 D e 2x is a solution of y 0 2y D 0. Look for solutions of the nonlinear equation of the xe 2x x .1 u/2 1 form y D ue 2x . Then u0 e 2x D ; u0 D ; .1 u/u0 D x; D .x 2 c/; 1 u 1 u 2 2 p p u D 1 ˙ c x 2 ; y D e 2x 1 ˙ c x 2 . 2.3 EXISTENCE AND UNIQUENESS OF SOLUTIONS OF NONLINEAR EQUATIONS
ex C y 1 2y.e x C y/ and fy .x; y/ D 2 are both continuous at all .x; y/ ¤ 2 2 2 x Cy x Cy .x 2 C y 2 /2 .0; 0/. Hence, Theorem 2.3.1 implies that if .x0 ; y0 / ¤ .0; 0/, then the initial value problem has a a unique solution on some open interval containing x0. Theorem 2.3.1 does not apply if .x0 ; y0 / D .0; 0/. 2.3.2. f .x; y/ D
x2 C y2 2y x2 C y2 and fy .x; y/ D are both continuous at all .x; y/ such ln xy ln xy x.ln xy/2 that xy > 0 and xy ¤ 1. Hence, Theorem 2.3.1 implies that if x0 y0 > 0 and x0 y0 ¤ 1, then the initial value problem has unique solution on an open interval containing x0 . Theorem 2.3.1 does not apply if x0 y0 0 or x0 y0 D 1. 2.3.4. f .x; y/ D
2.3.6. f .x; y/ D 2xy and fy .x; y/ D 2x are both continuous at all .x; y/. Hence, Theorem 2.3.1 implies that if .x0 ; y0 / is arbitrary, then the initial value problem has a unique solution on some open interval containing x0 . 3 2x C 3y 2x C 3y and fy .x; y/ D C4 are both continuous at all .x; y/ such x 4y x 4y .x 4y/2 that x ¤ 4y. Hence, Theorem 2.3.1 implies that if x0 ¤ 4y0 , then the initial value problem has a unique solution on some open interval containing x0 . Theorem 2.3.1 does not apply if x0 D 4y0 . 2.3.8. f .x; y/ D
4 xy.y 2 1/1=3 is continuous 3 at .x; y/ if and only if y ¤ ˙1. Hence, Theorem 2.3.1 implies that if y0 ¤ ˙1, then the initial value problem has a unique solution on some open interval containing x0, while if y0 D ˙1, then the initial value problem has at least one solution (possibly not unique on any open interval containing x0 ).
2.3.10. f .x; y/ D x.y 2
1/2=3 is continuous at all .x; y/, but fy .x; y/ D
1 are both continuous at all .x; y/ such that 2.x C y/1=2 x C y > 0 Hence, Theorem 2.3.1 implies that if x0 C y0 > 0, then the initial value problem has a unique solution on some open interval containing x0 . Theorem 2.3.1 does not apply if x0 C y0 0. 2.3.12. f .x; y/ D .x C y/1=2 and fy .x; y/ D
2.3.14. To apply Theorem 2.3.1, rewrite the given initial value problem as (A) y 0 D f .x; y/; y.x0 / D y0 , where f .x; y/ D p.x/y C q.x/ and fy .x; y/ D p.x/. If p and f are continuous on some open interval .a; b/ containing x0 , then f and fy are continuous on some open rectangle containing .x0 ; y0 /, so Theorem 2.3.1 implies that (A) has a unique solution on some open interval containing x0. The conclusion of Theorem 2.1.2 is more specific: the solution of (A) exists and is unique on .a; b/. For example, in the extreme case where .a; b/ D . 1; 1/, Theorem 2.3.1 still implies only existence and uniqueness on some open interval containing x0 , while Theorem 2.1.2 implies that the solution exists and is unique on . 1; 1/. 2.3.16. First find solutions of (A) y 0 D y 2=5 . Obviously y 0 is a solution. If y 6 0, then we 5 can separate variables on any open interval where y has no zeros: y 2=5 y 0 D 1; y 3=5 D x C c; 3 3 5=3 y D .x C c/ . (Note that this solution is also defined at x D c, even though y. c/ D 0. 5
12 Chapter 2 First Order Equations To satisfy the initial condition, let c D 1. Thus, y D
3 .x C 1/5=3 5
is a solution of the initial value
2 3=5 y are both continuous at 5 all .x; y/ such that y ¤ 0, this is the only solution on . 5=3; 1/, by an argument similar to that given in Example 2.3.7, the function ( 0; 1 < x 35 yD 5=3 3 5 xC1 ; 0, so (H) reduces to (D), or ˙ D C if y0 y0 x x0 < 0, so (H) reduces to (E). y0 x (d) Differentiating (A) yields 2x C 2yy 0 D 0, so y 0 D on either semicircle. Since (D) and (E) y xy x both reduce to y 0 D D (since x 2 C y 2 D 1) on both semicircles, the conclusion follows. 1 x2 y p 25 ˙ 49 0 (e) From (D) and (E) the slopes of tangent lines from (5,5) tangent to the circle are y D D 24 3 1 C 3x=5 4 1 4x=5 3 4 ; . Therefore, tangent lines are y D 5 C .x 5/ D and y D 5 C .x 5/ D , 4 3 4 4=5 3 3=5 which intersect the circle at . 3=5; 4=5/ .4=5; 3=5/, respectively. (See (B)). 4.5.16. (a) If .x0 ; y0 / is any point on the parabola such that x0 > 0 (and therefore y0 ¤ 0), then 1 differentiating (A) yields 1 D 2y0 y00 , so y00 D . Therefore,the equation of the tangent line is y D 2y0 1 y0 C .x x0 /. Since x0 D y02 , this is equivalent to (B). 2y0 1 y0 (b) Since y 0 D on the tangent line, we can rewrite (B) as D y xy 0 . Substituting this into (B) 2y0 2 x yields y D .y xy 0 / C , which implies (C). 4.y xy 0 / (c) Using the quadratic formula to solve (C) for y 0 yields p y ˙ y2 x 0 y D .F/ 2x y0 x 1 x 2 0 2 if .x; y/ is on a tangent line with slope y . If y D C , then y x D y0 so (F) 2 2y0 4 y0
48 Chapter 4 Applications of First Order Equations ˇ ˇ ˇ x x ˇˇ ˙ ˇˇy0 1 y0 y0 ˇ is equivalent to D which holds if and only if we choose the “˙" so that 2y 4x ˇ ˇ 0 ˇ x ˇˇ x D y0 . Therefore,we must choose ˙ D C if x > y02 D x0, so (F) reduces to (D), ˙ˇˇy0 ˇ y0 y0 or ˙ D if x < y02 D x0, so (F) reduces to (E). 1 (d) Differentiating (A) yields 1 D 2yy 0 , so y 0 D on either half of the parabola. Since (D) and (E) 2y both reduce to this if x D y 2 , the conclusion follows. y0 C
4.5.18. The equation of the line tangent to the curve at .x0 ; y.x0 // is y D y.x0 / C y 0 .x0 /.x x0 /. y 0 .x0 /x0 2y y0 2 D 0. Since x0 is arbitrary, it follows that y 0 D , so D , Since y.x0 =2/ D 0, y.x0 / 2 x y x ln jyj D 2 ln jxj C k, and y D cx 2 . Since .1; 2/ is on the curve, c D 2. Therefore,y D 2x 2 . 4.5.20. The equation of the line tangent to the curve at .x0 ; y.x0 // is y D y.x0 / C y 0 .x0 /.x x0 /. Since .x1 ; y1 / is on the line, y.x0 / C y 0 .x0 /.x1 x0 / D y1 . Since x0 is arbitrary, it follows that 1 y0 y C y 0 .x1 x/ D y1 , so D , ln jy y1 j D ln jx x1 j C k, and y y1 D c.x x1/. y y1 x x1 4.5.22. The equation of the line tangent to the curve at .x0 ; y.x0 // is y D y.x0 / C y 0 .x0 /.x y.0/ D x0 , x0 D y.x0 / y 0 .x0 /x0 . Since x0 is arbitrary, it follows that x D y xy 0 , so (A) y 0 1 The solutions of (A) are of the form y D ux, where u0 x D 1, so u0 D . Therefore,u D x and y D x ln jxj C cx.
x0 /. Since y D 1. x ln jxj C c
x x0 . Since y.0/ D y 0 .x0 / x0 y2 x2 c 2y.x0 /, y.x0 / C 0 D 2y.x0 /. Since x0 is arbitrary, it follows that y 0 y D x, so (A) D C y .x0 / 2 2 2 p and y 2 D x 2 C c. Now y.2/ D 1 $ c D 3. Therefore,y D x 2 3. 4.5.24. The equation of the line normal to the curve at .x0 ; y0 / is y D y.x0 /
4.5.26. Differentiating the given equation yields 2x C 4y C 4xy 0 C 2yy 0 D 0, so y 0 D
x C 2y 2x C y
2x C y is a differential equation for the x C 2y 2Cu 2.u2 1/ orthogonal trajectories. Substituting y D ux in (A) yields u0 x C u D , so u0 x D and 1 C 2u 1 C 2u 1 C 2u 2 3 1 4 u0 D , or C u0 D . Therefore, 3 ln ju 1jCln juC1j D 4 ln jxjC .u 1/.u C 1/ x u 1 uC1 x k y K, so .u 1/3 .u C 1/ D 4 . Substituting u D yields the orthogonal trajectories .y x/3 .y C x/ D k. x x
is a differential equation for the given family, and (A) y 0 D
y.1 C 2x 2 / 2 2 4.5.28. Differentiating yields ye x .1C2x 2 /Cxe x y 0 D 0, so y 0 D is a differential equation x x for the given family. Therefore,(A) y 0 D is a differential equation for the orthogonal y.1 C 2x 2 / x y2 1 k trajectories. From (A), yy 0 D , so D ln.1 C 2x 2/ C , and the orthogonal trajectories 2 1 C 2x 2 4 2 1 2 2 are given by y D ln.1 C 2x / C k. 2
Section 4.5 Applications to Curves 4.5.30. Differentiating (A) y D 1 C cx 2 yields (B) y 0 D 2cx. From (C), c D
y
1 x2
49
. Substituting this
2.y 1/ into (B) yields the differential equation y 0 D for the given family of parabolas. Therefore,y 0 D x x is a differential equation for the orthogonal trajectories. Separating variables yields 2.y 2.y 1/ x2 9 x2 9 C k. Now y. 1/ D 3 $ k D , so .y 1/2 D C . Therefore,(D) 1/y 0 D x, so .y 1/2 D 2 2 2 2 r 9 x2 . This curve interesects the parabola (A) if and only if the equation (C) cx 2 D y D 1C 2 r 9 x2 has a solution x 2 in .0; 9/. Therefore,c > 0 is a necessaary condition for intersection. We will 2 show that it is also sufficient. Squaring both sides of (C) and simplifying yields 2c 2 x 4 Cx 2 9 D 0. Using p 1 C 1 C 72c 2 the quadratic formula to solve this for x 2 yields x 2 D . The condition x 2 < 9 holds if 4c 2 p and only if 1C 1 C 72c 2 < 36c 4, which is equivalent to 1C72c 2 < .1C36c 2/2 D 1C72c 2 C1296c 4, which holds for all c > 0. 4.5.32. The angles and 1 from the x-axis to the tangents to C and C1 satisfy tan D f .x0 ; y0 / and f .x0 ; y0 / C tan ˛ tan C tan ˛ tan 1 D D D tan. C ˛/. Therefore, assuming and 1 are both in 1 f .x0 ; y0 / tan ˛ 1 tan tan ˛ Œ0; 2 /, 1 D C ˛. 4.5.34. Circles centered at the origin are given by x 2 C y 2 D r 2 . Differentiating yields 2x C 2yy 0 D 0, .x=y/ C tan ˛ x so y 0 D is a differential equation for the given family, and y 0 D is a differential y 1 C .x=y/ tan ˛ 1=u C tan ˛ 1 C u tan ˛ equation for the desired family. Substituting y D ux yields u0 x C u D D . 1 C .1=u/ tan ˛ u C tan ˛ 1 C u2 u C tan ˛ 0 1 1 Therefore,u0 x D , u D and ln.1 C u2 / C tan ˛ tan 1 u D ln jxj C k. u C tan ˛ 1 C u2 x 2 y 1 2 2 1 y Substituting u D yields ln.x C y / C .tan ˛/ tan D k. x 2 x
CHAPTER 5 Linear Second Order Equations
5.1 HOMOGENEOUS LINEAR EQUATIONS 5.1.2. (a) If y1 D e x cos x, then y10 D e x .cos x sin x/ and y100 D e x .cos x sin x sin x cos x/ D 2e x sin x, so y100 2y10 C 2y1 D e x . 2 sin x 2 cos x C 2 sin x C 2 cos x/ D 0. If y2 D e x sin x, then 0 y2 D e x .sin x C cos x/ and y200 D e x .sin x C cos x C cos x sin x/ D 2e x cos x, so y200 2y20 C 2y2 D e x .2 cos x 2 sin x 2 cos x C 2 sin x/ D 0. (b) If (B) y D e x .c1 cos x C c2 sin x/, then y 0 D e x .c1.cos x
sin x/ C c2 .sin x C cos x//
.C/
and y 00
D D
c1e x .cos x sin x sin x cos x/ Cc2e x .sin x C cos x C cos x sin x/ 2e x . c1 sin x C c2 cos x/;
so y 00
2y 0 C 2y
D c1 e x . 2 sin x
2 cos x C 2 sin x C 2 cos x/
Cc2 e x .2 cos x
2 cos x C 2 sin x/ D 0:
2 sin x
(c) We must choose c1 and c2 in (B) so that y.0/ D 3 and y 0 .0/ D 2. Setting x D 0 in (B) and (C) shows that c1 D 3 and c1 C c2 D 2, so c2 D 5. Therefore, y D e x .3 cos x 5 sin x/. (d) We must choose c1 and c2 in (B) so that y.0/ D k0 and y 0 .0/ D k1 . Setting x D 0 in (B) and (C) shows that c1 D k0 and c1 C c2 D k1 , so c2 D k1 k0 . Therefore, y D e x .k0 cos x C .k1 k0 / sin x/. 5.1.4. (a) If y1 D
1 x .x 2
1
, then y10 D
1 .x
1/2
1/y100 C 4xy10 C 2y1
and y100 D
2 .x
1/3
, so
D
2.x 2 1/ .x 1/3
D
2.x C 1/ 4x C 2.x .x 1/2
4x 2 C 2 .x 1/ x 1 1/
D 0:
1/y200 C 4xy20 C 2y2 D 0. The general solution on each of the c1 c2 intervals . 1; 1/, . 1; 1/, and .1; 1/ is (B) y D C . x 1 xC1
Similar manipulations show that .x 2
51
52 Chapter 5 Linear Second Order Equations c1 c2 . We must choose c1 and c2 in (B) so .x 1/2 .x C 1/2 0 that y.0/ D 5 and y .0/ D 1. Setting x D 0 in (B) and (C) shows that c1 C c2 D 5; c1 c2 D 1. 2 3 Therefore,c1 D 2 and c2 D 3, so y D on . 1; 1/. x 1 xC1 (d) The Wronskian of fy1 ; y2 g is (b) Differentiating (B) yields (C) y 0 D
1
ˇ 1 ˇ ˇ 2 x 1 xC1 ˇ ; ˇD 2 1 1 ˇ .x 1/2 ˇ .x 1/2 .x C 1/2 Z x Z x 4x 4t , so p.t/ dt D dt D ln.x 2 so W .0/ D 2. Since p.x/ D 2 2 x 1 t 1 0 0 2 2 2 implies that W .x/ D W .0/e ln.x 1/ D 2 , consistent with (D). .x 1/2 ˇ ˇ ˇ ˇ W .x/ D ˇ ˇ ˇ
Rx
Rx
2
.D/
1/2 , Abel’s formula
2
5.1.6. From Abel’s formula, W .x/ D W . /e 3 .t C1/ dt D 0 e 3 .t C1/ dt D 0. Z x Z x dt 1 5.1.8. p.x/ D ; therefore p.t/ dt D D ln x, so Abel’s formula yields W .x/ D W .1/e x t 1 1 1 . x 5.1.10. p.x/ D uD
K e 4
4x
2; P .x/ D
2x; y2 D uy1 D ue 3x ; u0 D
. Choose K D 4; then y2 D e
5.1.12. p.x/ D
2a; P .x/ D
4x 3x
e
De
x
Ke P.x/ Ke 2x D D Ke 2 e 6x y1 .x/
ln x
4x
D
;
.
2ax; y2 D uy1 D ue ax ; u0 D
Choose K D 1; then y2 D xe ax .
Ke P.x/ Ke 2ax D D K; u D Kx. e 2ax y12 .x/
Ke P.x/ K 1 Kx ; P .x/ D ln x; y2 D uy1 D ux; u0 D D 2 D ; u D K ln x. x x x y12 .x/ Choose K D 1; then y2 D x ln x. 5.1.14. p.x/ D
5.1.16. p.x/ D uD
1 ; P .x/ D x
ln jxj; y2 D uy1 D ux 1=2e 2x ; u0 D
Ke 4x . Choose K D 4; then y2 D e 4
4x
.x 1=2e 2x / D x 1=2e
2x
Ke P.x/ Kx D De 2 xe 4x y1 .x/
4x
;
.
2 Ke P.x/ Kx 2 ; P .x/ D 2 ln jxj; y2 D uy1 D ux cos x; u0 D D D x x 2 cos2 x y12 .x/ K sec2 x; u D K tan x. Choose K D 1; then y2 D tan x.x cos x/ D x sin x.
5.1.18. p.x/ D
3x C 2 3 D 1 ; P .x/ D x 3x 1 3x 1 x P.x/ Ke K.3x 1/e u0 D D D K.3x 1/e 3x ; u D 2 e 4x y1 .x/ y2 D xe 3x e 2x D xe x . 5.1.20. p.x/ D
1j; y2 D uy1 D ue 2x ;
ln j3x Kxe
3x
. Choose K D
1; then
Section 5.1 Homogeneous Linear Equations
53
2.2x 2 1/ 2 2 D 2 C ; P .x/ D 2x ln j2x C 1j C 2 ln jxj; y2 D x.2x C 1/ 2x C 1 x u Ke P.x/ K.2x C 1/e 2x 2 D x D K.2x C 1/e 2x ; u D Kxe 2x . Choose K D 1; then uy1 D ; u0 D x x2 y12 .x/ xe 2x y2 D D e 2x . x 5.1.22. p.x/ D
5.1.24. Suppose that y 0 on .a; b/. Then y 0 0 and y 00 0 on .a; b/, so y is a solution of (A) y 00 C p.x/y 0 C q.x/y D 0; y.x0 / D 0; y 0 .x0 / D 0 on .a; b/. Since Theorem 5.1.1 implies that (A) has only one solution on .a; b/, the conclusion follows. 5.1.26. If f´1 ; ´2 g is a fundamental set of solutions of (A) on .a; b/, then every solution y of (A) on .a; b/ is a linear combination of f´1 ; ´2 g; that is, y D c1 ´1 C c2 ´2 D c1.˛y1 C ˇy2 / C c2 . y1 C ıy2 / D .c1 ˛ C c2 /y1 C .c1 ˇ C c2 ı/y2 , which shows that every solution of (A) on .a; b/ can be written as a linear combination of fy1 ; y2 g. Therefore,fy1 ; y2 g is a fundamental set of solutions of (A) on .a; b/. 5.1.28. The Wronskian of fy1 ; y2 g is ˇ ˇ y1 y2 ˇ W Dˇ ˇ y0 y0 1 2 00
ˇ ˇ ˇ ˇ y1 ˇ ˇ ˇDˇ 0 ˇ ˇ y 1 0
ˇ ky1 ˇˇ ˇ D k.y1 y10 ky 0 ˇ
y10 y1 / D 0:
1
nor y2 can be a solution of y C p.x/y C q.x/y D 0 on .a; b/. 5.1.30. W .x0 / D y1 .x0 /y20 .x0 / y10 .x0 /y2 .x0 / D 0 if either y1 .x0 / D y2 .x0 / D 0 or y10 .x0 / D y20 .x0 / D 0, and Theorem 5.1.6 implies that fy1 ; y2 g is linearly dependent on .a; b/. 5.1.32. Let x0 be an arbitrary point in .a; b/. By the motivating argument preceding Theorem 5.1.4, (B) W .x0 / D y1 .x0 /y20 .x0 / y10 .x0 /y2 .x0 / ¤ 0. Now let y be the solution of y 00 C p.x/y 0 C q.x/y D 0; y.x0 / D y1 .x0 /; y 0 .x0 / D y10 .x0 /. By assumption, y is a linear combination of fy1 ; y2g on .a; b/; that is, y D c1 y1 C c2 y2 , where c1 y1 .x0 / C c2 y2 .x0 /
c1 y10 .x0 / C c2 y20 .x0 /
D y1 .x0 /
D y10 .x0 /:
Solving this system by Cramers’ rule yields ˇ ˇ 1 1 ˇˇ y1 .x0 / y2 .x0 / ˇˇ c1 D ˇ 0 ˇ D 1 and c2 D 0 ˇ ˇ W .x0 / y1 .x0 / y2 .x0 / W .x0 /
ˇ ˇ y1 .x0 / ˇ ˇ 0 ˇ y .x0 / 1
ˇ y1 .x0 / ˇˇ ˇ D 0: y1 .x0 / ˇ
Therefore,y D y1 , which shows that y1 is a solution of (A). A similar argument shows that y2 is a solution of (A).
5.1.34. Expanding the determinant by cofactors of its exercise can be written as ˇ 0 ˇ ˇ 0 y ˇˇ y1 y2 ˇˇ y 0 ˇˇ y1 y2 ˇ ˇ ˇ W ˇ y100 y200 ˇ W ˇ y100 y200 which is of the form (A) with
pD
1 W
ˇ ˇ y1 ˇ ˇ 00 ˇ y 1
ˇ y2 ˇˇ ˇ y 00 ˇ 2
and
first column shows that the first equation in the ˇ ˇ ˇ y 00 ˇ y1 ˇ ˇ ˇC ˇ ˇ W ˇ y0 1 1 qD W
ˇ 0 ˇ y ˇ 1 ˇ 00 ˇ y 1
ˇ y2 ˇˇ ˇ D 0; y20 ˇ ˇ y20 ˇˇ ˇ: y 00 ˇ 2
54 Chapter 5 Linear Second Order Equations 5.1.36. Theorem 5.1.6 implies that that there are constants c1 and c2 such that (B) y D c1y1 C c2 y2 on .a; b/. To see that c1 and c2 are unique, assume that (B) holds, and let x0 be a point in .a; b/. Then (C) y 0 D c1 y10 C c2y20 . Setting x D x0 in (B) and (C) yields c1 y1 .x0 / C c2 y2 .x0 / c1 y10 .x0 / C c2 y20 .x0 /
D y.x0 / D y 0 .x0 /:
Since Theorem 5.1.6 implies that y1 .x0 /y20 .x0 / y10 .x0 /y2 .x0 / ¤ 0, the argument preceding Theorem 5.1.4 implies that c1 and c2 are given uniquely by c1 D
y20 .x0 /y.x0 / y1 .x0 /y20 .x0 /
y2 .x0 /y 0 .x0 / y1 .x0 /y 0 .x0 / c D 2 y10 .x0 /y2 .x0 / y1 .x0 /y20 .x0 /
y10 .x0 /y.x0 / : y10 .x0 /y2 .x0 /
5.1.38. The general solution of y 00 D 0 is y D c1 C c2 x, so y 0 D c2 . Imposing the stated initial conditions on y1 D c1 C c2 x yields c1 C c2 x0 D 1 and c2 D 0; therefore c1 D 1, so y1 D 1. Imposing the stated initial conditions on y2 D c1 C c2x yields c1 C c2x0 D 0 and c2 D 1; therefore c1 D x0 , so y2 D x x0. The solution of the general initial value problem is y D k0 C k1 .x x0 /. 5.1.40. Let y1 D a1 cos !x C a2 sin !x and y2 D b1 cos !x C b2 sin !x. Then a1 cos !x0 C a2 sin !x0
D 1
!. a1 sin !x0 C a2 cos !x0 /
D 0
b1 cos !x0 C b2 sin !x0 !. b1 sin !x0 C b2 cos !x0 /
D 0 D 1:
and
sin !x0 cos !x0 , and b2 D . ! ! 1 Therefore, y1 D cos !x0 cos !x C sin !x0 sin !x D cos !.x x0 / and y2 D . sin !x0 cos !x C ! 1 cos !x0 sin !x/ D sin !.x x0 /. The solution of the general initial value problem is y D k0 cos !.x ! k1 x0 / C sin !.x x0 /. !
Solving these systems yields a1 D cos !x0 , a2 D sin !x0 , b1 D
5.1.42. (a) If y1 D x 2 , then y10 D 2x and y100 D 2, so x 2y100 4xy10 C 6y1 D x 2 .2/ 4x.2x/ C 6x 2 D 0 for x in . 1; 1/. If y2 D x 3 , then y20 D 3x 2 and y200 D 6x, so x 2 y200 4xy20 C 6y2 D x 2 .6x/ 4x.3x 2 / C 6x 3 D 0 for x in . 1; 1/. If x ¤ 0, then y2 .x/=y1 .x/ D x, which is nonconstant on . 1; 0/ and .0; 1/, so Theorem 5.1.6 implies that fy1 ; y2 g is a fundamental set of solutions of (A) on each of these intervals. 5.1.6 and (a) imply that y satisfies (A) on . 1; 0/ and on .0; 1/ if and only if y D (b) Theorem a1 x 2 C a2 x 3; x > 0; Since y.0/ D 0 we can complete the proof that y is a solution of (A) on b1 x 2 C b2 x 3; x < 0: . 1; 1/ by showing that y 0 .0/ and y 00 .0/ both exist if and only if a1 D b1 . Since y.x/ y.0/ a1 x C a2 x 2 ; if x > 0; D b1 x C b2 x 2 ; if x < 0; x 0 y.x/ y.0/ 2a1 x C 3a2 x 2 ; x 0; 0 0 it follows that y .0/ D lim D 0. Therefore, y D Since 2b1x C 3b2 x 2 ; x < 0: x!0 x 0 0 0 0 0 y .x/ y .0/ y .x/ y .0/ 2a1 C 3a2 x; if x > 0; D it follows that y 00 .0/ D lim exists if and 2b C 3b x; if x < 0; x!0 x 0 x 0 1 2
Section 5.2 Constant Coefficient Homogeneous Equations
55
only if a1 D b1 . By renaming a1 D b1 D c1, a2 D c2, and b2 D c3 we see that y is a solution of (A) on c1 x 2 C c2x 3 ; x 0; . 1; 1/ if and only if y D c1 x 2 C c3x 3 ; x < 0: (c) We have shown that y.0/ D y 0 .0/ D 0 for any choice of c1 and c2 in (C). Therefore,the given initial value problem has a solution if and only if k0 D k1 D 0, in which case every function of the form (C) is a solution. (d) If x0 > 0, then c1 and c2 in (C) are uniquely determined by k0 and k1 , but c3 can be chosen arbitrarily. Therefore,(B) has a unique solution on .0; 1/, but infinitely many solutions on . 1; 1/. If x0 < 0, then c1 and c3 in (C) are uniquely determined by k0 and k1 , but c2 can be chosen arbitrarily. Therefore,(B) has a unique solution on . 1; 0/, but infinitely many solutions on . 1; 1/. 5.1.44. (a) If y1 D x 3 , then y10 D 3x 2 and y100 D 6x, so x 2 y100 6xy10 C 12y1 D x 2.6x/ 6x.3x 2/ C 12x 3 D 0 for x in . 1; 1/. If y2 D x 4 , then y20 D 4x 3 and y200 D 12x 2 , so x 2 y200 6xy20 C 12y2 D x 2 .12x 2 / 6x.4x 3 / C 12x 4 D 0 for x in . 1; 1/. If x ¤ 0, then y2 .x/=y1 .x/ D x, which is nonconstant on . 1; 0/ and .0; 1/, so Theorem 5.1.6 implies that fy1 ; y2 g is a fundamental set of solutions of (A) on each of these intervals. (b) Theorem 5.1.2 and (a) imply that y satisfies (A) on . 1; 0/ and on .0; 1/ if and only if (C) a1 x 3 C a2 x 4 ; x > 0; y D Since y.0/ D 0 we can complete the proof that y is a solution of b1 x 3 C b2 x 4 ; x < 0: (A) on . 1; 1/ by showing that y 0 .0/ and y 00 .0/ both exist for any choice of a1 , a2 , b1 , and b2 . 2 y.x/ y.0/ y.x/ y.0/ a1 x C a2 x 3 ; if x > 0; Since D it follows that y 0 .0/ D lim D 0. b1x 2 C b2 x 3; if x < 0; x!0 x 0 x 0 y 0 .x/ y 0 .0/ 3a1 x 2 C 4a2 x 3 ; x 0; 3a1 x C 4a2 x 2 ; if x > 0; Therefore, y 0 D Since D 2 3 3b1x C 4b2 x ; x < 0: 3b1x C 4b2 x 2 ; if x < 0; x 0 0 0 y .x/ y .0/ it follows that y 00 .0/ D lim D 0. Therefore,(B) is a solution of (A) on . 1; 1/. x!0 x 0 0 (c) We have shown that y.0/ D y .0/ D 0 for any choice of a1 , a2 , b1 , and b2 in (B). Therefore,the given initial value problem has a solution if and only if k0 D k1 D 0, in which case every function of the form (B) is a solution. (d) If x0 > 0, then a1 and a2 in (B) are uniquely determined by k0 and k1 , but b1 and b2 can be chosen arbitrarily. Therefore,(C) has a unique solution on .0; 1/, but infinitely many solutions on . 1; 1/. If x0 < 0, then b1 and b2 in (B) are uniquely determined by k0 and k1 , but a1 and a2 can be chosen arbitrarily. Therefore,(C) has a unique solution on . 1; 0/, but infinitely many solutions on . 1; 1/. 5.2 CONSTANT COEFFICIENT HOMOGENEOUS EQUATIONS 5.2.2. p.r / D r 2
4r C 5 D .r
2/2 C 1; y D e 2x .c1 cos x C c2 sin x/.
5.2.4. p.r / D r 2
4r C 4 D .r
2/2 ; y D e 2x .c1 C c2x/.
5.2.6. p.r / D r 2 C 6r C 10 D .r C 3/2 C 1; y D e 5.2.8. p.r / D r 2 C r D r .r C 1/; y D c1 C c2 e
x
3x
.
5.2.10. p.r / D r 2 C 6r C 13y D .r C 3/2 C 4; y D e 5.2.12. p.r / D 10r 2
3r
1 D .2r
.c1 cos x C c2 sin x/. 3x
1/.5r C 1/ D 10.r
.c1 cos 2x C c2 sin 2x/. 1=2/.r C 1=5/; y D c1e
x=5
C c2 e x=2 .
5.2.14. p.r / D 6r 2 r 1 D .2r 1/.3r C 1/ D 6.r 1=2/.r C 1=3/; y D c1e x=3 C c2 e x=2; c1 x=3 c2 x=2 c1 c2 y0 D e C e ; y.0/ D 10 ) c1 C c2 D 10; y 0 .0/ D 0 ) C D 0; c1 D 6; c2 D 4; 3 2 3 2 x=2 x=3 y D 4e C 6e .
56 Chapter 5 Linear Second Order Equations 5.2.16. p.r / D 4r 2 4r 3 D .2r 3/.2r C 1/ D 4.r 3=2/.r C 1=2/; y D c1 e x=2 C c2e 3x=2 ; c1 x=2 3c2 3x=2 13 13 0 23 c1 3c2 23 y0 D e C e ; y.0/ D ) c1 C c2 D ; y .0/ D ) C D ; 2 2 12 12 24 2 2 24 1 3 e x=2 3e 3x=2 c1 D ; c2 D ; y D C . 3 4 3 4 5.2.18. p.r / D r 2 C 7r C 12 D .r C 3/.r C 4/; y D c1e 4x C c2 e 3x ; y 0 D 4c1e 4x 3c2 e y.0/ D 1 ) c1 C c2 D 1; y 0 .0/ D 0 ) 4c1 3c2 D 0; c1 D 3, c2 D 4; y D 3e 4x 4e
3x 3x
; .
e x=6 1/2 D 36.r 1=6/2 ; y D e x=6 .c1 C c2 x/; y 0 D .c1 C 6 5 c 5 1 c2 x/ C c2 e x=6 ; y.0/ D 3 ) c1 D 3; y 0 .0/ D ) C c2 D ) c2 D 2; y D e x=6 .3 C 2x/. 2 6 2
5.2.20. p.r / D 36r 2
12r C 1 D .6r
5.2.22. (a) From (A), ay 00 .x/ C by 0 .x/ C cy.x/ D 0 for all x. Replacing x by x x0 yields (C) ay 00 .x x0 / C by 0 .x x0 / C cy.x x0 / D 0. If ´.x/ D y.x x0 /, then the chain rule implies that ´0 .x/ D y 0 .x x0 / and ´00 .x/ D y 00 .x x0 /, so (C) is equivalent to a´00 C b´0 C c´ D 0. (b) If fy1 ; y2 g is a fundamental set of solutions of (A) then Theorem 5.1.6 implies that y2 =y1 is ´2 .x/ y2 .x x0 / nonconstant. Therefore, D is also nonconstant, so Theorem 5.1.6 implies that f´1 ; ´2 g ´1 .x/ y1 .x x0 / is a fundamental set of solutions of (A). (c) Let p.r / D ar 2 C br C c be the characteristic polynomial of (A). Then: If p.r / D 0 has distinct real roots r1 and r2 , then the general solution of (A) is y D c1 e r1.x
x0 /
C c2 e r2.x
x0 /
:
If p.r / D 0 has a repeated root r1 , then the general solution of (A) is y D e r1.x
x0 /
.c1 C c2 .x
x0 //:
If p.r / D 0 has complex conjugate roots r1 D C i ! and r2 D general solution of (A) is y D e .x 5.2.24. p.r / D r 2
7 D .r
6r
.c1 cos !.x
x0 / C c2 sin !.x
x0 //:
7/.r C 1/; y y0
1 y.2/ D ) c1 C c2 D 3 1 .x 2/ 2 7.x 2/ e e . 3 3
x0 /
i ! .where ! > 0/, then the
D c1 e .x 2/ C c2 e 7.x 2/ I D c1 e .x 2/ C 7c2e 7.x 2/ I
1 ; y 0 .2/ D 3
5 )
c1 C 7c2 D
5; c1 D
5.2.26. p.r / D 9r 2 C 6r C 1 D .3r C 1/2 D 9.r C 1=3/2 ; y y0
D e D
.x 2/=3
1 e 3
.c1 C c2 .x
.x 2/=3
2// I
.c1 C c2 .x
2// C c2 e
.x 2/=3
I
1 ; c2 D 3
2 ; y D 3
Section 5.2 Constant Coefficient Homogeneous Equations y.2/ D 2 ) c1 D 2; y 0 .2/ D 5.2.28. p.r / D r 2 C 3; y y0
14 ) 3
c1 C c2 D 3
14 ) c2 D 3
4; y D e
p p D c1cos 3 x C c2 sin 3 x I 3 3 p p p p 3c1sin 3 x C 3c2 cos 3 x D 3
y D 2 cos
p 3 x
p 1 p sin 3 x 3
3
.2
4.x
2//.
I 3
1 p ; 3
y.=3/ D 2 ) c1 D 2; y 0 .=3/ D 1 ) c2 D
.x 2/=3
57
: 3
5.2.30. y is a solution of ay 00 C by 0 C cy D 0 if and only if y y0
D D
c1 e r1.x r1c1 e r1.x
x0 /
C e r2.x x0 / C r2 e r2 .x
x0 /
x0 /:
Now y1 .x0 / D k0 and y10 .x0 / D k1 ) c1 C c2 D k0 ; r1c1 C r2 c2 D k1 . Therefore,c1 D and c2 D
k1 r1 k0 . Substituting c1 and c2 into the above equations for y and y’ yields r2 r1 y
D D
r2 k0 k1 r1.x x0 / k1 r1 k0 r2 .x x0 / e C e r2 r1 r2 r1 k0 r1.x x0 / k1 r2.x r2 e r1 e r2.x x0 / C e r2 r1 r2 r1
x0 /
e r1.x
x0 /
r2 k0 k1 r2 r1
:
5.2.32. y is a solution of ay 00 C by 0 C cy D 0 if and only if y D e .x
x0 /
.c1 cos !.x
x0 / C c2 sin !.x
x0 //
(A)
and y0
D
e .x x0 / .c1 cos !.x x0 / C c2 sin !.x x0 // C!e .x x0 / . c1 sin !.x x0/ C c2 cos !.x x0 // :
k1 k0 Now y1 .x0 / D k0 ) c1 D k0 and y10 .x0 / D k1 ) c1 C !c2 D k1 , so c2 D . Substituting ! c1 and c2 into (A) yields k1 k0 y D e .x x0 / k0 cos !.x x0 / C sin !.x x0 / : ! 5.2.34. (b) e i1 e i2
D .cos 1 C i sin 1 /.cos 2 C i sin 2 / D .cos 1 cos 2 sin 1 sin 2 / C i.sin 1 cos 2 C cos 1 sin 2 / D cos.1 C 2 / C i sin.1 C 2 / D e i.1 C2 / :
58 Chapter 5 Linear Second Order Equations (c) e ´1C´2
D e .˛1 Ciˇ1 /C.˛2 Ciˇ2 / D e .˛1 C˛2 /Ci.ˇ1 Cˇ2 /
D D D D
e .˛1 C˛2 / e i.ˇ1Cˇ2 / (from (F) with ˛ D ˛1 C ˛2 and ˇ D ˇ1 C ˇ2 ) e ˛1 e ˛2 e i.ˇ1 Cˇ2 / (property of the real–valued exponential function) e ˛1 e ˛2 e iˇ1 e iˇ2 (from (b)) e ˛1 e iˇ1 e ˛2 e iˇ2 D e ˛1Ciˇ1 e ˛2 Ciˇ2 D e ´1 e ´2 :
(d) The real and imaginary parts of ´1 D e .Ci !/x are u1 D e x cos !x and v1 D e x sin !x, which are both solutions of ay 00 C by 0 C cy D 0, by Theorem 5.2.1(c). Similarly, the real and imaginary parts of ´2 D e . i !/x are u2 D e x cos. !x/ D e x cos !x and v1 D e x sin. !x/ D e x sin !x, which are both solutions of ay 00 C by 0 C cy D 0, by Theorem 5.2.1,(c). 5.3 NONHOMOGENEOUS LINEAR EQUATIONS 5.3.2. The characteristic polynomial of the complementary equation is p.r / D r 2 4r C5 D .r 2/2 C1, so fe 2x cos x; e 2x sin xg is a fundamental set of solutions for the complementary equation. Let yp D A C Bx; then yp00 4yp0 C 5yp D 4B C 5.A C Bx/ D 1 C 5x. Therefore,5B D 5; 4B C 5A D 1, so B D 1, A D 1. Therefore,yp D 1 C x is a particular solution and y D 1 C x C e 2x .c1 cos x C c2 sin x/ is the general solution. 5.3.4. The characteristic polynomial of the complementary equation is p.r / D r 2 4r C 4 D .r 2/2 , so fe 2x ; xe 2x g is a fundamental set of solutions for the complementary equation. Let yp D A C Bx C Cx 2 ; then yp00 4yp0 C 4yp D 2C 4.B C 2Cx/ C 4.A C Bx C Cx 2 / D .2C 4B C 4A/ C . 8C C 4B/x C 4Cx 2 D 2 C 8x 4x 2. Therefore,4C D 4; 8C C 4B D 8; 2C 4B C 4A D 2, so C D 1, B D 0, and A D 1. Therefore,yp D 1 x 2 is a particular solution and y D 1 x 2 C e 2x .c1 C c2 x/ is the general solution. 5.3.6. The characteristic polynomial of the complementary equation is p.r / D r 2 C 6r C 10 D .r C 3/2 C 1, so fe 3x cos x; e 3x sin xg is a fundamental set of solutions for the complementary equation. Let yp D A C Bx; then yp00 C 6yp0 C 10yp D 6B C 10.A C Bx/ D 22 C 20x. Therefore,10B D 20; 6B C 10A D 22, so B D 2, A D 1. Therefore,yp D 1 C 2x is a particular solution and (A) y D 1 C 2x C e 3x.c1 cos x C c2 sin x/ is the general solution. Now y.0/ D 2 ) 2 D 1 C c1 ) c1 D 1. Differentiating (A) yields y 0 D 2 3e 3x .c1 cos x C c2 sin x/ C e 3x . c1 sin x C c2 cos x/, so y 0 .0/ D 2 ) 2 D 2 3c1 C c2 ) c2 D 1. y D 1 C 2x C e 3x .cos x sin x/ is the solution of the initial value problem. A 2 1 8 3A 6 2 2 00 0 5.3.8. If yp D , then x yp C 7xyp C 8yp D A x C 7x C D D if 3 2 x x x x x x 2 A D 2. Therefore,yp D is a particular solution. x 5.3.10. If yp D Ax 3 , then x 2 yp00 Therefore,yp D
1 x.3x 2 / C x 3 D 4Ax 3 D 2x 3 if A D . 2
xyp0 C yp D A x 2 .6x/
x3 is a particular solution. 2
5.3.12. If yp D Ax
1=3
, then
x 2 yp00
C
xyp0
C yp D A x
2
2x 9
5=3
!
10A 1=3 x D 10x 1=3 if A D 9. Therefore,yp D 9x 1=3 is a particular solution. 9
Cx
x
2=3
3
!
Cx
1=3
!
D
Section 5.3 Nonhomogeneous Linear Equations
59
A 3 3 2 12 2 00 0 , then x y C 3xy 3y D A x C 3x C D 0. Therefore,yp p p p x3 x5 x4 x3 is not a solution of the given equation for any choice of A.
5.3.14. If yp D
5.3.16. The characteristic polynomial of the complementary equation is p.r / D r 2 C5r 6 D .r C6/.r 1/, so fe 6x ; e x g is a fundamental set of solutions for the complementary equation. Let yp D Ae 3x ; then 1 e 3x yp00 C 5yp0 6yp D p.3/Ae 3x D 18Ae 3x D 6e 3x if A D . Therefore,yp D is a particular solution 3 3 e 3x C c1 e 6x C c2 e x is the general solution. and y D 3 5.3.18. The characteristic polynomial of the complementary equation is p.r / D r 2 C8r C7 D .r C1/.r C 7/, so fe 7x ; e x g is a fundamental set of solutions for the complementary equation. Let yp D Ae 2x ; then yp00 C 8yp0 C 7yp D p. 2/Ae 2x D 5Ae 2x D 10e 2x if A D 2. Therefore,yp D 2e 2x is a particular solution and (A) y D 2e 2x C c1 e 7x C c2 e x is the general solution. Differentiating (A) yields y 0 D 4e 2x 7c1 e 7x c2 e x . Now y.0/ D 2 ) 2 D 2 C c1 C c2 and y 0 .0/ D 10 ) 10 D 4 7c1 c2 . Therefore,c1 D 1 and c2 D 1, so y D 2e 2x e 7x C e x is the solution of the initial value problem. 5.3.20. The characteristic polynomial of the complementary equation is p.r / D r 2 C 2r C 10 D .r C 1/2 C 9, so fe x cos 3x; e x sin 3xg is a fundamental set of solutions for the complementary equation. 4 45 x=2 If yp D Ae x=2 , then yp00 C 2yp0 C 10yp D p.1=2/Ae x=2 D Ae D e x=2 if A D . Therefore, 4 45 4 x=2 4 x=2 yp D e is a particular solution and y D e Ce x .c1 cos 3x Cc2 sin 3x/ is the general solution. 45 45 5.3.22. The characteristic polynomial of the complementary equation is p.r / D r 2 7r C 12 D .r 4/.r 3/. If yp D Ae 4x , then yp00 7yp0 C 12yp D p.4/Ae 4x D 0 e 4x D 0, so yp00 7yp0 C 12yp ¤ 5e 4x for any choice of A. 5.3.24. The characteristic polynomial of the complementary equation is p.r / D r 2 8r C 16 D .r 4/2 , so fe 4x ; xe 4x g is a fundamental set of solutions for the complementary equation. If yp D A cos x C B sin x, then yp00 8yp0 C16yp D .A cos x CB sin x/ 8. A sin x CB cos x/C16.A cos x CB sin x/ D .15A 8B/ cos x C .8A C 15B/ sin x, so 15A 8B D 23; 8A C 15B D 7, which implies that A D 1 and B D 1. Hence yp D cos x sin x and y D cos x sin x C e 4x .c1 C c2 x/ is the general solution. 2 2 5.3.26. Thepcharacteristic p polynomial of the complementary equation is p.r / D r 2r C3 D .r 1/ C2, x x so fe cos 2x; e sin 2xg is a fundamental set of solutions for the complementary equation. If yp D A cos 3x C B sin 3x, then yp00 2yp0 C 3yp D 9.A cos 3x C B sin 3x/ 6. A sin 3x C B cos 3x/ C 3.A cos 3x C B sin 3x/ D .6A C 6B/ cos 3x C .6A 6B/ sin 3x, so 6A 6B D 6; 6A 6B D 6, which implies p that A Dp1 and B D 0. Hence yp D cos 3x is a particular solution and y D cos 3x C e x .c1 cos 2x C c2 sin 2x/ is the general solution.
5.3.28. The characteristic polynomial of the complementary equation is p.r / D r 2 C 7r C 12 D .r C 3/.r C 4/, so fe 4x ; e 3x g is a fundamental set of solutions for the complementary equation. If yp D A cos 2x C B sin 2x, then yp00 C 7yp0 C 12yp D 4.A cos 2x C B sin 2x/ C 14. A sin 2x C B cos 2x/ C 12.A cos x CB sin x/ D .8AC14B/ cos 2x C.8B 14a/ sin 2x, so 8AC14B D 2; 14AC8B D 36, which implies that A D 2 and B D 1. Hence yp D 2 cos 2x C sin 2x is a particular solution and (A) y D 2 cos 2x C sin 2x C c1 e 4x C c2e 3x is the general solution. Differentiating (A) yields y 0 D 2 sin 2x C 2 cos 2x 4c1e 4x 3c2e 3x . Now y.0/ D 3 ) 3 D 2 C c1 C c2 and y 0 .0/ D 3 ) 3 D 2 4c1 3c2 . Therefore, c1 D 2 and c2 D 3, so y D 2 cos 2x C sin 2x C 2e 4x 3e 3x is the solution of the initial value problem.
60 Chapter 5 Linear Second Order Equations I 5.3.30. fcos !0 x; sin !0 xg is a fundamental set of solutions of the complementary equation. If yp D A cos !x C B sin !x, then yp00 C !02 yp D ! 2 .A cos !x C B sin !x/ C !02 .A cos !x C B sin !x/ D M N .!02 ! 2 /.A cos !x C B sin !x/ D M cos !x C N sin !x if A D 2 and B D 2 . 2 !0 ! !0 ! 2 Therefore, 1 .M cos !x C N sin !x/ yp D 2 !0 ! 2 is a particular solution of the given equation and yD
1 !02
!2
.M cos !x C N sin !x/ C c1 cos !0 x C c2 sin !0 x
is the general solution. 5.3.32. If yp D A cos !x C B sin !x, then ayp00 C byp0 C cyp D a! 2 .A cos !x C B sin !x/ C b!. A sin !xCB cos !x/Cc.A cos !xCB sin !x/ D .c a! 2 /A C b!B cos !xC b!A C .c a! 2 /B sin !x. Therefore, yp is a solution of (A) if and only if the set of equations (B) .c a! 2 /A C b!B D 2 2 2 2 M; b!A C .c a! /B D N has a solution. If .c a! / C .b!/ ¤ 0, then (B) has the so.c a! 2 /M b!N .c a! 2 /N C b!M lution A D , B D , and yp D A cos !x C B sin !x is a .c a! 2 /2 C .b!/2 .c a! 2 /2 C .b!/2 solution of (A). If .c a! 2 /2 C .b!/2 D 0 (which is true if and only if the left side of (A) is of the form a.y 00 C ! 2 y/, then the coefficients of A and B in (B) are all zero, so (B) does not have a solution, so (A) does not have a solution of the form yp D A cos !x C B sin !x. 5.3.34. From Exercises 5.3.2 and 5.3.17, yp1 D 1 C x and yp2 D e 2x are particular solutions of y 00 4y 0 C 5y D 1 C 5x and y 00 4y 0 C 5y D e 2x respectively, and fe 2x cos x; e 2x sin xg is a fundamental set of solutions of the complementary equation. Therefore,yp D yp1 C yp2 D 1 C x C e 2x is a particular solution of the given equation, and y D 1 C x C e 2x .1 C c1 cos x C c2 sin x/ is the general solution. 5.3.36. From Exercises 5.3.4 and 5.3.19, yp1 D 1 x 2 and yp2 D e x are particular solutions of y 00 4y 0 C 4y D 2 C 8x 4x 2 and y 00 4y 0 C 4y D e x respectively, and fe 2x ; xe 2x g is a fundamental set of solutions of the complementary equation. Therefore,yp D yp1 C yp2 D 1 x 2 C e x is a particular solution of the given equation, and y D 1 x 2 C e x C e 2x .c1 C c2x/ is the general solution. 5.3.38. From Exercises 5.3.6 and 5.3.21, yp1 D 1 C 2x and yp2 D e 3x are particular solutions of y 00 C 6y 0 C 10y D 22 C 20x and y 00 C 6y 0 C 10y D e 3x respectively, and fe 3x cos x; e 3x sin xg is a fundamental set of solutions of the complementary equation. Therefore,yp D yp1 Cyp2 D 1C2x Ce 3x is a particular solution of the given equation, and y D 1 C 2x C e 3x .1 C c1 cos x C c2 sin x/ is the general solution. 5.3.40. Letting c1 D c2 D 0 shows that (A) yp00 C p.x/yp0 C q.x/yp D f . Letting c1 D 1 and c2 D 0 shows that (B) .y1 C yp /00 C p.x/.y1 C yp /0 C q.x/.y1 C yp / D f . Now subtract (A) from (B) to see that y100 C p.x/y10 C q.x/y1 D 0. Letting c1 D 0 and c2 D 1 shows that (C) .y2 C yp /00 C p.x/.y2 C yp /0 C q.x/.y2 C yp / D f . Now subtract (A) from (C) to see that y200 C p.x/y20 C q.x/y2 D 0. 5.4 THE METHOD OF UNDETERMINED COEFFICIENTS I 5.4.2. If y D ue 3x , then y 00 6y 0 C5y D e 3x Œ.u00 6u0 C 9u/ 6.u0 3u/ C 5u D e 3x .35 8x/, so u00 12u0 C32u D 35 8x and up D ACBx, where 12B C32.ACBx/ D 35 8x. Therefore,32B D 1 x x 8, 32A 12B D 35, so B D , A D 1, and up D 1 . Therefore, yp D e 3x 1 . 4 4 4
Section 5.4 The Method of Undetermined Coefficients I
61
5.4.4. If y D ue 2x , then y 00 C 2y 0 C y D e 2x Œ.u00 C 4u0 C 4u/ C 2.u0 C 2u/ C u D e 2x . 7 15x C 9x 2 / so u00 C 6u0 C 9u D 7 15x C 9x 2 and up D A C Bx C Cx 2 , where 2C C 6.B C 2Cx/ C 9.A C Bx C Cx 2 / D 7 15x C 9x 2. Therefore,9C D 9, 9B C 12C D 15, 9A C 6B C 2C D 7, so C D 1, B D 3, A D 1, and up D 1 3x C x 2 . Therefore,yp D e 2x .1 3x C x 2 /. 5.4.6. If y D ue x , then y 00 y 0 2y D e x Œ.u00 C 2u0 C u/ .u0 C u/ 2u D e x .9 C 2x 4x 2 / so u00 C u0 2u D 9 C 2x 4x 2, and up D A C Bx C Cx 2, where 2C C .B C 2Cx/ 2.A C Bx C Cx 2/ D 9 C 2x 4x 2 . Therefore, 2C D 4, 2B C 2C D 2, 2A C B C 2C D 9, so C D 2, B D 1, A D 2, and up D 2 C x C 2x 2 . Therefore, yp D e x . 2 C x C 2x 2/. 5.4.8. If y D ue x , then y 00 3y 0 C 2y D e x Œ.u00 C 2u0 C u/ 3.u0 C u/ C 2u D e x .3 4x/, so u00 u0 D 3 4x and up D Ax C Bx 2 , where 2B .A C 2Bx/ D 3 4x. Therefore, 2B D 4, A C 2B D 3, so B D 2, A D 1, and up D x.1 C 2x/. Therefore,yp D xe x .1 C 2x/. 5.4.10. If y D ue 2x , then 2y 00 3y 0 2y D e 2x Œ2.u00 C 4u0 C 4u/ 3.u0 C 2u/ 2u D e 2x . 6 C 10x/, so 2u00 C 5u0 D 6 C 10x and up D Ax C Bx 2 , where 2.2B/ C 5.A C 2Bx/ D 6 C 10x. Therefore,10B D 10, 5A C 4B D 6, so B D 1, A D 2, and up D x. 2 C x/. Therefore,yp D xe 2x . 2 C x/. 5.4.12. If y D ue x , then y 00
2y 0 C y D e x Œ.u00 C 2u0 C u/
2.u0 C u/ C u D e x .1 6x/, so 1 u00 D 1 6x Integrating twice and taking the constants of integration to be zero yields up D x 2 x . 2 1 Therefore, yp D x 2 e x x . 2 2u0 u u 5.4.14. If y D ue x=3 , then 9y 00 C 6y 0 C y D e x=3 9 u00 C C 6 u0 Cu D 3 9 3 1 x=3 2 2 00 2 00 e .2 4x C 4x /, so 9u D 2 4x C 4x , or u D .2 4x C 4x /. Integrating twice 9 x2 and taking the constants of integration to be zero yields up D .3 2x C x 2 /. Therefore, yp D 27 x 2 e x=3 .3 2x C x 2 /. 27 5.4.16. If y D ue x , then y 00 6y 0 C 8y D e x Œ.u00 C 2u0 C u/ 6.u0 C u/ C 8u D e x .11 6x/, so u00 4u0 C 3u D 11 6x and up D A C Bx, where 4B C 3.A C Bx/ D 11 6x. Therefore,3B D 6, 3A 4B D 11, so B D 2, A D 1 and up D 1 2x. Therefore,yp D e x .1 2x/. The characteristic polynomial of the complementary equation is p.r / D r 2 6r C 8 D .r 2/.r 4/, so fe 2x ; e 4x g is a fundamental set of solutions of the complementary equation. Therefore, y D e x .1 2x/C c1 e 2x C c2 e 4x is the general solution of the nonhomogeneous equation. 5.4.18. If y D ue x , then y 00 C 2y 0 3y D e x Œ.u00 C 2u0 C u/ C 2.u0 C u/ 3u D 16xe x , so u00 C 4u0 D 16x and up D Ax C Bx 2 , where 2B C 4.A C 2Bx/ D 16x. Therefore,8B D 16, 4AC 2B D 0, so B D 2, A D 1, and up D x.1 2x/. Therefore,yp D xe x .1 2x/. The characteristic polynomial of the complementary equation is p.r / D r 2 C 2r 3 D .r C 3/.r 1/, so fe x ; e 3x g is a fundamental set of solutions of the complementary equation. Therefore, y D xe x .1 2x/Cc1e x Cc2e 3x is the general solution of the nonhomogeneous equation. 5.4.20. If y D ue 2x , then y 00 4y 0 5y D e 2x Œ.u00 C 4u0 C 4u/ 4.u0 C 2u/ 5u D 9e 2x .1 C x/, so u00 9u D 9 C 9x and up D A C Bx, where 9.A C Bx/ D 9 C 9x. Therefore, 9B D 9, 9A D 9, so B D 1, A D 1, and up D 1 x. Therefore,yp D e 2x .1 C x/. The characteristic polynomial of the complementary equation is p.r / D r 2 4r 5 D .r 5/.r C 1/, so fe x ; e 5x g is a fundamental
62 Chapter 5 Linear Second Order Equations I set of solutions of the complementary equation. Therefore,(A) y D e 2x .1 C x/ C c1 e x C c2 e 5x is the general solution of the nonhomogeneous equation. Differentiating (A) yields y 0 D 2e 2x .1 C x/ e 2x c1 e x C 5c2e 5x . Now y.0/ D 0; y 0 .0/ D 10 ) 0 D 1 C c1 C c2 ; 10 D 3 c1 C 5c2, so c1 D 2, c2 D 1. Therefore,y D e 2x .1 C x/ C 2e x e 5x is the solution of the initial value problem. 5.4.22. If y D ue x , then y 00 C 4y 0 C 3y D e x Œ.u00 2u0 C u/ C 4.u0 u/ C 3u D e x .2 C 8x/, so u00 C 2u0 D 2 8x and up D Ax C Bx 2 , where 2B C 2.A C 2Bx/ D 2 8x. Therefore,4B D 8, 2A C 2B D 2, so B D 2, A D 1, and up D x.1 2x/. Therefore,yp D xe x .1 2x/. The characteristic polynomial of the complementary equation is p.r / D r 2 C 4r C 3 D .r C 3/.r C 1/, so fe x ; e 3x g is a fundamental set of solutions of the complementary equation. Therefore,(A) y D xe x .1 2x/ C c1e x C c2 e 3x is the general solution of the nonhomogeneous equation. Differentiating (A) yields y 0 D xe x .1 2x/ C e x .1 4x/ c1 e x 3c2e 3x . Now y.0/ D 1; y 0 .0/ D 2 ) 1 D c1 C c2 ; 2 D 1 c1 3c2 , so c1 D 2, c2 D 1. Therefore,y D e x .2 C x 2x 2 / e 3x is the solution of the initial value problem. 5.4.24. We must find particular solutions yp1 and yp2 of (A) y 00 C y 0 C y D xe x and (B) y 00 C y 0 C y D e x .1 C 2x/, respectively. To find a particular solution of (A) we write y D ue x . Then y 00 C y 0 C y D e x Œ.u00 C 2u0 C u/ C .u0 C u/ C u D xe x so u00 C 3u0 C 3u D x and up D A C Bx, 1 1 where 3B C 3.A C Bx/ D x. Therefore,3B D 1, 3A C 3B D 0, so B D , A D , and up D 3 3 x e 1 .1 x/, so yp1 D .1 x/. To find a particular solution of (B) we write y D ue x . Then 3 3 0 00 0 x 00 y C y C y D e Œ.u 2u C u/ C .u0 u/ C u D e x .1 C 2x/, so u00 u0 C u D 1 C 2x and up D A C Bx, where B C .A C Bx/ D 1 C 2x. Therefore, B D 2, A B D 1, so A D 3, and ex up D 2 C 3x, so yp2 D e x .3 C 2x/. Now yp D yp1 C yp2 D .1 x/ C e x .3 C 2x/. 3 5.4.26. We must find particular solutions yp1 and yp2 of (A) y 00 8y 0 C 16y D 6xe 4x and (B) y 00 8y 0 C 16y D 2 C 16x C 16x 2, respectively. To find a particular solution of (A) we write y D ue 4x . Then y 00 8y 0 C 16y D e x Œ.u00 C 8u0 C 16u/ 8.u0 C 4u/ C 16u D 6xe 4x , so u00 D 6x, up D x 3. and yp1 D x 3 e 4x . To find a particular solution of (B) we write yp D ACBxCCx 2 . Then yp00 8yp0 C16yp D 2C 8.B C2Cx/C16.ACBx CCx 2 / D .16A 8B C2C /C.16B 16C /x C16Cx 2 D 2C16x C16x 2 if 16C D 16, 16B 16C D 16, 16A 8B C 2C D 2. Therefore,C D 1, B D 2, A D 1, and yp2 D 1 C 2x C x 2 . Now yp D yp1 C yp2 D x 3 e 4x C 1 C 2x C x 2 . 5.4.28. We must find particular solutions yp1 and yp2 of (A) y 00 2y 0 C 2y D e x .1 C x/ and (B) y 00 2y 0 C 2y D e x .2 8x C 5x 2/, respectively. To find a particular solution of (A) we write y D ue x . Then y 00 2y 0 C 2y D e x Œ.u00 C 2u0 C u/ 2.u0 C u/ C 2u D e x .1 C x/, so u00 C u D 1 C x and up D 1 C x, so yp1 D e x .1 C x/. To find a particular solution of (B) we write y D ue x . Then y 00 2y 0 C 2y D e x Œ.u00 2u0 C u/ 2.u0 u/ C 2u D e x .2 8x C 5x 2 /, so u00 4u0 C 5u D 2 8x C 5x 2 and up D A C Bx C Cx 2 , where 2C 4.B C 2Cx/ C 5.A C Bx C Cx 2 / D 2 8x C 5x 2. Therefore,5C D 5, 5B 8C D 8, 5A 4B C 2C D 2, so C D 1, B D 0, A D 0, and up D x 2 . Therefore,yp2 D x 2 e x . Now yp D yp1 C yp2 D e x .1 C x/ C x 2 e x . 5.4.30. (a) If y D ue ˛x , then ay 00 C by 0 C cy D e ˛x a.u00 C 2˛u0 C ˛ 2 u/ C b.u0 C ˛u/ C cu D e ˛x au00 C .2a˛ C b//u0 C .a˛ 2 C b˛ C c/u D e ˛x .au00 Cp 0 .˛/u0 Cp.˛/u/. Therefore,ay 00 Cby 0 C cy D e ˛x G.x/ if and only if au00 C p 0 .˛/u0 C p.˛/u D G.x/. (b) Substituting up D A C Bx C Cx 2 C Dx 3 into (B) yields a.2C C 6Dx/ C p 0 .˛/.B C 2Cx C 3Dx 2 / C p.˛/.A C Bx C Cx 2 C Dx 3 / D Œp.˛/A C p 0 .˛/B C 2aC C Œp.˛/B C 2p 0 .˛/C C 6aDx CŒp.˛/C C 3p 0.˛/Dx 2 C p.˛/Dx 3 D g0 C g1 x C g2 x 2 C g3 x 3
Section 5.4 The Method of Undetermined Coefficients I if
D g3 D g2 D g1 D g0 :
p.˛/D p.˛/C C 3p 0 .˛/D p.˛/B C 2p 0 .˛/C C 6aD p.˛/A C p 0 .˛/B C 2aC
63
.C/
Since e ˛x is not a solution of the complementary equation, p.˛/ ¤ 0. Therefore,the triangular system (C) can be solved successively for D, C , B and A. (c) Since e ˛x is a solution of the complementary equation while xe ˛x is not, p.˛/ D 0 and p 0 .˛/ ¤ 0. Therefore, (B) reduces to (D) au00 C p 0 .˛/u D G.x/. Substituting up D Ax C Bx 2 C Cx 3 C Dx 4 into (D) yields a.2B C 6Cx C 12Dx 2/ C p 0 .˛/.A C 2Bx C 3Cx 2 C 4Dx 3 / D .p 0 .˛/A C 2aB/ C .2p 0 .˛/B C 6aC /x C .3p 0 .˛/C C 12aD/x 2 C4p 0 .˛/Dx 3 D g0 C g1 x C g2 x 2 C g3 x 3 if
4p 0 .˛/D 3p .˛/C C 12aD 2p 0 .˛/B C 6aC p 0 .˛/A C 2aB 0
D g3 D g2 D g1 D g0 :
Since p 0 .˛/ ¤ 0 this triangular system can be solved successively for D, C , B and A. (d) Since e ˛x and xe ˛x are solutions of the complementary equation, p.˛/ D 0 and p 0 .˛/ D 0. G.x/ Therefore, (B) reduces to (D) au00 D G.x/, so u00 D . Integrating this twice and taking the a g g1 g2 g3 3 0 constants of integration yields the particular solution up D x 2 C x C x2 C x . 2 6 12 20
5.4.32. If yp D Axe 4x , then yp00 if A D 1, so yp D 5xe 4x .
7yp0 C 12yp D Œ.8 C 16x/
7.1 C 4x/ C 12xAe 4x D Ae 4x D 5e 4x
5.4.34. If yp D e 3x .A C Bx C Cx 2 /, then yp00
3yp0 C 2yp
D
e 3x Œ.9A C 6B C 2C / C .9B C 12C /x C 9Cx 2
D D
e 3x Œ.2A C 3B C 2C / C .2B C 6C /x C 2Cx 2 e 3x . 1 C 2x C x 2 /
3e 3x Œ.3A C B/ C .3B C 2C /x C 3Cx 2 C2e 3x .A C Bx C Cx 2 /
if 2C D 1; 2B C 6C D 2; 2A C 3B C 2C D yp D
e 3x .1 C 2x 4
5.4.36. If yp D e
x=2
1. Therefore,C D
1 ,B D 2
1 ,A D 2
2x 2 /. .Ax 2 C Bx 3 C Cx 4 /, then
4yp00 C 4yp0 C yp
D e
x=2
Œ8A
.8A
24B/x C .A
12B C 48C /x 2
Ce x=2 Œ.B 16C /x 3 C Cx 4 Ce x=2Œ8Ax .2A 12B/x 2 .2B 16C /x 3 2Cx 4 Ce x=2 .Ax 2 C Bx 3 C Cx 4 / D e x=2 .8A C 24Bx C 48Cx 2 / D e x=2. 8 C 48x C 144x 2/
1 , and 4
64 Chapter 5 Linear Second Order Equations if 48C D 144, 24B D 48, and 8A D 8. Therefore,C D 3, B D 2, A D 1, and yp D x 2 e x=2. 1 C 2x C 3x 2 /. R 5.4.38. If y D e ˛x P .x/ dx, then y 0 D e ˛x P .x/. Let y D ue ˛x ; then .u0 C˛u/e ˛x D e ˛x P .x/, which implies (A). We must show that it is possible to choose A0 ; : : : ; Ak so that (B) .A0 C A1 x C Ak x k /0 C ˛.A0 C A1 x C Ak x k / D p0 C p1 x C C pk x k . By equating the coefficients of x k ; x k 1; : : : ; 1 (in that order) on the two sides of (B), we see that (B) holds if and only if ˛Ak D pk and .k j C 1/Ak j C1 C ˛Ak D pk j ; 1 j k. R 5.4.40. If y D x k e ˛x dx, then y 0 D x k e ˛x . Let y D ue ˛x ; then .u0 C ˛u/e ˛x D x k e ˛x , so u0 C ˛u D x k . This equation has a particular solution up D A0 C A1 x C Ak x k , where (A) .A0 C A1 x C Ak x k /0 C ˛.A0 C A1 x C Ak x k / D x k . By equating the coefficients of x k ; x k 1; : : : ; 1 on the two sides of (A), we see that (A) holds if and only if ˛Ak D 1 and .k j C 1/Ak j C1 C 1 k k.k 1/ ˛Ak j D 0; 1 j k. Therefore, Ak D , Ak 1 D , Ak 2 D , and, in general, 2 ˛ ˛ ˛3 j . 1/ kŠ k.k 1/ .k j C 1/ Ak j D . 1/j D j C1 ; 1 j k. By introducing the index j C1 ˛ ˛ .k j /Š k . 1/k kŠ X . ˛x/r . 1/k r kŠ r D k j we can rewrite this as Ar D k r C1 ; 0 r k. Therefore, up D rŠ ˛ rŠ ˛ kC1 k
and y D
. 1/ kŠe ˛ kC1
˛x
r D0
k X . ˛x/r C c. rŠ r D0
5.5 THE METHOD OF UNDETERMINED COEFFICIENTS II 5.5.2. Let yp yp0 yp00
D .A0 C A1 x/ cos x C .B0 C B1x/ sin xI then
D .A1 C B0 C B1x/ cos x C .B1 A0 A1 x/ sin x D .2B1 A0 A1 x/ cos x .2A1 C B0 C B1 x/ sin x; so yp00 C 3yp0 C yp
D .3A1 C 3B0 C 2B1 C 3B1 x/ cos x C.3B1 3A0 2A1 3A1 x/ sin x
D .2
6x/ cos x
9 sin x
if 3B1 D 6, 3A1 D 0, 3B0 C 3A1 C 2B1 D 2, 3A0 C 3B1 C 2A1 D B1 D 2, A0 D 1, B0 D 2, and yp D cos x C .2 2x/ sin x.
9. Therefore,A1 D 0,
5.5.4. Let y D ue 2x . Then y 00 C 3y 0
2y
if u00 C 7u0 C 8u D 5 cos 2x u00p C 7u0p C 8up
D e 2x .u00 C 4u0 C 4u/ C 3.u0 C 2u/ D e 2x .u00 C 7u0 C 8u/ D
2u
e 2x .5 cos 2x C 9 sin 2x/
9 sin 2x. Now let up D A cos 2x C B sin 2x. Then D
4.A cos 2x C B sin 2x/ C 14. A sin 2x C B cos 2x/ C8.A cos 2x C B sin 2x/ D .4A C 14B/ cos 2x .14A 4B/ sin 2x D 5 cos 2x 9 sin 2x
Section 5.5 The Method of Undetermined Coefficients II 1 ,B D 2
if 4A C 14B D 5, 14A C 4B D 9. Therefore,A D 1 up D .cos 2x 2 2x
5.5.6. Let y D ue
u0
2y
D e
D e D e
2x 2x 2x
4u D .4 C 20x/ cos 3x C .26
e 2x .cos 2x 2
sin 2x/:
00 .u
4u0 C 4u/ C 3.u0
.u00 u0 4u/ Œ.4 C 20x/ cos 3x C .26
2u/
2u
32x/ sin 3x
32x/ sin 3x. Let
D .A0 C A1 x/ cos 3x C .B0 C B1 x/ sin 3xI then D .A1 C 3B0 C 3B1x/ cos 3x C .B1 3A0 3A1x/ sin 3x
up u0p u00p
u00p
1 , 2
. Then
y 00 C 3y 0
if u00
sin 2x/; and yp D
65
D .6B1
u0p
4up
13A1 3A1
3B1 13B1
9A0
9A1 x/ cos 3x
.2A1 C 9B0 C 9B1 x/ sin 3x; so
D
Œ13A0 C A1 C 3B0 6B1 C .13A1 C 3B1 /x cos 3x Œ13B0 C B1 3A0 C 6A1 C .13B1 3A1 /x sin 3x D .4 C 20x/ cos 3x C .26 32x/ sin 3x if D D
20 and 32
13A0 3A0
3B0 13B0
A1 C 6B1 6A1 B1
D 4 D 26:
From the first two equations, A1 D 2, B1 D 2. Substituting these in the last two equations yields 13A0 3B0 D 10, 3A0 13B0 D 16. Solving this pair yields A0 D 1, B0 D 1. Therefore, up D .1
2x/.cos 3x
sin 3x/ and yp D e
2x
.1
2x/.cos 3x
sin 3x/:
5.5.8. Let yp yp0 yp00
D .A0 x C A1 x 2 / cos x C .B0 x C B1 x 2 / sin xI then D A0 C .2A1 C B0 /x C B1 x 2 cos x C B0 C .2B1 D 2A1 C 2B0 .A0 4B1/x A1 x 2 cos x C 2B1 2A0 .B0 C 4A1 /x B1x 2 sin x; so
yp00 C yp
D .2A1 C 2B0 C 4B1 x/ cos x C .2B1
D . 4 C 8x/ cos x C .8
A0 /x
2A0
B1 x 2 sin x
4A1 x/ sin x
4x/ sin x
if 4B1 D 8, 4A1 D 4, 2B0 C 2A1 D 4, 2A0 C 2B1 D 8. Therefore, A1 D 1, B1 D 2, A0 D 2, B0 D 3, and yp D x Œ.2 x/ cos x C .3 2x/ sin x. 5.5.10. Let y D ue
x
. Then
y 00 C 2y 0 C 2y
D e D e
x
.u00 2u0 C u/ C 2.u0 x 00 .u C u/ D e x .8 cos x
u/ C 2u 6 sin x/
66 Chapter 5 Linear Second Order Equations if u00 C u D 8 cos x
6 sin x. Now let up u0p u00p
D Ax cos x C Bx sin xI then D .A C Bx/ cos x C .B Ax/ sin x
D .2B
Ax/ cos x
u00p C up D 2B cos x
.2A C Bx/ sin x; so
2A sin x D 8 cos x
6 sin x
if 2B D 8, 2A D 6. Therefore,A D 3, B D 4, up D x.3 cos x C 4 sin x/, and yp D xe 4 sin x/.
x
.3 cos x C
5.5.12. Let yp yp0
D D
yp00
D
yp00 C 2yp0 C yp
.A0 C A1 x C A2 x 2 / cos x C .B0 C B1 x C B2 x 2 / sin xI then A1 C B0 C .2A2 C B1/x C B2 x 2 cos x C B1 A0 C .2B2 A1 /x A2 x 2 sin x; A0 C 2A2 C 2B1 .A1 4B2/x A2 x 2 cos x C B0 C 2B2 2A1 .B1 C 4A2 /x B2 x 2 sin x; so
D 2 A1 C A2 C B0 C B1 C .2A2 C B1 C 2B2/x C B2 x 2 cos x C2 B1 C B2 A0 A1 C .2B2 A1 2A2 /x A2 x 2 sin x
D 8x 2 cos x (i)
2B2 2A2
4x sin x if
D 8 ; (ii) D 0
2B1 C 4A2 C 4B2 2A1 4A2 C 4B2
2B0 C 2A1 C 2B1 C 2A2 2A0 2A1 C 2B1 C 2B2
(iii)
D D
D D
0 ; 4
0 : 0
From (i), A2 D 0, B2 D 4. Substituting these into (ii) and solving for A1 and B1 yields A1 D 10, B1 D 8. Substituting the known coefficients into (iii) and solving for A0 and B0 yields A0 D 14, B0 D 2. Therefore,yp D .14 10x/ cos x .2 C 8x 4x 2/ sin x. 5.5.14. Let yp yp0 yp00
D .A0 C A1 x C A2 x 2 / cos 2x C .B0 C B1 x C B2 x 2 / sin 2xI then D A1 C 2B0 C .2A2 C 2B1/x C 2B2x 2 cos 2x C B1 2A0 C .2B2 2A1 /x 2A2 x 2 sin 2x D 4A0 C 2A2 C 4B1 .4A1 8B2 /x 4A2x 2 cos 2x C 4B0 2B2 4A1 .4B1 C 8A2 /x 4B2 x 2 sin 2x; so
yp00 C 3yp0 C 2yp
D Π2A0 C 3A1 C 4A2 C 6B0 C 4B1 .2A1 6A2 6B1 8B2 /x C Π2B0 C 3B1 C 4B2 6A0 .2B1
D .1
x
6B2 C 6A1 C 8A2 /x
4x 2 / cos 2x
.2A2 4A1
6B2 /x 2 cos 2x
.2B2 C 6A2 /x 2 sin 2x
.1 C 7x C 2x 2 / sin 2x if
Section 5.5 The Method of Undetermined Coefficients II 2A2 C 6B2 6A2 2B2
(i)
D D
4 ; (ii) 2
2A1 C 6B1 C 6A2 C 8B2 6A1 2B1 8A2 C 6B2
2A0 C 6B0 C 3A1 C 4B1 C 2A2 6A0 2B0 4A0 C 3B1 C 2B2
(iii)
D D
D D
67
1 ; 7
1 : 1
1 1 . Substituting these into (ii) and solving for A1 and B1 yields A1 D 0, From (i), A2 D , B2 D 2 2 B1 D 0. Substituting the known coefficients into (iii) and solving for A0 and B0 yields A0 D 0, B0 D 0. x2 .cos 2x sin 2x/. Therefore, yp D 2 5.5.16. Let y D ue x . Then y 00
if u00 D .3 C 4x up u0p u00p
2y 0 C y
D e x .u00 C 2u0 C u/ 2.u0 C u/ C u D e x u00 D e x .3 C 4x x 2 / cos x C .3 4x x 2 / sin x
x 2 / cos x
.3
x 2 / sin x. Now let
4x
D .A0 C A1 x C A2 x 2 / cos x C .B0 C B1 x C B2x 2 / sin xI then D A1 C B0 C .2A2 C B1 /x C B2 x 2 cos x C B1 A0 C .2B2 A1 /x A2 x 2 sin x; D A0 C 2A2 C 2B1 .A1 4B2 /x A2 x 2 cos x C B0 C 2B2 2A1 .B1 C 4A2 /x B2x 2 sin x
D
.3 C 4x
A2 B2
(i)
x 2/ cos x D D
(iii)
1 ; (ii) 1
.3
4x
x 2 / sin x if
A1 C 4B2 B1 4A2
A0 C 2B1 C 2A2 B0 2A1 C 2B2
D D
D D
4 ; 4
3 : 3
From (i), A2 D 1, B2 D 1. Substituting these into (ii) and solving for A1 and B1 yields A1 D 0, B1 D 0. Substituting the known coefficients into (iii) and solving for A0 and B0 yields A0 D 1, B0 D 1. Therefore,up D .1 x 2 /.cos x C sin x/ and yp D e x .1 x 2 /.cos x C sin x/. 5.5.18. Let y D ue
x
. Then
y 00 C 2y 0 C y if u00 D .5
2x/ cos x up u0p u00p
D e D e
x
.u00 2u0 C u/ C 2.u0 u/ C u x 00 u D e x Œ.5 2x/ cos x .3 C 3x/ sin x
.3 C 3x/ sin x. Let D
D D D
.A0 C A1 x/ cos x C .B0 C B1 x/ sin xI then
.A1 C B0 C B1 x/ cos x C .B1 A0 A1 x/ sin x .2B1 A0 A1 x/ cos x .2A1 C B0 C B1 x/ sin x .5 2x/ cos x .3 C 3x/ sin x
if A1 D 2, B1 D 3, A0 C 2B1 D 5, B0 2A1 =-3. Therefore, A1 D 2, B1 D 3, A0 D 1, B0 D 1, up D e x Œ.1 C 2x/ cos x .1 3x/ sin x, and yp D e x Œ.1 C 2x/ cos x .1 3x/ sin x.
68 Chapter 5 Linear Second Order Equations 5.5.20. Let D .A0 x C A1 x 2 C A2 x 3/ cos x C .B0 x C B1 x 2 C B2 x 3 / sin xI then D A0 C .2A1 C B0 /x C .3A2 C B1/x 2 C B2 x 3 cos x C B0 C .2B1 A0 /x C .3B2 A1 /x 2 A2 x 3 sin x D 2A1 C 2B0 .A0 6A2 4B1/x .A1 6B2/x 2 A2 x 3 cos x C 2B1 2A0 .B0 C 6B2 C 4A1 /x .B1 C 6A2 /x 2 B2 x 3 sin x; so
yp yp0 yp00
yp00 C yp
2A1 C 2B0 C .6A2 C 4B1 /x C 6B2 x 2 cos x C 2B1 2A0 C .6B2 4A1 /x 6A2 x 2 sin x
D
.2 C 2x/ cos x C .4 C 6x 2 / sin x if
D 6B2 6A2
(i)
D D
0 ; (ii) 6
4B1 C 6A2 4A1 C 6B2
D 2 ; (iii) D 0
2B0 C 2A1 2A0 C 2B1
D 2 : D 4
From (i), A2 D 1, B2 D 0. Substituting these into (ii) and solving for A1 and B1 yields A1 D 0, B1 D 2. Substituting the known coefficients into (iii) and solving for A0 and B0 yields A0 D 0, B0 D 2. Therefore,yp D x 3 cos x C .x C 2x 2 / sin x. 5.5.22. Let y D ue x . Then y 00
if u00
7y 0 C 6y
D e x .u00 C 2u0 C u/ 7.u0 C u/ C 6u D e x .u00 5u0 / D e x .17 cos x 7 sin x/
5u0 D 17 cos x C 7 sin x. Now let up D A cos x C B sin x. Then u00p
5u0p
D
.A cos x C B sin x/
D . A
5B/ cos x
.B
5. A sin x C B cos x/
5A/ sin x D 17 cos x C 7 sin x
if A 5B D 17, 5A B D 7. Therefore,A D 2, B D 3, up D 2 cos x C 3 sin x, and yp D e x .2 cos x C 3 sin x/. The characteristic polynomial of the complementary equation is p.r / D r 2 7r C 6 D .r 1/.r 6/, so fe x ; e 6x g is a fundamental set of solutions of the complementary equation. Therefore, (A) y D e x .2 cos x C 3 sin x/ C c1 e x C c2 e 6x is the general solution of the nonhomogeneous equation. Differentiating (A) yields y 0 D e x .2 cos x C 3 sin x/C e x . 2 sin x C 3 cos x/C c1 e x C 6c2e 6x , so y.0/ D 4; y 0 .0/ D 2 ) 4 D 2 C c1 C c2; 2 D 2 C 3 C c1 C 6c2 ) c1 C c2 D 2; c1 C 6c2 D 3, so c1 D 3, c2 D 1, and y D e x .2 cos x C 3 sin x/ C 3e x e 6x . 5.5.24. Let y D ue x . Then y 00 C 6y 0 C 10y
D e x .u00 C 2u0 C u/ C 6.u0 C u/ C 10u D e x .u00 C 8u0 C 17u/ D 40e x sin x
if u00 C 8u0 C 17u D 40 sin x. Let up D A cos x C B sin x. Then u00p C 6u0p C 17up
D
.A cos x C B sin x/ C 8. A sin x C B cos x/
C17.A cos x C B sin x/ D .16A C 8B/ cos x .8A
16B/ sin x D 40 sin x
Section 5.5 The Method of Undetermined Coefficients II
69
if 16A C 8B D 0, 8A C 16B D 40. Therefore,A D 1, B D 2, and yp D e x .cos x 2 sin x/. The characteristic polynomial of the complementary equation is p.r / D r 2 C 6r C 10 D .r C 3/2 C 1, so fe 3x cos x; e 3x sin xg is a fundamental set of solutions of the complementary equation, and (A) y D e x .cos x 2 sin x/ C e 3x .c1 cos x C c2 sin x/ is the general solution of the nonhomogeneous equation. Therefore,y.0/ D 2 ) 2 D 1 C c1 , so c1 D 1. Differentiating (A) yields y 0 D e x .cos x 2 sin x/ e x .sin x C2 cos x/ 3e 3x .c1 cos x Cc2 sin x/Ce 3x . c1 sin x Cc2 cos x/. Therefore,y 0 .0/ D 3 ) 3 D 1 2 3c1 C c2, so c2 D 1, and y D e x .cos x 2 sin x/ C e 3x .cos x C sin x/. 5.5.26. Let y D ue 3x . Then y 00
3y 0 C 2y
D e 3x .u00 C 6u0 C 9u/
D e 3x .u00 C 3u0 C 2u/ D e 3x Œ21 cos x
if u00 C 3u0 C 2u D 21 cos x up u0p u00p
D D D
3.u0 C 3u/ C 2u
.11 C 10x/ sin x
.11 C 10x/ sin x. Now let .A0 C A1 x/ cos x C .B0 C B1 x/ sin xI then .A1 C B0 C B1 x/ cos x C .B1 A0 A1 x/ sin x .2B1 A0 A1 x/ cos x .2A1 C B0 C B1 x/ sin x; so
u00 C 3u0 C 2u D ŒA0 C 3A1 C 3B0 C 2B1 C .A1 C 3B1/x cos x C ŒB0 C 3B1 3A0 2A1 C .B1 3A1 /x sin x D 21 cos x .11 C 10x/ sin x if A1 C 3B1 3A1 C B1
D D
0 and 10
A0 C 3B0 C 3A1 C 2B1 3A0 C B0 2A1 C 3B1
D D
21 : 11
From the first two equations A1 D 3, B1 D 1. Substituting these in last two equations yields and solving for A0 and B0 yields A0 D 2, B0 D 4. Therefore, up D .2 C 3x/ cos x C .4 x/ sin x and yp D e 3x Œ.2 C 3x/ cos x C .4 x/ sin x. The characteristic polynomial of the complementary equation is p.r / D r 2 3r C2 D .r 1/.r 2/, so fe x ; e 2x g is a fundamental set of solutions of the complementary equation, and (A) y D e 3x Œ.2 C 3x/ cos x C .4 x/ sin x C c1 e x C c2 e 2x is the general solution of the nonhomogeneous equation. Differentiating (A) yields y0
D 3e 3x Œ.2 C 3x/ cos x C .4 x/ sin x Ce 3x Œ.7 x/ cos x .3 C 3x/ sin x C c1 e x C 2c2e 2x :
Therefore,y.0/ D 0; y 0 .0/ D 6 ) 0 D 2 C c1 C c2 ; 6 D 6 C 7 C c1 C 2c2, so c1 C c2 D 2; c1 C 2c2 D 7. Therefore, c1 D 3, c2 D 5, and y D e 3x Œ.2 C 3x/ cos x C .4 x/ sin x C 3e x 5e 2x . 5.5.28. We must find particular solutions yp1 , yp2 , and yp3 of (A) y 00 C y D 4 cos x 2 sin x and (B) y 00 C y D xe x , and (C) y 00 C y D e x , respectively. To find a particular solution of (A) we write yp1 yp0 1 yp001
D D D
Ax cos x C Bx sin xI then .A C Bx/ cos x C .B Ax/ sin x .2B Ax/ cos x .2A C Bx/ sin x; so
yp001 C yp1 D 2B cos x 2A sin x D 4 cos x 2 sin x if 2B D 4, 2A D 2. Therefore, A D 1, B D 2, and yp1 D x.cos x C 2 sin x/. To find a particular solution of (B) we write y D ue x . Then y 00 C y D e x .u00 C 2u0 C u/ C u D e x .u00 C 2u0 C 2u/ D xe x
70 Chapter 5 Linear Second Order Equations if u00 C 2u0 C 2u D x. Now up D A C Bx, where 2B C 2.A C Bx/ D x. Therefore, 2B D 1, 1 1 1 ex 2A C 2B D 0, so B D , A D , up D .1 x/, and yp2 D .1 x/. To find a particular 2 2 2 2 1 solution of (C) we write yp3 D Ae x . Then yp003 C yp3 D 2Ae x D e x if 2A D 1, so A D and 2 e x ex e x Now yp D yp1 C yp2 C yp3 D x.cos x C 2 sin x/ .1 x/ C . yp3 D 2 2 2 5.5.30. We must find particular solutions yp1 , yp2 and yp3 of (A) y 00 2y 0 C 2y D 4xe x cos x, (B) y 00 2y 0 C 2y D xe x , and (C) y 00 2y 0 C 2y D 1 C x 2 , respectively. To find a particular solution of (A) we write y D ue x . Then y 00 2y 0 C 2y D e x Œ.u00 C 2u0 C u/ 2.u0 C u/ C 2u D e x .u00 C u/ D 4xe x cos x if u00 C u D 4x cos x. Now let up u0p u00p
D .A0 x C A1 x 2 / cos x C .B0 x C B1 x 2 / sin xI then D A0 C .2A1 C B0 /x C B1 x 2 cos x C B0 C .2B1 D 2A1 C 2B0 .A0 4B1 /x A1 x 2 cos x C 2B1 2A0 .B0 C 4A1 /x B1 x 2 sin x; so
u00p C up
D .2A1 C 2B0 C 4B1 x/ cos x C .2B1
A0 /x
2A0
B1x 2 sin x
4A1 x/ sin x
D 4x cos x
if 4B1 D 4, 4A1 D 0, 2B0 C 2A1 D 0, 2A0 C 2B1 D 0. Therefore, A1 D 0, B1 D 1, A0 D 1, B0 D 0, up D x.cos x C x sin x/, and yp1 D xe x .cos x C x sin x/. To find a particular solution of (B) we write y D ue x . Then y 00 2y 0 C 2y D e x .u00 2u0 C u/ 2.u0 u/ C 2u D e x .u00 4u0 C 5u/ D xe x
if u00
4u0 C 5u D x. Now up D A C Bx where 4B C 5.A C Bx/ D x. Therefore, 5B D 1, 4 1 e x 1 5A 4B D 0, B D , A D , up D .4 C 5x/, and yp2 D .4 C 5x/. To find a particular 5 25 25 25 2 solution of (C) we write yp3 D A C Bx C Cx . Then yp003
2yp0 3 C 2yp3
D 2C
D .2A
2.B C 2Cx/ C 2.A C Bx C Cx 2 / 2B C 2C / C .2B
4C /x C 2Cx 2 D 1 C x 2
1 x2 , B D 1, A D 1, and yp3 D 1 C x C . 2 2 x 2 e x .4 C 5x/ C 1 C x C . D xe x .cos x C x sin x/ C 25 2
if 2A 2B C 2C D 1, 2B 4C D 0, 2C D 1. Therefore,C D Now yp D yp1 C yp2 C yp3
5.5.32. We must find particular solutions yp1 and yp2 of (A) y 00 4y 0 C4y D 6e 2x and (B) y 00 4y 0 C4y D 25 sin x, respectively. To find a particular solution of (A), let y D ue 2x . Then y 00 4y 0 C 4y D e 2x .u00 C 4u0 C 4u/ 4.u0 C 2u/ C 4u D e 2x u00 D 6e 2x
if u00 D 6. Integrating twice and taking the constants of integration to be zero yields up D 3x 2 , so yp1 D 3x 2e 2x . To find a particular solution of (B), let yp2 D A cos x C B sin x. Then yp002
4yp0 2 C 4yp2
D
.A cos x C B sin x/ 4. A sin x C B cos x/ C4.A cos x C B sin x/ D .3A 4B/ cos x C .4A C 3B/ sin x D 25 sin x
Section 5.5 The Method of Undetermined Coefficients II
71
if 3A 4B D 0, 4A C 3B D 25. Therefore,A D 4, B D 3, and yp2 D 4 cos x C 3 sin x. Now yp D yp1 C yp2 D 3x 2 e 2x C 4 cos x C 3 sin x. The characteristic polynomial of the complementary equation is p.r / D r 2 4r C 4 D .r 2/2 , so fe 2x ; xe 2x g is a fundamental set of solutions of the complementary equation. Therefore,(C) y D 3x 2 e 2x C 4 cos x C 3 sin x C e 2x .c1 C c2 x/ is the general solution of the nonhomogeneous equation. Now y.0/ D 5 ) 5 D 4 C c1 , so c1 D 1. Differentiating (C) yields y 0 D 6e 2x .x C x 2/ 4 sin x C 3 cos x C 2e 2x .c1 C c2x/ C c2 e 2x , so y 0 .0/ D 3 ) 3 D 3 C 2 C c2. Therefore,c2 D 2, and y D .1 2x C 3x 2/e 2x C 4 cos x C 3 sin x. 5.5.34. We must find particular solutions yp1 and yp2 of (A) y 00 C 4y 0 C 4y D 2 cos 2x C 3 sin 2x and (B) y 00 C 4y 0 C 4y D e x , respectively. To find a particular solution of (A) we write yp1 D A cos 2x C B sin 2x. Then yp001 C 4yp0 1 C 4yp1
D
4.A cos 2x C B sin 2x/ C 8. A sin 2x C B cos 2x/ C4.A cos 2x C B sin 2x/ D 8A sin 2x C 8B cos 2x
D 2 cos 2x C 3 sin 2x
1 3 1 3 , B D , and yp1 D cos 2x C sin 2x. To find a 8 x 4 8 4 particular solution of (B) we write yp2 D Ae . Then yp002 C 4yp0 2 C 4yp2 D A.1 4 C 4/e x D 3 1 Ae x D e x if A D 1. Therefore,yp2 D e x . Now yp D yp1 C yp2 D cos 2x C sin 2x C e x . 8 4 The characteristic polynomial of the complementary equation is p.r / D r 2 C 4r C 4 D .r 2/2 , so fe 2x ; xe 2x g is a fundamental set of solutions of the complementary equation. Therefore,(C) y D 3 1 cos 2x C sin 2x C e x C e 2x .c1 C c2 x/ is the general solution of the nonhomogeneous equation. 8 4 13 3 3 Now y.0/ D 1 ) 1 D C 1 C c1 , so c1 D . Differentiating (C) yields y 0 D sin 2x C 8 8 4 1 1 3 x 2x 2x 0 cos 2x e 2e .c1 C c2x/C c2 e , so y .0/ D 2 ) 2 D 1 2c1 C c2. Therefore,c2 D , 2 2 4 1 13 2x 3 3 and y D cos 2x C sin 2x C e x e xe 2x . 8 4 8 4
if 8B D 2,
8A D 3. Therefore,A D
5.5.36. (a), (b), and (c) require only routine manipulations. (d) The coefficients of sin !x in yp0 , yp00 , ayp00 Cbyp0 Ccyp , and yp00 C! 2 yp can be obtained by replacing A by B and B by A in the corresponding coefficients of cos !x. 5.5.38. Let y D ue x . Then ay 00 C by 0 C cy
D e x a.u00 C 2u0 C 2 u/ C b.u0 C u/ C cu D e x au00 C .2a C b/u0 C .a2 C b C c/u D e x au00 C p 0 ./u0 C p./u D e x .P .x/ cos !x C Q.x/ sin !x/ if
(A) au00 C p 0 ./u0 C p./uP .x/ cos !x C Q.x/ sin !x, where p.r / D ar C br C c is that characteristic polynomial of the complementary equation (B) ay 00 C by 0 C cy D 0. If e x cos !x and e x sin !x are not solutions of (B), then cos !x and sin !x are not solutions of the complementary equation for (A). Then Theorem 5.5.1 implies that (A) has a particular solution up D .A0 C A1 x C C Ak x k / cos !x C .B0 C B1 x C C Bk x k / sin !x; and yp D up e x is a particular solution of the stated form for the given equation. If e x cos !x and e x sin !x are solutions of (B), then cos !x and sin !x are solutions of the complementary equation for
72 Chapter 5 Linear Second Order Equations (A). Then Theorem 5.5.1 implies that (A) has a particular solution up D .A0 x C A1 x 2 C C Ak x kC1 / cos !x C .B0 x C B1 x 2 C C Bk x kC1 / sin !x; and yp D up e x is a particular solution of the stated form for the given equation. R 5.5.40. (a) Let y D x 2 cos x dx; then y 0 D x 2 cos x Now let
(i)
.A0 C A1 x C A2 x 2 / cos x C .B0 C B1 x C B2 x 2 / sin xI then A1 C B0 C .2A2 C B1/x C B2 x 2 cos x C B1 A0 C .2B2 A1 /x A2 x 2 sin x D x 2 cos x if
D D
yp yp0
D 1 ; (ii) D 0
B2 A2
B1 C 2A2 A1 C 2B2
D 0 ; (iii) D 0
B0 C A1 A0 C B1
D D
0 : 0
Solving these equations yields A2 D 0, B2 D 1, A1 D 2, B1 D 0, A0 D 0, B0 D 2. Therefore,yp D 2x cos x .2 Rx 2 / sin x and y D 2x cos x .2 x 2 / sin x C c. (b) Let y D x 2 e x cos x dx D ue x ; then y 0 D .u0 C u/e x D x 2 e x cos x if u0 C u D x 2 cos x. Now let D .A0 C A1 x C A2 x 2 / cos x C .B0 C B1 x C B2x 2 / sin xI then D A1 C B0 C .2A2 C B1 /x C B2 x 2 cos x C B1 A0 C .2B2 A1 /x A2 x 2 sin x; so
up u0p
u00p C up
D
A0 C A1 C B0 C .A1 C 2A2 C B1 /x C .A2 C B2 /x 2 cos x C B0 C B1 A0 C .B1 C 2B2 A1 /x C .B2 A2 /x 2 sin x
D x 2 cos x if (i)
A2 C B2 A2 C B2
D 1 ; (ii) D 0
(iii)
A1 C B1 C 2A2 A1 C B1 C 2B2
A0 C B0 C A1 A0 C B0 C B1
D 0 ; D 0
D 0 : D 0
1 1 , B2 D . Substituting these into (ii) and solving for A1 and B1 yields A1 D 0, 2 2 1 1 B1 D 1. Substituting these into (iii) and solving for A0 and B0 yields A0 D , B0 D . Therefore, 2 2 1 ex up D .1 x 2 / cos x .1 x/2 sin x and y D .1 x 2 / cos x .1 x/2 sin x . 2 2 R (c) Let y D xe x sin 2x dx D ue x ; then y 0 D .u0 u/e x D xe x sin 2x if u0 u D x sin 2x. Now let
From (i), A2 D
up
D .A0 C A1 x/ cos 2x C .B0 C B1x/ sin 2xI then
u0p
D Œ.A1 C 2B0/ C 2B1x cos 2x C Œ.B1
u00p
up
2A0 /
2A1 x sin 2x; so
D ΠA0 C A1 C 2B0 .A1 2B1/x cos 2x C ΠB0 C B1 2A0 .B1 C 2A1 /x sin 2x D x sin 2x if
Section 5.5 The Method of Undetermined Coefficients II A1 C 2B1 2A1 B1
(i)
2 From (i), A1 D , B1 D 5 3 . Therefore, B0 D 25
D 0 ; (ii) D 1
A0 C 2B0 C A1 2A0 B0 C B1
D 0 : D 0
1 . Substituting these into (ii) and solving for A0 and B0 yields A0 D 5
1 Œ.4 C 10x/ cos 2x 25
up D
.3
73
4 , 25
5x/ sin 2x C c and
e x Œ.4 C 10x/ cos 2x .3 5x/ sin 2x C c: 25 R (d) Let y D x 2 e x sin x dx D ue x ; then y 0 D .u0 u/e x D x 2 e x sin x if u0 Now let yp D
up u0p
u00p
up
u D x 2 sin x.
D .A0 C A1 x C A2 x 2 / cos x C .B0 C B1 x C B2x 2 / sin xI then D A1 C B0 C .2A2 C B1 /x C B2 x 2 cos x C B1 A0 C .2B2 A1 /x A2 x 2 sin x; so D
A0 C A1 C B0 .A1 2A2 B1/x .A2 B2 /x 2 cos x C B0 C B1 A0 .B1 2B2 C A1 /x .B2 C A2 /x 2 sin x
D x 2 sin x if (i)
A2 C B2 A2 B2
D 0 ; (ii) D 1
(iii)
A1 C B1 C 2A2 A1 B1 C 2B2
A0 C B0 C A1 A0 B0 C B1
D 0 ; D 0
D 0 : D 0
1 1 , B2 D . Substituting these into (ii) and solving for A1 and B1 yields A1 D 1, 2 2 1 1 B1 D 0. Substituting these into (iii) and solving for A0 and B0 yields A0 D , B0 D . Therefore, 2 2
From (i), A2 D
up D yD
(e) Let y D up u0p
R
x
e 2 e
x
.1 C x/2 cos x
.1 C x/2 cos x
.1 .1
x 2 / sin x and
x 2 / sin x C c:
2 x 3 e x sin x dx D ue x ; then y 0 D .u0 C u/e x D x 3 e x sin x if u0 C u D x 3 sin x. Now let
D .A0 C A1 x C A2 x 2 C A3 x 3 / cos x C .B0 C B1 x C B2 x 2 C B3x 3 / sin xI then D A1 C B0 C .2A2 C B1 /x C .3A3 C B2 /x 2 C B3 x 3 cos x C B1 A0 C .2B2 A1 /x C .3B3 A2 /x 2 A3 x 3 sin x; so u00p C up
D ŒA0 C A1 C B0 C .A1 C 2A2 C B1 /x
C.A2 C 3A3 C B2 /x 2 C .A3 C B3 /x 3 cos x C ŒB0 C B1 A0 C .B1 C 2B2 A1 /x C.B2 C 3B3 A2 /x 2 C .B3 A3 /x 3 sin x D x 3 sin x if
74 Chapter 5 Linear Second Order Equations A3 C B3 A3 C B3
(i)
D 0 ; (ii) D 1
A1 C B1 C 2A2 A1 C B1 C 2B2
(iii)
A2 C B2 C 3A3 A2 C B2 C 3B3
D 0 ; (iv) D 0
D 0 ; D 0
A0 C B0 C A1 A0 C B0 C B1
D 0 : D 0
1 1 3 , B3 D . Substituting these into (ii) and solving for A2 and B2 yields A2 D , 2 2 2 3 3 , B1 D . Substituting B2 D 0. Substituting these into (iii) and solving for A1 and B1 yields A1 D 2 2 3 these into (iv) and solving for A0 and B0 yields A0 D 0, B0 D . Therefore, 2 From (i), A3 D
1 x.3 2
up D
3x C x 2 / cos x
.3
3x C x 3 / sin x and
ex x.3 3x C x 2 / cos x .3 3x C x 3 / sin x C c: 2 R (f) Let y D e x Œx cos x .1 C 3x/ sin x dx D ue x ; then y 0 D .u0 Cu/e x D e x Œx cos x if u0 C u D x cos x .1 C 3x/ sin x. Now let yD
D D
up u0p
.A0 C A1 x/ cos x C .B0 C B1 x/ sin xI then ŒA1 C B0 C B1 x cos x C ŒB1 A0 A1 x sin x; so
u00p C up
(i)
.1 C 3x/ sin x
D ŒA0 C A1 C B0 C .A1 C B1 /x cos x
C ŒB0 C B1 A0 C .B1 A1 /x sin x D x cos x .1 C 3x/ sin x if
A1 C B1 A1 C B1
D D
1 ; (ii) 3
A0 C B0 C A1 A0 C B0 C B1
D D
0 : 1
From (i), A1 D 2, B1 D 1. Substituting these into (ii) and solving for A0 and B0 yields A0 D 1, B0 D 1. Therefore, up D Œ.1 2x/ cos x C .1 C x/ sin x and y D e x Œ.1 2x/ cos x C .1 C x/ sin xC c. R (g) Let y D e x .1 C x 2/ cos x C .1 x 2 / sin x dx D ue x ; then y 0 D .u0 u/e x D e x .1 C x 2 / cos x C .1 x 2 / sin x
if u0
u D .1 C x 2 / cos x C .1 up u0p
u00p
up
x 2 / sin x. Now let
D .A0 C A1 x C A2 x 2 / cos x C .B0 C B1 x C B2x 2 / sin xI then D A1 C B0 C .2A2 C B1 /x C B2 x 2 cos x C B1 A0 C .2B2 A1 /x A2 x 2 sin x; so D
A0 C A1 C B0 .A1 2A2 B1/x .A2 B2 /x 2 cos x C B0 C B1 A0 .B1 2B2 C A1 /x .B2 C A2 /x 2 sin x
D .1 C x 2 / cos x C .1 (i)
A2 C B2 A2 B2
D D
x 2 / sin x if
1 ; (ii) 1
A1 C B1 C 2A2 A1 B1 C 2B2
D 0 ; D 0
Section 5.6 Reduction of Order (iii)
A0 C B0 C A1 A0 B0 C B1
75
D 1 : D 1
From (i), A2 D 0, B2 D 1. Substituting these into (ii) and solving for A1 and B1 yields A1 D 1, B1 D 1. Substituting these into (iii) and solving for A0 and B0 yields A0 D 0, B0 D 0. Therefore,up D x cos x C x.1 C x/ sin x and y D e x Œx cos x C x.1 C x/ sin x C c. 5.6 REDUCTION OF ORDER (N OTE: The term uy100 is indicated by “ " in some of the following solutions, where y100 is complicated. Since this term always drops out of the differential equation for u, it is not necessary to include it.) 5.6.2. If y D ux, then y 0 D u0 x C u and y 00 D u00Zx C 2u0 , so x 2 y 00 C xy 0 y D x 3 u00 C 3x 2u0 D 3 4 3 1 4 if u0 D ´, where (A) ´0 C ´ D 5 . Since dx D 3 ln jxj, ´1 D 3 is a solution of the x2 x x x x v complementary equation for (A). Therefore,the solutions of (A) are of the form (B) ´ D 3 , where x v0 4 4 4 4 C1 4 C1 0 0 D 5 , so v D 2 . Hence, v D C C1 ; u D ´ D C 3 (see (B)); u D C C2 ; x3 x x x x4 x 3x 3 2x 2 C1 4 c2 4 y D ux D C C2 x, or y D C c1 x C . As a byproduct, fx; 1=xg is a fundamental set 3x 2 2x 3x 2 x of solutions of the complementary equation. 5.6.4. If y D ue 2x , then y 0 D .u0 C 2u/e 2x and y 00 D .u00 C 4u0 C 4u/e 2x , so y 00 3y 0 C 2y D 1 e 2x 0 0 .u00 C u0 /e 2x D if u D ´, where (A) ´ C ´ D . Since ´1 D e x is a solution of 1Ce x 1Ce x the complementary equation for (A), the solutions of (A) are of the form (B) ´ D ve x , where v 0 e x D e 2x e x , so v 0 D . Hence, v D ln.1 C e x / C C1 ; u0 D ´ D e x ln.1 C e x / C C1 e x x 1Ce 1Ce x (see (B)); u D .1 C e x / ln.1 C e x / 1 e x C1 e x C C2 ; y D ue 2x D .e 2x C e x / ln.1 C e x / .C1 C 1/e x C .C2 1/e 2x , or y D .e 2x C e x / ln.1 C e x / C c1 e 2x C c2e x . As a byproduct, fe 2x ; e x g is a fundamental set of solutions of the complementary equation. ! ! x 1=2 x x 1=2 x 1=2 1=2 1=2 x 0 0 1=2 x 00 00 1=2 x 0 5.6.6. If y D ux e , then y D u x e Cu x C e and y D u x e C2u x C e C 2 2 so 4x 2 y 00 C .4x
8x 2 /y 0 C .4x 2 4x 1/yZD e x .4x 5=2 u00 C 8x 3=2 u0 / D 4x 1=2e x .1 C 4x/ if 2 1 C 4x 2 2 u0 D ´, where (A) ´0 C ´ D . Since dx D 2 ln jxj, ´1 D 2 is a solution of the x x2 x x v complementary equation for (A). Therefore,the solutions of (A) are of the form (B) ´ D 2 , where x v0 1 C 4x 1 C1 0 2 0 D , so v D 1 C 4x. Hence, v D x C 2x C C1 ; u D ´ D C 2 C 2 (see x2 x2 x x C1 1=2 x x 3=2 1=2 1=2 (B)); u D ln x C 2x C C2 ; y D ux e D e .2x C x ln x C1 x C C2 x 1=2/, or x y D e x .2x 3=2 C x 1=2 ln x C c1x 1=2 C c2x 1=2 /. As a byproduct, fx 1=2 e x ; x 1=2e x g is a fundamental set of solutions of the complementary equation. 2
2
2
2
2
5.6.8. If y D ue x , then y 0 D u0 e x 2xue x and y 00 D u00 e x 4xu0 e x C , so y 00 C 4xy 0 C 2 2 .4x 2 C 2/y D u00 e x D 8e x.xC2/ D 8e x e 2x if u00 D 8e 2x . Therefore,u0 D 4e 2x C C1 ; 2 2 2 u D 2e 2x C C1 x C C2 , and y D ue x D e x .2e 2x C C1 x C C2 /, or y D e x .2e 2x C c1 C c2x/. 2 2 As a byproduct, fe x ; xe x g is a fundamental set of solutions of the complementary equation.
76 Chapter 5 Linear Second Order Equations 5.6.10. If y D uxe
x
, then y 0 D u0 xe
x
x
1/ and y 00 D u00 xe
x
2u0 e
x
1/ C , 3x e so x 2 y 00 C 2x.x 1/y 0 C .x 2 2x C 2/y D x 3u00 D x 3 e 2x if u00 D e 3x . Therefore,u0 D C C1 ; 3 3x 2x 2x e xe xe C C1 x C C2 , and y D uxe x D C xe x .C1 x C C2 /, or y D C xe x .c1 C c2 x/. uD 9 9 9 As a byproduct, fxe x ; x 2 e x g is a fundamental set of solutions of the complementary equation. ue
.x
.x
5.6.12. If y D ue x , then y 0 D .u0 C u/e x and y 00 D .u00 C 2u0 C u/e x , so .1
2x/y 00 C 2y 0 C 4 4x .2x 3/y D e x Œ.1 2x/u00 C .4 4x/u0 D .1 4x C 4x 2/e x if u0 D ´, where (A) ´0 C ´D 1 2x Z Z 4 4x 2 1 2x. Since dx D 2C dx D 2x ln j1 2xj, ´1 D .1 2x/e 2x is 1 2x 1 2x a solution of the complementary equation for (A). Therefore,the solutions of (A) are of the form (B) e 2x ´ D v.1 2x/e 2x , where v 0 .1 2x/e 2x D .1 2x/, so v 0 D e 2x . Hence, v D C C1 ; 2 1 .2x 1/2 u0 D ´ D C C1 e 2x .1 2x/ (see (B)); u D C C1 xe 2x C C2 ; y D ue x D 2 8 .2x 1/2 e x .2x 1/2 e x C C1 xe x C C2 e x , or y D C c1 e x C c2 xe x . As a byproduct, fe x ; xe x g 8 8 is a fundamental set of solutions of the complementary equation. , so 2xy 00 C .4x C 1 3 1/y 0 C .2x C 1/y D e x .2xu00 C u0 / D 3x 1=2 e x if u0 D ´, where (A) ´0 C ´ D x 1=2 . Since 2x 2 Z 1 1 dx D ln jxj, ´1 D x 1=2 is a solution of the complementary equation for (A). Therefore,the 2x 2 3 3 solutions of (A) are of the form (B) ´ D vx 1=2, where v 0 x 1=2 D x 1=2, so v 0 D . Hence, 2 2 3x 3 1=2 1=2 0 3=2 1=2 v D C C1 ; u D ´ D x C C1 x (see (B)); u D x C 2C1 x C C2 ; y D ue x D 2 2 e x .x 3=2 C 2C1 x 1=2 C C2 /, or y D e x .x 3=2 C c1 C c2x 1=2 / As a byproduct, is a fe x ; x 1=2e x g fundamental set of solutions of the complementary equation. 5.6.14. If y D ue
x
, then y 0 D .u0
u/e
x
and y 00 D .u00
2u0 C u/e
x
u u0 00 00 1=2 and y D u x C C so 4x 2 y 00 2x 1=2 x 1=2 4x.x C 1/y 0 C .2x C 3/y D 4x 5=2.u00 u0 / D 4x 5=2e 2x if u0 D ´, where (A) ´0 ´ D e 2x . Since ´1 D e x is a solution of the complementary equation for (A), the solutions of (A) are of the form (B) x 0 2x ´ D ve x , where v 0 e x D e 2x , so v 0 D e x . Hence, C1 e x (see (B)); 2x v D e C C1; u D ´ D e C 2x 2x e e e uD C C1 e x C C2 ; y D ux 1=2 D x 1=2 C C1 e x C C2 , or y D x 1=2 C c1 C c2 e x . 2 2 2 As a byproduct, fx 1=2; x 1=2e x g is a fundamental set of solutions of the complementary equation. 5.6.16. If y D ux 1=2, then y 0 D u0 x 1=2 C
5.6.18. If y D ue x , then y 0 D .u0 C u/e x and y 00 D .u00 C 2u0 C u/e x , so xy 00 C .2 2x/y 0 C u00 2 C1 C1 .x 2/y D e x .xu00 C 2u0 / D 0 if 0 D ; ln ju0 j D 2 ln jxj C k; u0 D 2 ; u D C C2 . u x x x C1 Therefore,y D ue x D e x C C2 is the general solution, and fe x ; e x =xg is a fundamental set of x solutions. u 2u0 and y 00 D u00 ln jxj C , so x 2 .ln jxj/2 y 00 x x .2x ln jxj/y 0 C .2 C ln jxj/y D x 2 .ln jxj/3 u00 D 0 if u00 D 0; u0 D C1 ; u D C1 x C C2 . Therefore,y D u ln jxj D .C1 x C C2 / ln jxj is the general solution, and fln jxj; x ln jxjg is a fundamental set of solutions. 5.6.20. If y D u ln jxj, then y 0 D u0 ln jxj C
Section 5.6 Reduction of Order
77
5.6.22. If y D ue x , then y 0 D u0 e x Cue x and y 00 D u00 e x C2u0 e x Cue x , so xy 00 .2xC2/y 0 C.xC2/y D u00 2 C1 x 3 e x .xu00 2u0 / D 0 if 0 D ; ln ju0 j D 2 ln jxj C k; u0 D C1 x 2 ; u D C C2 . Therefore, u x 3 3 C1 x C C2 e x is the general solution, and fe x ; x 3 e x g is a fundamental set of solutions. y D ue x D 3 5.6.24. If y D ux sin x, then y 0 D u0 x sin x C u.x cos x C sin x/ and y 00 D u00 x sin x C 2u0 .x cos x C u00 2 cos x sin x/C , so x 2 y 00 2xy 0 C.x 2 C2/y D .x 3 sin x/u00 C2.x 3 cos x/u0 D 0 if 0 D ; ln ju0 j D u sin x C1 2 ln j sin xjCk; u0 D ; u D C1 cot x CC2. Therefore, y D ux sin x D x. C1 cos x CC2 sin x/ sin2 x is the general solution, and fx sin x; x cos xg is a fundamental set of solutions. u u0 00 00 1=2 and y D u x C C so 4x 2 .sin x/y 00 2x 1=2 x 1=2 u00 cos x 4x.x cos x C sin x/y 0 C .2x cos x C 3 sin x/y D 4x 5=2.u00 sin x u0 cos x/ D 0 if 0 D ; ln ju0 j D u sin x ln j sin xj C k; u0 D C1 sin x; u D C1 cos x C C2 . Therefore,y D ux 1=2 D . C1 cos x C C2 /x 1=2 is the general solution, and fx 1=2; x 1=2 cos xg is a fundamental set of solutions. 5.6.26. If y D ux 1=2, then y 0 D u0 x 1=2 C
u u00 2u0 and y 00 D C , so .2x C 1/xy 00 2.2x 2 1/y 0 4.x C 2 x x x2 u00 4x C 4 2 1/y D .2x C 1/u00 .4x C 4/u0 D 0 if 0 D D 2C ; ln ju0 j D 2x C ln j2x C 1j C k; u 2x C 1 2x C 1 u C2 u0 D C1 .2x C 1/e 2x ; u D C1 xe 2x C C2 . Therefore,y D D C1 e 2x C is the general solution, and x x 2x f1=x; e g is a fundamental set of solutions. 5.6.28. If y D
u0 u , then y 0 D x x
5.6.30. If y D ue 2x , then y 0 D .u0 C 2u/e 2x and y 00 D .u00 C 4u0 C 4u/e 2x , so xy 00 .4x C 1/y 0 C 1 C1 x 2 u00 C C2 . .4x C 2/y D e 2x .xu00 u0 / D 0 if 0 D ; ln ju0 j D ln jxj C k; u0 D C1 x; u D u x 2 2 C1 x Therefore, y D ue 2x D e 2x C C2 is the general solution, and fe 2x ; x 2 e 2x g is a fundamental 2 set of solutions. 5.6.32. If y D ue 2x , then y 0 D .u0 C 2u/e 2x and y 00 D .u00 C 4u0 C 4u/e 2x , so .3x 1/y 00 u00 9x 6 3 .3x C 2/y 0 .6x 8/y D e 2x Œ.3x 1/u00 C .9x 6/u0 D 0 if 0 D D 3C . u 3x 1 3x 1 0 0 3x 3x Therefore,ln ju j D 3x C ln j3x 1j C k, so u D C1 .3x 1/e , u D C1 xe C C2 . Therefore,the general solution is y D ue 2x D C1 xe x C C2 e 2x , or (A) y D c1 e 2x C c2 xe x . Now y.0/ D 2 ) c1 D 2. Differentiating (A) yields y 0 D 2c1e 2x C c2 .e x xe x /. Now y 0 .0/ D 3 ) 3 D 2c1 C c2, so c2 D 1 and y D 2e 2x xe x . 5.6.34. If y D ux, then y 0 D u0 x Cu andZy 00 D u00 x C2u0 , so x 2 y 00 C2xy 0 2y D x 3 u00 C4x 2u0 D x 2 if 4 1 4 1 u0 D ´, where (A) ´0 C ´ D . Since dx D 4 ln jxj, ´1 D 4 is a solution of the complementary x x x x v v0 1 equation for (A). Therefore,the solutions of (A) are of the form (B) ´ D 4 , where 4 D , so v 0 D x 3 . x x x x4 1 C1 x C1 Hence, v D CC1; u0 D ´ D C 4 (see (B)); u D CC2 . Therefore,the general solution is 4 4 x 4 3x 3 2 2 x C1 x c2 x c2 y D ux D C C2 x, or (C) y D C c1 x C 2 . Differentiating (C) yields y 0 D C c1 2 3 : 4 3x 2 4 x 2 x
78 Chapter 5 Linear Second Order Equations Now y.1/ D
5 0 3 ; y .1/ D ) c1 C c2 D 1; c1 4 2
2c2 D 1, so c1 D 1, c2 D 0 and y D
x2 C x. 4
5.6.36. If y D uy1 , then y 0 D u0 y1 C uy10 and y 00 D u00 y1 C 2u0 y10 C uy100 , so y 00 C p1 .x/y 0 C p2 .x/y D u00 y0 y1 u00 C .2y10 C p1 y1 /u0 D 0 if u is any function such that (B) 0 D 2 1 p1 . If ln ju0 .x/j D u y1 Z x Z x 1 0 2 ln jy1 .x/j p1 .t/ dt, then u satisfies (B); therefore, if (C) u .x/ D 2 exp p1 .s/ ds , y1 .x/ x0 x0 Z t Z x 1 exp p1 .s/ds satisfies (C), y2 D uy1 is a solution then u satisfies (B). Since u.x/ D 2 x0 y1 .t/ x0 y2 D u is nonconstant, Theorem 5.1.6 implies that fy1 ; y2 g is a fundamental set of (A) on .a; b/. Since y1 of solutions of (A) on .a; b/. 5.6.38. (a) The associated linear equation is (A) ´00 C k 2 ´ D 0, with characteristic polynomial p.r / D r 2 C k 2 . The general solution of (A) is ´ D c1 cos kx C c2 sin kx. Since ´0 D kc1 sin kx C kc2 cos kx, kc1 sin kx C kc2 cos kx ´0 yD D . ´ c1 cos kx C c2 sin kx (b) The associated linear equation is (A) ´00 3´0 C 2´ D 0, with characteristic polynomial p.r / D 2 r 3r C 2 D .r 1/.r 2/. The general solution of (A) is ´ D c1 e x C c2 e 2x . Since ´0 D c1 e x C c2e 2x , ´0 c1 C 2c2 e x yD D . ´ c1 C c2 e x (c) The associated linear equation is (A) ´00 C5´0 6´ D 0, with characteristic polynomial p.r / D r 2 C 5r 6 D .r C 6/.r 1/. The general solution of (A) is ´ D c1 e 6x C c2e x . Since ´0 D 6c1 e 6x C c2e x , ´0 6c1 C c2 e 7x yD D . ´ c1 C c2 e 7x (d) The associated linear equation is (A) ´00 C 8´0 C 7´ D 0, with characteristic polynomial p.r / D 2 r C 8r C 7 D .r C 7/.r C 1/. The general solution of (A) is ´ D c1 e 7x C c2e x . Since ´0 D 7c1 C c2 e 6x ´0 7c1 e 7x 2c2e x , y D D . ´ c1 C c2 e 6x (e) The associated linear equation is (A) ´00 C 14´0 C 50´ D 0, with characteristic polynomial p.r / D r 2 C 14r C 50 D .r C 7/2 C 1. The general solution of (A) is ´ D e 7x .c1 cos x C c2 sin x/. Since ´0 D 7e 7x .c1 cos x C c2 sin x/ C e 7x . c1 sin x C c2 cos x/ D .7c1 c2 / cos x .c1 C 7c2 / sin x, ´0 .7c1 c2 / cos x C .c1 C 7c2/ sin x yD D . ´ c1 cos x C c2 sin x 1 1 (f) The given equation is equivalent to (A) y 0 C y 2 y D 0. The associated linear equation is 6 6 1 1 1 1 1 1 (B) ´00 ´0 ´ D 0, with characteristic polynomial p.r / D r 2 r D rC r . 6 6 6 6 3 2 c1 x=3 c2 ´0 The general solution of (B) is ´ D c1 e x=3 C c2 e x=2. Since ´0 D e C e x=2, y D D 3 2 ´ 2c1 C 3c2e 5x=6 . 6.c1 C c2 e 5x=6 / 1 1 (g) The given equation is equivalent to (A) y 0 C y 2 yC D 0. The associated linear equation is 3 36 1 1 1 2 00 1 0 2 1 (B) ´ ´ C ´ D 0, with characteristic polynomial p.r / D r rC D r . The general 3 36 3 36 6 e x=6 e x=6 solution of (B) is ´ D e x=6 .c1 C c2x/. Since ´0 D .c1 C c2 x/ C c2 e x=6 D .c1 C c2.x C 6//, 6 6
Section 5.7 Variation of Parameters yD
79
´0 c1 C c2 .x C 6/ D . ´ 6.c1 C c2 x/
´0 ´00 r 0 .x/ 5.6.40. (a) Suppose that ´ is a solution of (B) and let y D . Then (D) C p.x/ yCq.x/ D r´ r´ r .x/ 2 1 ´0 r 0 ´0 ´00 r0 ´00 r0 ´00 D ry 2 y, so D y 0 C ry 2 C y. Therefore, (D) 0 and y 0 D 2 r´ r ´ r ´ r´ r r´ r implies that y satisfies (A). Now suppose that y is a solution of (A) and let ´ be any function such that r0 r0 ´0 D ry´. Then ´00 D r 0 y´ C ry 0 ´ C ry´0 D ´0 C .y 0 C ry 2 /r ´ D ´0 .p.x/y C q.x//r ´, so r r r0 0 00 ´ ´ C p.x/ry´ C q.x/r ´ D 0, which implies that ´ satisfies (B), since ry´ D ´0 . r (b) If f´1 ; ´2 g is a fundamental set of solutions of (B) on .a; b/, then ´ D c1 ´1 C c2 ´2 is the general solution of (B) on .a; b/. This and (a) imply that (C) is the general solution of (A) on .a; b/. 5.7 VARIATION OF PARAMETERS 5.7.2. (A) yp D u1 cos 2x C u2 sin 2x; u01 cos 2x C u02 sin 2x 2u01 sin 2x C 2u02 cos 2x
D 0 (B) D sin 2x sec2 x:
(C)
Multiplying (B) by 2 sin 2x and (C) by cos 2x and adding the resulting equations yields 2u02 D tan 2x, tan2 2x 1 sec2 2x tan 2x so u02 D . Then (B) implies that u01 D u02 tan.2x/ D D . Therefore,u1 D 2 2 2 tan 2x ln j cos 2xj sin 2x ln j cos 2xj x cos 2x sin 2x x and u2 D . Now (A) yields yp D C . 2 4 4 4 2 4 sin 2x ln j cos 2xj x cos 2x C . Since sin 2x satisfies the complementary equation we redefine yp D 4 2 5.7.4. (A) yp D u1 e x cos x C u2 e x sin x; u01 .e x
cos x
u01 e x cos x C u02 e x sin x e x sin x/ C u02 .e x sin x C e x cos x/
D 0 (B) x D 3e sec x:
(C)
Subtracting (B) from (C) and cancelling e x from the resulting equations yields u01 cos x C u02 sin x u01 sin x C u02 cos x
D 0 (D) D 3 sec x:
(E)
Multiplying (D) by sin x and (E) by cos x and adding the results yields u02 D 3. From (D), u01 D u02 tan x D 3 tan x. Therefore u1 D 3 ln j cos xj, u2 D 3x. Now (A) yields yp D 3e x .cos x ln j cos xjC x sin x/. 5.7.6. (A) yp D u1 e x C u2 e
x
; u01 e x C u02 e
x
D 0
u01 e x
x
D
Adding (B) to (C) yields 2u01 e x D 1
2 e
2x
D
u02 e
(B)
4e x : 1 C e 2x
(C)
4e x 2e 2x 0 , so u D . From (B), u02 D 1 1 C e 2x 1 e 2x
2e 2x . Using the substitution v D e 1 e 2x
2x
e 2x u01 D
we integrate u01 to obtain u1 D ln.1
e
2x
/.
80 Chapter 5 Linear Second Order Equations Using the substitution v D e 2x we integrate u02 to obtain u1 D ln.1 e x ln.1 e 2x / e x ln.e 2x 1/. 5.7.8. (A) yp D u1 e x C u2
ex ; x ex u01 e x C u02 D x x e ex D u01 e x C u02 x x2
u02 e x e 2x D , so u02 D x2 x e 2x . u1 D e x , u2 D xe x C e x . Now (A) yields yp D x
Subtracting (B) from (C) yields
5.7.10. (A) yp D u1 e
x2
e 2x /. Now (A) yields yp D
C u2 xe
2xu01 e
x2
x2
0
(B)
e 2x : x
(C)
xe x . From (B), u01 D
u02 D e x . Therefore x
;
u01 e
x2
C u02 .e
x2
C u02 xe 2x 2 e
x2 x2
/
D 0 D 4e
(B) x.xC2/
:
(C)
2
Multiplying (B) by 2x and adding the result to (C) yields u02 e x D 4e x.xC2/ , so u02 D 4e 2x . From (B), u01 D u02 x D 4xe 2x . Therefore u1 D .2x C 1/e 2x , u2 D 2e 2x . Now (A) yields yp D e x.xC2/ . 5.7.12. (A) yp D u1 x C u2 x 3; u01 x C u02 x 3
D
u01 C 3u02 x 2
D
0
(B) 4
2x sin x D 2x 2 sin x x2
(C):
1 and subtracting the result from (C) yields 2x 2 u02 D 2x 2 sin x, so u02 D sin x. From x (B), u01 D u02 x 2 D x 2 sin x. Therefore u1 D .x 2 2/ cos x 2x sin x, u2 D cos x. Now (A) yields yp D 2x 2 sin x 2x cos x. p p 5.7.14. (A) yp D u1 cos x C u2 sin x; p p u01 cos x C u02 sin x D 0 (B) p p p x cos x sin x sin u01 p C u02 p D (C): 4x 2 x 2 x p p sin x u0 Multiplying (B) by p and (C) by cos x and adding the resulting equations yields p2 D 2 x 2 x p p p p 2p p sin x cos x sin x cos x sin x , so u02 D p . From (B), u01 D u02 tan x D p . Therefore, 4x p 2 x 2p x p p p p 2p sin x cos x x sin x sin x x cos x u1 D , u2 D . Now (A) yields yp D . Since 2 2 2 2 p p2 p x cos x sin x satisfies the complementary equation we redefine yp D . 2
Multiplying (B) by
Section 5.7 Variation of Parameters
81
5.7.16. (A) yp D u1 x a C u2 x a ln x; u01 x a C u02 x a ln x au01 x a
1
C u02 .ax a
1
ln x C x a
1
D 0
x D xa x2
D
/
(B) aC1
1
(C):
a and subtracting the result from (C) yields u02 x a 1 D x a 1, so u02 D 1. From (B), x ln x. Therefore, u1 D x ln x, u2 D x. Now (A) yields yp D x aC1.
Multiplying (B) by u01 D u02 ln x D 2
5.7.18. yp D u1 e x C u2 e
x2
; 2
u01 e x C u02 e 2u01 xe
x2
2u02 xe
x2
D
x2
0 (B) 8x 5 D 8x 4 : x
D
(B)
2
2
Multiplying (B) by 2x and adding the result to (C) yields 4u01xe x D 8x 4 , so u01 D 2x 3 e x . From (B), 2 2 2 2 u02 D u01 e 2x D 2x 3e x . Therefore u1 D e x .x 2 C 1/, u2 D e x .x 2 1/. Now (A) yields 2 yp D 2x . p p 5.7.20. (A) yp D u1 xe 2x C u2 xe 2x ; p p (B) u01 xe 2x C u02 xe 2x D 0 5=2 p p p 1 1 8x u01 e 2x 2 x C p u02 e 2x 2 x p D (C): D2 x 4x 2 2 x 2 x 1 Multiplying (B) by , subtracting the result from (C), and cancelling common factors from the resulting 2x equations yields u01 e 2x C u02 e u01 e 2x u02 e Adding (D) to (E) yields 2u01 e 2x D 1, so u01 D u1 D
e
2x
4
, u2 D
2x
e
e 2x . Now (A) yields yp D 4
5.7.22. (A)yp D u1 xe x C u2 xe
x
u01 .x C 1/e x
u02 .x
1/e
2x
. From (D), u02 D
2p x . 2
u01 e 4x D
e 2x . Therefore, 2
x
D
x
D
0 (B) 4 3x D 3x 2 x2
(C):
1 , subtracting the resulting equation from (C), and cancelling common factors yields x u01 e x C u02 e u01 e x u02 e
Adding (D) to (E) yields 2u01e x D 3x, so u01 D u1 D
(D) (E)
; u01 xe x C u02 xe
Multiplying (B) by
D 0 D 1:
2x
3e x .x C 1/ , u2 D 2
3e x .x 2
1/
x x
3xe 2
D 0 D 3x:
(D) (E)
x
. From (D), u02 D
. Now (A) yields yp D 3x 2 .
u01 e 2x D
3xe x . Therefore 2
82 Chapter 5 Linear Second Order Equations 5.7.24. (A) yp D
u1 C u2 x 3 ; x u01 C u02 x 3 x u01 C 3u02 x 2 x2
D 0 D
(B)
x 3=2 Dx x2
1=2
:
(C)
1 x 5=2 and adding the result to (C) yields 4u02x 2 D x 1=2 , so u02 D . From (B), x 4 x 3=2 x 5=2 x 3=2 4x 3=2 . Therefore u1 D , u2 D . Now (A) yields yp D . 4 10 6 15
Multiplying (B) by u01 D u02 x 4 D
5.7.26. (A) yp D u1 x 2 e x C u2 x 3 e x ; u01 x 2 e x C u02 x 3 e x
D
u01 .x 2 e x C 2xe x / C u02 .x 3 e x C 3x 2e x / D
0 (B) 2xe x 2e x D : x2 x
(C)
Subtracting (B) from (C) and cancelling common factors in the resulting equations yields u01 C u02 x 2u01 x C 3u02 x 2
D 0 2 D : x
(D) (E)
2 2 Multiplying (D) by 2x and subtracting the result from (E) yields x 2u02 D , so u02 D 3 . From (D), x x 2 1 2 u01 D u02 x D . Therefore u1 D , u2 D . Now (A) yields yp D xe x . 2 x x x2 5.7.28. (A) yp D u1 x C u2 e x ; u01 x C u02 e x u01 C u02 e x
D 0 (B) 2.x 1/2 e x D 2.x D x 1
1/e x :
(C)
Subtracting (B) from (C) yields u01 .1 x/ D 2.x 1/e x , so u01 D 2e x . From (B), u02 D Therefore, u1 D 2e x , u2 D x 2 . Now (A) yields yp D xe x .x 2/. 5.7.30. (A) yp D u1 e 2x C u2 xe
x
x
x
D 2x.
;
u01 e 2x C u02 xe 2u01 e 2x C u02 .e
u01 xe
xe
x x
/
D 0 (B) .3x 1/2 e 2x D D .3x 3x 1
1/e 2x :
(C)
Multiplying (B) by 2 and subtracting the result from (C) yields u02 .1 3x/e x D .3x 1/e 2x , so x2 e 3x u02 D e 3x . From (B), u01 D u02 xe 3x D x. Therefore u1 D , u2 D . Now (A) yields 2 3 2x 2x xe .3x 2/ xe .3x 2/ yp D . The general solution of the given equation is y D Cc1 e 2x Cc2xe x . 3 3 e 2x .3x 2 C x 1/ Differentiating this yields y 0 D C 2c1e 2x C c2.1 x/e x . Now y.0/ D 1; y 0 .0/ D 3 1 1 e 2x .3x 2 2x C 6/ xe x 2 ) c1 D 1; 2 D C 2c1 C c2, so c2 D , and y D C . 3 3 6 3
Section 5.7 Variation of Parameters 5.7.32. (A) yp D u1 .x
u01 .x
1/e x C u2 .x
1/e x C u02 .x
83
1/; 1/
u01 xe x C u02
D 0 (B) .x 1/3 e x D D .x .x 1/2
1/e x :
(C)
From (B), u01 D u02 e x . Substituting this into (C) yields u02 .x 1/ D .x 1/e x , so u02 D e x , u01 D 1. Therefore u1 D x, u2 D e x . Now (A) yields yp D e x .x 1/2 . The general solution of the given equation is y D .x 1/2 e x C c1.x 1/e x C c2.x 1/. Differentiating this yields y 0 D .x 2 1/e x C c1 xe x C c2. Now y.0/ D 4; y 0 .0/ D 6 ) 4 D 1 c1 c2 ; 6 D 1 C c2 , so c1 D 2; c2 D 5 and y D .x 2 1/e x 5.x 1/. u2 5.7.34. (A) yp D u1 x C 2 ; x u0 u01 x C 22 D 0 (B) x 2x 2 2u02 u01 D D 2: (C) x3 x2 2 2 Multiplying (B) by and adding the result to (C) yields 3u01 D 2, so u01 D . From (B), u02 D x 3 2x 3 2x x4 x2 u01 x 3 D . Therefore u1 D , u2 D . Now (A) yields yp D . The general solution 3 3 6 2 2 c2 2c2 x of the given equation is y D C c1 x C 2 . Differentiating this yields y 0 D x C c1 . Now 2 x x3 1 1 y.1/ D 1; y 0 .1/ D 1 ) 1 D C c1 C c2 ; 1 D 1 C c1 2c2, so c1 D 1; c2 D , and 2 2 x2 1 yD C x C 2. 2 2x 5.7.36. Since y D yp
a1 y1
a2 y2 ,
00
P0 .x/y C P1 .x/y 0 C P2 .x/y
D P0 .x/.yp
a1 y1
a2 y2 /00
CP1 .x/.yp a1 y1 a2 y2 /0 CP2 .x/.yp a1 y1 a2 y2 / D .P0 .x/yp00 C P1 .x/yp0 C P2 .x/yp / a1 P0 .x/y100 C P1 .x/y10 C P2 .x/y1 a2 P0 .x/y200 C P1 .x/y20 C P2 .x/y2 D F .x/ a1 0 a2 0 D F .x/I
hence y is a particular solution of (A). is a solution of (A) on .a; 1/ if u01 e x C u02 e x D 0 and u01 e x e xf 0 ex f u02 e x D f .x/. Solving these two equations yields u01 D , u2 D . The functions u1 .x/ D 2 2 Z x Z x 1 1 e t f .t/ dt and u2 .x/ D e t f .t/ dt satisfy these conditions. Therefore, 2 0 2 0 Z Z ex x t e x x t yp .x/ D e f .t/ dt e f .t/ dt 2 0 2 0 Z Z x 1 x .x t/ D f .t/ e .x t / e dt D f .t/ sinh.x t/ dt: 2 0 0
5.7.38.
(a) yp D u1 e x C u2 e
x
84 Chapter 5 Linear Second Order Equations is a particular solution of y 00 yp0 .x/
D D D
y D f .x/. Differentiating yp yields Z Z ex x t ex e x x t e x x e f .t/ dt C e x C e f .t/ dt e 2 0 2 2 0 2 Z Z e x x t ex x t e f .t/ dt C e f .t/ dt 2 0 2 0 Z x Z x 1 .x t / .x t / f .t/ e Ce dt D f .t/ cosh.x t/ dt: 2 0 0
Since yp .x0 / D yp0 .x0 / D 0, the solution of the initial value problem is y
D yp C k0 cosh x C k1 sinh x Z x D k0 cosh x C k1 sinh x C sinh.x
t/f .t/ dt:
0
The derivative of the solution is y0
D yp0 C k0 sinh x C k1 cosh x Z x D k0 sinh x C k1 cosh x C cosh.x 0
t/f .t/ dt:
CHAPTER 6 Applications of Linear Second Order Equations
6.1 SPRING PROBLEMS I g 32 k D D D 320 the equation of motion is (A) y 00 C 320y D 0. The general m l :1 p p p p p solution of (A) is y D c1 cos 8 5t C c2 sin 8 5t, so y 0 D 8 5. c1 sin 8 5t C c2 cos 8 5t/. Now p p 1 1 1 1 1 y.0/ D ) c1 D and y 0 .0/ D 2 ) c2 D p , so y D cos 8 5t p sin 8 5t ft. 4 4 4 4 5 4 5 6.1.2. Since
g 32 k D D D 64 the equation of motion is (A) y 00 C 64y D 0. The general solution m l :5 1 1 of (A) is y D c1 cos 8t C c2 sin 8t, so y 0 D 8. c1 sin 8t C c2 cos 8t/. Now y.0/ D ) c1 D and 4 4 p 1 1 1 17 1 0 y .0/ D ) c2 D , so y D cos 8t sin 8t ft; R D ft; !0 D 8 rad/s; T D =4 s; 2 16 4 16 16 ı :245 rad 14:04 . 6.1.4. Since
.9:8/10 mg D D 140, the equation of motion of the 2 kg mass is (A) y 00 C 70y D l :7 p p p p 0. The general solution of (A) is y D c1 cos 70t C c2 sin 70t, so y 0 D 70. c1 sin 70t C p p p 1 2 1 1 2 c2 cos 70t/. Now y.0/ D ) c1 D and y 0 .0/ D 2 ) c2 D p , so y D cos 70 t C p sin 70 t 4 4 4 70 70 r p p 1 67 m; R D m; !0 D 70 rad/s; T D 2= 70 s; 2:38 rad 136:28ı. 4 35 6.1.6. Since k D
k g 32 D D D 64 the equation of motion is (A) y 00 C 64y D 0. The general solution m l 1=2 1 1 of (A) is y D c1 cos 8t C c2 sin 8t, so y 0 D 8. c1 sin 8t C c2 cos 8t/. Now y.0/ D ) c1 D and 2 2 3 1 3 0 y .0/ D 3 ) c2 D , so y D cos 8t sin 8t ft. 8 2 8 6.1.8. Since
64 D 2, so the equation of motion is 2y 00 C 8y D 2 sin t, or (A) y 00 C 4y D sin t. Let 32 yp D A cos t C B sin t; then yp00 D A cos t B sin t, so yp00 C 4yp D 3A cos t C 3B sin t D sin t if 1 1 3A D 0, 3B D 1. Therefore,A D 0, B D , and yp D sin t. The general solution of 3 3 6.1.10. m D
85
86 Chapter 6 Applications of Linear Second Order Equations 1 1 1 (A) is (B) y D sin t C c1 cos 2t C c2 sin 2t, so y.0/ D ) c1 D . Differentiating (B) yields 3 2 2 1 1 5 0 0 y D cos t 2c1 sin 2t C 2c2 cos 2t, so y .0/ D 2 ) 2 D C 2c2, so c2 D . Therefore, 3 3 6 1 1 5 y D sin t C cos 2t C sin 2t ft. 3 2 6 4 1 mg 1 1 D and k D D 4, so the equation of motion is y 00 C 4y D sin 8t, or (A) 6.1.12. m D 32 8 l 8 4 y 00 C 32y D 2 sin 8t. Let yp D A cos 8t C B sin 8t; then yp00 D 64A cos t 64B sin 8t, so yp00 C 32yp D 1 32A cos 8t 32B sin 8t D 2 sin 8t if 32A D 0, 32B D 2. Therefore,A D 0, B D , and 16 p p 1 1 yp D sin 8t. The general solution of (A) is (B) y D sin 8t C c1 cos 4 2t C c2 sin 4 2t, so 16 16 p p p 1 1 1 y.0/ D ) c1 D . Differentiating (B) yields y 0 D cos 8t C4 2. c1 sin 4 2t Cc2 cos 4 2t/, so 3 3 2 p p 1 p 1 1 1 1 0 y .0/ D 1 ) 1 D C4 2c2 , so c2 D p . Therefore,y D sin 8t C cos 4 2t p sin 4 2t ft. 2 16 3 8 2 8 2 r 2 m 6.1.14. Since T D D 2 the period is proportional to the square root of the mass. Therefore, !0 k p doubling p the mass mutiplies the period by 2; hence the period of the system with the 20 gm mass is T D 4 2 s.
3 mg 6 3 00 6 D and k D D D 18 so the equation of motion is y C 18y D 32 16 l 1=3 16 64 4 sin !t 6 cos !t, or (A) y 00 C 96y D sin !t 32 cos !t. The displacement will be unbounded if 3 p p p p 64 ! D 96 D 4 6, in which case (A) becomes (B) y 00 C 96y D sin 4 6t 32 cos 4 6t. Let 3 p p yp D At cos 4 6t C Bt sin 4 6tI then p p p p yp0 D .A C 4 6Bt/ cos 4 6t C .B 4 6At/ sin 4 6t p p p p yp00 D .8 6B 96At/ cos 4 6t .8 6A C 96Bt/ sin 4 6t; so p p p p p p 64 yp00 C 96yp D 8 6B cos 4 6t 8 6A sin 4 6t D sin 4 6t 32 cos 4 6t 3 p p p p 8 4 t 8 64 if 8 6B D 32, 8 6A D . Therefore, A D cos 4 6t C 4 sin 4 6t . p , B D p , and yp D p 3 3 6 6 6 3 The general solution of (B) is p p p p t 8 yD p cos 4 6t C 4 sin 4 6t C c1 cos 4 6t C c2 sin 4 6t; .C/ 6 3 6.1.16. m D
so y.0/ D 0 ) c1 D 0. Differentiating (C) yields p p p p 8 4 8 y0 D p cos 4 6t C p sin 4 6t 4t sin 4 6t C 4 cos 4 6t 3 3 6 6 p p p C4 6. c1 sin 4 6t C c2 cos 6t/; p 8 1 so y 0 .0/ D 0 ) 0 D p C 4 6c2, and c2 D . Therefore, 9 3 6 p 1 p p t 8 yD p cos 4 6t C 4 sin 4 6t C sin 4 6t ft: 9 6 3
Section 6.2 Spring Problems II
87
6.1.18. The equation of motion is (A) y 00 C !02 y D 0. The general solution of (A) is y D c1 cos !0 t C c2 sin !0 t. Now y.0/ D y0 ) c1 D y0 . Since y 0 D !0 . c1 sin !0 t C c2 cos !0 t/, y 0 .0/ D v0 ) c2 D v0 v0 . Therefore, y D y0 cos !0 t C sin !0 t; !0 !0 1 p v0 y0 !0 RD I sin D p : .!0 y0 /2 C .v0 /2 I cos D p !0 .!0 y0 /2 C .v0 /2 .!0 y0 /2 C .v0 /2
Discussion 6.1.1 In Exercises 19, 20, and 21 we use the r fact that in a spring–mass system with mass m m and spring constant k the period of the motion is T D 2 . Therefore, if we have two systems with k s T2 m2 k1 masses m1 and m2 and spring constants k1 and k2 , then the periods are related by D . We T1 m1 k2 will use this formula in the solutions of these exercises. s p p T2 2m1 6.1.20. Let m2 D 2m1 . Since k1 D k2 , D D 2, so T2 D 2T1 . T1 m1 s k1 6.1.21. Suppose that T2 D 3T1 . Since m1 D m2 , D 3, k1 D 9k2 . k2 6.2 SPRING PROBLEMS II 16 1 mg D D 5 the equation of motion is y 00 Cy 0 C5y D 0, or (A) y 00 C2y 0 C10y D 0. l 3:2 2 The characteristic polynomial of (A) is p.r / D r 2 C2r C10 D .r C1/2 C9. Therefore,the general solution of (A) is y D e t .c1 cos 3t C c2 sin 3t/, so y 0 D y C 3e t . c1 sin 3t C c2 cos 3t/. Now y.0/ D 3 1 1 t 0 and y .0/ D 2 ) c1 D 3 and 2 D 3 C 3c2, or c2 D . Therefore,y D e 3 cos 3t C sin 3t 3 3 p 82 t ft. The time–varying amplitude is e ft. 3 96 mg 6.2.4. Since k D D D 30 the equation of motion is 3y 00 C 18y 0 C 30y D 0, or (A) y 00 C 6y 0 C l 3:2 10y D 0. The characteristic polynomial of (A) is p.r / D r 2 C 6r C 10 D .r C 3/2 C 1. Therefore,the general solution of (A) is y D e 3t .c1 cos t C c2 sin t/, so y 0 D 3y C e 3t . c1 sin t C c2 cos t/. 5 5 15 63 Now y.0/ D and y 0 .0/ D 12 ) c1 D and 12 D C c2 , or c2 D . Therefore, 4 4 4 4 3t e yD .5 cos t C 63 sin t/ ft. 4 mg 8 1 3 6.2.6. Since k D D D 25 the equation of motion is y 00 C y 0 C 25y D 0, or (A) y 00 C 6y 0 C l :32 4 2 2 2 100y D 0. The characteristic polynomial of (A) is p.r / D r C 6r C 100 D .r C p p p 3/ C3t91. Therefore,the p 3t 0 general solution of (A) is y D e .c1 cos 91t Cc2 sin 91t/, so y D 3yC 91e . c1 sin 91t C p p 1 1 3 11 c2 cos 91t/. Now y.0/ D and y 0 .0/ D 4 ) c1 D and 4 D C 91c2, or c2 D p . 2 2 2 2 91 p p 1 3t 11 Therefore,y D e cos 91t C p sin 91t ft. 2 91 6.2.2. Since k D
mg 20 980 D D 3920 the equation of motion is 20y 00 C 400y 0 C 3920y D 0, l 5 or (A) y 00 C 20y 0 C 196y D 0. The characteristic polynomial of (A) is p.r / D r 2 C 20r C 196 D
6.2.8. Since k D
88 Chapter 6 Applications of Linear Second Order Equations p p .r C 10/2 C 96.pTherefore,the general solution of (A) is y D e 10t .c1 cos 4 6t C c2 sin 4 6t/, so p p y 0 D 10y C 4 6e 10t . c1 sin 4 6t C c2 cos 4 6t/. Now y.0/ D 9 and y 0 .0/ D 0 ) c1 D 9 and p p p 45 45 0 D 90 C 4 6c2 , or c2 D p . Therefore, y D e 10t 9 cos 4 6t C p sin 4 6t cm. 2 6 2 6 mg 32 D D 32 the equation of motion is (A) y 00 C 3y 0 C 32y D 0. The characterisl 1 3 2 119 tic polynomial of (A) is p.r / D r 2 C3r C32 D r C C . Therefore,the general solution of (A) is 2 4 ! ! p p p p p 3 119 119 119 3t =2 119 119 3t =2 0 t C c2 sin t , so y D yC e c1 sin t C c2 cos t . yDe c1 cos 2 2 2 2 2 2 p 1 119 1 3 9 Now y.0/ D and y 0 .0/ D 3 ) c1 D and 3 D C c2 , or c2 D p . Therefore, 2 2 ! 4 2 2 119 p p 3 1 119 9 119 y D e 2t cos t p sin t ft. 2 2 2 2 119 6.2.10. Since k D
2 25 1 00 1 25 mg D D the equation of motion is y C y0 C y D 0, or (A) l :32 4 16 8 4 2 y 00 C 2y 0 C 100y D 0. The characteristic polynomial of (A) is p.r C 100 D .r C p/ D r C 2r p 1/2 C 99. Therefore,the general solution of (A) is y D e t .c1 cos 3 11t C c2 sin 3 11t/, so y 0 D p p p 1 1 y C 3 11e t . c1 sin 3 11t C c2 cos 3 11t/. Now y.0/ D and y 0 .0/ D 5 ) c1 D and 3 3 p p p 1 14 1 14 5 D C 3 11c2 , or c2 D p . Therefore,y D e t cos 3 11t C p sin 3 11t ft. 3 3 9 11 9 11 6.2.12. Since k D
mg D 32 the equation of motion is (A) y 00 C 12y 0 C 32y D 0. The characteristic l polynomial of (A) is p.r / D r 2 C 12r C 32 D .r C 8/.r C 4/. Therefore, the general solution of (A) is 2 2 and y 0 .0/ D 0 ) c1 C c2 D ;y D c1e 8t C c2 e 4t , so y 0 D 8c1e 8t 4c2e 4t . Now y.0/ D 3 3 2 4 2 8t 4t 8c1 4c2 D 0, so c1 D , c2 D , and y D .e 2e /. 3 3 3 6.2.14. Since k D
100 980 mg D D 100 the equation of motion is 100y 00 C 600y 0 C 1000y D 0, or l 98 (A) y 00 C 6y 0 C 10y D 0. The characteristic polynomial of (A) is p.r / D r 2 C 6r C 10 D .r C 3/2 C 1. Therefore, the general solution of (A) is y D e 3t .c1 cos t C c2 sin t/, so y 0 D 3y C e t . c1 sin t C c2 cos t/. Now y.0/ D 10 and y 0 .0/ D 100 ) c1 D 10 and 100 D 30 C c2 , or c2 D 70. Therefore, y D e 3t .10 cos t 70 sin t/ cm. 6.2.16. Since k D
6.2.18. The equation of motion is (A) 2y 00 C 4y 0 C 20y D 3 cos 4t 5 sin 4t. The steady state component of the solution of (A) is of the form yp D A cos 4t CB sin 4t; therefore yp0 D 4A sin 4t C4B cos 4t and yp00 D 16A cos 4t 16B sin 4t, so 2yp00 C 4yp0 C 20yp D . 12A C 16B/ cos 4t .16A C 12B/ sin 4t D 11 27 3 cos 4t 5 sin 4t if 12A C 16B D 3, 16A 12B D 5; therefore A D ,B D , and 100 100 11 27 yp D cos 4t C sin 4t cm. 100 100 mg 9:8 6.2.20. Since k D D D 20 the equation of motion is (A) y 00 C 4y 0 C 20y D 8 sin 2t 6 cos 2t. l :49 The steady state component of the solution of (A) is of the form yp D A cos 2t C B sin 2t; therefore
Section 6.3 The RLC Circuit
89
yp0 D 2A sin 2t C 2B cos 2t and yp00 D 4A cos 2t 4B sin 2t, so yp00 C 4yp0 C 20yp D .16A C 8B/ cos 2t .8A 16B/ sin 2t D 8 sin 2t 6 cos 2t if 16A C 8B D 6, 8A C 16B D 8; therefore 1 1 1 1 , B D , and y D cos 2t C sin 2t m. AD 2 4 2 4 6.2.22. If e r1t .c1 C c2 t/ D 0, then (A) c1 C c2 t D 0. If c2 D 0, then c1 ¤ 0 (by assumption), so (A) is impossible. If c1 ¤ 0, then the left side of (A) is strictly monotonic and therefore cannot have the same value for two distinct values of t. c y C !1 e ct =2m. c1 sin !1 t C 6.2.24. If y D e ct =2m.c1 cos !1 t C c2 sin !1 t/, then y 0 D 2m cy0 1 cy0 c2 cos !1 t/, so y.0/ D y0 and y 0 .0/ D v0 ) c1 D y0 and v0 D Cc2 !1 , so c2 D v0 C t 2m !1 2m 1 cy0 and y D e ct =2m y0 cos !1 t C v0 C sin !1 t . !1 2m
6.2.26. If y D e r1t .c1 C c2t/, then y 0 D r1 y C c2 e r1t , so y.0/ D y0 and y 0 .0/ D v0 ) c1 D y0 and v0 D r1 y0 C c2, so c2 D v0 r1 y0 . Therefore, y D e r1t .y0 C .v0 r1 y0 /t/.
6.3 THE RLC CIRCUIT 1 00 6.3.2. Q C 2Q0 C 100Q D 0; Q00 C 40Q0 C 2000Q D 0; r 2 C 40r C 2000 D .r C 20/2 C 20 1600 D 0; r D 20 ˙ 40i ; Q D e 20t .2 cos 40t C c2 sin 40t/ (since Q0 D 2); I D Q0 D 21 e 20t ..40c2 40/ cos 40t .20c2 C 80/ sin 40t/; I0 D 2 ) 40c2 40 D 2 ) c2 D , so 20 20t 20c2 C 80 D 101; I D e .2 cos 40t 101 sin 40t/. 1 00 Q C 6Q0 C 250Q D 0; Q00 C 60Q0 C 2500Q D 0; r 2 C 60r C 2500 D .r C 30/2 C 10 1600 D 0; r D 30 ˙ 40i ; Q D e 30t .3 cos 40t C c2 sin 40t/ (since Q0 D 3); I D Q0 D e 30t ..40c2 90/ cos 40t .30c2 C 120/ sin 40t/; I0 D 10 ) 40c2 90 D 10 ) c2 D 2, so 30c2 120 D 180; I D 10e 30t .cos 40t C 18 sin 40t/. 6.3.4.
6.3.6. Qp D A cos 10t C B sin 10t; Qp0 D 10B cos 10t 10A sin 10t; Qp00 D 100A cos 10t 1 00 Q C 3Qp0 C 100Q9 D .90A C 30B/ cos 10t .30A 90B/ sin 10t D 5 cos 10t 100B sin 10t; 10 p 5 sin 10t, so 90A C 30B D 5, 30A C 90B D 5. Therefore, A D 1=15, B D 1=30, Qp D 1 cos 10t sin 10t , and Ip D .cos 10t C 2 sin 10t/. 15 30 3 6.3.8. Qp D A cos 50t C B sin 50t; Qp0 D 50B cos 50t 50A sin 50t; Qp00 D 2500A cos 50t 1 2500B sin 50t; Qp00 C 2Qp0 C 100Qp . 150AC 100B/ cos 50t .100AC 150B/ sin 50t D 3 cos 50t 10 6 sin 50t, so 150A C 100B D 3, 100A C 1500B D 6. Therefore,A D 3=650, B D 12=325, 3 3 Qp D .cos 50t C 8 sin 50t/, and Ip D .8 cos 50t sin 50t/. 650 13 6.3.10. Qp D A cos 30t C B sin 30t; Qp0 D 30B cos 30t 30A sin 30t; Qp00 D 900A cos 30t 1 900B sin 30t; Qp00 C 4Qp0 C 125Qp D .80A C 120B/ cos 30t .120A 80B/ sin 30t D 15 cos 30t 20 3 30 sin 30t, so 80AC120B D 15, 120AC80B D 30, A D 3=13, B D 3=104, Qp D .8 cos 30t 104 45 sin 30t/, and Ip D .cos 30t C 8 sin 30t/. 52
90 Chapter 6 Applications of Linear Second Order Equations 6.3.12. Let D .!/ be the amplitude of Ip . From the solution of Exercise 6.3.11, Qp D A cos !t C .1=C L! 2 /U R!V R!U C .1=C L! 2 /V , B D , and D .1=C B cos !t, where A D 2 2 2 2 0 2 L! / C R ! . Since Ip D Qp D !. A sin !t C B cos !t/, it follows that .!/ D ! 2 .A2 C B 2 / D U2 CV2 , with .!/ D 2 D .1=C! L!/2 C R2 , which attains it mininmum value R2 when .!/ ! p U2 CV2 1 ! D !0 D p . The maximum amplitude of Ip is .!/ D . R LC 6.4 MOTION UNDER A CENTRAL FORCE 6.4.2. Let h D r0200 ; then D
h2 . Since r D k 1 C e cos.
/
, it follows that (A) e cos. 0
/ D
r 1. Differentiating this with respect to t yields e sin. / 0 D , so (B) e sin. / D r 0 r2 r , since r 2 0 h. Squaring and adding (A) and (B) and setting t D 0 in the result yields e D "h 2 0 2 #1=2 r0 1 C . If e D 0, then 0 is undefined, but also irrelevant; if e ¤ 0, then set t D 0 r0 h r 0 1 in (A) and (B) to see that D 0 ˛, where ˛ < , cos ˛ D 1 and sin ˛ D 0 e r0 eh 1 1 1 d 2u 6 f .1=u/. Let u D D ; then D D 6cu2 . 2 2 2 2 mh u r c d c 4 6c 1 1 6cu2 C u D f .1=u/, so f .1=u/ D mh2 .6cu4 C u3 / and f .r / D mh2 C 3 . mh2 u2 r4 r mk d 2u k 6.4.6. (a) With f .r / D , Eqn. 6.4.11 becomes (A) C 1 u D 0. The initial conditions r3 d 2 h2 1 du r00 imply that u.0 / D and .0 / D (see Eqn. (6.4.) r d h ˇ 0 ˇ1=2 ˇ k ˇˇ d 2u (b) Let D ˇˇ1 . (i) If h2 < k, then (A) becomes
2 u D 0, and the solution ˇ 2 h d 2 r00 1 of the initial value problem for u is u D cosh . 0 / sinh . 0 /; therefore r D r0
h 1 r0 r00 d 2u r0 cosh . 0 / sinh . 0 / . (ii) If h2 D k, then (A) becomes D 0, and the solu h d 2 1 1 r00 r0r00 tion of the initial value problem for u is u D . 0 /; therefore r D r0 1 . 0 / . r0 h h d 2u (iii) If h2 > k, then (A) becomes C 2 u D 0, and the solution of the initial value problem for u is d 2 1 1 r00 r0r00 uD cos . 0 / sin . 0 /; therefore r D r0 cos . 0 / sin . 0 / . r0
h
h 6.4.4. Recall that (A)
d 2u 2 D d
CHAPTER 7 Series Solutions of Linear Second Equations
7.1 REVIEW OF POWER SERIES P m 7.1.2. From Theorem 7.1.3, 1 j´j > 1=L. Therefore, mD0 bm ´ converges p if j´j < 1=L and diverges if p P1 2 x0 / converges if jx x0 j < 1= L and diverges if jx x0 j > 1= L. mD0 bm .x P m 7.1.4. From Theorem 7.1.3, 1 if j´j < 1=L and diverges if j´jp> 1=L. Therefore, mD0 bm ´ converges p P1 k k km b .x x / converges if jx x j < 1= L and diverges if jx x0 j > 1= L. m 0 0 mD0 2
00
2 0
7.1.12. .1 C 3x /y C 3x y 2
1 X
nD0
an x n D 2a0 C
2a2
1 X
2y D
1 X
.n C 2/.n C 1/anC2 x n C 3
nD0
1 X
2 3
nD0 1 X
nD0
C
n
an x C 4
nan x n C 3
D
nD0 1 X
nD0
n.n
nD1
1 X
nD0 1 X
nD0
an x n D
1 X
nD0
1 X
1 X
C3
1 X
2/an C 3.n
1/
n.n 1/an x n
nD2
.n C 2/.n C 1/anC2 x n C 2
nD0
2
C2
1 X
n
1/an x C 3
n.n
nD2
1/nan x n C 3
n.n
nD1
an x n C 4
nD1 1 X
1 X
Œ.n C 2/.n C 1/anC2 C .3n.n
1 X
.n
1/an
1 X
nD2
n.n
nD0
1/an
an x n D
1/an x n C 2
1 X
nD0
1 X
1 X
nD2
1 x
2
C
1 X
1 X
nD0
nan x n C 3
n.n
nD2
.n C 2/.n C 1/anC2 C 2.n C 1/anC1 C .n2 91
1 X
n
an x n
nD0
2n C 3/an x n .
2
1 X
an x n D
.
1/an x n C 2
.n C 2/.n C 1/anC2 x n
.n C 1/anC1 x n
nD0
1/an x n
n.n
nan x nC1
nD0
1 X an x .n C 2/.n C 1/anC2 C 2.n C 1/anC1 C .2n2
x/y 0 C 3y D
n 1x
n.n 1/an x n C 2
nD0
1 X
nD1
nD1
n
7.1.14. .1 C x 2 /y 00 C .2 1 X
1/an x
nD2
7.1.13. .1 C 2x 2 /y 00 C .2 3x/y 0 C 4y D 1 X
n.n
n 2
1 X
nD0
1 X
nan x n
1
nD1
.n C 1/anC1 x n
5n C 4/an x n .
1/an x n C 2
1 X
nD1
nan x n
1
92 Chapter 7 Series Solutions of Linear Second Equations 1 X
7.1.16. Let t D xC1; then xy 00 C.4C2x/y 0 C.2Cx/y D . 1Ct/y 00 C.2C2t/y 0 C.1Ct/y D 1/an t n
2
C
1 X
1/an t n
n.n
nD2
2/.nC 1/anC2 t n C
1 X
C2
1 X
nan t n
nD1
.nC 1/nanC1 t n C 2
nD0 1 X
. 2a2 C2a1 Ca0 /C
1
nD1
1 X
nD0
1
C2
1 X
nD1
nan t n C
.nC 1/anC1 t n C 2
1 X
nD0 1 X
nD0
an t n C
nan t n C
1 X
nD0 1 X
nD0
an t nC1 D an t n C
1 X
an
1
1t
n
D
n 1 .xC2/ .
1/x r
C .r
.n C
nD1
Π.n C 2/.n C 1/anC2 C .n C 1/.n C 2/anC1 C .2n C 1/an C an
1 1 X X nan x n 1 C r x r 1 an x n D .n C r /x nCr 1 nD0 " nD0 # nD0 1 1 X X d d y 00 D y 0 .x/ D xr 1 .n C r /an x n D x r 1 .n C r /nan x n dx dx nD0 nD0 1 X r /an x n D .n C r /.n C r 1/an x nCr 2 .
nD0
1 X
7.1.20. y 0 .x/ D x r
n.n
nD2 1 X
2
1 X
nD0
.n C
nD0
7.1.22. x 2 .1 C x/y 00 C x.1 C 2x/y 0 .4 C 6x/y D .x 2 y 00 C xy 0 1 1 X X Œ.n C r /.n C r 1/ C .n C r / 4an x nCr C Œ.n C r /.n C r
nD0 1 X
nD0
.n C r
2/.n C r C 2/an x
r C 2/an x nCr C bn D .n C r
1 X
nD1
nCr
C
1 X
nD0
.n C r C 2/.n C r
nD0
nCr D xr 1x
2/.n C r C 2/an C .n C r C 2/.n C r
1 X
nD0
3/an
1,
1 X
6y/ D
6an x nCr C1 D
1/ C 2.n C r /
2/an x nCr C1 D
.n C r C 3/.n C r
3/an
4y/ C x.x 2 y 00 C 2xy 0
.n C r
nD0
bn x n with b0 D .r
2/.n C
2/.r C 2/a0 and
n 1.
7.1.24. x 2 .1 C 3x/y 00 C x.2 C 12x C x 2 /y 0 C 2x.3 C x/y D .x 2 y 00 C 2xy 0 / C x.3x 2 y 00 C 12xy 0 C 1 1 X X 6y/ C x 2 .xy 0 C 2y/ D Œ.n C r /.n C r 1/ C 2.n C r /anx nCr C Œ3.n C r /.n C r 1/ C 12.n C r / C 6an x
nCr C1
C
r C 2/an x nCr C1 C 1/an
1x
nCr
C
1 X
nD2
1 X
nD0 1 X
nD0
nD0
Œ.n C r / C 2an x
nCr C2
D
.n C r C 2/an x nCr C2 D
.n C r /an
2x
nCr
D xr
1 X
nD0
1 X
nD0 1 X
nD0
.n C r /.n C r C 1/an x nCr C 3
.n C r /.n C r C 1/an x nCr C 3
nD0
1 X
.n C r C 1/.n C
nD0 1 X
.n C r /.n C r C
nD1
bn x n with b0 D r .r C 1/a0 , b1 D .r C 1/.r C 2/a1 C
3.r C 1/.r C 2/a0 , bn D .n C r /.n C r C 1/an C 3.n C r /.n C r C 1/an
1
C .n C r /an
2,
n 2.
7.1.26. x 2 .2 C x 2 /y 00 C 2x.5 C x 2/y 0 C 2.3 x 2 /y D .2x 2 y 00 C 10xy 0 C 6y/C x 2 .x 2 y 00 C 2xy 0 2y/ D 1 1 X X Œ2.n C r /.n C r 1/ C 10.n C r / C 6an x nCr C Œ.n C r /.n C r 1/ C 2.n C r / 2an x nCr C2 D
nD0 1 X
2
nD0
.n C r C 1/.n C r C 3/an x nCr C
1 X
nD0
nD0
.n C r
1/.n C r C 2/an x nCr C2 D 2
1 X
nD0
.n C r C 1/.n C
Section 7.2 Series Solutions Near an Ordinary Point I r C 3/an x nCr C
1 X
nD2
.n C r
3/.n C r /an
2x
nCr
1 X
D xr
nD0
93
bn x n with b0 D 2.r C 1/.r C 3/a0 ,
b1 D 2.r C 2/.r C 4/a1 bn D 2.n C r C 1/.n C r C 3/an C .n C r
3/.n C r /an
2,
n 2.
7.2 SERIES SOLUTIONS NEAR AN ORDINARY POINT I 7.2.2. p.n/ D n.n a2m D
1/ C 2n
. 1/m a0 ; a2mC3 D 2m 1
2 D .n C 2/.n
1/; anC2 D
n 1 a ; nC1 n 1 X
a2mC2 D
2m 1 a2m , so 2m C 1
m x 2m a2mC1 D 0 if m 0; y D a0 . 1/mC1 C a1 x. mC2 2m 1 mD0
7.2.4. p.n/ D n.n 1/ 8n 12 D .n C 3/.n C 4/; anC2 D .nC3/.nC4/ a ; a2mC2 D .nC2/.nC1/ n .m C 2/.2m C 3/ .m C 2/.2m C 5/ a2m , so a2m D .mC1/.2mC1/a0 ; a2mC3 D a2mC1 so a2mC1 D .m C 1/.2m C 1/ .m C 1/.2m C 3/ 1 1 X .m C 1/.2m C 3/ a1 X a1 ; y D a0 .m C 1/.2m C 1/x 2m C .m C 1/.2m C 3/x 2mC1 . 3 3 mD0 mD0 .4m C 1/2 1 .2n C 1/2 .2nC1/2 an ; a2mC2 D 7.2.6. p.n/ D n.n 1/C2nC D ; anC2 D 4.nC2/.nC1/ a2m , 4 8.m C 1/.2m C 1/ 3 4 2 m Y1 .4j C 1/2 .4m C 3/2 m4 5 1 a0 ; a2mC3 D so a2m D . 1/ a2mC1 so a2mC1 D 2j C 1 8m mŠ 8.2m C 3/.m C 1/ j D0 3 2 3 2 3 2 1 m 1 m m X Y1 .4j C 1/2 x 2m X Y1 .4j C 3/2 x 2mC1 Y1 .4j C 3/2 1 5 5 5 . 1/m 4 a1 ; y D a0 . 1/m 4 C a1 . 1/m 4 m mŠ 2j C 3 8m mŠ 2j C 1 8 2j C 3 8m mŠ mD0 mD0 j D0
j D0
7.2.8. p.n/ D n.n 1/ 10nC28 D .n 7/.n 4/; anC2 D
j D0
5 35 a2 D a0 , a2m D 0 if m 3; a2mC3 D 6 3 1 3 1 1 a3 D 3a1 , a5 D a3 D a1 , a7 D a5 D a1 ; 5 35 5 21 3 5 1 7 35 4 3 2 y D a0 1 14x C x C a1 x 3x C x C x . 3 5 35
so a2 D
2.2m 7/.m 2/ a2m , 2.m C 1/.2m C 1/ .m 3/.2m 3/ a2mC1 , so .2m C 3/.m C 1/
.n 7/.n 4/ .nC2/.nC1/ an ; a2mC2
14a0 , a4 D
D
2n C 3 4m C 3 7.2.10. p.n/ D 2n C 3; anC2 D an ; a2mC2 D a2m , so a2m D .n C 2/.n C 1/ 2.m C 1/.2m 3 2 C 1/ 3 2 m m 1 Y Y1 4j C 3 . 1/m 4m C 5 4j C 5 . 1/m 5 5 4 a0 ; a2mC3 D a2mC1 so a2mC1 D 4 a1 ; m 2j C 1 2 mŠ 2.2m C 3/.m C 1/ 2j C 3 2m mŠ j D0 j D0 2 3 2 3 1 m 1 m X Y1 4j C 3 x 2m X Y1 4j C 5 x 2mC1 m4 m 5 5 y D a0 . 1/ C a1 . 1/ 4 2j C 1 2m mŠ 2j C 3 2m mŠ mD0
7.2.12. p.n/ D 2n.n
a1 D y 0 .0/ D 1.
7.2.13. p.n/ D 8n.n
y 0 .0/ D 1.
mD0
j D0
1/
9n
6 D .n
1/ C 2 D 2.2n
j D0
6/.2n C 1/; anC2 D 1/2 ; anC2 D
.n 6/.2n C 1/ an ; a0 D y.0/ D 1; .n C 2/.n C 1/
2.2n 1/2 an ; a0 D y.0/ D 2; a1 D .n C 2/.n C 1/
94 Chapter 7 Series Solutions of Linear Second Order Equations 1 1 1 an ; a2mC2 D a2m , so a2m D a0 ; .n C 2/.n C 1/ .2m C 2/.2m C 1/ .2m/Š 1 1 X X 1 1 .x 3/2m .x 3/2mC1 D a2mC1 , so a2mC1 D a1 ; y D a0 C a1 . .2m C 3/.2m C 1/ .2m C 1/Š .2m/Š .2m C 1/Š mD0 mD0
7.2.16. p.n/ D a2mC3
1; anC2 D
7.2.18. Let t D x
2t 2/y 00
10ty 0
2n.n 2 1/ 10n 36 D m 1 2.n C 3/ 2m C 3 1 4Y 2.n C 1/.n C 3/; anC2 D an ; a2mC2 D a2m , so a2m D .2j C 3/5 a0 ; nC2 mC1 mŠ j D0 2 3 1 m 1 m X Y 4.m C 2/ 4 .m C 1/Š .x 1/2m 4 a2mC1 , so a2mC1 D Qm 1 a1 ; y D a0 .2j C 3/5 a2mC3 D 2m C 3 mŠ j D0 .2j C 3/ mD0 1; then .1
6y D 0; p.n/ D
j D0
1 X
4m .m C 1/Š C a1 .x Qm 1 j D0 .2j C 3/ mD0
1/2mC1 .
3t 2 7.2.20. Let t D x C 1; then 1 C y 00 3 1/2 ; anC2 D
3.n C 1/ an ; a2mC2 D 2.n C 2/
9t 0 3 3 9 3 3 y C y D 0; p.n/ D n.n 1/ C n C D .n C 2 2 2 2 2 2 3 2 m 1 Y 3.2m C 1/ 3m a2m , so a2m D . 1/m 4 .2j C 1/5 m a0 ; 4.m C 1/ 4 mŠ j D0
3.m C 1/ 3m mŠ a2mC3 D a2mC1 , so a2mC1 D . 1/m Qm 1 a1 ; 2m C 3 j D0 .2j C 3/ 2 3 1 m 1 X Y1 X 3m 3m mŠ y D a0 . 1/m 4 .2j C 1/5 m .x C 1/2m C a1 . 1/m Qm 1 .x C 1/2mC1 . 4 mŠ .2j C 3/ j D0 mD0
mD0
j D0
7.2.22. p.n/ D n C 3; anC2 D
nC3 an ; a0 D y.3/ D 2; a1 D y 0 .3/ D 3. .n C 2/.n C 1/
7.2.24. Let t D x 3; .1 C 4t 2 /y 00 C y D 0; p.n/ D .4n.n 1/ C 1 D 2n 1/2 ; anC2 D .2n 1/2 an ; a0 D y.3/ D 4; a1 D y 0 .3/ D 6. .n C 2/.n C 1/ 20 0 2 20 2t 2 7.2.26. Let t D x C 1; 1 C y 00 ty C 20y D 0; p.n/ D n.n 1/ n C 20 D 3 3 3 3 2.n 6/.n 5/ 2.n 6/.n 5/ ; anC2 D an ; a0 D y. 1/ D 3; a1 D y 0 . 1/ D 3. 3 3.n C 2/.n C 1/ p.n/ p.2m/ 7.2.28. From Theorem 7.2.2, anC2 D an ; a2mC2 D a2m , so .n C 2/.n C 1/ .2m C22/.2m C 1/ 2 3 3 m m Y1 Y1 . 1/m p.2m C 1/ . 1/m a2m D 4 p.2j /5 a0 ; a2mC3 D a2m , so a2mC1 D 4 p.2j C 1/5 a1 . .2m/Š .2m C 3/.2m C 2/ .2m C 1/Š j D0
j D0
7.2.30. (a) Here p.n/ D Œn.n 1/ C 2bn ˛.˛ C 2b 1/ D .n ˛/.n C ˛ C 2b Exercise 7.2.28 implies that y1 and y2 have the stated forms. If ˛ D 2k, then 2 3 1 m X Y1 x 2m 4 y1 D .2j 2k/.2j C 2k C 2b 1/5 .2m/Š mD0
j D0
1/, so
.C/:
Section 7.2 Series Solutions Near an Ordinary Point I
95
If ˛ D 2k C 1, then y2 D
1 X
mD0
2
m Y1
4
.2j
j D0
3
2k/.2j C 2k C 2b/5
x 2mC1 : .2m C 1/Š
.D/
Q Since 2b is not a negative integer and jmD01 .2j 2k/ D 0 if m > k, y1 in (C) and y2 in (D) have the stated properties. This implies the conclusions regarding Pn . (b) Multiplying (A) through by .1 x 2 /b 1 yields x 2 /b Pn0 0 D n.n C 2b
1/.1
x 2 /b
x 2 /b Pm0 0 D m.m C 2b
1/.1
x 2 /b
Œ.1
1
Pn :
.E/
(c) Therefore, Œ.1
1
Pm :
.F/
Subtract Pn times (F) from Pm times (E) to obtain (B). (d) Integrating the left side of (B) by parts over Œ 1; 1 yields zero, which implies the conclusion. 7.2.32. (a) Let Ly D .1 C ˛x 3/y 00 C ˇx 2 y 0 C xy. If y D 1 X
p.n/an x
nC1
nD0
D 2a2 C
1 X
nD0
1 X
nD0
1 X
an x n , then Ly D
n.n
1/an x n
2
nD2
C
Œ.n C 3/.n C 2/anC3 C p.n/an x nC1 D 0 if and only if a2 D 0 and
p.n/ an for n 0. .n C 3/.n C 2/
anC3 D
7.2.34. p.r / D . 1/m 2m
2r .r
m Y1
m Y1 . 2/.3j C 2/2 p.3j / D D 3j C 2 3j C 2
10r
m Y1
m Y1 . 2/.3j C 3/2 . 1/m 2m .mŠ/2 p.3j C 1/ D D Qm 1 . Substituting (A) 3j C 4 3j C 4 j D0 .3j C 4/
m Y1 j D0
2.r C 2/2 ; (A)
8 D
1/
.3j C 2/; (B)
j D0
j D0
j D0
j D0
and (B) into the result 2 of Exercise 7.2.32(c) yields 3 1 m m 1 1 3m X Y X 2 x 6m mŠ 4 y D a0 .3j C 2/5 C a1 x 3mC1 . Qm 1 3 mŠ .3j C 4/ j D0 mD0 mD0 j D0
7.2.36. p.r / D 2r .r (B)
m Y1 j D0
1/ C 6r C 24 D 2.r
m Y1 . 6/.j C 1/.3j p.3j C 1/ D 3j C 4 3j C 4 j D0
6/.r C 2/; (A) 5/
m Y1
m Y1 p.3j / D . 6/.j 3j C 2
j D0 m Y1 m m
D . 1/ 6 m
(B) into the result of Exercise 7.2.32(c) yields 2 3 1 m X Y1 3j 5 5 x 3mC1 . y D a0 .1 4x 3 C 4x 6/ C a1 2m 4 3j C 4 mD0
j D0
2/.
j D0
3j 5 . Substituting (A) and 3j C 4
j D0
7.2.38. (a) Let Ly D .1 C ˛x kC2/y 00 C ˇx kC1 y 0 C x k y. If y D 1/an x n
2
C
1 X
nD0
p.n/an x nCk D
1 X
nD k
.n C k C 2/.n C k
1 X
nD0
an x n , then Ly D
1/anCkC2 x nCk C
1 X
nD0
1 X
n.n
nD2
Œ.n C k C 2/.n C
96 Chapter 7 Series Solutions of Linear Second Order Equations k C 1/anCkC2 C p.n/an x nCk D 0 if and only if ak D 0 for 2 n k C 1 and (A) anCkC1 D p.n/ an for n 0. .n C k C 2/.n C k C 1/ (b) If an D 0 the anC.kC2/m D 0 for all m 0, from (A). m Y1
m Y1 p.4j C 1/ p.4j / 1 1 D Qm 1 ; (B) D Qm 1 . 7.2.40. k D 2 and p.r / D 1; (A) 4j C 3 .4j C 5/ j D0 .4j C 3/ j D0 4j C 5 j D0 j D0 Substituting (A) and (B) into the result of Exercise t.2.38(c) yields 1 1 X X x 4m x 4mC1 y D a0 . 1/m C a . 1/m . Q Q 1 m 1 4m mŠ j D0 .4j C 3/ 4m mŠ jmD01 .4j C 5/ mD0 mD0
7.2.42. k D 6 and p.r / D r .r 1/ 16r C72 D .r 9/.r 8/; (A) (B)
m Y1 j D0
m Y1 8.j 1/.8j p.8j C 1/ D .8j C 9/ 8j C 9
7/
m Y1 j D0
m Y1 8.j 1/.8j p.8j / D 8j C 7 8j C 7
9/
;
j D0
;
j D0
Substituting (A) and (B) into the result of Exercise 7.2.38(c) yields 9 8 7 9 y D a0 1 x C a1 x x . 7 9 7.2.44. k D 4 and p.r / D r C 6; (A) (B)
m Y1 j D0
p.6j C 1/ D 1; .6j C 7/
m Y1
m Y1 6.j C 1/ 6m mŠ p.6j / D D Qm 1 ; 6j C 5 6j C 5 j D0 .6j C 5/
j D0
j D0
Substituting (A) and (B) into the result of Exercise 7.2.38(c) yields 1 1 X X x 6m x 6mC1 m y D a0 . 1/ Qm 1 C a1 . 1/m m . 6 mŠ j D0 .6j C 5/ mD0 mD0
7.3 SERIES SOLUTIONS NEAR AN ORDINARY POINT II 1 1 X X n 2 00 0 7.3.2. If y D an x , then .1 C x C 2x /y C .2 C 8x/y C 4y D n.n 1/an x n
1
nD0 1 X
C2
1/an x
n 2
n.n
nD2
1/an x n C 2
1 X
nan x n
1
nD1
C8
1 X
nD1
nan x n C 4
1 X
nD0
an x n D
1 X
nD0
1/.anC2 C anC1 C 2an /x n D 0 if anC2 D anC1 2an , an 0. Starting with a0 D yields y D 1 C 2x 4x 3 C 4x 4 C 4x 5 12x 6 C 4x 7 C . 7.3.4. If y D 1/an x n
1
1 X
an x n , then .1 C x C 3x 2/y 00 C .2 C 15x/y 0 C 12y D
nD0 1 X
C3
nD2
C
nD2
n.n
1/an x n C 2
1 X
nan x n
nD1
1
C 15
1 X
nD1
1 X
x2
nan x n C 12
7 3 15 4 45 5 x C x C x 2 2 8
1/an x n
nD2
1/anC2 C .n C 1/.n C 2/anC1 C 3.n C 2/2an x n D 0 if anC2 D anC1 with a0 D 0 and a1 D 1 yields y D x
n.n
1 X
nD0
an x n D
1 X
n.n
nD2
.n C 2/.n C
1 and a1 D 2
2
1 X
nD0
C
1 X
n.n
nD2
Œ.n C 2/.n C
3.n C 2/ an , an 0. Starting nC1 261 6 207 7 x C x C . 8 16
Section 7.3 Series Solutions Near an Ordinary Point II
7.3.6. If y D 3
1 X
1 X
nD0
an x n , then .3 C 3x C x 2 /y 00 C .6 C 4x/y 0 C 2y D 3
1/an x n
n.n
1
nD2
C
1 X
1/an x n C 6
n.n
nD2
1 X
nan x n
1
C4
nD1
1 X
nD1
1 X
n.n
nD2 1 X
nan x n C 2
nD0
97
1/an x n
an x n D
1 X
nD0
2
C
.n C
2/.n C 1/Œ3anC2 C 3anC1 C an x n D 0 if anC2 D anC1 an =3, an 0. Starting with a0 D 7 and 16 2 13 3 23 4 10 5 7 6 1 7 a1 D 3 yields y D 7 C 3x x C x x C x x x C . 3 3 9 9 27 9 7.3.8. The equation is equivalent to .1 C t C 2t 2 /y 00 C .2 C 6t/y 0 C 2y D 0 with t D x 1. If 1 1 1 X X X yD an t n , then .1 C t C 2t 2 /y 00 C .2 C 6t/y 0 C 2y D n.n 1/an t n 2 C n.n 1/an t n 1 C nD0
2
1 X
n
1/an t C 2
n.n
nD2
1 X
nan t
n 1
nD1
C6
1 X
nD1
nD2
nan t C 2
2/anC1 C2.nC1/an t n D 0 if anC2 D anC1 yields y D 1
4 1/ C .x 3
.x
4 .x 3
1/3
n
1 X
nD0
n
an t D
1 X
nD0
nD2
.n C 1/Œ.n C 2/anC2 C .n C
2.n C 1/ an , an 0. Starting with a0 D 1 and a1 D 1 nC2 4 136 104 1/4 .x 1/5 C .x 1/6 .x 1/7 C . 5 45 63
7.3.10. The equation is equivalent to .1 C t C t 2 /y 00 C .3 C 4t/y 0 C 2y D 0 with t D x 1. If 1 1 1 X X X yD an t n , then .1 C t C t 2 /y 00 C .3 C 4t/y 0 C 2y D n.n 1/an t n 2 C n.n 1/an t n 1 C nD0
1 X
n.n 1/an t n C3
2
.x
nD2
1 X
nan t n
nD1
1
C4
nD1
nan t n C2
1 X
nD0
nD2
an t n D
1 X
nD0
nD2
.nC1/Œ.nC2/anC2 C.nC3/anC1 C
nC3 anC1 an , an 0. Starting with a0 D 2 and a1 D 1 yields y D nC2 5 19 7 59 1091 1/2 C .x 1/3 .x 1/4 C .x 1/5 C .x 1/6 .x 1/7 C 3 12 30 45 630
.n C 2/an t n D 0 if anC2 D 1 .x 2
1/
1 X
7.3.12. The equation is equivalent to .1 C 2t C t 2 /y 00 C .1 C 7t/y 0 C 8y D 0 with t D x 1. If y D 1 1 1 1 X X X X an t n , then .1C2t Ct 2 /y 00 C.1C7t/y 0 C8y D n.n 1/an t n 2 C2 n.n 1/an t n 1 C n.n
nD0
1/an t n C
1 X
nan t n
nD1
1
C7
1 X
nD1
nan t n C 8
7.3.16. If y D 2
1 X
nD1
nan x
2.x 1 X
nD1
0 if anC2 D anC1
3.x
an x n , then .1 x/y 00
nD0 1 X n 1
C
1/
nan x n C
1 X
nD0
nD2
an t n D
1 X
nD0
nD2
nD2
Œ.n C 2/.n C 1/anC2 C .n C 1/.2n C 1/anC1 C
2n C 1 nC4 anC1 an , an 0. Starting with a0 D 1 and a1 D 2 nC2 nC1 42 604 1/2 C 8.x 1/3 4.x 1/4 .x 1/5 C 19.x 1/6 .x 1/7 C 5 35
.nC2/.nC4/an t n D 0 if anC2 D yields y D 1
1 X
nD0
.2 x/y 0 C y D
an x n D
an ,an 0. nC2
1 X
nD0
1 X
nD2
n.n 1/an x n
2
1 X
n.n 1/an x n
1
nD2
Œ.nC2/.nC1/anC2 .nC2/.nC1/anC1 C.nC1/an x n D
98 Chapter 7 Series Solutions of Linear Second Order Equations
7.3.18. If y D 1 X
nan x n
1
nD1
anC2 D
1 X
an x n , then .1 C x 2 /y 00 C y 0 C 2y D
nD0 1 X
C2
nD0
an x n D
1 X
nD0
1 X
1/an x n
n.n
nD2
2
C
1 X
1/an x n C
n.n
nD2
Œ.n C 2/.n C 1/anC2 C .n C 1/anC1 C .n2
n C 2/an x n D 0 if
n2 n C 2 an . .n C 2/.n C 1/
1 anC1 nC2
7.3.20. The equation is equivalent to .3 C 2t/y 00 C .1 C 2t/y 0 .1 2t/y D 0 with t D x 1. If 1 1 1 X X X y D an t n , then .3 C 2t/y 00 C .1 C 2t/y 0 .1 2t/y D 3 n.n 1/an t n 2 C 2 n.n nD0
1/an t
n 1
C
1 X
nan t
n 1
nD1
C2
1 X
nan t
n
nD1
1 X
nD0
n
an t C 2
1 X
nD2
an t
nC1
nD0
nD2
D .6a2 C a1
1 X
a0 / C
nD1
Œ3.n C
a1 a0 2/.n C 1/anC2 C .n C 1/.2n C 1/anC1 C .2n 1/an C 2an 1 t n D 0 if a2 D and anC2 D 6 2n 1 2 2n C 1 anC1 an an 1 , n 1. Starting with a0 D 1 and 3.n C 2/ 3.n C 2/.n C 1/ 3.n C 2/.n C 1/ 1 1 5 73 a1 D 2 yields y D 1 2.x 1/ C .x 1/2 .x 1/3 C .x 1/4 .x 1/5 C . 2 6 36 1080 7.3.22. The equation is equivalent to .1Ct/y 00 C.2 2t/y 0 C.3Ct/y D 0 with t D xC3. If y D then .1 C t/y 00 C .2 2
1 X
nD1
n
nan t C 3
1/anC1
.2n
1 X
nD0
2t/y 0 C .3 C t/y D n
an t C
3/an C an
1 X
an t
nC1
nD0 1 t
n
1 X
n.n
1/an t n
nD2
2
C
D .2a2 C 2a1 C 3a0 / C
D 0 if a2 D
1 X
nD2 1 X
nD1
n.n
1/an t n
2.x C 3/
C2
2a1 C 3a0 and anC2 D 2
anC1 C
2 yields 11 67 .x C 3/ C .x C 3/3 .x C 3/4 C .x C 3/5 C . 12 60
1 X
n.n 1/an t n
nD2
an t nC1 D .2a2 C 3a1 C a0 / C
if a2 D
3a1 C a0 and anC2 D 2
a1 D 3 yields y D 2
nan t n
1
nD1
2
.1 C 2t/y 00 C 3y 0 C .1 t/y D
nD0
nD0
.2n 3/an an 1 , .n C 2/.n C 1/
7.3.24. The equation is equivalent to .1C2t/y 00 C3y 0 C.1 t/y D 0 with t D xC1. If y D
1 X
an t n ,
Œ.n C 2/.n C 1/anC2 C .n C 2/.n C
n 1. Starting with a0 D 2 and a1 D yD2
1
1 X
1 X
1 X
nD1
2
C2
1 X
nD2
n.n 1/an t n
1
C3
1 X
nan t n
nD1
Œ.n C 2/.n C 1/anC2 C .2n C 3/.n C 1/anC1 C an
1 X
an t n , then
nD0 1
C
1 X
an t n
nD0
an
1 t
n
D0
2n C 3 an an 1 anC1 , n 1. Starting with a0 D 2 and nC2 .n C 2/.n C 1/ 7 197 287 3.x C 1/ C .x C 1/2 5.x C 1/3 C .x C 1/4 .x C 1/5 C . 2 24 20
7.3.26. The equation is equivalent to .6
2t/y 00 C .3 C t/y D 0 with t D x
2. If y D
1 X
nD0
an t n , then
Section 7.3 Series Solutions Near an Ordinary Point II 2t/y 00 C .3 C t/y D 6
.6
1 X
.12a2 C 3a0 / C
nD1
1 X
1/an t n
n.n
2
2
nD2
1 X
1/an t n
n.n
1
C3
nD2
Œ6.n C 2/.n C 1/anC2
1 X
an t n C
nD0
2.n C 1/nanC1 C 3an C an
1 t
n
1 X
nD0
an t nC1 D
D 0 if a2 D
n 3an C an 1 anC1 , n 1. Starting with a0 D 2 and a1 D 3.n C 2/ 6.n C 2/.n C 1/ 1 2 49 23 4.x 2/ .x 2/2 C .x 2/3 C .x 2/4 C .x 2/5 C . 2 9 432 1080
and anC2 D yD2
7.3.28. The equation is equivalent to .2 C 4t/y 00 .2 C 4t/y 00
a0 / C
.4a2
2t/y D 2
.1 1 X
nD1
1 X
n.n
1/an t n
nD2
2
2t/y D 0 with t D x C 4. If y D
.1 C4
1 X
1/an t n
n.n
1 X
1
nD2
nD0
Œ2.n C 2/.n C 1/anC2 C 4.n C 1/nanC1
an C 2an
1 t
an t n C 2
n
1 X
nD0 1 X
nD0
99
a0 4
4 yields
an t n , then an t nC1 D
D 0 if a2 D
a0 and 4
an 2an 1 2n anC2 D anC1 C , n 1. Starting with a0 D 1 and a1 D 2 yields y D nC2 2.n C 2/.n C 1/ 1 1 65 67 1 C 2.x C 1/ .x C 1/2 C .x C 1/3 .x C 1/4 C .x C 1/5 C . 4 2 96 80 N=5; b=zeros(N,1); b(1)=-1;b(2)=2; b(3)=b(1)/4; for n=1:N-2 b(n+3)=-2*n*b(n+2)/(n+2)+(b(n+1)2*b(n))/(2*(n+2)*(n+1)); end 7.3.29. Let Ly D .1 C ˛x C ˇx 2 /y 00 C . C ıx/y 0 C y. If y D 1/an x n 1 X
2
C˛
1 X
n.n 1/an x n
nD2
.n C 2/.n C 1/anC2 x n C ˛
nD0 1 X
ı
nD0
nan x n C
Œˇn.n
1 X
nD0
an x n D
1
Cˇ
1 X
nD0
1 X
nD0
1 X
nD2
n.n 1/an x n C
.n C 1/nanC1 x n C ˇ
1 X
nD0
nD0
nan x n
nD1
1 X
1 X
n.n
1
an x n , then Ly D
Cı
1 X
nD1
nan x n C
1/an x n C
1 X
nD0
1 X
n.n
nD2 1 X
nD0
an x n D
.n C 1/anC1 x n C
bn x n , where bn D .n C 1/.n C 2/anC2 C .n C 1/.˛ n C /anC1 C
1/ C ın C an , which implies the conclusion.
7.3.30. (a) Let D 2˛, ı D 4ˇ, and D 2ˇ in Exercise 7.3.29 to obtain (B). (b) If an D c1 r1n C c2 r2n , then anC2 C ˛anC1 C ˇan D c1 r1n .r12 C ˛r C ˇ/ C c2 r2n .r22 C ˛r2 C ˇ/ D c1 r1n P0 .r1 / C c2 r2n PP 0 .r2 / D 0, so fan g satisfies (B). Since 1=r1 and 1=r2 are the zeros of P0 , n n n Theorem 7.2.1 implies that 1 nD0 .c1 r1 C c2 r2 /x is a solution of (A) on . ; /. 1 X 1 (c) If jxj < , then jr1 xj < and jr2 xj < 1, so rin x n D D yi , i D 1; 2. Therefore, (b) 1 ri x nD0 implies that fy1 ; y2g is a fundamental set of solutions of (A) on . ; /. (d) (A) can written as P0 y 00 C 2P00 y 0 C P000 y D .P0 y/00 D 0. Therefore,P0 y D a C bx where a and b are arbitrary constants, and a partial fraction expansion shows that the general solution of (A) on any a C bx c1 c2 interval not containing 1=r1 or 1=r2 is y D D C D c1 y1 C c2 y2 . P0 .x/ 1 r1 x 1 r2 x (e) If an D c1 r1n C c2r2n , then anC2 C ˛anC1 C ˇan D c1r1n .r12 C ˛r C ˇ/ C c2r1n Œ.n C 2/r12 C ˛.n C 1/r2 C ˇn/ D .c1 C nc2 /r1n P0 .r1 / C c2 r1n P00 .r1 / D 0, so fan g satisfies (B). Since 1=r1 is the only zero
100 Chapter 7 Series Solutions of Linear Second Order Equations of P0 , Theorem 7.2.1 implies that (f) If jxj < , then jr1 xj < , 1 X
P1
n n nD0 .c1 C c2 n/r1 /x is 1 X 1 so r1n x n D 1 r1 x nD0
a solution of (A) on . ; /. D y1 . Differentiating this and multiplying
r1 x D r1 y2 . Therefore, (e) implies that fy1 ; y2g is a .1 r1 x/2 nD0 fundamental set of solutions of (A) on . ; /. (g) The argument is the same as in (c), but now the partial fraction expansion can be written as y D a C bx c1 c2 x D C D c1y1 C c2y2 . P0 .x/ 1 r1 x .1 r2 x/2 the result by x shows that
7.3.32. If y D 3
1 X
nD0
2an
n
an x C 2
2 x
n
1 X
n
nD0
1 X
nr1n x n D
00
0
an x , then y C 2xy C .3 C 2x /y D
an x
nD0
D 0 if a2 D
nC2
D .2a2 C 3a0 / C .6a3 C 5a1 /x C
3a0 =2, a3 D
with a0 D 1 and a1 D 2 yields y D 1 7.3.34. If y D 1 X
1 X
nD0
3a0 / C .6a3 C 2a1 /x C a1 =3, and anC2 D
a1 D 2 yields y D 6
2 2x C 9x 2 C x 3 3
2
1 X
nD0
4an
an x n C 4
2 x
n
1 X
an x n , then y 00
nD0
1 X
nD0
4
an x n
2 x
nD0
1 X
nD0
nD0
an
1 X
n
1 X
1 X
nD1
nan x n C
Œ.n C 2/.n C 1/anC2 C .2n C 3/an C
n.n 1/an x n
nD2
2
2
C5
1 X
, n 2. Starting
nan x n 3
nD1
1 X
nD0
3/an C an
an x n C
2 x
n
D0
.5n 3/an C an 2 , n 2. Starting with a0 D 6 and .n C 2/.n C 1/ 23 4 3 5 x x C . 4 10
a1 /x C
2x 4
x3
1/an x n
n.n
2
3
nD2
1 X
nD2
x 2 /y D 3
nD1
Œ.n C 2/.n C 1/anC2
1 X
nD2
1 X
1 X
n.n
2
.3n
nan x n C 2/an C
, n 2. Starting with a0 D 3
1/an x n
2
C2
1 X
nD1
nan x n C
Œ3.n C 2/.n C 1/anC2 C .2n C 4/an
nD2
.2n C 4/an an 2 , n 2. Starting with 3.n C 2/.n C 1/ 19 4 13 5 x C x C . 54 60
a1 =3, and anC2 D
4 a0 D 2 and a1 D 3 yields y D 2 C 3x C x 2 3
1 X
.3n 2/an 4an .n C 2/.n C 1/ 17 5 x C . 20
an x nC2 D .6a2 C 4a0 / C .18a3 C 6a1 /x C 2a0 =3, a3 D
nD2
3xy 0 C .2 C 4x 2/y D
3x 2 C x 3
C2
nD2
1 X
nD2
an x n , then 3y 00 C 2xy 0 C .4
D 0 if a2 D
1/an x
Œ.n C 2/.n C 1/anC2 C .5n
D 0 if a2 D a0 , a3 D a1 =6, and anC2 D
7.3.38. If y D 1 X
1 X
an x nC2 D .2a2 C 2a0 / C .6a3
and a1 D 6 yields y D 3 C 6x
n.n
n 2
.2n C 3/an C 2an .n C 2/.n C 1/ 3 2 5 3 17 4 11 5 x C x C x x C . 2 3 24 20
2x
if a2 D 3a0 =2, a3 D
7.3.36. If y D
1 X
5a1 =6, and anC2 D
an x n , then y 00 C5xy 0 .3 x 2 /y D
an x nC2 D .2a2
nD0
2
Section 7.3 Series Solutions Near an Ordinary Point II
7.3.40. If y D 1/an x n
1
C
1 X
nD0
1 X
nD1
an x n , then .1 C x/y 00 C x 2 y 0 C .1 C 2x/y D
nan x nC1 C
1 X
nD0
an x n C 2
1 X
1 X
nD2 1 X
an x nC1 D .2a2 C a0 / C
nD0
n.n
nD1
1/an x n
2
C
1 X
101
n.n
nD2
Œ.n C 2/.n C 1/anC2 C .n C
.n C 1/nanC1 C an C .n C 1/an .n C 2/.n C 1/ 1 3 3 4 31 5 2 n 1. Starting with a0 D 2 and a1 D 3 yields y D 2 C 3x C x x x C x C . 6 4 120
1/nanC1 C an C .nC 1/an
7.3.42. If y D 2
1 X
nan x
nD1
1 X
n
D 0 if a2 D
nD1
nan x nC1 C
1 X
nD1
1
a0 =2 and anC2 D
an x n , then .1Cx 2 /y 00 C.2Cx 2 /y 0 Cxy D
nD0 1 X n 1
C
1 x
1 X
n.n 1/an x n
nD2 1 X
an x nC1 D .2a2 C2a1 /C
nD1
2
C
1 X
nD2
,
n.n 1/an x n C
Œ.nC2/.nC1/anC2 C2.nC1/anC1 C
Œ2.n C 1/anC1 C n.n 1/an C nan 1 , n 1. .n C 2/.n C 1/ 23 23 4 11 5 Starting with a0 D 3 and a1 D 5 yields y D 3 C 5x 5x 2 C x 3 x C x C . 6 12 30
n.n 1/an C nan
1 x
n
D 0 if a2 D a1 and anC2 D
7.3.44. The equation is equivalent to y 00 C .1 C 3t 2/y 0 C .1 C 2t/y D 0 with t D x then y 00 C .1 C 3t 2 /y 0 C .1 C 2t/y D 2
1 X
nD0
an t nC1 D .2a2 C a1 C a0 / C
if a2 D .a1 C a0 /=2 and anC2 D and a1 D 3 yields
1 3.x C 2/ C .x C 2/2 2
yD2
1 X
n.n 1/an t n
2
nD2
1 X
nD1
C
1 X
nan t n
1
nD1
C3
1 X
nD1
2. If y D
nan t nC1 C
Œ.n C 2/.n C 1/anC2 C .n C 1/anC1 C an C .3n
Œ.n C 1/anC1 C an C .3n .n C 2/.n C 1/
1/an
1 31 .x C 2/3 C .x C 2/4 3 24
53 .x C 2/5 C . 120
1
nD0
1/an
t 2 /y 00
then .1 8
1 X
nan t n
nD1
9/an
1 X
nD1
.n
2/an
8t C t 2 /y 0 C ty D
.7
nan t nC1 C 1 t
n
1 X
nD0
n.n
1/an t n
nD2
an t nC1 D .2a2 7a1 /C
D 0 if a2 D 7a1 =2 and anC2 D
n 1. Starting with a0 D 2 and a1 D 1 yields 7 43 y D 2 .x C 2/ .x C 2/2 .x C 2/3 2 6
1 X
2
1 X
n.n
an t n ,
nD0
1 X
an t n C
1 t
n
D0
, n 1. Starting with a0 D 2
7.3.46. The equation is equivalent to .1 t 2 /y 00 .7 8t Ct 2 /y 0 Cty D 0 with t D xC2. If y D 1 X
1 X
1/an t n
nD2
7
1 X
1 X
an t n ,
nD0
nan t n
nD1
1
C
Œ.nC2/.nC1/anC2 7.nC1/anC1 n.n
nD1
Œ7.n C 1/anC1 C n.n 9/an C .n .n C 2/.n C 1/
203 .x C 2/4 24
2/an
1
,
167 .x C 2/5 C . 30
7.3.48. The equation is equivalent to .1 C 3t C 2t 2/y 00 .3 C t t 2 /y 0 .3 C t/y D 0 with t D x 1. If 1 1 1 X X X yD an t n , then .1 C 3t C 2t 2 /y 00 .3 C t t 2 /y 0 .3 C t/y D n.n 1/an t n 2 C 3 n.n nD0
nD2
nD2
102 Chapter 7 Series Solutions of Linear Second Order Equations 1/an t n
1
C2
1 X
n.n 1/an t n 3
nD2
.2a2 3a1 3a0 /C
1 X
nD1
1 X
nan t n
1
nD1
1 X
nD1
nan t n C
1 X
nan t nC1 3
nD1
1 X
an t n
nD0
1 X
nD0
an t nC1 D
Œ.nC2/.nC1/anC2 C3.n2 1/anC1 C.2n2 3n 3/.nC1/an C.n 2/an
1 t
n
D0
1/anC1 C .2n2 3n 3/.n C 1/an C .n 2/an 1 ,n .n C 2/.n C 1/ 3 1 1 1. Starting with a0 D 1 and a1 D 0 yields y D 1 C .x 1/2 C .x 1/3 .x 1/5 C . 2 6 8 7.4 REGULAR SINGULAR POINTS; EULER EQUATIONS
if a2 D 3.a1 C a0 /=2 and anC2 D
Œ3.n2
7r C 7 D .r
1/; y D c1 x C c2x 7 .
7.4.2. p.r / D r .r
1/
7.4.4. p.r / D r .r
1/ C 5r C 4 D .r C 2/2 ;y D x
7.4.6. p.r / D r .r
1/
7/.r
3r C 13 D .r
7.4.8. p.r / D 12r .r
1/
5r C 6 D .3r
7.4.10. p.r / D 3r .r
1/
r C 1 D .r
2
.c1 C c2 ln x/
2/2 C 9; y D x 2 Œc1 cos.3 ln x/ C c2 sin.3 ln x/. 2/.4r 1/.3r
3/; y D c1x 2=3 C c2 x 3=4 . 1/; y D c1x C c2 x 1=3 .
7.4.12. p.r / D r .r
1/ C 3r C 5 D .r C 1/2 C 4; y D
7.4.14. p.r / D r .r
1/
r C 10 D .r
1 Œc1 cos.2 ln x/ C c2 sin.2 ln x x
1/2 C 9; y D x Œc1 cos.3 ln x/ C c2 sin.3 ln x/.
c1 C c2x 1=2 . x 1 1 1 2 7.4.18. p.r / D 2r .r 1/C 10r C 9 D 2.r C 2/ C 1; y D 2 c1 cos p ln x C c2 sin p ln x . x 2 2 7.4.16. p.r / D 2r .r
1/ C 3r
1 D .r C 1/.2r
1/; y D
7.4.20. If p.r / D ar .r 1/ C br C c D a.r r1 /2 , then (A) p.r1 / D p 0 .r1 / D 0. If y D ux r1 , then y 0 D u0 x r1 C r1 ux r1 1 and y 00 D u00 x r1 C 2r1u0 x1r1 1 C r1 .r1 1/x1r1 2 , so ax 2 y 00 C bxy 0 C cy
D D
ax r1C2 u00 C .2ar1 C b/x r1 C1 u0 C .ar1 .r1 1/ C br1 C c/ x r1 u r C1 ax r1C2 u00 C p 0 .r1 /x11 u0 C p.r /x r1 u D ax r1 C2 u00 ;
from (A). Therefore,u00 D 0, so u D c1 C c2x and y D x r1 .c1 C c2 x/. 7.4.22. (a) If t D x 1 and Y .t/ D y.t C 1/ D y.x/, then .1 x 2 /y 00 2xy 0 C ˛.˛ C 1/y D d 2Y dY t.2 C t/ 2 2.1 C t/ C ˛.˛ C 1/Y D 0, so y satisfies Legendre’s equation if and only if Y dt dt 2 d Y dY d 2Y satisfies (A) t.2 C t/ 2 C 2.1 C t/ ˛.˛ C 1/Y D 0. Since (A) can be rewritten as t 2 .2 C t/ 2 C dt dt dt dY 2t.1 C t/ ˛.˛ C 1/tY D 0, (A) has a regular singular point at t D 00 . dt d2Y (b) If t D x C 1 and Y .t/ D y.t 1/ D y.x/, then .1 x 2 /y 00 2xy 0 C ˛.˛ C 1/y D t.2 t/ 2 C dt dY d2Y 2.1 t/ C ˛.˛ C 1/Y , so y satisfies Legendre’s equation if and only if Y satisfies (B) t.2 t/ 2 C dt dt
Section 7.5 The Method of Frobenius I
103
dY d 2Y dY C ˛.˛ C 1/Y , Since (B) can be rewritten as (B) t 2 .2 t/ 2 C 2t.1 t/ C ˛.˛ C 1/tY , 2.1 t/ dt dt dt (B) has a regular singular point at t D 00 . 7.5 The Method of Frobenius I 7.5.2. p0 .r / D r .3r 1/; p1 .r / D 2.r C 1/; p2 .r / D 4.r C 2/. 2 2an 1 .r / 4an 2 .r / a1 .r / D ; an .r / D , n 1. 3r C 2 3n C 3r 1 2an 1 .1=3/ 4an 2 .1=3/ , n 1; r1 D 1=3; a1 .1=3/ D 2=3; an .1=3/ D 3n 2 8 40 y1 D x 1=3 1 x C x2 x3 C . 3 9 81 2an 1 .0/ 4an 2 .0/ , n 1; r2 D 0; a1 .0/ D 1; an .0/ D 3n 1 6 4 3 y2 D 1 x C x 2 x C . 5 5 7.5.4. p0 .r / D .r C 1/.4r 1/; p1 .r / D 2.r C 2/; p2 .r / D 4r C 7. 2 2 1 a1 .r / D ; an .r / D an 1 .r / an 2 .r /, n 1. 4r C 3 4n C 4r 1 nCr C1 1 4 r1 D 1=4; a1 .1=4/ D 1=2; an .1=4/ D an 1 .1=4/ an 2 .1=4/, n 1; 2n 4n C5 19 2 1571 3 1 y1 D x 1=4 1 x x C x C . 2 104 10608 2 1 r2 D 1; a1 . 1/ D 2; an . 1/ D an 1 . 1/ an 2 . 1/, n 1; 4n 5 n 11 2 1 3 y2 D x 1 1 C 2x x x C . 6 7 7.5.6. p0 .r / D r .5r 1/; p1 .r / D .r C 1/2 ; p2 .r / D 2.r C 2/.5r C 9/. r C1 nCr a1 .r / D ; an .r / D an 1 .r / 2an 2 .r /, n 1. 5r C 4 5n C 5r 1 5n C 1 an 1 .1=5/ 2an 2 .1=5/, n 1; r1 D 1=5; a1 .1=5/ D 6=25; an .1=5/ D 25n 6 1217 41972 y1 D x 1=5 1 x x2 C x3 C . 25 625 46875 n r2 D 0; a1 .0/ D 1=4; an .0/ D an 1 .0/ 2an 2 .0/, n 1; 5n 1 1 2 35 3 11 4 y2 D x x x C x C . 4 18 12 7.5.8. p0 .r / D .3r 1/.6r C 1/; p1 .r / D .3r C 2/.6r C 1/; p2 .r / D 3r C 5. 6r C 1 6n C 6r 5 1 ; an .r / D an 1 .r / an 2 .r /, n 1. a1 .r / D 6r C 7 6n C 6r C 1 6n C 6r C 1 2n 1 1 r1 D 1=3; a1 .1=3/ D 1=3; an .1=3/ D an 1 .1=3/ an 2 .1=3/, n 1; 2n 6n C 3 C1 1 2 5 3 y1 D x 1=3 1 x C x2 x C . 3 15 63 n 1 1 r2 D 1=6; a1 . 1=6/ D 0; an . 1=6/ D an 1 . 1=6/ an 2 . 1=6/, n 1; n 6n 1 2 1 y2 D x 1=6 1 x C x3 C . 12 18
104 Chapter 7 Series Solutions of Linear Second Order Equations 7.5.10. p0 .r / D .2r C 1/.5r 1/; p1 .r / D .2r 1/.5r C 4/; p2 .r / D 2.2r C 5/.5r 1/. 2r 1 2n C 2r 3 10n C 10r 22 a1 .r / D ; an .r / D an 1 .r / an 2 .r /, n 1. 2r C 3 2n C 2r C 1 5n C 5r 1 10n 13 2n 4 r1 D 1=5; a1 .1=5/ D 3=17; an .1=5/ D an 1 .1=5/ an 2 .1=5/, n 1; 10n C7 n 7 2 547 3 3 x x C . y1 D x 1=5 1 C x 17 153 5661 n 2 20n 54 r2 D 1=2; a1 . 1=2/ D 1; an . 1=2/ D an 1 . 1=2/ an 2 . 1=2/, n 1; 10n 7 n 14 556 3 y2 D x 1=2 1 C x C x 2 x C . 13 897 1 an 1 .r /. nCr C1 1 X 2 . 2/n Qn an 1 .1=2/; y1 D x 1=2 r1 D 1=2; an .1=2/ D x n. 2n C 3 .2j C 3/ j D1 nD0 1 n X 1 . 1/ r2 D 1; an . 1/ D an 1 . 1/; y2 D x 1 xn. n nŠ nD0
7.5.14. p0 .r / D .r C 1/.2r
1/; p1 .r / D 2r C 1; an .r / D
1 an 2n C 2r 1 1 n X . 1/ 1 r1 D 1=2; an .1=2/ D an 1 .1=2/; y1 D x 1=2 x n. 2n 2n nŠ nD0 1 1 X 1 . 1/n Qn r2 D 2; an . 2/ D an 1 . 2/; y2 D 2 xn. 2n 5 x nD0 j D1 .2j 5/
7.5.16. p0 .r / D .r C 2/.2r
1/; p1 .r / D r C 3; an .r / D
1 .r /.
2 an .n C r 1/.2n C 2r 1/ 1 X 2 2n Qn xn. r1 D 1; an .1/ D /an 1 .1/; y1 D x n.2n C 1 nŠ .2j C 1/ j D1 nD0 1 X 2 2n Qn r2 D 1=2; an .1=2/ D x n. /an 1 .1=2/; y2 D x 1=2 n.2n 1 nŠ .2j 1/ j D1 nD0
7.5.18. p0 .r / D .r
1/.2r
1/; p1 .r / D 2; an .r / D
1 .r /.
nCr 4 3; an .r / D an 1 .r /. .n Cr 1/.3n C 3r C 1/ n 3 2 1 an 1 .1/; y1 D x 1 C x C x 2 . r1 D; an .1/ D n.3n C 4/ 7 70 0 1 1 n n X Y 3n 13 . 1/ 3j 13 @ A x n. r2 D 1=3; an . 1=3/ D an 1 . 1=3/; y2 D x 1=3 n nŠ 3n.3n 4/ 3 3j 4 nD0
7.5.20. p0 .r / D .r
1/.3r C 1/; p1 .r / D r
j D1
nCr C1 an 1 .r /. 4n C 4r 1 1 X nC2 . 1/n .n C 2/Š n Q r1 D 1; an .1/ D an 1 .1/; y1 D x x . 4n C 3 2 njD1 .4j C 3/ nD0 1 n X 4n C 5 . 1/n Y r2 D 1=4; an .1=4/ D an 1 .1=4/; y2 D x 1=4 .4j C 5/x n n nŠ 16n 16 nD0
7.5.22. p0 .r / D .r
1/.4r
1/; p1 .r / D r .r C 2/; an .r / D
j D1
Section 7.5 The Method of Frobenius I
105
2n C 2r C 1 an 1 .r /. 3n0C 3r 1 1 n 1 X 6n C 5 . 1/n 2 n Y @ r1 D 1=3; an .1=3/ D 2 an 1 .1=3/; y1 D x 1=3 .6j C 5/A x n ; 9n nŠ 9 nD0 j D1 0 1 1 n X Y 2n 1 2j 1 A xn an 1 . 1/; y2 D x 1 . 1/n 2n @ r2 D 1; an . 1/ D 2 3n 4 3j 4
7.5.24. p0 .r / D .r C 1/.3r
1/; p1 .r / D 2.r C 2/.2r C 3/; an .r / D 2
nD0
7.5.28. p0 .r / D .2r
j D1
.n C r /2 an 1 .r /. .2n C 2r 1/.4n C 4r 1/ 4n2 C 4n C 1 9 5 2 245 3 an 1 .1=2/; y1 D x 1=2 1 xC x x C . 8n.4n C 1/ 128 39936 40 16n2 C 8n C 1 25 675 38025 3 1=4 2 an 1 .1=4/; y2 D x 1 xC x x C . 32n.4n 1/ 96 14336 5046272
1/.4r
r1 D 1=2; an .1=2/ D r2 D 1=4; an .1=4/ D
1/; p1 .r / D .r C 1/2 ; an .r / D
.3n C 3r 2/ 1/.2r C 1/; p1 .r / D .2r C 1/.3r C 1/; an .r / D an .r /. .2n C 2r C 1/ 6n 1 5 55 935 3 r1 D 1=2; an .1=2/ D an 1 .1=2/; y1 D x 1=2 1 x C x2 x C . 4.n C 1/ 96 1536 8 1 5 2 55 3 6n 7 1=2 r2 D 1=2; an . 1=2/ D an 1 . 1=2/; y2 D x 1C x x x C . 4n 4 32 384
7.5.30. p0 .r / D .2r
.n C r /.n C r C 1/ an .r /. .2n C2r C 1/.3n C 3r C 1/ 10 200 2 17600 3 .3n 1/.3n C 2/ 1=3 an 1 . 1=3/; y1 D x 1 xC x x C . 9n.6n C 1/ 63 7371 3781323 .2n 1/.2n C 1/ 3 9 2 105 3 1=2 an 1 . 1=2/; y2 D x 1 xC x x C . 4n.6n 1/ 20 352 23936
7.5.32. p0 .r / D .2r C 1/.3r C 1/; p1 .r / D .r C 1/.r C 2/; an .r / D r1 D
1=3; an . 1=3/ D
r2 D
1=2; an . 1=2/ D
7.5.34. p0 .r / D .2r
1/.4r
r1 D 1=2; a2m .1=2/ D
8m 3 a2m 8m C 1
r2 D 1=4; a2m .1=4/ D
2m 1 a2m 2m
7.5.36. p0 .r / D r .3r
1/; p2 .r / D .r
r1 D 1=3; a2m .1=3/ D r2 D 0; a2m .0/ D 7.5.38. p0 .r / D .2r
8m C 4r 5 .2r C 3/.4r C 3/; a2m .r / D a2m 8m 0 1 C 4r 1 1 m X Y 8j 3 A 2m 1=2 @ .1=2/; y D x x . 2 1 8j C1 mD0 j D1 0 1 1 m X Y 1 1=4 @ .2j 1/A x 2m 2 .1=4/; y2 D x m mŠ 2 mD0
1/; p2 .r / D
6m 17 a2m 18m
2m 6m 1/.3r
6 a2m 1
2 .r /.
j D1
2m C r 6 a2m 6m C 3r0 1 1 m X . 1/m Y 1=3 @ .6j 2 .1=3/; y1 D x 18m mŠ mD0
4/.r C 2/; a2m .r / D
4 2 8 4 x 2 .0/; y2 D 1 C x C 5 55
1/; p2 .r / D .r C 1/.3r C 5/; a2m .r / D
j D1
2 .r /.
1
17/A x 2m .
2m C r 1 a2m 4m C 2r 1
2 .r /.
106 Chapter 7 Series Solutions of Linear Second Order Equations 0 1 m Y 1 1=2 @ .4j 1/A x 2m . 2 .1=2/; y1 D x m mŠ 8 mD0 0 j D1 1 1 m X Y 6m 2 3j 1 A x 2m . a2m 2 .1=3/; y2 D x 1=3 2m @ r2 D 1=3; a2m .1=3/ D 12m 1 12j 1 mD0 1 X
4m 1 r1 D 1=2; a2m .1=2/ D a2m 8m
j D1
7.5.40. p0 .r / D .2r
4m 1 a2m 4.2m C 1/
r1 D 1=2; a2m .1=2/ D r2 D
4m 3 a2m 8m
1=2; a2m . 1=2/ D
7.5.42. p0 .r / D .r C 1/.3r r1 D 1=3; a2m .1=3/ D r2 D
2m C r 1 a2m 2 .r /. 4m 0 C 2r C 1 1 1 m X . 1/m @ Y 4j 1 A 2m 1=2 x . 2 .1=2/; y1 D x 4m 2j C 1 mD0 j D1 0 1 1 m m X Y . 1/ 1=2 @ .4j 3/A x 2m 2 . 1=2/; y2 D x m mŠ 8 mD0
1/.2r C 1/; p1 .r / D .r C 1/.2r C 3/; a2m .r / D
1; a2m . 1/ D
2m C r 3 a2m 2 .r /. 02m C r C 11 1 m X Y 3j 4 A 2m 1=3 m@ .1=3/; y D x . 1/ x . 2 1 3j C2 mD0
1/; p1 .r / D .r 3m 4 a2m 3m C 2
m
2
m
a2m
2.
j D1
1/.3r C 5/; a2m .r / D
j D1
1/; y2 D x
1
.1 C x 2 /
.2m C r 2/2 a2m 2 .r /. .2m C r C 1/.4m 0 C 2r 1/ 1 1 m m 2 X Y .4m 3/2 . 1/ .4j 3/ @ A x 2m . r1 D 1=2; a2m .1=2/ D a2m 2 .1=2/; y1 D x 1=2 m mŠ 8m.4m C 3/ 8 4j C 3 mD0 0 j D1 1 1 m 2 m 2 X Y .2m 3/ . 1/ .2j 3/ @ A x 2m . r2 D 1; a2m . 1/ D a2m 2 . 1/; y2 D x 1 m mŠ 2m.4m 3/ 2 4j 3 mD0
7.5.44. p0 .r / D .r C 1/.2r
1/; p1 .r / D r 2; a2m .r / D
j D1
1 a2m 2 .r /. 6m C 3r C 1 1 X 1 . 1/m Q r1 D 1=3; a2m .1=3/ D a2m 2 .1=3/; y1 D x 1=3 x 2m . m m 2.3m C 1/ 2 .3j C 1/ j D1 mD0 1 m X 1 . 1/ r2 D 1=3; a2m . 1=3/ D a2m 2 . 1=3/; y2 D x 1=3 x 2m m mŠ 6m 6 mD0
7.5.46. p0 .r / D .3r
1/.3r C 1/; p1 .r / D 3r C 5; a2m .r / D
2m C r C 1 1/; p2 .r / D .r C 3/2 ; a2m .r / D a2m 2 .r /. 2.8m C 4r 1/ 8m C 5 13 2 273 4 2639 6 r1 D 1=4; a2m .1=4/ D a2m 2 .1=4/; y1 D x 1=4 1 x C x x C . 64m 64 8192 524288 m 1 2 2 4 2 6 1 r2 D 1; a2m . 1/ D a2m 2 . 1/; y2 D x 1 x C x x C . 8m 5 3 33 209
7.5.48. p0 .r / D 2.r C 1/.4r
7.5.50. p0 .r / D .2r
1/.2r C 1/; p2 .r / D .2r C 5/2 ; a2m .r / D
4m C 2r C 1 a2m 4m C 2r 1
2 .r /.
Section 7.5 The Method of Frobenius I
107
2m C 1 3 2 15 4 35 6 a2m 2 .1=2/; y1 D x 1=2 1 x C x x C . 2m 2 8 16 2m 8 16 6 1=2 2 4 r2 D 1=2; a2m . 1=2/ D a2m 2 . 1=2/; y2 D x 1 2x C x x C . 2m 1 3 5 P P1 n n 7.5.52. (a) Multiplying (A) c1 y1 C c2y2 0 by x r2 yields c1x r1 r2 1 nD0 an x C c2 nD0 bn x D 0, 0 < x < . Letting x ! 0C shows that c2 D 0, since b0 D 1. Now (A) reduces to c1y1 0, so c1 D 0. Therefore,y1 and y2 P are linearly independent on .0; /. P1 1 n n (b) Since y1 D a .r /x and y D 2 nD0 n 1 nD0 an .r2 /x are linearly independent solutions of Ly D 0 .0; /, fy1 ; y2 g is a fundamental set of solutions of Ly D 0 on .0; /, by Theorem 5.1.6. r1 D 1=2; a2m .1=2/ D
7.5.54. (a) If x > 0, then jxjr x n D x nCr , so the assertions are obvious. If x < 0, then jxjr D . x/r , so r . x/r r jxjr d r jxjr n d jxjr D r . x/r 1 D D . Therefore,(A) .jxjr x n / D x C jxjr .nx n 1 / D dx x x dx x d2 d .n C r /jxjr x n 1 and .jxjr x n / D .n C r / .jxjr x n 1 / D .n C r /.n C r 1/jxjr x n 2 , from (A) dx 2 dx with n replaced by n 1. p2 .n C r 2/ an 2 .r /, p0 .n C r / r 0, which implies that a2mC1 .r / D 0 for m D 1; 2; 3; : : : . Therefore,Eqn. (7.5.12) actually reduces p2 .2m C r 2/ to a0 .r / D 1, a2m .r / D , which holds because of condition (A). p0 .2m C r / (b) Similar to the proof of Exercise 7.5.55(a). (c) p0 .2m C r1/ D 2m˛0.2m C r1 r2 /, which is nonzero if m > 0, since r1 r2 0. Therefore, the assumptions of Theorem 7.5.2 hold with r D r1 , and Ly1 D p0 .r1 /x r1 D 0. If r1 r2 is not an even integer, then p0 .2m C r2/ D 2m˛0 .2m r1 C r2/ ¤ 0, m D 1; 2; . Hence, the assumptions of Theorem 7.5.2 hold with r D r2 and Ly2 D p0 .r2 /x r2 D 0. From Exercise 7.5.52, fy1 ; y2 g is a fundamental set of solutions. (d) Similar to the proof of Exercise 7.5.55(c). 7.5.56. (a) Here p1 0, so Eqn. (7.5.12) reduces to a0 .r / D 1, a1 .r / D 0, an .r / D
7.5.58. (a) From Exercise 7.5.57, bn D 0 for n 1. 7.5.60. (a) .˛0 C˛1 x C˛2 x 2/ 1 X
1 X
nD0
an x n D ˛0 a0 C.˛0 a1 C˛1a0 /x C
1 X
nD2
.˛0 an C˛1 an
1 C˛2 an 2 /x
n
D
˛0 a0 . 2 ˛ C ˛ 0 1 x C ˛2 x nD0 p1 .r 1/ ˛1 p2 .r 2/ ˛2 D and D , then Eqn. (7.5.12) is equivalent to a0 .r / D 1, ˛0a1 .r /C (b) If p0 .r / ˛0 p0 .r / ˛0 ˛1 a0 .r / D 0, ˛0an .r / C ˛1 an 1 .r / C ˛2 an 2 .r / D 0, n 2. Therefore,Theorem 7.5.2 implies the conclusion.
1, so
an x n D
7.5.62. p0 .r / D .2r
1/.3r
1/; p1 .r / D 0; p2 .r / D 2.2r C 3/.3r C 5/;
p1 .r 1/ ˛1 D0D ; p0 .r / ˛0
p2 .r 2/ ˛2 x 1=3 x 1=2 D2D ; y1 D ; y2 D . 2 p0 .r / ˛0 1 C 2x 1 C 2x 2 7.5.64. p0 .r / D 5.3r
1/.3r C 1/; p1 .r / D .3r C 2/.3r C 4/; p2 .r / D 0;
p2 .r 2/ ˛2 x 1=3 x 1=3 D0D ; y1 D ; y2 D . p0 .r / ˛0 5Cx 5Cx
p1 .r 1/ 1 ˛1 D D ; p0 .r / 5 ˛0
108 Chapter 7 Series Solutions of Linear Second Order Equations 7.5.66. p0 .r / D .2r 3D
3/.2r
1/; p1 .r / D 3.2r
1/.2r C 1/; p2 .r / D .2r C 1/.2r C 3/;
p1 .r 1/ D p0 .r /
˛2 x 1=2 x 3=2 ˛1 p2 .r 2/ ; D1D ; y1 D ; y D . 2 ˛0 p0 .r / ˛0 1 C 3x C x 2 1 C 3x C x 2
7.5.68. p0 .r / D 3.r
1/.4r
1/; p1 .r / D 2r .4r C 3/; p2 .r / D .r C 1/.4r C 7/;
p1 .r 1/ 2 ˛1 D D ; p0 .r / 3 ˛0
p2 .r 2/ 1 ˛2 x x 1=4 D D ; y1 D ; y2 D . 2 p0 .r / 3 ˛0 3 C 2x C x 3 C 2x C x 2 7.6 THE METHOD OF FROBENIUS II 7.6.2. p0 .r / D .r C 1/2 ; p1 .r / D .r C 2/.r C 3/; p2 .r / D .r C 3/.2r 1/; r C3 nCr C2 2n C 2r 5 a1 .r / D ; an .r / D an 1 .r / an 2 .r /, n 2. r C2 nCr C1 nCr C1 1 nCr C2 0 2n C 2r 5 0 1 a10 .r / D ; a0 .r / D a .r / a .r / C an 1 .r / .r C 2/2 n nCr C1 n 1 nCr C1 n 2 .n C r C 1/2 7 an 2 .r /, n 2. .n C r C 1/2 nC1 2n 7 r1 D 1; a1 . 1/ D 2; an . 1/ D an 1 . 1/ an 2 . 1/, n 2; n n 9 20 3 y1 D x 1 1 2x C x 2 x C ; 2 3 nC1 2n 7 1 7 a10 . 1/ D 1; an0 . 1/ D an 1 . 1/ an 2 . 1/ C 2 an 1 . 1/ an 2 . 1/, n 2; n n n n2 15 133 2 y2 D y1 ln x C 1 xC x C . 4 18 7.6.4. p0 .r / D .2r 1/2 ; p1 .r / D .2r C 1/.2r C 3/; p2 .r / D .2r C 1/.2r C 3/; .2n C 2r C 1/an 1 .r / .2n C 2r 3/an 2 .r / 2r C 3 a1 .r / D ; an .r / D , n 2. 2r C 1 2n C 2r 1 0 0 4 .2n C 2r C 1/an 1 .r / .2n C 2r 3/an 2 .r / 4.an 1 .r / an 2 .r // a10 .r / D ; a0 .r / D C ; .2r C 1/2 n 2n C 2r 1 .2n C 2r 1/2 n 2. .n C 1/an 1 .1=2/ C .n 1/an 2 .1=2/ ; n 2; r1 D 1=2; a1 .1=2/ D 2; an .1=2/ D n 5 y1 D x 1=2 1 2x C x 2 2x 3 C ; 2 .n C 1/an0 1 .1=2/ C .n 1/an0 2 .1=2/ an 1 .1=2/ an 2 .1=2/ a10 .1=2/ D 1; an0 .1=2/ D C ,n n n2 2; 9 17 2 3=2 y2 D y1 ln x C x 1 x C x C . 4 6 7.6.6. p0 .r / D .3r C 1/2 ; p1 .r / D 3.3r C 4/; p2 .r / D 2.3r C 7/; 3 3an 1 .r / C 2an 2 .r / a1 .r / D ; an .r / D ; n 2; 3r C 4 3n C 3r C 1 9 3an0 1 .r / C 2an0 2 .r / 9an 1 .r / 6an 2 .r / a10 .r / D ; an0 .r / D C ; n 2. 2 .3r C 4/ 3n C 3r C 1 .3n C 3r C 1/2 3an 1 . 1=3/ C 2an 2 . 1=3/ r1 D 1=3; a1 . 1=3/ D 1; an . 1=3/ D , n 2; 3n 5 1 3 y1 D x 1=3 1 x C x 2 x C ; 6 2
Section 7.6 The Method of Frobenius II 3an0 1 .r / C 2an0 2 .r / 3an a10 . 1=3/ D 1; an0 . 1=3/ D C 3n 11 25 x C x2 C . y2 D y1 ln x C x 2=3 1 12 36
1 .r /
2an 3n2
2 .r /
109
; n 2;
7.6.8. p0 .r / D .r C 2/2 ; p1 .r / D 2.r C 3/2 ; p2 .r / D 3.r C 4/; 3an 2 .r / a1 .r / D 2; an .r / D 2an 1 .r / ; n 2; nCr C2 0 3an 2 .r / 3an 2 .r / C , n 2. a10 .r / D 0; an0 .r / D 2an0 1 .r / nCr C2 .n C r C 2/2 3an 2 . 2/ r1 D 2; a1 . 2/ D 2; an . 2/ D 2an 1 . 2/ , n 2; n 5 y1 D x 2 1 2x C x 2 3x 3 C ; 2 3an 2 . 2/ 3an 2 . 2/ a10 . 2/ D 0; an0 . 2/ D 2an 1 . 2/ C ; n 2; n n2 3 13 y2 D y1 ln x C x C . 4 6 7.6.10. p0 .r / D .4r C 1/2 ; p1 .r / D 4r C 5; p2 .r / D 2.4r C 9/; 1 an 1 .r / C 2an 2 .r / a1 .r / D ; an .r / D ; n 2; 4r C 5 4n C 4r C 1 0 0 an 1 .r / C 2an 2 .r / 4an 1 .r / C 8an 2 .r / 4 a10 .r / D ; a0 .r / D C ; n 2. .4r C 5/2 n 4n C 4r C 1 .4n C 4r C 1/2 an 1 . 1=4/ C 2an 2 . 1=4/ r1 D 1=4; a1 . 1=4/ D 1=4; an . 1=4/ D ; n 2; 4n 1 7 2 23 3 y1 D x 1=4 1 x x C x C ; 4 32 384 an0 1 . 1=4/ C 2an0 2 . 1=4/ an 1 . 1=4/ C 2an 2 . 1=4/ a10 . 1=4/ D 1=4; an0 . 1=4/ D C ;n 4n 4n2 2; 5 157 2 1 C x x C . y2 D y1 ln x C x 3=4 4 64 2304 7.6.12. p0 .r / D .2r
1/2 ; p1 .r / D 4;
4 an 1 .r /; .2n C 2r 1/2 n . 4/ an .r / D Qn . 1/ j D1 .2j C 2r
an .r / D
By logarithmic differentiation, an0 .r / D an .r / . 1/n ; r1 D 1=2; an .1=2/ D 2 0 .nŠ/ 1 n X 1 A; an0 .1=2/ D an .1=2/ @ 2 j 1 X . 1/n n y1 D x 1=2 x ; .nŠ/2 nD0
j D1
n X
j D1
2 2j C 2r
1
;
110 Chapter 7 Series Solutions of Linear Second Order Equations
y2 D y1 ln x
0 1 1 n n X X . 1/ 1 @ A xn; 2x 1=2 2 .nŠ/ j nD1 j D1
2/2 ; p1 .r / D r 2 ; an .r / D
7.6.14. p0 .r / D .r
.n C r .n C r
1/2 an 2/2
1 .r /;
an .r / D . 1/n
.n C r 1/2 ; .r 1/2
2n.r C n 1/ ; r1 D 2; an .2/ D . 1/n .n C 1/2 ; an0 .2/ D . 1/nC1 2n.n C 1/; .r 1/3 1 1 X X y1 D x 2 . 1/n .n C 1/2 x n ; y2 D y1 ln x 2x 2 . 1/n n.n C 1/x n . an0 .r / D . 1/nC1 nD0
nD1
7.6.16. p0 .r / D .5r 1/2 ; p1 .r / D r C 1; .n C r / an 1 .r /; an .r / D .5n C 5r 1/2 n Y .j C r / an .r / D . 1/n ; .5j C 5r 1/2 j D1
By logarithmic differentiation, n X .5j C 5r C 1/ an0 .r / D an .r / ; .j C r /.5j C 5r 1/ j D1
n Y .5j C 1/ r1 D 1=5; an .1=5/ D . 1/ ; 125n .nŠ/2 n
j D1
n X 5j C 2 ; an0 .1=5/ D an .1=5/ j.5j C 1/ j D1 Q 1 X . 1/n njD1 .5j C 1/ n 1=5 y1 D x x ; 125n .nŠ/2 nD0 0 1 Qn 1 n n X X . 1/ .5j C 1/ 5j C 2 j D1 @ A x n. y2 D y1 ln x x 1=5 125n.nŠ/2 j.5j C 1/ nD1
7.6.18. p0 .r / D .3r 1/2 ; p1 .r / D .2r .2n C 2r 3/2 an .r / D an 1 .r /; .3n C 3r 1/2 n Y .2j C 2r 3/2 an .r / D . 1/n ; .3j C 3r 1/2 j D1
j D1
1/2 ;
By logarithmic differentiation, n X 1 an0 .r / D 14an .r / ; .2j C 2r 3/.3j C 3r 1/ j D1 Q . 1/n njD1 .6j 7/2 r1 D 1=3; an .1=3/ D ; 81n .nŠ/2 n X 1 an0 .1=3/ D 14an .1=3/ /; j.6j 7/ j D1 Q 1 X . 1/n njD1 .6j 7/2 n 1=3 y1 D x x ; 81n .nŠ/2 nD0
Section 7.6 The Method of Frobenius II
y2 D y1 ln x C 14x
1 X . 1/n 1=3 nD1
Qn
j D1 .6j 81n .nŠ/2
7/2
0 @
n X
j D1
1 1 /A x n . j.6j 7/
7.6.20. p0 .r / D .r C 1/2 ; p1 .r / D 2.r C 2/.2r C 3/; n Y 2.2n C 2r C 1/ 2j C 2r C 1 an .r / D an 1 .r /, n 1; an .r / D 2n ; nCr C1 j Cr C1 j D1
By logarithmic differentiation, n X 1 ; an0 .r / D an .r / .j C r C 1/.2j C 2r C 1/ j D1 Q 2n njD1 .2j 1/ ; r1 D 1; an . 1/ D nŠ n X 1 an0 . 1/ D an . 1/ ; j.2j 1/ j D1 Qn 1 n 1/ n 1X2 j D1 .2j y1 D x ; x nD0 nŠ 0 1 Qn 1 n 1 X 2n j D1 .2j 1/ X 1 @ A x n. y2 D y1 ln x C x nD1 nŠ j.2j 1/ j D1
7.6.22. p0 .r / D 2.r 2n C 2r an .r / D 2.n C r n . 1/n Y an .r / D 2n
j D1
2/2 ; p1 .r / D .r 1 an 1 .r /; 2/ 2j C 2r 1 ; j Cr 2
1/.2r C 1/;
By logarithmic differentiation, n X 1 an0 .r / D 3an .r / ; .j C r 2/.2j C 2r 1/ j D1 Q . 1/n njD1 .2j C 3/ r1 D 2; an .2/ D ; 2n nŠ n X 1 an0 .2/ D 3an .2/ ; j.2j C 3/ j D1 Q 1 X . 1/n njD1 .2j C 3/ n y1 D x 2 x ; 2n nŠ nD0 0 1 Qn 1 n n X X . 1/ .2j C 3/ 1 j D1 @ A x n. y2 D y1 ln x 3x 2 n nŠ 2 j.2j C 3/ nD0 j D1
7.6.24. p0 .r / D .r 3/2 ; p1 .r / D 2.r 1/.r C 2/; 2.n C r 2/.n C r C 1/ an .r / D an 1 .r /; .n C r 3/2 2.n C r 2/.n C r C 1/ 0 2.5n C 5r 7/ an0 .r / D an 1 .r / an 2 .n C r 3/ .n C r 3/3 2.n C 1/.n C 4/ r1 D 3; an .3/ D an 1 .3/; n2
1 .r /;
111
112 Chapter 7 Series Solutions of Linear Second Order Equations y1 D x 3 .1 C 20x C 180x 2 C 1120x 3 C ; 2.n C 1/.n C 4/ 0 2.5n C 8/ an0 .3/ D an 1 .3/ an n2 n3 6968 y2 D y1 ln x x 4 26 C 324x C x2 C 3
1 .3/;
7.6.26. p0 .r / D r 2; p1 .r / D r 2 C r C 1; .n2 C n.2r 1/ C r 2 r C 1/ an 1 .r /; an .r / D .n C r /2 2 2 .n C n.2r 1/ C r r C 1/ 0 .n C r 2/ an0 .r / D an 1 .r / an 2 .n C r / .n C r /3 .n2 n C 1/ r1 D 0; an .0/ D an 1 .0/; n2 3 7 3 x C ; y1 D 1 x C x 2 4 12 2 .n 2/ .n n C 1/ 0 an0 .0/ D an 1 .0/ an 1 .0/; 2 n n3 3 5 y2 D y1 ln x C x 1 x C x2 C . 4 9 7.6.28. p0 .r / D .r
1/2 ; p2 .r / D r C 1;
. 1/m 1 a2m 2 .r /, n 1; a2m .r / D Qm 2m C r 1 j D1 .2j C r By logarithmic differentiation, m X 0 .r / D a2m .r / a2m ;
a2m .r / D
j D1
. 1/m ; 2m mŠ m X1 1 a2m .1/ ; 2 j
r1 D 1; a2m .1/ D 0 a2m .1/ D
j D1
1 X . 1/m 2m y1 D x x ; 2m mŠ mD0
y2 D y1 ln x
0 1 1 m x X . 1/m @ X 1 A 2m x . 2 2m mŠ j
7.6.30. p0 .r / D .2r
mD1
j D1
1/2 ; p2 .r / D 2r C 3;
1 a2m 2 .r /; 4m C 2r 1 m . 1/ a2m .r / D Qm ; 1/ j D1 .4j C 2r By logarithmic differentiation, m X 1 0 a2m .r / D 2a2m .r / ; 4j C 2r 1
a2m .r / D
1 .r /;
j D1
r1 D 1=2; a2m .1=2/ D
. 1/m ; 4m mŠ
1/
;
Section 7.6 The Method of Frobenius II m
0 a2m .1=2/ D
y1 D x 1=2
X1 1 a2m .1=2/ ; 2 j j D1
1 X . 1/m 2m x ; 4m mŠ
mD0
y2 D y1 ln x
0 1 1 m x 1=2 X . 1/m @ X 1 A 2m x . 2 mD1 4m mŠ j j D1
7.6.32. p0 .r / D .2r a2m .r / D . 1/m
1/2 ; p2 .r / D .r C 1/.2r C 3/; a2m .r / D
m Y
2j C r 1 ; 4j C 2r 1
j D1
By logarithmic differentiation, m X 0 a2m .r / D a2m .r / .2j C r j D1
r1 D 1=2; a2m .1=2/ D 0 a2m .1=2/ D a2m .1=2/
y1 D x 1=2
1 X . 1/m
mD0
y2 D y1 ln x C
x
1=2
2
. 1/m
m X
j D1 Q m
1 1/.4j C 2r
Qm
j D1 .4j 8m mŠ
1 2j.4j
j D1 .4j
1/ 1/
8m mŠ
1 X . 1/m
mD1
Qm
1/
1/
;
;
;
x 2m ;
j D1 .4j
1/
8m mŠ
7.6.34. p0 .r / D .4r C 1/2 ; p2 .r / D .r 2m C r 3 a2m .r / D a2m 2 .r /; 8m C 4r C 1 m Y 2j C r 3 a2m .r / D . 1/m ; 8j C 4r C 1
2m C r 1 a2m 4m C 2r 1
0 @
m X
j D1
1/.4r C 9/;
1
1 A x 2m . j.4j 1/
j D1
By logarithmic differentiation, m X 0 a2m .r / D a2m .r / .2j C r
13 ; 3/.8j C 4r C 1/ j D1 Q . 1/m m 13/ j D1 .8j r1 D 1=4; a2m . 1=4/ D ; .32/m mŠ m X 13 0 a2m . 1=4/ D a2m . 1=4/ ; 2j.8j 13/ j D1 Q 1 X . 1/m m 13/ 2m j D1 .8j 1=4 y1 D x x ; .32/m mŠ mD0 0 1 Qm 1 m 13 1=4 X . 1/m j D1 .8j 13/ X 1 @ A x 2m . y2 D y1 ln x C x m mŠ 2 .32/ j.8j 13/ mD1 j D1
7.6.36. p0 .r / D .2r
1/2 ; p2 .r / D 16r .r C 1/;
2 .r /;
113
114 Chapter 7 Series Solutions of Linear Second Order Equations 16.2m C r 2/.2m C r 1/ a2m 2 .r /; .4m C 2r 1/2 m Y .2j C r 2/.2j C r 1/ ; a2m .r / D . 16/m .4j C 2r 1/2 a2m .r / D
j D1
By logarithmic differentiation, m X 0 a2m .r / D a2m .r / .2j C r j D1
r1 D 1=2; a2m .1=2/ D 0 a2m .1=2/ D a2m .1=2/
y1 D x 1=2
1 X .
mD0
. 1/m
m X
8j C 4r 5 2/.2j C r 1/.4j C 2r
Qm
j.4j
3/.4j j D1 .4j m 2 4 .mŠ/ 8j 3 3/.4j
j D1 Q 1/m m 3/.4j j D1 .4j m 2 4 .mŠ/
y2 D y1 ln x C x 1=2
1 X . 1/m
mD1
1/ 1/
Qm
j D1 .4j
1/
;
; x 2m ;
3/.4j
4m .mŠ/2
7.6.38. p0 .r / D .r C 1/2 ; p2 .r / D .r C 3/.2r 4m C 2r 5 a2m .r / D a2m 2 .r /; 2m C r C 1 m Y 4j C 2r 5 a2m .r / D . 1/m ; 2j C r C 1
;
1/
1/;
1/
0 @
m X
j D1
j.4j
j D1
8j 3 3/.4j
By logarithmic differentiation, m X 7 0 a2m .r / D a2m .r / ; .2j C r C 1/.4j C 2r 5/ j D1 Q . 1/m m 7/ j D1 .4j r1 D 1; a2m . 1/ D ; m 2 mŠ m X 7 0 a2m . 1/ D a2m . 1/ ; 2j.4j 7/ j D1 Qm 1 1 X . 1/m j D1 .4j 7/ 2m y1 D x ; x mD0 2m mŠ 0 1 Qm 1 m 7 X . 1/m j D1 .4j 7/ X 1 @ A x 2m . y2 D y1 ln x C 2x mD1 2m mŠ j.4j 7/ j D1
1/2 ; p2 .r / D r C 1; 1 a2m .r / D a2m 2 .r /; 2m C r 1 1 1 0 a2m .r / D a0 .r / C 2m C r 1 2m 2 .2m C r 1 r1 D 1; a2m .1/ D a2m 2 .1/; 2m 1 2 1 4 1 6 y1 D x 1 x C x x C ; 2 8 48
7.6.40. p0 .r / D .r
1/2
a2m
2 .r /;
1/
1
A x 2m .
Section 7.6 The Method of Frobenius II
115
1 1 0 a2m 2 .1/ C a2m 2 .1/, m 1; 2m 4m2 3 2 11 4 3 1 y2 D y1 ln x C x x C x C . 4 32 576 0 a2m .1/ D
7.6.42. p0 .r / D 2.r C 3/2 ; p2 .r / D r 2 2r C 2; 4m2 C 4m.r 3/ C r 2 6r C 10 a2m .r / D a2m 2 .r /; 2.2m C r C 3/2 4m2 C 4m.r 3/ C r 2 6r C 10 0 12m C 6r 19 0 a2m 2 .r / a2m a2m .r / D 2.2m C r C 3/2 .2m C r C 3/3 2 4m 24m C 37 r1 D 3; a2m . 3/ D a2m 2 . 3/; 8m2 17 2 85 4 85 6 3 x C x x C ; y1 D x 1 8 256 18432 2 4m 24m C 37 0 37 12m 0 a2m . 3/ D a2m 2 . 3/ C a2m 2 . 3/, m 1; 2 8m3 8m 25 471 2 1583 4 y2 D y1 ln x C x 1 x C x C . 8 512 110592
2 .r /;
7.6.44. p0 .r / D .r C 1/2 ; p1 .r / D 2.2 r /.r C 1/; r1 D 1. n Y 2.n C r /.n C r 3/ .j C r /.j C r 3/ n an .r / D a .r /; a .r / D 2 , n 0. Therefore,an . 1/ D n 1 n .n C r C 1/2 .j C r C 1/2 j D1
0 if n 1 and y1 D 1=x. If n 4, then an .r / D .r C1/2 bn .r /, where bn0 . 1/ exists; therefore an0 . 1/ D 4.r 2/.r 1/ 2.r 2/ 0 if n 4. For r D 1; 2; 3, an .r / D .r C 1/cn .r /, where c1 .r / D , c2 .r / D , 2 .r C 2/ .r C 2/.r C 3/2 8r .r 2/.r 1/ c3 .r / D . Hence, a10 . 1/ D c1. 1/ D 6, a20 . 1/ D c2. 1/ D 6, a30 . 1/ D .r C 2/.r C 3/.r C 4/2 8 2 c3 . 1/ D 8=3, and y2 D y1 ln x 6 C 6x x . 3 7.6.46. p0 .r / D .r C 1/2 ; p1 .r / D .r 1/.r C 2/; r1 D 1. m Y nCr 2 j Cr 2 an .r / D an 1 .r /; an .r / D , n 0. Therefore,a1 . 1/ D nCr C1 j Cr C1 and an . 1/ D 0 if n 3, so y1 D
j D1 2
.x
1/
x
2, a2 . 1/ D 1,
.
r 1 0 3 r .r 1/ 6.r 2 C 2r 1/ 0 0 , a1 .r / D , a . 1/ D 3; a .r / D , a .r / D , 2 r C2 .r C 2/2 1 .r C 2/.r C 3/ 2 .r C 2/2 .r C 3/2 r .r 1/ a20 . 1/ D 3; if n 3 an .r / D .r C 1/cn .r / where cn .r / D , so .n C r /.n C r 1/.n C r C 1/ 1 X 2 1 an0 . 1/ D cn . 1/ D and y2 D y1 ln x C 3 3x C 2 x n. 2 n.n 2/.n 1/ n.n 1/ nD2 a1 .r / D
7.6.48. p0 .r / D .r 2/2 ; p1 .r / D .r 5/.r 1/; r1 D 2. nCr 6 an .r / D an 1 .r /; nCr 2 m Y j Cr 6 an .r / D , n 0. Therefore,a1 .2/ D 3, a2 .2/ D 3, a3 .2/ D j Cr 2 j D1
n 4, so y1 D x 2 .1
x/3 .
1, and an .2/ D 0 if
116 Chapter 7 Series Solutions of Linear Second Order Equations r 5 0 4 , a1 .r / D , a0 .2/ D 4; r 1 .r 1/2 1 .r 5/.r 4/ 0 4.2r 2 10r C 5/ 0 a2 .r / D , a2 .r / D , a2 .2/ D 7; r .r 1/ r 2 .r 1/2 4 .r 5/.r 4/.r 3/ 0 12.r 8r 3 C 16r 2 5/ 0 , a3 .r / D , a3 .2/ D 11=3; if n 4, then a3 .r / D r .r 1/.r C 1/ r 2.r 1/2 .r C 1/2 .r 5/.r 4/.r 3/ an .r / D .r 2/cn .r / where cn .r / D , so an0 .2/ D .n C r 5/.n C r 4/.n C r 3/.n C r 2/ 6 cn .2/ D and n.n 2/.n2 1/ ! 1 X 11 2 1 3 n y2 D y1 ln x C x 4 7x C x 6 x . 3 n.n 2/.n2 1/ a1 .r / D
nD3
7.6.50. p0 .r / D .3r 1/2 ; p2 .r / D 7 3r ; r1 D 1=3. 6m C 3r 13 a2m 2 .r /; a2m .r / D .6m C 3r 1/2 m Y 6j C 3r 13 a2m .r / D , m 0. Therefore,a2 .1=3/ D 1=6 and a2m .1=3/ D 0 if m 2, so .6j C 3r 1/2 j D1 1 2 y1 D x 1=3 1 x . 6 3r 7 3.19 3r / 0 a2 .r / D ; a0 .r / D ; a .1=3/ D 1=4. If m 2, then a2m .r / D .r 1=3/c2m .r / .3r C 5/2 2 .3r C 5/3 2 3.3r 7/ 0 Q where c2m .r / D , so a2m .1=3/ D c2m .1=3/ D .6m C 3r 7/.6m C 3r 1/ m 1/ j D1 .6j C 3r 1 1 , and m 1 12 6 .m 1/m mŠ ! 1 1 X 1 7=3 1 2m y2 D y1 ln x C x x . 4 12 6m m.m C 1/.m C 1/Š mD1
7.6.52. p0 .r / D .2r C 1/2 ; p2 .r / D 7 2r ; r1 D 1=2. 4m C 2r 11 a2m .r / D a2m 2 .r /; .4m C 2r C 1/2 m Y 4j C 2r 11 a2m .r / D , m 0. Therefore,a2 . 1=2/ D 1=2, a4 . 1=2/ D 1=32, and .4j C 2r C 1/2 j D1 1 2 1 a2m . 1=2/ D 0 if m 3, so y1 D x 1=2 1 x C x4 . 2 32 2r 7 2.19 2r / 0 0 a2 .r / D , a .r / D , a . 1=2/ D 5=8, .2r C 5/2 2 .2r C 5/3 2 .2r 7/.2r 3/ 4.8r 3 60r 2 146r C 519/ 0 0 a4 .r / D , a .r / D , a4 . 1=2/ D 9=128; if 4 .2r C 5/2 .2r C 9/2 .2r C 5/3 .2r C 9/3 m 3, then a2m .r / D .r C 1=2/c2m .r / where 2.2r 7/.2r 3/ 0 Q c2m .r / D , so a2m . 1=2/ D c2m . 1=2/ D .4m C 2r 7/.4m C 2r 3/.4m C 2r C 1/ m .4j C 2r C 1/ j D1 1 , and m 4 .m 2/.m 1/m mŠ
Section 7.6 The Method of Frobenius II
y2 D y1 ln x C x
3=2
5 8
1 9 2 X x C 128 4mC1 .m mD2
n . 1/n Y p1 .j C r1 ˛0n .nŠ/2 j D1
j D1
n . 1/n Y p1 .j C r 1/ . Therefore, an .r1 / D ˛0n .j C r r1 /2 j D1
1/. Theorem 7.6.2 implies Ly1 D 0.
(b) From (A), ln jan .r /j D n 0 X p1 .j C r an .r / p1 .j C r
! 1 2m x . 1/m.m C 1/.m C 1/Š
r1 /2 , then (A) an .r / D
7.6.54. (a) If p0 .r / D ˛0 .r
117
1/ 1/
n ln j˛0 j C
2 j Cr
r1
n X
j D1
and
.ln jp1 .j C r
an0 .r1 /
1/j
2 ln jj C r
n 0 X p1 .j C r1 D an .r1 / p1 .j C r1 j D1
r1 j/, so an0 .r / D 1/ 1/
2 . Theoj
rem 7.6.2 implies that Ly2 D 0. (c) Since p1 .r / D 1 , y1 and y2 reduce to the stated forms. If 1 D 0, then y1 D x r1 and y2 D r1 x ln x, which are solutions of the Euler equation ˛0 x 2 y 00 C ˇ0 xy 0 C 0 y. Q 7.6.54. (a) Ly1 D p0 .r1 /x r1 D 0. Now use the fact that p0 .j C r1 / D ˛0 j 2 , so njD1 p0 .j C r1 / D ˛0n .nŠ/2 . 1 X (b) From Theorem 7.6.2, y2 D y1 ln x C x r1 an0 .r1 /x n is a second solution of Ly D 0. Since nD1
n n X . 1/n Y p1 .j C r 1/ an .r / D , (A) ln ja .r /j D n ln j˛ j C ln jp1 .j C r n 0 n ˛0 .j C r r1 /2 j D1
j D1
1/j
2
n X
j D1
ln jj C
r r1 j, provided that p1 .j Cr 1/ and j Cr r1 are nonzero for all positive integers j . Differentiating (A) n n X X p10 .j C r1 1/ 1 a0 .r1 / and then setting r D r1 yields n D 2 , which implies the conclusion. an .r1 / p2 .j C r1 1/ j j D1 j D1 n X 1 . 1/n 1 n 0 and Jn D 2 . If (c) In this case p1 .r / D 1 and p1 .r / D 0, so an .r1 / D .nŠ/2 ˛0 j j D1
1 D 0, then y1 D x r1 and y2 D x r1 ln x, while the differential equation is an Euler equation with indicial polynomial ˛0 .r r12 /. See Theorem 7.4.3. 7.6.56. p0 .r / D r 2; p1 .r / D 1; r1 D 0. a2m .r / D
a2m 1 .r / . 1/m Q , m 1; a .r / D , 2m m 2 .2m C r /2 j D1 .2j C r /
1 X . 1/m . 1/m 2m , so y D x . 1 m 2 4 .mŠ/ 4m .mŠ/2 mD0 m X 1 0 0 By logarithmic differentiation, a2m .r / D 2a2m .r / , so a2m .0/ D 2m C r j D1 0 1 1 m X X . 1/m 1 @ A x 2m . y2 D y1 ln x m .mŠ/2 4 j mD1
m 0. Therefore,a2m .0/ D
a2m .0/
m X 1 and j
j D1
j D1
7.6.58. p0 .r / D .2r 1/2 ; p1 .r / D .2r C1/2; p2 .r / D 0; y1 D
x 1=2 x 1=2 ln x ; y2 D . 1Cx 1Cx
p1 .r 1/ ˛1 p2 .r 2/ ˛2 D1D ; D0D ; p0 .r / ˛0 p0 .r / ˛0
118 Chapter 7 Series Solutions of Linear Second Order Equations 1/2 ; p1 .r / D 0; p2 .r / D
7.6.60. p0 .r / D 2.r 1=2 D
˛2 x x ln x ; y1 D ; y2 D . 2 ˛0 2 x 2 x2 1/2 ; p1 .r / D 3r 2; p2 .r / D 0;
7.6.62. p0 .r / D 4.r y1 D
.r C 1/2 ;
x x ln x ; y2 D . 4 C 3x 4 C 3x
p1 .r 1/ ˛1 p2 .r 2/ ˛2 D 3=4 D ; D0D ; p0 .r / ˛0 p0 .r / ˛0
1/2 ; p1 .r / D 2r 2; p2 .r / D .r C 1/2;
7.6.64. p0 .r / D .r
p1 .r 1/ ˛1 p2 .r 2/ D 0 D ; D p0 .r / ˛0 p0 .r /
p1 .r 1/ ˛1 p2 .r 2/ D 2D ; D1D p0 .r / ˛0 p0 .r /
˛2 x x ln x ; y1 D ; y2 D . 2 ˛0 .1 x/ .1 x/2 7.6.66. See the proofs of Theorems 7.6.1 and 7.6.2. 7.7 THE METHOD OF FROBENIUS III 7.7.2. p0 .r / D r .r
1/; p1 .r / D 1; r1 D 1; r2 D 0; k D r1 1 an .r / D an 1 .r /; .n C r /.n C r 1/ . 1/n an .r / D Qn ; 1/ j D1 .j C r /.j C r . 1/n an .1/ D ; nŠ.n C 1/Š 1 X . 1/n y1 D x x n; nŠ.n C 1/Š nD0 ´ D 1; C D p1 .0/a0 .0/ D 1. By logarithmic differentiation, n X 2n C 2r 1 an0 .r / D an .r / ; .n C r /.n C r 1/ an0 .1/ D an .1/ y2 D 1
j D1 n X
j D1
y1 ln x C x
r2 D 1;
2j C 1 ; j.j C 1/ 1 X
nD1
0 1 n . 1/n @ X 2j C 1 A n x . nŠ.n C 1/Š j.j C 1/ j D1
7.7.4. p0 .r / D r .r 1/; p1 .r / D r C1; r1 D 1; r2 D 0; k D r1 r2 D 1; an .r / D . 1/n j D1 .j C r
Qn
1/
; an .1/ D
1 X . 1/n . 1/n n ; y1 D x x D xe nŠ nŠ nD0
By logarithmic differentiation, an0 .r / D an .r /
n X
j D1
1 j Cr
x
an 1 .r / ; an .r / D nCr 1
; ´ D 1; C D
p1 .0/a0 .0/ D
n X 1 ; an0 .1/ D an .1/ ; y2 D 1 1 j j D1
1.
0 1 1 n X . 1/n X 1 @ A y1 ln x C x nŠ j nD1
7.7.6. p0 .r / D .r 1/.r C 2/; p1 .r / D r C 3; r1 D 1; r2 D 2; k D r1 r2 D 3. an .r / D 1 X 1 . 1/n . 1/n . 1/n n an 1 .r /; an .r / D Qn ; an .1/ D ; y1 D x x D xe x ; nCr 1 nŠ nŠ .j C r 1/ j D1 nD0
j D1
Section 7.7 The Method of Frobenius III 1 1 p1 .0/ a2 . 2/ D 1=2. 1 C x C x2 ; C D 2 2 3 n n X X 1 1 an .r / ; an0 .1/ D an .1/ ; y2 D x 2 1 C j Cr 1 j
´ D x
2
j D1
j D1
119
By logarithmic differentiation, an0 .r / D 0 0 1 1 1 n X 1 1 2 1 . 1/n X 1 @y1 ln x x @ A x n A; xC x 2 2 2 nŠ j nD1 j D1
7.7.8. p0 .r / D .r C 2/.r C 7/; p1 .r / D 1; r1 D 2; r2 D 7; k D r1 r2 D 5; an .r / D n Y . 1/n an 1 .r / . 1/n ; an .r / D Qn ; an . 2/ D 120 ; .n C r C 2/.n C r C 7/ nŠ.n C 5/Š j D1 .j C r C 2/.j C r C 7/ j D1
1 120 X . 1/n x n; x2 nŠ.n C 5/Š nD0 1 1 1 3 1 4 p1 . 3/ 7 x C x ;C D a4 . ´Dx 1 C x C x2 C 4 24 144 576 5 n X 2j C 2r C 9 mic differentiation, an0 .r / D an .r / ; a0 . .j C r C 2/.j C r C 7/ n j D1 0 1 1 1 1 1 @y1 ln x y2 D x 7 1 C x C x 2 C x3 C x4 4 24 144 576 2880
y1 D
7/ D
1=2880. By logarith-
n X 2j C 5 ; j.j C 5/ j D1 0 1 1 1 n 120 X . 1/n @ X 2j C 5 A n A x . x2 nŠ.n C 5/Š j.j C 5/
2/ D an . 2/
nD1
j D1
7.7.10. p0 .r / D r .r 4/; p1 .r / D .r 6/.r 5/; r1 D 4; r2 D 0; k D r1 r2 D 4; an .r / D n Y .j C r 7/.j C r 6/ .n C r 7/.n C r 6/ an 1 .r /; an .r / D . 1/n . Setting r D 4 yields .n C r /.n C r 4/ .j C r /.j C r 4/ j D1 2 p1 .3/ .r 6/.r 5/ 4 y1 D x 1 x . ´ D 1C10xC50x 2C200x 3; C D a3 .0/ D 300. a1 .r / D ; 5 4 .r 3/.r C 1/ 2 2 3.3r 22r C 31/ 0 .r 6/.r 5/ a10 .r / D ; a1 .4/ D 27=25. a2 .r / D .r 4/c2 .r /, with c2 .r / D , 2 2 .r 3/ .r C 1/ .r 3/.r 2/.r C 1/.r C 2/ 2 0 0 so a20 .4/ D c2 .4/ D 1=30. If n 3, then an .r / D .r 4/ bn .r /where bn .4/ exists, so an .4/ D 0 and 1 6 27 y2 D 1 C 10x C 50x 2 C 200x 3 300 y1 ln x C x 5 x . 25 30 7.7.12. p0 .r / D .r 2/.r C 2/; p1 .r / D 2r 1; r1 D 2; r2 D 2; k D r1 r2 D 4; an .r / D n Q Y 2j C 2r 1 2n C 2r 1 n 2j C3 1 an 1 .r /; an .r / D ; an .2/ D nŠ j D1 j C4 ; .j C r 2/.j C r C 2/ .n C r 2/.n C r C 2/ j D1 0 1 1 n X 1 Y 2j C 3 1 2 1 3 p1 .1/ 2 n 2 @ A y1 D x x ;´ D x 1CxC x x ;C D a3 . 2/ D 1=16. nŠ j C4 4 12 4 nD0 j D1
By logarithmic differentiation, n n X X j 2 C j.2r 1/ C r 2 r C 4 .j 2 C 3j C 6/ an0 .r / D 2an .r / ; an0 .2/ D 2an .2/ ; .j C r 2/.j C r C 2/.2j C 2r 1/ j.j C 4/.2j C 3/ j D1 j D1 0 10 1 1 n n 2 X 2 Y X 1 1 1 x 1 2j C 3 .j C 3j C 6/ @ A@ A x n. y2 D x 2 1 C x C x 2 x3 y1 ln x C 4 12 16 8 nD1 nŠ j C4 j.j C 4/.2j C 3/ j D1
j D1
7.7.14. p0 .r / D .r C 1/.r C 7/; p1 .r / D .r C 5/.2r C 1/; r1 D 1; r2 D 7; k D r1 n Y .n C r C 4/.2n C 2r 1/ .j C r C 4/.2j C 2r 1/ r2 D 6; an .r / D an 1 .r /; an .r / D . 1/n ; .n C r C 1/.n C r C 7/ .j C r C 1/.j C r C 7/ j D1
120 Chapter 7 Series Solutions of Linear Second Order Equations 0 n . 1/n Y .j C 3/.2j @ an . 1/ D nŠ j C6 C D
0 1 n 1 X . 1/n Y .j C 3/.2j A; y1 D @ x nD0 nŠ j C6 j D1 j D1 p1 . 2/ 26 143 2 a5 . 7/ D 0; y2 D x 7 1 C x C x . 6 5 20 3/
1
3/
1
A x n; ´ D x
7
26 143 2 1C xC x ; 5 20
7.7.16. p0 .r / D .3r .n C r an .r / D .3n C 3r
10/.3r C 2/; p1 .r / D r .3r 4/; r1 D 10=3; r2 D 2=3; k D r1 r2 D 4; n Y .j C r 1/.3j C 3r 7/ 1/.3n C 3r 7/ an 1 .r /; an .r / D . 1/n ; 10/.3n C 3r C 2/ .3j C 3r 10/.3j C 3r C 2/ j D1 0 1 0 1 n 1 n n X Y . 1/n .n C 1/ Y 3j C 7 . 1/ .n C 1/ 3j C 7 @ A; y1 D x 10=3 @ A x n; ´ D an .10=3/ D n 9n j C4 9 j C 4 nD0 j D1 j D1 4 1 p .7=3/ 4 1 2 1 2=3 2 2=3 x 1C x x ;C D a3 . 2=3/ D 0; y2 D x 1C x x . 27 243 36 27 243 7.7.18. p0 .r / D .r 3/.r C 2/; p1 .r / D .r C 1/2 ; r1 D 3; r2 D 2; k D r1 r2 D 5; n Y .n C r /2 .j C r /2 an 1 .r /; an .r / D . 1/n ; an .3/ D an .r / D .n C r 3/.n C r C 2/ .j C r 3/.j C r C 2/ j D1 0 1 n 1 Q 2 n Y X n 1 p1 .2/ 2 . 1/ .j C 3/ n . 1/ .j C3/ 3 n 2 @ A ; y1 D x x ;´ D x 1C x ;C D a4 . 2/ D j D1 j C5 nŠ nŠ j C5 4 5 nD0 j D1 1 2 0; y2 D x 1C x . 4 7.7.20. p0 .r / D .r 6/.r 1/; p1 .r / D .r 8/.r 4/; r1 D D 6; r2 D 1; k D r1 r2 D 5; an .r / .n C r 9/.n C r 5/ 2 1 2 21 21 2 35 3 6 an 1 .r /; y1 D x 1 C x C x ; ´ D x 1 C x C x C x ; .n C r 6/.n C r 1/ 3 7 4 2 4 21 2 35 3 p1 .5/ 21 C D a5 .1/ D 0; y2 D x 1 C x C x C x . 6 4 2 4 7.7.22. p0 .r / D r .r 10/; p1 .r / D 2.r 6/.r C 1/; r1 D 10; r2 D 0; k D r1 r2 D 10; an .r / D .n C r 9/.n C r 8/.n C r 7/ . 1/n 2n .n C 1/.n C 2/.n C 3/ 2.n C r 7/ an 1 .r /; an .r / D . 2/n ; an .10/ D ; n C r 10 .r 9/.r 8/.r 7/ 6 1 x 10 X 4 5 40 3 40 4 32 5 16 6 y1 D . 1/n 2n .n C 1/.n C 2/.n C 3/x n ; ´ D 1 x C x2 x C x x C x ; 6 nD0 3 3 21 21 21 21 p1 .9/ 4 5 40 3 40 4 32 5 16 6 C D a9 .0/ D 0; y2 D 1 x C x2 x C x x C x . 10 3 3 21 21 21 21 Pk 1 r2 2m Note: in the solutions to Exercises 7.7.23–7.7.40, ´ D x . mD0 a2m .r2 /x
7.7.24. p0 .r / D .r 6/.r 2/; p2 .r / D r ; r1 D 6; r2 D 2; k D .r1 r2/=2 D 2; a2m .r / D 1 X a2m 2 .r / . 1/m . 1/m . 1/m 2m 2 ; a2m .r / D Qm x D x 6 e x =2 ; ; a2m .6/ D m ; y1 D x 6 m 2m C r 6 2 mŠ 2 mŠ 6/ j D1 .2j C r mD0 m X 1 2 p2 .4/ 1 2 0 ´ D x 1C x ;C D a2 .2/ D 1=2. By logarithmic differentiation, a2m .r / D a2m .r / 2 4 2j C r j D1 0 1 m 1 m 6 X m X X 1 1 1 x . 1/ 1 0 @ A x 2m . a2m .6/ D a2m .6/ ; y2 D x 2 1 C x 2 y1 ln x C 2j 2 2 4 mD1 2m mŠ j j D1
j D1
6
;
Section 7.7 The Method of Frobenius III
121
7.7.26. p0 .r / D .r 1/.r C 1/; p2 .r / D 2r C 10; r1 D 1; r2 D 1; k D .r1 r2 /=2 D 1; m Y 2.2m C r C 3/ 2j C r C 3 a2m .r / D a2m 2 .r /; a2m .r / D . 2/m ; .2m C r 1/.2m C r C 1/ .2j C r 1/.2j C r C 1/ 1 . 1/ .m C 2/ x X . 1/m .m C 2/ 2m ; y1 D x ;´Dx a2m .1/ D 2 mŠ 2 mD0 mŠ By logarithmic differentiation, m X .4j 2 C 4j.r C 3/ C r 2 C 6r C 1/ 0 ; a2m .r / D a2m .r / .2j C r 1/.2j C r C 1/.2j C r C 3/
j D1
m
0 a2m .1/ D a2m .1/
y2 D x
1
j D1 m X
j D1
1
;C D
p2 . 1/ a0 . 1/ D 2
4.
j 2 C 4j C 2 ; 2j.j C 1/.j C 2/
0 1 m 1 X . 1/m .m C 2/ @ X j 2 C 4j C 2 A 2m 4y1 ln x C x x . mŠ j.j C 1/.j C 2/ mD1
j D1
7.7.28. p0 .r / D .2r C 1/.2r C 5/; p2 .r / D 2r C 3; r1 D 1=2; r2 D 5=2; k D .r1 r2 /=2 D 1; m Y .4m C 2r 1/ .4j C 2r 1/ a2m 2 .r /; a2m .r / D . 1/m ; a2m .r / D .4m C 2r C 1/.4m C 2r C 5/ .4j C 2r C 1/.4j C 2r C 5/ j D1 Q Q 1 X . 1/m m 1/ . 1/m m 1/ 2m j D1 .2j j D1 .2j 1=2 a2m . 1=2/ D ; y1 D x x ; ´ D x 5=2 ; C D m m 8 mŠ.m C 1/Š 8 mŠ.m C 1/Š mD0 p2 . 5=2/ a0 . 5=2/ D 1=4. By logarithmic differentiation, 8 m m X X .16j 2 C 8j.2r 1/ C 4r 2 4r 11/ 2j 2 2j 1 0 0 a2m .r / D 2a2m .r / ; a2m . 1=2/ D a2m . 1=2/ ; .4j C 2r 1/.4j C 2r C 1/.4j C 2r C 5/ 2j.j C 1/.2j 1/ j D1 j D1 0 1 Qm 1 m m 2 X X . 1/ .2j 1/ 1 2j 1 2j j D1 @ A x 2m . y2 D x 5=2 C y1 ln x x 1=2 mC1 mŠ.m C 1/Š 4 8 j.j C 1/.2j 1/ mD1 j D1
7.7.30. p0 .r / D .r 2/.r C 2/; p2 .r / D 2.r C 4/; r1 D 2; r2 D 2; k D .r1 r2 /=2 D 2; a2m .r / D 1 X 2 2m 1 2m 1 2 a2m 2 .r /; a2m .r / D Qm ; a2m .2/ D ; y1 D x 2 x D x2ex ; 2m C r 2 mŠ mŠ 2/ j D1 .2j C r mD0 m X p2 .0/ 1 0 ´ D x 2 .1 x 2 /; C D a2 . 2/ D 2. By logarithmic differentiation, a2m .r / D a2m .r / 4 2j C r j D1 0 1 m 1 m X X 1 1 X1 0 @ A x 2m . a2m .2/ D a2m .2/ ; y2 D x 2 .1 x 2 / 2y1 ln x C x 2 2j mŠ j mD1 j D1
2
;
j D1
7.7.32. p0 .r / D .3r 13/.3r 1/; p2 .r / D 2.5 3r /; r1 D 13=3; r2 D 1=3; k D .r1 r2 /=2 D 2; m Y 2.6m C 3r 11/ .6j C 3r 11/ a2m .r / D a2m 2 .r /; a2m .r / D 2m ; .6m C 3r 13/.6m C 3r 1/ .6j C 3r 13/.6j C 3r 1/ j D1 Qm 1 Qm X 2 2 j D1 .3j C 1/ j D1 .3j C 1/ 2m 13=3 1=3 a2m .13=3/ D m ; y1 D 2x x ; ´ D x 1C x ; C D 9 mŠ.m C 2/Š 9m mŠ.m C 2/Š 9 mD0 m X p2 .7=3/ .12j 2 C 4j.3r 11/ C 3r 2 22r C 47/ 0 a2 .1=3/ D 2=81. By logarithmic differentiation, a2m .r / D 9a2m .r / ; 36 .6j C 3r 13/.6j C 3r 11/.6j C 3r 1/ j D1
122 Chapter 7 Series Solutions of Linear Second Order Equations m X
3j 2 C 2j C 2 ; 2j.j C 2/.3j C 1/ j D1 0 0 1 1 1 Qm m 2 X X .3j C 1/ 2 2 3j C 2j C 2 j D1 @y1 ln x x 13=3 @ A x 2m A. y2 D x 1=3 1 C x 2 C m mŠ.m C 2/Š 9 81 9 j.j C 2/.3j C 1/ mD0
0 a2m .13=3/ D a2m .13=3/
j D1
7.7.34. p0 .r / D .r 2/.r C 2/; p2 .r / D 3.r 4/; r1D 2; r2 D 2; k D .r1 r2 /=2 D 2; 3.2m C r 6/ 1 2 9 2 2 2 a2m .r / D a2m 2 .r /; y1 D x 1 x ; ´ D x 1C x ; C D .2m C r 2/.2m C r C 2/ 2 2 p2 .0/ 3.r 4/ 0 3.r 2 8r 16/ 0 a2 . 2/ D 27=2; a2 .r / D , a .r / D , a2 .2/ D 7=12. If m 2, 4 r .r C 4/ 2 r 2 .r C 4/2 m 3 .r 4/ Q , so then a2m .r / D .r 2/c2m .r / where c2m .r / D .2m C r 4/.2m C r 2/ m j D1 .2j C r C 2/ 1 3 m X 9 27 7 2 0 a2m .2/ D c2m .2/ D ; y2 D x 2 1 C x 2 y1 ln x C x 4 x 2 m.m 1/.m C 2/Š 2 2 12 m.m mD2
3 m 2
1/.m C 2/Š
x
2m
7.7.36. p0 .r / D .2r 5/.2r C 7/; p2 .r / D .2r 1/2 ; r1 D 5=2; r2 D 7=2; k D .r1 r2 /=2 D 3; .2r 1/.2r C 3/.2r C 7/ 4m C 2r 5 a2m .r / D a2m 2 .r /; a2m .r / D ; a2m .5=2/ D 4m C 2r C 7 .4m C 2r 1/.4m C 2r C 3/.4m C 2r C 7/ 1 X . 1/m . 1/m ; y1 D x 5=2 x 2m ; ´ D x 7=2 .1 C x 2 /2 C D .m C 1/.m C 2/.m C 3/ .m C 1/.m C 2/.m C 3/ mD0 p2 .1=2/ 7=2 a4 . 7=2/ D 0; y2 D x .1 C x 2 /2 . 24 7.7.38. p0 .r / D .r 3/.r C 7/; p2 .r / D r .r C 1/; r1 D 3; r2 D 7; k D .r1 r2 /=2 D 5; a2m .r / D m Y .2j C r 2/.2j C r 1/ .2m C r 2/.2m C r 1/ a2m 2 .r /; a2m .r / D . 1/m ; a2m .3/ D .2m C r 3/.2m C r C 7/ .2j C r 3/.2j C r C 7/ j D1 0 1 0 1 m 1 m Y X Y mC1 2j C 1 mC1 2j C 1 21 35 35 A; y1 D x 3 A x 2m ; ´ D x 7 1 C x 2 C x 4 C x 6 . 1/m m @ . 1/m m @ 2 j C5 2 j C5 8 16 64 mD0 j D1 j D1 p2 .1/ 21 35 35 C D a8 . 7/ D 0; y2 D x 7 1 C x 2 C x 4 C x 6 . 10 8 16 64 7.7.40. p0 .r / D .2r
3/.2r C 5/; p2 .r / D .2r 1/.2r C 1/; r1 D 3=2; r2 D 5=2; k D m Y 4m C 2r 5 4j C 2r 5 .r1 r2 /=2 D 2; a2m .r / D a2m 2 .r /; a2m .r / D . 1/m ; a2m .3=2/ D 4m C 2r C 5 4j C 2r C 5 j D1 Q Q 1 X . 1/m m 1/ . 1/m m 1/ 2m 3 2 p2 . 1=2/ j D1 .2j j D1 .2j 3=2 5=2 ; y1 D x x ;´ D x 1C x C D a2 . 5=2/ D m 1 m 1 2 .m C 2/Š 2 .m C 2/Š 2 16 mD0 3 0; y2 D x 5=2 1 C x 2 . 2 7.7.42. p0 .r / D r 2 2 ; p2 .r / D 1; r1 D ; r2 D ; k D .r1 r2 /=2 D ; a2m .r / D a2m 2 .r / . 1/m . 1/m ; a2m .r / D Qm ; a2m ./ D m Qm ; .2m C r C /.2m C r / / 4 mŠ j D1 .j C / j D1 .2j C r C /.2j C r 1 X . 1/m . 1/m a2m . / D m Qm , j D 0; : : : ; 1; y1 D x Q x 2m ; ´ D m m mŠ 4 mŠ j D1 .j / 4 .j C / j D1 mD0
!
.
Section 7.7 The Method of Frobenius III 1 X
. 1/m Qm j D1 .j
123
p2 . 2/ a2 2 . / 2 a2 2 . / D D . By Š. 2 2 4 1/Š mD0 m m X X 2j C 2j C 0 0 logarithmic differentiation, a2m .r / D 2a2m .r / ; a2m ./ D a2m ./ ; .2j C r C /.2j C r / 2j.j C / j D1 j D1 0 0 1 1 m 1 1 m X m X X 2 2j C . 1/ x . 1/ @ @y1 ln x A x 2m A. Qm x 2m Qm y2 D x Š. m mŠ m mŠ 4 1/Š 2 j.j C / 4 .j / 4 .j C / j D1 j D1 mD0 mD1 x
4m mŠ
/
x 2m ; C D
j D1
7.7.44. Since an .r2 / D
p1 .n C r2 1/ an p0 .n C r2 /
p1 .r1 1/ ak k˛0
But C D and only if
k Y
j D1
p1 .r2 C j
1 .r2 /
1 .r2 /, 1 n k 1, ak
p1 .r2 C k k˛0
D
1/
ak
1 .r2 /
k 1 .r2 / D . 1/
D . /k
1/ D 0.
Qk
1
kY 1
j D1
p1 .r2 C j 1/ . p0 .r2 C j /
p1 .r2 C j 1/ D 0 if Qk 1 k˛0 j D1 p0 .r2 C j / j D1
1 n Qn 7.7.46. Since p1 .r / D 1 , an .r / D an 1 .r / and (A) an .r / D . 1/ ˛0 .n C r r2 / ˛0 j D1 .j C r . 1/n 1 n 1 1 . 1/n 1 n Therefore, an .r1 / D Qn for n 0 (so Ly1 D 0) and an .r2 / D Qn nŠ ˛0 nŠ ˛0 k/ j D1 .j C k/ j D1 .j ! k 1 1 X X for n D 0; : : : ; k 1. Ly2 D 0 if y2 D x r2 an .r2 /x n C C y1 ln x C x r1 an0 .r1 /x n if C D
1 r1 /.n C r
n
nD0
nD1
k
1
1 . 1/k 1 1 k 1 . 1/k 1 1
1 ak 1 .r2 / D D . From (A), ln jan .r /j D k˛0 k˛0 .k 1/Š ˛0 .k 1/Š kŠ.k 1/Š ˛0 ˇ ˇ n n X ˇ 1 ˇ X 1 1 n ln ˇˇ ˇˇ ..ln jj C r r1 j C ln jj C r r2 j/, so an0 .r / D an .r / C ˛0 j C r r1 j C r r2 j D1
j D1
n X 2j C k and an0 .r1 / D an .r1 / . j.j C k/ j D1
@y 7.7.48. (a) From Exercise 7.6.66(a) of Section 7.6, L .x; r / D p00 .r /x r Cx r p0 .r / ln x. Setting r D @r ! 1 X r1 yields L y1 ln x C x r1 an0 .r1 / D p00 .r1 /x r1 . Since p00 .r / D ˛0.2r r1 r2 /, p00 .r1 / D k˛0 . nD1 ! 1 1 X X r2 n (b) From Exercise 7.5.57 of Section 7.5, L x an .r2 /x D x r2 bn x n , where b0 D p0 .r2 / D nD0
0 and bn D while bk D
n X
j D0 k X
j D0
pj .n C r2
pk .k C r2
j /an j /ak
j .r2 /
nD0
if n 1. From the definition of fan .r2 /g, bn D 0 if n ¤ k,
j .r2 / D
k X
j D1
pj .r1
j /ak
j .r2 /.
(d) Let faQ n .r2 /g be the coefficients that would obtained if aQ k .r2 / D 0. Then an .r2 / D aQ n .r2 / if n D
1 r1 /.j C
124 Chapter 7 Series Solutions of Linear Second Order Equations
0; : : : ; k
1, and (A) an .r2 /
aQn .r2 / D
n Xk 1 pj .n C r2 p0 .n C r2 / j D0
j /.an
j .r2 /
aQ n
j r2 /
if n > k.
Now let cm D akCm .r2 / aQ kCm .r2 /. Setting n D m C k in (A) and recalling the k C r2 D r1 yields (B) m X 1 cm D pj .m C r1 j /cm j . Since ck D ak .r2 /, (B) implies that cm D ak .r2 /am .r1 / p0 .m C r1 / j D0
for all m 0, which implies the conclusion.
CHAPTER 8 Laplace Transforms
8.1 INTRODUCTION TO THE LAPLACE TRANSFORM 1 1 1 s2 C 2 1 t t 8.1.2. (a) cosh t sin t D e sin t C e sin t $ C D . 2 2 .s C 1/2 C 1 Œ.s 1/2 C 1Œ.s C 1/2 C 1 2 .s 1/ C 1 1 cos 2t 1 1 s 2 (b) sin2 t D $ D . 2 2 2 s .s C 4/ s.s C 4/ 2 s s2 C 8 1 1 C 2 D . (c) cos2 2t D 2 s s C 16 s.s 2 C 16/ .e t C e t /2 / .e 2t C 2 C e 2t / 1 1 2 1 s2 2 2 (d) cosh t D D $ C C D . 4 4 s 2 s sC2 s.s 2 4/ 4 2t 2t 1 1 1 te te 4s $ D 2 . (e) t sinh 2t D 2 2 2 2 .s 2/ .s C 2/ .s 4/2 sin 2t 1 (f) sin t cos t D $ 2 . 2 s C4 1 sC1 (g) sin.t C =4/ D sin t cos.=4/ C cos t cos.=4/ $ p 2 . 2 s C1 s s 5s (h) cos 2t cos 3t $ 2 D 2 . 2 s C4 s C9 .s C 4/.s 2 C 9/ 2 s s 3 C 2s 2 C 4s C 32 (i) sin 2t C cos 4t $ 2 C 2 D . s C4 s C 16 .s 2 C 4/.s 2 C 16/ Z 1 Z 1 Z 1 st st 0 8.1.6. If F .s/ D e f .t/ dt, then F .s/ D . te /f .t/ dt D e st .tf .t// dt. Apply0
0
ing this argument repeatedly yields the assertion.
0
8.1.8. Let f .t/ D 1 and F .s/ D 1=s. From Exercise 8.1.6, t n $ . 1/n F .n/ .s/ D nŠ=s nC1 . st .s s0 /t 8.1.10. If jf .t/j Me s0 t for t t0 , then for t t0 . Let g.t/ D e st f .t/, R 1 jf .t/e j Me .s s0 /t w.t/ D Me , and D t0 . Since t0 w.t/ dt converges if s > s0, F .s/ is defined for s > s0 .
Z
8.1.12. Z 1 T e s 0
T
e
0 st
st
Z
t
f . / d 0
dt D
e
st
s
Z
t
0
ˇT Z ˇ 1 ˇ f . / d ˇ C s 0
T
e 0
st
f .t/ dt D
e
sT
s
Z
T
f . / d C 0
f .t/ dt. Since f is of exponential order s0 , the second integral on the right converges to 125
126 Chapter 8 Laplace Transforms Z T 1 sT L.f / as T ! 1 (Exercise 8.1.10). Now it suffices to show that (A) lim e f . / d D 0 T !1 s 0 s0 t that jf .t/j Me if t t0 ˇand jf .t/j K if 0 t t0 , and let T > t0 . Then ˇifZ s > s0 . Suppose ˇ ˇ Z t0 ˇ ˇˇZ T Z T ˇ T ˇ ˇ ˇ ˇ Me s0 T ˇ ˇ ˇ ˇ f . / d ˇ ˇˇ f . / d ˇˇ C ˇ f . / d ˇ < Kt0 C M e s0 d < Kt0 C , which ˇ ˇ 0 ˇ ˇ t0 ˇ s0 t0 0 proves (A). Z T Z T 8.1.14. (a) If T > 0, then e st f .t/ dt D e .s s0 /t .e s0 t f .t// dt. Use integration by parts with 0 0 Z T .s s0 /t s0 t .s s0 /t uDe , dv D e f .t/ dt, du D .s s0 /e , and v D g t obtain e st f .t/ dt D 0 ˇT Z T Z T ˇ e .s s0 /t g.t/ ˇˇ C .s s0 / e .s s0 /t g.t/ dt. Since g.0/ D 0 this reduces to e st f .t/ dt D 0 0 Z T0 .s s0 /T .s s0 /t e g.T / C .s s0 / e g.t/ dt. Since jg.t/j M for all t 0, we can let t ! 1 to 0 Z 1 Z 1 conclude that e st f .t/ dt D .s s0 / e .s s0 /t g.t/ dt if s > s0 . 0
0
(b) If F .s0 / exists, then g.t/ is bounded on Œ0; 1/. Now apply (a). ˇZ t ˇ 2 ˇ ˇ 1 d j sin.e t / sin.1/j 2 (c) Since f .t/ D sin.e t /, ˇˇ f . / d ˇˇ D 1 for all t 0. Now apply (a) 2 dt 2 0 with s0 D 0. ˇ Z 1 Z 1 1 ˛ x .˛ C 1/ x ˛ e x ˇˇ1 8.1.16. (a) .˛/ D x ˛ 1 e x dx D C x e dx D . ˇ ˛ ˛ ˛ 0 0 0 Z 1 (b) Use induction. .1/ D e x dx D 1. If (A) .n C 1/ D nŠ, then .n C 2/ D .n C 1/.n C 1/ 0
(from (a)) D .n CZ1/nŠ (from (A)) D .n C 1/Š. 1
(c) .˛ C 1/ D
x˛e
x
0
.˛ C 1/ . ˛
2
dt. Let x D st. Then .˛ C 1/ D
1
Z
2
1 0
.st/˛ e
st
s dt, so
Z
1 0
e
st ˛
t dt D
s e s .s C 1/ e .s 1/ e 2s 1 C C D s2 s2 s2 s2 0 0 1 2e s e 2s 1 .1 e s /2 .1 e s /2 1 e s 1 s C C D . Therefore,F .s/ D D D 2 tanh . s2 s2 s2 s2 s 2 .1 e 2s / s 2.1 C e s / s 2 Z 1 Z 1=2 Z 1 1 e s=2 e s e s=2 2e s=2 (b) e st f .t/ dt D e st dt e st dt D C D C s s s s s 0 0 1=2 e s 1 .1 e s=2 /2 .1 e s=2 /2 1 e s=2 1 s C D . Therefore,F .s/ D D D tanh . s s=2 s Zs s s.1 e / s 4 s.1 C e / Z 1 C e s 1 C e s 1 s st st (c) e f .t/ dt D e sin t dt D 2 . Therefore,F .s/ D 2 coth . s / s 2 C 1 .s C 1/ .s C 1/.1 e 2 0Z Z0 2 1 C e s 1 C e s . Therefore,F .s/ D D (d) e st f .t/ dt D e st sin t dt D .s 2 C 1/ .s 2 C 1/1 C e 2s 0 0 1 . 2 .s C 1/.1 e s /
8.1.18. (a)
Z
e
st
f .t/ dt D
Z
e
st
t dtC
Z
e
st
.2 t/ dt D
Section 8.2 The Inverse Laplace Transform
127
8.2 THE INVERSE LAPLACE TRANSFORM 2s C 3 2.s 7/ C 17 2 17 2Š 17 3Š 17 3 7t 2 8.2.2. (a) D D C D C $ e t C t . .s 7/4 .s 7/4 .s 7/3 .s 7/4 .s 7/3 6 .s 7/4 6 s2 1 Œ.s 2/ C 22 1 .s 2/2 C 4.s 2/ C 3 1 4 3 D D D C C D (b) 6 6 6 4 5 .s 2/ .s 2/ .s 2/ .s 2/6 .s 2/ .s 2/ 1 3Š 1 4Š 1 5Š 1 3 1 4 1 C C $ t C t C t 5 e 2t . 4 5 6 6 .s 2/ 6 .s 2/ 40 .s 2/ 6 6 40 sC5 .s C 3/ 2 3 2 3t (c) 2 D C $ e cos 3t C sin 3t . s C 6s C 18 .s C 3/2 C 9 3 .s C 3/2 C 9 3 2s C 1 s 1 3 1 (d) 2 D2 2 C $ 2 cos 3t C sin 3t. s C9 s C9 3 s2 C 9 3 s .s C 1/ 1 1 1 D D $ .1 t/e t . (e) 2 s C 2s C 1 .s C 1/2 s C 1 .s C 1/2 sC1 s 1 3 1 (f) 2 D 2 C $ cosh 3t C sinh 3t. 2 s 9 s 9 3s 9 3 (g) Expand the numerator in powers of s C 1: s 3 C 2s 2 s 3 D Œ.s C 1/ 13 C 2Œ.s C 1/ 12 s 3 C 2s 2 s 3 1 1 Œ.s C 1/ 1 3 D .s C 1/3 .s C 1/2 2.s C 1/ 1; therefore D 4 .s C 1/ s C 1 .s C 1/2 1 6 2 1 $ 1 t t2 t3 e t. .s C 1/3 6 .s C 1/4 6 .s 1/ 5 2 5 2s C 3 t (h) D 2 C $ e 2 cos 2t C sin 2t . .s 1/2 C 4 .s 1/2 C 4 2 .s 1/2 C 4 2 1 s (i) $ 1 cos t. 2 s s C1 3s C 4 3s 4 3s C 4 3s C 4 (j) 2 D 2 C 2 $ 3 cosh t C 4 sinh t. Alternatively, 2 D D s 1 s 1 s 1 s 1 .s 1/.s C 1/ t t 7 1 7e e 1 $ . 2 s 1 sC1 2 3 4s C 1 1 s 1 3 1 (k) C 2 D3 C4 2 C $ 3e t C 4 cos 3t C sin 3t. 2 s 1 s C9 s 1 s C9 3s C9 3 3 2s C 6 1 s 2 2t (l) D3 2 2 3 2 $ 3te 2 cos 2t 3 sin 2t. .s C 2/2 s 2 C 4 .s C 2/2 s C4 s C4 8.2.4. (a) .s 2 where
A B Cs C D 2 C 3s D C C 2 ; C 1/.s C 2/.s C 1/ sC2 s C1 s C1
A.s 2 C 1/.s C 1/ C B.s 2 C 1/.s C 2/ C .C s C D/.s C 2/.s C 1/ D 2 C 3s: 5A 2B A C 2B C 2D ACB CC
Solving this system yields A D .s 2
4 ,B D 5
2 C 3s C 1/.s C 2/.s C 1/
D D D D
4 1 2 0
1 ,C D 2 D $
.set s D 2/I .set s D 1/I .set s D 0/I .equate coefficients of s 3 /: 3 11 ,D D . Therefore, 10 10
4 1 1 1 1 3s 11 5 s C 2 2 s C 1 10 s 2 C 1 4 2t 1 t 3 11 e e cos t C sin t: 5 2 10 10
128 Chapter 8 Laplace Transforms (b) 3s 2 C 2s C 1 As C B C.s C 1/ C D D 2 C ; 2 2 .s C 1/.s C 2s C 2/ s C1 .s C 1/2 C 1
where
.As C B/..s C 1/2 C 1/ C .C.s C 1/ C D/.s 2 C 1/ D 3s 2 C 2s C 1: 2B C C C D A C B C 2D 2B C C C D ACC
D D D D
1 2 1 0
.set s D 0/I .set s D 1/I .set s D 0/I .equate coefficients of s 3/:
Solving this system yields A D 6=5, B D 2=5, C D 6=5, D D 7=5. Therefore, 3s 2 C 2s C 1 1 6s C 2 6.s C 1/ 7 D .s 2 C 1/.s 2 C 2s C 2/ 5 s 2 C 1 .s C 1/2 C 1 2 6 t 7 6 cos t C sin t e cos t C e $ 5 5 5 5
t
sin t:
(c) s 2 C 2s C 5 D .s C 1/2 C 4; .s
3s C 2 A B.s C 1/ C C D C ; 2 2/..s C 1/ C 4/ s 2 .s C 1/2 C 4
where A .s C 1/2 / C 4 C .B.s C 1/ C C / .s 13A D 4A 3C D ACB D
Solving this system yields A D
8 ,B D 13
3s C 2 2/..s C 1/2 C 4/
.s
8 1 0
2/ D 3s C 2:
.set s D 2/I .set s D 1/I .equate coefficients of s 2 /:
8 15 ,C D . Therefore, 13 13 1 8 8.s 1/ 15 D 13 s 2 .s C 1/2 C 4 8 2t 8 t 15 $ e e cos 2t C e 13 13 26
t
sin 2t:
(d) .s where
3s 2 C 2s C 1 A B C D D C C C ; 2 2 1/ .s C 2/.s C 3/ s 1 .s 1/ sC2 sC3
1/ C B/.s C 2/.s C 3/ C .C.s C 3/ C D.s C 2//.s
.A.s
12B 9C 16D ACC CD
D D D D
3s 2 C 2s C 1 1/2 .s C 2/.s C 3/
D
6 9 22 0
1/2 D 3s 2 C 2s C 1:
.set s D 1/I .set s D 2/I .set s D 3/I .equate coefficients of s 3 /:
Solving this system yields A D 3=8, B D 1=2, C D 1, D D 11=8. Therefore, .s
$
3 1 1 1 1 C C 2 8s 1 2 .s 1/ sC2 3 t 1 t 11 e C te C e 2t e 3t : 8 2 8
11 1 8 s C3
Section 8.2 The Inverse Laplace Transform
129
(e) 2s 2 C s C 3 A B C D D C C C ; 2 2 2 .s 1/ .s C 2/ s 1 .s 1/ sC2 .s C 2/2
where
1/ C B/.s C 2/2 C .C.s C 2/ C D/.s
.A.s
9B 9D 4A C 4B C 2C C D ACC
D D D D
1/2 D 2s 2 C s C 3:
.set s D 1/I .set s D 2/I .set s D 0/I .equate coefficients of s 3 /:
6 9 3 0
Solving this system yields A D 1=9, B D 2=3, C D 1=9, D D 1. Therefore, 2s 2 C s C 3 .s 1/2 .s C 2/2
D $
(f)
1 1 2 1 1 1 1 C C 9s 1 3 .s 1/2 9 s C 2 .s C 2/2 1 t 2 t 1 2t e C te e C te 2t : 9 3 9
3s C 2 A B Cs C D D C C 2 ; .s 2 C 1/.s 1/2 s 1 .s 1/2 s C1
where A.s
1/.s 2 C 1/ C B.s 2 C 1/ C .C s C D/.s
Setting s D 1 yields 2B D 5, so B D A.s
1/2 D 3s C 2:
.A/
5 . Substituting this into (A) shows that 2
1/.s 2 C 1/ C .C s C D/.s
1/2
D 3s C 2 5s 2
D
5 2 .s C 1/ 2 6s C 1 .s D 2
Therefore, A.s 2 C 1/ C .C s C D/.s 2A D A D D ACC D
1
5s 2
1/
:
2 .set s D 1/I 1=2 .set s D 0/I 0 .equate coefficients of s 2 /: 3 . Therefore, 2
Solving this system yields A D 1, C D 1, D D 3s C 2 .s 2 C 1/.s 1/2
1/ D
1/.5s 2
D $
1
5 1 s 3=2 C 2 2 .s 1/2 s C1 5 3 e t C te t C cos t sin t: 2 2 s
1
C
8.2.6. (a) .s 2 where .A.s
A.s 1/ C B C.s C 1/ C D 17s 15 D C 2s C 5/.s 2 C 2s C 10/ .s 1/2 C 4 .s C 1/2 C 9
1/ C B/..s C 1/2 C 9/ C .C.s C 1/ C D/..s
1/2 C 4/ D 17s
15:
:
130 Chapter 8 Laplace Transforms 13B C 8C C 4D 18A C 9B C 8D 10A C 10B C 5C C 5D ACC
D D D D
2 32 15 0
Solving this system yields A D 1, B D 2, C D 1, D D .s 2
17s 15 2s C 5/.s 2 C 2s C 10/
D $
.set s D 1/I .set s D 1/I .set s D 0/I .equate coefficients of s 3/: 4. Therefore,
.s 1/ C 2 .s 1/2 C 4
.s C 1/ C 4 .s C 1/2 C 9 4 e t .cos 2t C sin 2t/ e t cos 3t C sin 3t : 3
(b) .s 2 where .A.s
8s C 56 A.s 3/ C B C.s C 1/ C D D C 2 2 6s C 13/.s C 2s C 5/ .s 3/ C 4 .s C 1/2 C 4
3/ C B/..s C 1/2 C 4/ C .C.s C 1/ C D/..s
20B C 16C C 4D 16A C 4B C 20D 15A C 5B C 13C C 13D ACC
3/2 C 4/ D 8s C 56:
D 80 .set s D 3/I D 48 .set s D 1/I D 56 .set s D 0/I D 0 .equate coefficients of s 3 /:
Solving this system yields A D 1, B D 3, C D 1, D D 1. Therefore, .s 2
8s C 56 6s C 13/.s 2 C 2s C 5/
D $
.s 3/ C 3 .s C 1/ C 1 C 2 .s 3/ C 4 .s C 1/2 C 4 3 e 3t cos 2t C sin 2t C e 2
t
1 cos 2t C sin 2t : 2
(c) s C9 .s 2 C 4s C 5/.s 2
where
.A.s C 2/ C B/..s
4s C 13/
D
A.s C 2/ C B C.s 2/ C D C .s C 2/2 C 1 .s 2/2 C 9
2/2 C 9/ C .C.s
25B 4C C D 36A C 9B C 17D 26A C 13B 10C C 5D ACC
2/ C D/..s C 2/2 C 1/ D s C 9:
D 7 .set s D 2/I D 11 .set s D 2/I D 9 .set s D 0/I D 0 .equate coefficients of s 3 /:
Solving this system yields A D 1=8, B D 1=4, C D 1=8, D D 1=4. sC9 1 .s C 2/ C 2 .s D D 2 2 2 .s C 4s C 5/.s 4s C 13/ 8 .s C 2/ C 1 .s 1 1 2t $ e cos t C sin t 8 4
Therefore, 2/ 2 2/2 C 3 1 2t e cos 3t 8
1 sin 3t : 12
(d) .s 2 where .A.s
3s 2 A.s 2/ C B C.s 3/ C D D C 4s C 5/.s 2 6s C 13/ .s 2/2 C 1 .s 3/2 C 4
2/ C B/..s
3/2 C 4/ C .C.s
3/ C D/..s
2/2 C 1/ D 3s
2:
Section 8.2 The Inverse Laplace Transform 5B C C D 4A C 4B C 2D 26A C 13B 15C C 5D ACC
D D D D
.set s D 2/I .set s D 3/I .set s D 0/I .equate coefficients of s 3 /:
4 7 2 0
Solving this system yields A D 1, B D 1=2, C D 1, D D 1=2. Therefore, 3s 2 1 2.s 2/ C 1 2.s 3/ 1 D D .s 2 4s C 5/.s 2 6s C 13/ 2 .s 2/2 C 1 .s 3/2 C 4 1 1 $ e 2t cos t C sin t e 3t cos 2t sin 2t : 2 4 (e) .s 2 where
3s 1 A.s 1/ C B C.s C 1/ C D D C 2 2 2s C 2/.s C 2s C 5/ .s 1/ C 1 .s C 1/2 C 4
1/ C B/..s C 1/2 C 4/ C .C.s C 1/ C D/..s
.A.s
8B C 2C C D 8A C 4B C 5D 5A C 5B C 2C C 2D A C 5B C C
D D D D
1/2 C 1/ D 3s
1:
.set s D 1/I .set s D 1/I .set s D 0/I .equate coefficients of s 3/:
2 4 1 0
Solving this system yields A D 1=5, B D 2=5, C D 1=5, D D 4=5. Therefore, 1 .s 1/ C 2 .s C 1/ C 4 3s 1 D : .s 2 2s C 2/.s 2 C 2s C 5/ 5 .s 1/2 C 1 .s C 1/2 C 4 2 1 2 1 $ et cos t C sin t e t cos 2t C sin 2t : 5 5 5 5 (f) .4s 2 where .A.s
20s C 40 A.s 1=2/ C B C.s C 1=2/ C D DΠC 2 2 4s C 5/.4s C 4s C 5/ .s 1=2/ C 1 .s C 1=2/2 C 1
1=2/ C B/..s C 1=2/2 C 1/ C .C.s C 1=2/ C D/..s 2B C C C D A C B C 2D 5A C 10B C 5C C 10D ACC
D D D D
25=8 15=8 20 0
1=2/2 C 1/ D
5s C 10 : 4
.set s D 1=2/I .set s D 1=2/I .set s D 0/I .equate coefficients of s 3 /:
Solving this system yields A D 1, B D 9=8, C D 1, D D 1=8. Therefore, 20s C 40 1 8.s 1=2/ C 9 8.s C 1=2/ D C .4s 2 4s C 5/.4s 2 C 4s C 5/ 8 .s 1=2/2 C 1 .s C 1=2/2 C 1 9 1 $ e t =2 cos t C sin t C e t =2 cos t sin t : 8 8 8.2.8. (a) .s 2
2s C 1 C 1/.s 1/.s
3/
D
A s
1
C
B s
3
C
Cs C D s2 C 1
131
132 Chapter 8 Laplace Transforms where 1//.s 2 C 1/ C .C s C D/.s
3/ C B.s
.A.s
D D D D
4A 20B 3A B C 3D ACB CC
1/.s
3/ D 2s C 1:
.set s D 1/I .set s D 3/I .set s D 0/I .equate coefficients of s 3/:
3 7 1 0
Solving this system yields A D 3=4, B D 7=20, C D 2=5, D D 3=10. Therefore, .s 2
2s C 1 C 1/.s 1/.s
3/
3 1 7 1 2 s 3 1 C C 2 2 4 s 1 20 s 3 5 s C 1 10 s C 1 3 t 7 3t 2 3 e C e C cos t sin t: 4 20 5 10
D $
(b)
where
sC2 .s 2 C 2s C 2/.s 2 .A.s C 1/ C B.s
1/
A s
1
C
B C.s C 1/ C D C sC1 .s C 1/2 C 1
1//..s C 1/2 C 1/ C .C.s C 1/ C D/.s 2
10A 2B 2B C D ACB CC
2A
D
D D D D
1/ D s C 2:
.set s D 1/I .set s D 1/I .set s D 0/I .equate coefficients of s 3 /:
3 1 2 0
Solving this system yields A D 3=10, B D 1=2, C D 1=5, D D 3=5. Therefore, sC2 .s 2 C 2s C 2/.s 2
1/
D $
3 1 1 1 1 sC1 C 10 s 1 2 s C 1 5 .s C 1/2 C 1 1 3 t 1 t e e C e t cos te t sin t: 10 2 5
3 1 5 .s C 1/2 C 1
(c) .s 2 where
2s 1 2s C 2/.s C 1/.s
.A.s C 1/ C B.s
2A
2//..s
2/
D
A s
2
1/2 C 1/ C .C.s
6A 15B 4B C 2C 2D ACB CC
D D D D
3 3 1 0
C
B C.s 1/ C D C sC1 .s 1/2 C 1
1/ C D/.s
2/.s C 1/ D 2s
1:
.set s D 2/I .set s D 1/I .set s D 0/I .equate coefficients of s 3 /:
Solving this system yields A D 1=2, B D 1=5, C D 7=10, D D 1=10. Therefore, .s 2
2s 1 2s C 2/.s C 1/.s
2/
D $
1 1 1 1 7 s 1 C 2s 2 5 s C 1 10 .s 1/2 C 1 1 2t 1 7 t 1 t e C e t e cos t e sin t: 2 5 10 10
D
(d) .s 2
s 6 A B Cs C D D C C 2 2 1/.s C 4/ s 1 sC1 s C4
1 10 .s
1 1/2 C 1
Section 8.2 The Inverse Laplace Transform where 1//.s 2 C 4/ C .C s C D/.s 2
.A.s C 1/ C B.s
D D D D
10A 10B 4A 4B D ACB CC
5 7 6 0
1/ D s
6:
.set s D 1/I .set s D 1/I .set s D 0/I .equate coefficients of s 3 /:
Solving this system yields A D 1=2, B D 7=10, C D 1=5, D D 6=5. Therefore, .s 2
s 6 1/.s 2 C 4/
D
D
$
1 1 7 1 1 s 3 1 C C C 2 2s 1 10 s C 1 5 s 2 C 4 5 s C4 1 t 7 1 3 e C e t cos 2t C sin 2t: 2 10 5 5
(e) s.s where .A.s
A B C.s 1/ C D 2s 3 D C C 2 2/.s 2s C 5/ s s 2 .s 1/2 C 4 1/2 C 4/ C .C.s
2/ C Bs/..s
10A 10B 4A C 4B D ACB CC
D D D D
3 1 1 0
1/ C D/s.s
2/ D 2s
3:
.set s D 0/I .set s D 2/I .set s D 1/I .equate coefficients of s 3 /:
Solving this system yields A D 3=10, B D 1=10, C D 2=5, D D 1=5. Therefore, 2s 3 2/.s 2 2s C 5/
s.s
3 1 1 2 s 1 1 C C 2 10s 10 s 2 5 .s 1/ C 4 5 .s 3 1 2t 2 t 1 t C e e cos 2t C e sin 2t: 10 10 5 10
D
D
$
1 1/2 C 4
(f) .s 2 where .A.s
5s 15 4s C 13/.s 2/.s
2/ C B.s
26A
10A 9B 13B 4C C 2D ACB CC
5s 15 4s C 13/.s 2/.s
1/
A s
1
2/2 C 9/ C .C.s
1//..s
Solving this system yields A D 1, B D .s 2
1/
D
D D D D
5=9, C D
D
D
$
et
1 s
1 5 2t e 9
10 5 15 0
C
B s
2
C
C.s 2/ C D .s 2/2 C 9
2/ C D/.s
1/.s
2/ D 5s
15:
.set s D 1/I .set s D 2/I .set s D 0/I .equate coefficients of s 3/:
4=9, D D 1. Therefore, 5 1 4 9 s 2 9 .s 4 2t e cos 3t C 9
s
2 C C9 .s
2/2
1 2t e sin 3t: 3
1 2/2 C 9
133
134 Chapter 8 Laplace Transforms 8.2.10. (a) Let i D 1. (The proof for i D 2; : : : ; n) is similar. Multiplying the given equation through by s s1 yields P .s/ A2 An D A1 C .s s1/ CC ; .s s2/ .s sn / s s2 s sn and setting s D s1 yields A1 D
.s1
P .s1 / s2 / .s2
sn /
. P .s/ D s1 /Q1 .s/ P .s/ s1 shows that D AC Q1 .s/
(b) From calculus we know that F has a partial fraction expansion of the form
.s
A C G.s/ where G is continuous at s1 . Multiplying through by s s1 P .s1 / .s s1 /G.s/. Now set s D s1 to obtain A D . Q.s1 / (c) The result in (b) is generalization of the result in (a), since it shows that if s1 is a simple zero of the denominator of the rational function, then Heaviside’s method can be used to determine the coefficient of 1=.s s1/ in the partial fraction expansion even if some of the other zeros of the denominator are repeated or complex. s
8.3 SOLUTION OF INITIAL VALUE PROBLEMS 8.3.2. .s 2 Since .s 2
s
6/ D .s
s
2 C s.s 1/ D s.s 3/.s C 2/
1 8 4 C e 3t C e 3 15 5
2t
.s 2 4 D .s
2 C s.s s
1/
:
1 8 1 4 1 C C 3s 15 s 3 5sC2
4/Y .s/ D
2 s
3
C . 1 C s/ D
2 C .s
s
1/.s 3
3/
:
2/.s C 2/, Y .s/ D
and y D
1D
.
8.3.4.
Since s 2
2 Cs s
3/.s C 2/, Y .s/ D
and y D
6/Y .s/ D
1 2t 17 e C e 4 20
2t
2 C .s 1/.s 3/ D .s 2/.s C 2/.s 3/
1 1 17 1 2 1 C C 4s 2 20 s C 2 5s 3
2 C e 3t . 5
8.3.6. .s 2 C 3s C 2/Y .s/ D
6 s
1
C . 1 C s/ C 3 D
6 C .s
1/.s C 2/ : s 1
2
Since s C 3s C 2 D .s C 2/.s C 1/, Y .s/ D and y D e t C 2e
2t
2e t .
6 C .s 1/.s C 2/ 1 2 D C .s 1/.s C 2/.s C 1/ s 1 sC2
2 sC1
Section 8.3 Solution of Initial Value Problems 8.3.8. .s 2 Since s 2
3s C 2/Y .s/ D
3s C 2 D .s
1/.s
Y .s/ D and y D 4e t
2 s
2 C .s 3/.s 4/ 4 D .s 1/.s 2/.s 3/ s 1
3s C 2/Y .s/ D
3s C 2 D .s
1/.s
Y .s/ D 5 t e 2
1 s
3
C. 4
2/Y .s/ D
2/ D .s C 2/.s
e
2t
4 C .3 C 2s/ C 2 D s
4 C s.5 C 2s/ 2 D s.s C 2/.s 1/ s
C
2
:
1
.s
1 s
3
3/.s C 1/ : 3
s
4 s
2
C
1 1 2s 3
.s 2 6 D .s
s
1 1 C ; sC2 s 1
6/Y .s/ D
2 Cs s
1D
2 C s.s s
1/
:
3/.s C 2/, 2 C s.s 1/ D s.s 3/.s C 2/
Y .s/ D 8 4 1 C e 3t C e 3 15 5
2t
1 8 1 4 1 C C 3s 15 s 3 5sC2
.
8.3.16. .s 2 1 D .s
4 C s.5 C 2s/ : s
C et .
8.3.14.
Since s 2
4 s
1/,
Y .s/ D
and y D
4/
1 4e 2t C e 3t . 2
Since .s 2 C s
s
s/ C 3 D
5 1 1 .s 3/.s C 1/ D .s 1/.s 2/.s 3/ 2s 1
.s 2 C s
Since s 2
3/.s s 3
2/,
8.3.12.
and y D 2
2 C .s
4e 2t C e 3t . .s 2
and y D
3D
2/,
8.3.10.
Since s 2
3
C . 1 C s/
1/Y .s/ D
1 1 C s2 Cs D : s s
1/.s C 1/, Y .s/ D
and y D 1 C e t C e t .
1 C s2 D s.s 1/.s C 1/
1 1 1 C C s s 1 sC1
135
136 Chapter 8 Laplace Transforms 8.3.18. .s 2 C s/Y .s/ D
2 s
3
C .4
.s 3/2 : s 3
2
1D
s/
2
Since s C s D s.s C 1/, Y .s/ D and y D
7 3
7 e 2
t
2 .s 3/2 7 D s.s C 1/.s 3/ 3s
7 1 1 1 C 2sC1 6s 3
1 C e 3t . 6
8.3.20. .s 2 C 1/Y .s/ D
1 C 2; s2
so Y .s/ D
Substituting x D s 2 into 1 1 D .x C 1/x xC1 so Y .s/ D
1 x
.s 2 C 5s C 6/Y .s/ D Since s 2 C 5s C 6 D .s C 2/.s C 3/, Y .s/ D t
C 4e
2t
.s 2
4e
3t
s2
1 ; C1
2s
3 D .s
3/Y .s/ D
2s
2 2 C .s C 1/.s C 8/ C .3 C s/ C 5 D : sC1 sC1
3/.s C 1/,
10s C .7 C 2s/ C1
s2
4D
10s C .2s C 3/: C1
s2
10s 2s C 3 C : 3/.s C 1/.s 2 C 1/ .s 3/.s C 1/
.s
2s C 3 9 1 D .s 3/.s C 1/ 4s 3 .s
4 sC3
.
Y .s/ D
where
1 1 D 2 2 C 1/s s
1 4 2 C .s C 1/.s C 8/ D C .s C 1/.s C 2/.s C 3/ sC1 sC2
8.3.24.
Since s 2
.s 2
1 2 C 2 : 2 C 1/s s C1
1 1 C 2 and y D t C sin t. s2 s C1
8.3.22.
and y D e
yields
.s 2
1 1 9 $ e 3t 4sC1 4
1 t e : 4
10s A B Cs C D D C C 2 3/.s C 1/.s 2 C 1/ s 3 sC1 s C1
.A.s C 1/ C B.s
3//.s 2 C 1/ C .C s C D/.s
40A 8B A 3B 3D ACB CC
D D D D
30 10 0 0
3/.s C 1/ D 10s:
.set s D 3/I .set s D 1/I .set s D 0/I .equate coefficients of s 3 /:
.A/ .B/
Section 8.3 Solution of Initial Value Problems Solving this system yields A D 3=4, B D 5=4, C D 2, D D .s
10s 3/.s C 1/.s 2 C 1/
D $
From this, (A), and (B), y D 8.3.26.
137
1. Therefore,
3 1 5 1 2s C 1 C 4s 3 4 s C 1 s2 C 1 3 3t 5 t e C e 2 cos t sin t: 4 4
2 cos t C 3e 3t C e t .
sin t
16 9s C 2 C s; so C4 s C1 16 9s s C 2 C 2 : Y .s/ D 2 .s C 4/2 .s C 4/.s 2 C 1/ s C4 .s 2 C 4/Y .s/ D
s2
From the table of Laplace transforms, t cos 2t
$ D
s2 4 s2 C 4 D .s 2 C 4/2 .s 2 C 4/2 1 8 : 2 2 s C 4 .s C 4/2
.s 2
8 C 4/2
Therefore, 8 1 D 2 .s 2 C 4/2 s C4
Substituting x D s 2 into
and multiplying by s yields
L.t cos 2t/; so
16 $ sin 2t .s 2 C 4/2
3 9 D .x C 4/.x C 1/ xC1
9s 3s D 2 .s 2 C 4/.s 2 C 1/ s C1 Finally,
2t cos 2t:
3 xC4
3s $ 3 cos t s2 C 4
3 cos 2t:
s $ cos 2t: C4 .2t C 2/ cos 2t C sin 2t C 3 cos t. s2
Adding (A), (B), and (C) yields y D 28.
.s 2 C 2s C 2/Y .s/ D
.A/
.B/
.C/
2 C . 7 C 2s/ C 4: s2
Since .s 2 C 2s C 2/ D .s C 1/2 C 1, Y .s/ D
2 2s 3 C : s 2 ..s C 1/2 C 1/ .s C 1/2 C 1
2s 3 2.s C 1/ 5 D $ e t .2 cos t 2 .s C 1/ C 1 .s C 1/2 C 1
5 sin t/:
A B C.s C 1/ C D 2 D C 2C ; s 2..s C 1/2 C 1/ s s .s C 1/2 C 1
.A/
.B/
138 Chapter 8 Laplace Transforms where .As C B/ .s C 1/2 C 1 C s 2 .C.s C 1/ C D/ D 2. D D D D
2B ACB CD ACC 2A C B C C C D
.set s D 0/I .set s D 1/I .equate coefficients of s 3 /I .equate coefficients of s 2 /:
2 2 0 0
Solving this system yields A D 1, B D 1, C D 1, D D 0. Therefore, s 2..s
1 1 .s C 1/ C 2C $ s s .s C 1/2 C 1
2 D C 1/2 C 1/
From this, (A), and (B), y D 8.3.30. .s 2 C 4s C 5/Y .s/ D
1 C t C e t .cos t
1Ct Ce
t
cos t:
5 sin t/.
.s C 1/ C 3 C 4. Since .s 2 C 4s C 5/ D .s C 2/2 C 1, .s C 1/2 C 1
Y .s/ D
..s C
1/2
4 sC4 C : 2 C 1/..s C 2/ C 1/ .s C 2/2 C 1
4 $ 4e .s C 2/2 C 1
2t
.A/
sin t:
.B/
sC4 A.s C 1/ C B C.s C 2/ C D D C ; 2 2 ..s C C 1/..s C 2/ C 1/ .s C 1/ C 1 .s C 2/2 C 1 where .A.s C 1/ C B/ .s C 2/2 C 1 C .C.s C 2/ C D/ .s C 1/2 C 1 D 4 C s. 1/2
5A C 5B C 4C C 2D 2B C C C D A C B C 2D ACC
D D D D
.set s D 0/I .set s D 1/I .set s D 2/I .equate coefficients of s 3/:
4 3 2 0
Solving this system yields A D 1, B D 1, C D 1, D D 0. Therefore, ..s C
1/2
s C4 .s C 1/ C 1 s C2 D C ; $ e t . cos t C sin t/ C e 2 2 C 1/..s C 2/ C 1/ .s C 1/ C 1 .s C 2/2 C 1
From this, (A), and (B), y D e t . cos t C sin t/ C e 8.3.32. .2s 2 Since 2s
2
3s
3s
2 D .s
Y .s/ D and y D
1 2t e 5
2/Y .s/ D
s
1
.cos t C 4 sin t/.
C 2. 2 C s/
3D
4 C .2s 7/.s s 1
1/
2/.2s C 1/,
4 C .2s 2.s 2/.s
4 t 32 e C e 3 15
4
2t
t =2
7/.s 1/ 1 1 D 1/.s C 1=2/ 5s 2
4 1 32 1 C 3s 1 15 s C 1=2
.
8.3.34. .2s 2 C 2s C 1/Y .s/ D
2 2 C 2. 1 C s/ C 2 D 2 C 2s: s2 s
2t
cos t:
Section 8.3 Solution of Initial Value Problems
139
Since 2s 2 C 2s C 1 D 2..s C 1=2/2 C 1=4/, Y .s/ D
s 2..s
1 s C : 2 C 1=2/ C 1=4/ ..s C 1=2/2 C 1=4/
s $e ..s C 1=2/2 C 1=4/
where
t =2
.cos.t=2/
sin.t=2//:
.A/ .B/
A B C.s C 1=2/ C D 1 D C 2C s 2 ..s C 1=2/2 C 1=4/ s s ..s C 1=2/2 C 1=4/ .As C B/..s C 1=2/2 C 1=4/ C .C.s C 1=2/ C D/s 2 D 1: B A C 2B C 2D 5A C 10B C 2C C 2D ACC
D D D D
2 8 8 0
.set s D 0/I .set s D 1=2/I .set s D 1=2/I .equate coefficients of s 3 /:
Solving this system yields A D 4, B D 2, C D 4, D D 0. Therefore, 1 s 2 ..s C 1=2/2 C 1=4/ This, (A), and (B) imply that y D e
t =2
4 2 4.s C 1=2/ C 2C s s .s C 1=2/2 C 1=4
D $
4 C 2t C 4e
t =2
sin.t=2// C 2t
.5 cos.t=2/
cos.t=2/: 4.
8.3.36. .4s 2 C 4s C 1/Y .s/ D
3Cs 3Cs C 4. 1 C 2s/ C 8 D 2 C 4. 1 C 2s/ C 8s C 4: 2 s C1 s C1
Since 4s 2 C 4s C 1 D 4.s C 1=2/2 , Y .s/ D
where
3Cs 2 C : 4.s C 1=2/2.s 2 C 1/ s C 1=2
.A/
A B Cs C D 3Cs D C C 2 2 2 2 4.s C 1=2/ .s C 1/ s C 1=2 .s C 1=2/ s C1 .A.s C 1=2/ C B/.s 2 C 1/ C .C s C D/.s C 1=2/2 D 10B 2A C 4B C D 12A C 8B C 9C C 9D ACC
D D D D
5 3 4 0
3Cs : 4
.set s D 1=2/I .set s D 0/I .set s D 1/I .equate coefficients of s 3/:
Solving this system yields A D 3=5, B D 1=2, C D 3=5, D D 1=5. Therefore, 3Cs 4.s C 1=2/2.s 2 C 1/
D $
3 1 1 1 1 3s C 1 C : 2 5 s C 1=2 2 .s C 1=2/ 5 s2 C 1 3 t =2 1 1 e C te t =2 .3 cos t C sin t/: 5 2 5
140 Chapter 8 Laplace Transforms Since
2 $ 2e s C 1=2
t =2
e
t =2
, this and (A) imply that y D
10
1 .3 cos t C sin t/. 5
.5t C 26/
8.3.38. Transforming the initial value problem ay 00 C by 0 C cy D 0; y.0/ D 1; y 0 .0/ D 0 as C b . Therefore, y1 D L as 2 C bs C c satisfies the initial conditions y1 .0/ D 1, y10 .0/ D 0. Transforming the initial value problem
1
yields .as 2 C bs C c/Y .s/ D as C b, so Y .s/ D
as C b as 2 C bs C c
a 2 as C bs C c
ay 00 C by 0 C cy D 0; y.0/ D 0; y 0 .0/ D 1 a . Therefore, y2 D L 2 as C bs C c satisfies the initial conditions y1 .0/ D 0, y10 .0/ D 1.
1
yields .as 2 C bs C c/Y .s/ D a, so Y .s/ D 8.4 THE UNIT STEP FUNCTION 8.4.2. Z
L.f / D
1
st
e
0
Z
f .t/ dt D
1 st
e 0
t dt C
Z
Z
1
This and (B) imply that L.f / D .1 Alternatively, f .t/ D t
u.t
8.4.4. L.f / D
Z
1
Letting t D x C 1 in the last integral yields Z 1 e st .t 1/ dt D
Z
1 0
e
s
e
1
/L.t/ D
f .t/ dt D
s.xC1/
e
0
st
0
dt in (A) to obtain st
.t
1/ dt:
1
dt C
s
e
.
s2
Z
1
e
st
1
.t C 2/ dt:
R1 To relate the first term to a Laplace transform we add and subtract 1 e st dt in (A) to obtain Z 1 Z 1 Z 1 L.f / D e st dt C e st .t C 1/ dt D L.t/ C e st .t C 1/ dt: 0
1
1
Letting t D x C 1 in the last integral yields Z 1 Z e st .t C 1/ dt D
1
1
0
This and (B) imply that L.f / D L.1/ C e
s
e
s.xC1/
.B/
1
.
1
e
e
.A/
x dx D e s L.t/:
e s /L.t/ D
Z
1
dt:
s
e s2
1/ $ .1
1/.t
st
1
st
e
1 R 1 st t 1 e
To relate the first term to a Laplace transform we add and subtract Z 1 Z 1 L.f / D e st t dt C e st .1 t/ dt D L.t/ 0
1
.x C 2/ dx D e
1 L.t C 2/ D C e s
s
s
L.t C 2/:
1 2 C . s2 s
.A/
.B/
Section 8.4 The Unit Step Function
141
Alternatively, 1 1/.t C 1/ $ L.1/ C e L.t C 2/ D C e s s
f .t/ D 1 C u.t 8.4.6.
L.f / D
Z
1
st
e
0
f .t/ dt D
Z
s
1 2 C . s2 s
1
e
Z
st 2
0
t D L.t 2 /
1
t 2 dt:
.A/
1
Letting t D x C 1 in the last integral yields Z 1 Z 1 st 2 e t dt D e s.xC1/ .t 2 C 2t C 1/ dx D e s L.t 2 C 2t C 1/: 1
0
This and (A) imply that L.f / D L.t 2 / C e s L.t 2 C 2t C 1/ D
2 s3
e
s
2 2 1 C C : s3 s2 s
Alternatively, 2
f .t/ D t .1
u.t
2
1// $ L.t / C e
8.4.8. f .t/ D t 2 C 2 C u.t L u.t
t2
1/.t
2/
2 2 it follows that F .s/ D 3 C s s 8.4.10. f .t/ D e e
.sC2/
L.e
2t
/
t
t2
1/.t
s
e
2 L.t C 2t C 1/ D 3 s 2
1/.e
2 2 1 C 2C s3 s s
2 2 C and 3 s s e s L .t C 1/ .t C 1/2 2 1 2 2 e s L.t 2 C t C 2/ D e s 3 C 2 C ; s s s 1 2 C 2C . s s
D 2 s3
e t / $ L.e t / C e s L.e 1 e .sC1/ e .sC2/ e .sC1/ L.e t / D C . sC1 sC2 C u.t
e
s
2/. Since t 2 C 2 $ D
s
2t
8.4.12.f .t/ D Œu.t 1/ u.t 2/t $ e s L.t C 1/ 1 1 1 2 De s 2C e 2s C . s s s2 s
e
2s
2.t C1/
/
e
s
L.e
t 1
/ D L.e t / C
L.t C 2/
8.4.14. f .t/
D
t
D
1 s2
2u.t 2e s s2
1 1/ C u.t 2/.t C 4/ $ 2 s 1 6 C e 2s 2 C : s s
1/.t
8.4.16. f.t/ D 2 2u.t 1/t C u.t 3/.5t 2 2 2 5 13 e s 2C C e 3s 2 C . s s s s s
2/ $ L.2/
2e s L.t/ C e
2s
2e s L.t C 1/ C e
L.t C 6/
3s
L.5t C 13/ D
142 Chapter 8 Laplace Transforms 8.4.18. f .t/ D .t C 1/2 C u.t
1/ .t C 2/2 .t C 1/2 D t 2 C 2t C 1 C u.t 2 2 1 2 5 2 s s C . L.t C 2t C 1/ C e L.2t C 5/ D 3 C 2 C C e s s s s2 s 1 1 1 8.4.20. D $1 e s.s C 1/ s sC1
t
)e
s
1 $ u.t 1/ 1 s.s C 1/
e
.t 1/
D
1/.2t C 3/ $
(
0; 1
e
.t 1/
0 t < 1; ; t 1:
8.4.22. 3 1 t )e $ u.t 1/ .3 .t 1// D u.t 1/.4 t/I s s2 1 1 1 3s 1 C 2 $1Ct )e C 2 $ u.t 3/ .1 C .t 3// D u.t 3/.t 2/I s s s s 3 s
1 $3 s2
s
therefore
h.t/ D 2 C t C u.t
8.4.24.
t/ C u.t
1/.4
3/.t
1 2s 5 2.s C 2/ D $e s 2 C 4s C 5 .s C 2/2 C 1
therefore,
h.t/
D u.t D u.t (
D
/e /e
2.t / 2.t /
2.t /
2t
.5 sin t
2 cos t/I
.5 sin.t .2 cos t
0;
e
8 2 C t; 0 t < 1; ˆ ˆ < 6; 1 t < 3; 2/ D ˆ ˆ : t C 4; t 3:
.2 cos t
/ 2 cos.t // 5 sin t/ 0 t < ; : 5 sin t/; t :
s C1 3.s 3/ 4 1 3.s 3/ . Since D and .s C 1/.s 2/ .s 1/.s 2/ .s C 1/.s 2/ sC1 s 2 sC1 3 2 4 4 2 D , F .s/ D C $ 4e t 4e 2t C2e t . Therefore,e 2s F .s/ $ .s 1/.s 2/ s 2 s 1 s C(1 s 2 s 1 0; 0 t < 2; u.t 2/ 4e .t 2/ 4e 2.t 2/ C 2e .t 2/ D .t 2/ 2.t 2/ .t 2/ 4e 4e C 2e ; t 2: 8.4.26. Denote F .s/ D
8.4.28. 3 s
1 $3 s3
t2 )e 2
2s
3 s
1 s3
$ u.t
2/ 3
1 e 4s $ t ) $ u.t s2 s2
2/2 2
4/.t
4/I
.t
D u.t
2 t 2/ C 2t C 1 I 2
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions
143
therefore h.t/
2
t2 C 2t C 1 C u.t 2 0t M . Therefore, M X .fm .t/ fm 1 .t// D fM .t/ f .t/ D f0 .t/ C mD1
8.4.32. Since
1 X
e
Km
converges if > 0,
mD0
1 X
e
tm
converges if > 0, by the comparison test.
mD0
Therefore,(C) of Exercise 8.3.31 holds if s > s0 C if is any positive number. This implies that it holds if s > s0 . 8.4.34. Let tm D m and fm .t/ D . 1/m , m D 0; 1; 2; : : : . Then fm .t/ fm 1 .t/ D . 1/m 2, so ! 1 1 X X 1 f .t/ D 1 C 2 . 1/m u.t m/ and F .s/ D 1C2 . 1/m e ms Substituting x D e s in the s mD1 mD1 1 X 1 2e s 1 1 e s x m m identity . 1/ x D (jxj < 1) yields F .s/ D 1 D . 1Cx s 1Ce s s 1Ce s mD1
8.4.36. Let tm D m and fm .t/ D . 1/m m, m D 0; 1; 2; : : : . Then fm .t/ fm 1 .t/ D . 1/m .2m 1/, 1 1 X 1 X so f .t/ D . 1/m .2m 1/u.t m/and F .s/ D . 1/m .2m 1/e ms . Substituting x D e s s mD1 mD1 1 1 X X x x in the identities . 1/m x m D and . 1/m mx m D (jxj < 1) yields F .s/ D 1Cx .1 C x/2 mD1 mD1 1 e s 2e s 1 .1 e s / D . s s 2 s 1Ce .1 C e / s .1 C e s /2 8.5 CONSTANT COEEFFICIENT EQUATIONS WITH PIECEWISE CONTINUOUS FORCING FUNCTIONS
8.5.2. y 00 C y D 3 C u.t L .u.t
4/.2t 4/.2t
8/; y.0/ D 1; y 0 .0/ D 0. Since 8// D e
4s
L .2.t C 4/
.s 2 C 1/Y .s/ D
8/ D e
4s
3 2e 4s C C s: s s2
L.2t/ D
2e 4s ; s2
144 Chapter 8 Laplace Transforms 3 2e 4s s C C 2 s.s 2 C 1/ s 2.s 2 C 1/ s C1 1 s 1 1 s 4s D 3 C 2e C 2 s s2 C 1 s2 s2 C 1 s C1 3 2s 1 1 4s C 2e : D s s2 C 1 s2 s2 C 1
D
Y .s/
Since 1 s2 yD3
1 $t s2 C 1
2 cos t C 2u.t
8.5.4. y 00
4s
sin t ) e
4/ .t
y D e 2t C u.t
4
sin.t
1 s2
1 s2 C 1
$ u.t
4/ .t
4
sin.t
4// ;
4//.
e 2t /; y.0/ D 3; y 0 .0/ D 1. Since e4 1 L.u.t 2/.1 e 2t // D e 2s L.1 e 2.t C2// D e 2s ; s s 2 1 e4 1 .s 2 1/Y .s/ D C e 2s C . 1 C 3s/: s 2 s s 2 2/.1
Therefore, Y .s/
.s
D
1 3s 1 C .s 1/.s .s 1/.s C 1/ C 1/.s 2/ e4 1 2s Ce s.s 1/.s C 1/ .s 1/.s C 1/.s
1 1/.s C 1/.s
e 2s e 4 1/.s C 1/.s
.s
2/
:
1 1 1 1 1 1 C C 2s 1 6sC1 3s 2 1 t 1 t 1 2t $ e C e C e I 2 6 3 1 t C2 1 .t 6/ 1 2t $ u.t 2/ e C e C e I 2/ 2 6 3 2/
D
1 D 1/.s C 1/
1 1 1 1 1 1 1 C C $ 1 C et C e t I s.s s 2s 1 2sC1 2 2 2s 1 1 e $ u.t 2/ 1 C e t 2 C e .t 2/ I s.s 1/.s C 1/ 2 2 .s Therefore, 1 13 y D et C e 2 6
t
3s 1 1 2 D C $ e t C 2e t : 1/.s C 1/ s 1 sC1
1 C e 2t C u.t 3
2/
1 1 C et 2
2
1 C e 2
8.5.6. Note that j sin tj D sin t if 0 t < , while j sin tj D value problem as y 00 C 4y D sin t
2u.t
/ sin t C u.t
.t 2/
1 C e t C2 2
1 e 6
.t 6/
1 2t e : 3
sin t if t < 2 . Rewrite the initial
2 / sin t; y.0/ D 3; y 0 .0/ D 1:
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions
145
Since / sin t/ D e
L .u.t
s
L.sin.t C // D e
s
L.sin t/
and 2 / sin t/ D e
L .u.t .s 2 C 4/Y .s/ D
s
1 C 2e Ce 2 .s C 1/
2s
e s .s 2 C 1/.s 2 C 4/
L.sin.t C 2 // D e
C1
1 D .s 2 C 1/.s 2 C 4/
therefore
2s
3s; so Y .s/ D
1 s2 C 1
1 s2 C 4
$
2s
L.sin t/;
1 C 2e s C e 2s 1 3s C 2 : 2 2 .s C 1/.s C 4/ s C4 1 sin t 3
1 sin 2tI 6
1 1 u.t / sin.t / sin 2.t 3 6 1 1 u.t / sin t C sin 2t 3 6
$ D
/
and e 2s .s 2 C 1/.s 2 C 4/
$
u.t
D
u.t
1 2 / sin.t 3 1 2 / sin t 3
1 2 / sin 2.t 6 1 sin 2t I 6
2 /
therefore 1 y D sin 2t 3
1 3 cos 2t C sin t 3
2u.t
8.5.8. y 00 C 9y D cos t C u.t
3=2/.sin t
L .u.t
cos t//
3=2/.sin t
1 1 / sin t C sin 2t C u.t 3 6
1 2 / sin t 3
1 sin 2t : 6
cos t/; y.0/ D 0; y 0 .0/ D 0. Since D e
3s=2
e
L .sin.t C 3=2/
3s=2
L.cos t C sin t/;
cos.t C 3=2//
s C1 1 sC1 ; so Y .s/ D 2 e 3s=2 2 : 2 2 s C1 .s C 1/.s C 9/ .s C 1/.s 2 C 9/ 1 1 1 1 1 1 D $ sin t sin 3t and .s 2 C 1/.s 2 C 9/ 8 s2 C 1 s2 C 9 8 3 s 1 s s 1 D $ .cos t cos 3t/ : .s 2 C 1/.s 2 C 9/ 8 s2 C 1 s2 C 9 8
.s 2 C 9/Y .s/ D
s2
1 C1
e
3s=2
sC1 .s 2 C 1/.s 2 C 9/
D $
sC1 sC1 sC1 sC1 D .s 2 C 1/.s 2 C 9/ 8 s2 C 1 s2 C 9 1 1 cos t C sin t cos 3t sin 3t ; so 8 3
146 Chapter 8 Laplace Transforms 3s=2
e
sC1 .s 2 C 1/.s 2 C 9/
u.t
$
3=2/ .cos.t 8
3=2/ C sin.t
3=2/ =2/
1 sin 3.t 3=2/ 3 u.t 3=2/ 1 sin t cos t C sin 3t cos 3t : 8 3 1 3=2/ sin t cos t C sin 3t cos 3t . 3 cos 3.t
D Therefore, y D
1 .cos t 8
8.5.10. y 00 C y D t
cos 3t/
1 u.t 8
/t; y.0/ D 0; y 0 .0/ D 0. Since
2u.t
1 L .u.t /t/ D e L .t C / D e C s2 s 1 1 .s 2 C 1/Y .s/ D 2 2e s 2 C I s s s s
Y .s/
s
1 C s 2 .s 2 C 1/ s.s 2 C 1/ 1 1 s 2e 2 e s2 s2 C 1
1 D 2e 2 2 s .s C 1/ 1 1 D s2 s2 C 1
s
s
;
1 s
s : s2 C 1
Since 1 s2
1 s2 C 1
$
t
$
u.t
s
sin t ) e / .t
1 1 s2 s2 C 1 sin.t // D u.t
C sin t/
/.t
and 1 s yDt
sin t
2u.t
8.5.12. y 00 C y D t
s 2 s C1
$
1
$
u.t
cos t ) e
s
/ .1
cos.t
1 s
s s2 C 1 // D u.t /.1 C cos t/;
/.t C sin t C cos t/. 2 /t; y.0/ D 1; y 0 .0/ D 2I
3u.t
2 /t/ D e
L.u.t
.s 2 C 1/Y .s/ D Y .s/ D
e 2s $ u.t s 2.s 2 C 1/
2s
1
L.t C 2 / D e 3e s2
1 3e 2s s 2 .s 2 C 1/
1 1 D 2 s 2 .s 2 C 1/ s 2 /..t
2
2s
2 1 C 2 s s
I
2s
6 e 2s C 2 C sI s 6 e 2s 2Cs C 2 I 2 s.s C 1/ s C1
1 $t s2 C 1
sin.t
sin tI
2 // D u.t
2 /.t
2
sin t/I
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions 1 1 D C 1/ s
s.s 2 e 2s $ u.t s.s 2 C 1/
y D t C sin t C cos t
2 /.1
s2
s $1 C1
cos tI
2 // D u.t
cos.t
2 /.1
cos t/I
2Cs $ 2 sin t C cos tI s2 C 1 2 /.3t 3 sin t 6 cos t/.
u.t
8.5.14. y 00
4y 0 C 3y D 1 C 2u.t 1/; y.0/ D 0; y 0 .0/ D 0; 1 C 2e s 1 C 2e s .s 2 4s C 3/Y .s/ D ; Y .s/ D ; s s.s 1/.s 3/ 1 1 1 1 1 1 1 1 1 t D C $ C e 3t e ; s.s 1/.s 3/ 3s 6 s 3 2 s 1 3 6 2 s e 1 1 1 t 1 $ u.t 1/ C e 3.t 1/ e ; s.s 1/.s 3/ 3 6 2 1 1 3t 1 2 1 yD e C e t C u.t 1/ C e 3.t 1/ e t 1 . 3 6 2 3 3
8.5.16. y 00 C 2y 0 C y D 4e t
4u.t
L 4u.t
1/e t ; y.0/ D 0; y 0 .0/ D 0. Since 4e 1/e t D e s L 4e .t C1/ D s
.s 2 C 2s C 1/Y .s/ D Y .s/ D
.s
where
4 1/.s C 1/2
A.s C 1/2 C B.s A C ACB
1/.s C 1/ C C.s
D et
e
t
2te
t
eu.t
D et
e
t
2te
t
u.t
4y 0 C 4y D e 2t
2u.t L u.t
1/ D 4:
1, C D 2. Therefore,
1 1 D 1/.s C 1/2 s 1
.s
4e sC1 ; so 1 s 1 4e sC1 : .s 1/.s C 1/2
D 1 .set s D 1/I D 2 .set s D 1/I D 0 .equate coefficients of s 2 /:
Solving this system yields A D 1, B D
8.5.18. y 00
s
;
1
A B C 1 D C C ; 2 1/.s C 1/ s 1 sC1 .s C 1/2
.s
y
4
sC1
1 sC1
1/ e t 1 1/ e t e
2 .s C 1/2 e
.t 1/
.t 2/
2.t
and
2.t 1/e
2/e 2t ; y.0/ D 0; y 0 .0/ D 1. Since 2/e 2t D e
2s
e 2sC4 L e 2t C4 D ; s 2
1/e
.t 1/
.t 2/
:
147
148 Chapter 8 Laplace Transforms .s 2
4s C 4/Y .s/ D Y .s/ D
1 2/3
.s
therefore y D
$
t 2 e 2t 2
t 2 e 2t e ) 2 .s te 2t
u.t
8.5.20. y 00 C 2y 0 C 2y D 1 C u.t
1 s
2
1
2sC4
2/3
2e .s
2/3
.s $
2e s
e4 u.t 2
2sC4
2sC4
1
2/3
2/e 2.t
1;
2
2/2
.s 2/
.t
so :
2/2 D u.t
2/
.t
2/2 e 2t I 2
2/2 e 2t .
2/.t
3 /.t C 1/; y.0/ D 2; y 0 .0/ D 1; 1 2 1 2s 2s C I L.u.t 2 /.t 1// D e L..t C 2 1// D e s2 s 3 C 1 1 C I L.u.t 3 /.t C 1// D e 3s L..t C 3 C 1// D e 3s s2 s 1 1 2 1 1 3 C 1 2 2s 3s .s C 2s C 2/Y .s/ D C e C e C C . 1 C 2s/ C 4: s s2 s s2 s
Let G.s/ D
2 /.t
1/
u.t
1 1 , H.s/ D ; then s.s 2 C 2s C 2/ s.s 2 C 2s C 2/ Y .s/ D Y1 .s/ C e
where
2s
Y2 .s/
3s
Y3 .s/;
(A)
2s C 3 ; C 2s C 2 Y2.s/ D H.s/ C .2 1/G.s/; Y1 .s/ D G.s/ C
Let yi .t/ D L
1
1
(B)
s2
(C)
Y3 .s/ D H.s/ C .3 C 1/G.s/:
(D)
.Yi .s//, .i D 1; 2; 3/. From (A), y.t/ D y1 .t/ C u.t
Find L
e
2 /y2.t
2 /
u.t
3 /y3 .t
3 /:
.G.s//: G.s/ D
A B.s C 1/ C C C s .s C 1/2 C 1
where A..s C 1/2 C 1/ C .B.s C 1/ C C /s D 1. Setting s D 0 yields A D 1=2; setting s D A C D 1, so C D 1=2; since A C B D 0 (coefficient of x 2 ), B D 1=2. Therefore, 1 1 .s C 1/ C 1 1 1 t G.s/ D $ e .cos t C sin t/: 2 s .s C 1/2 C 1/ 2 2 Find L
1
(E)
.H.s//: H.s/ D
A B C.s C 1/ C D C C s sC2 .s C 1/2 C 1
1 yields
(F)
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions
149
where .As C B/..s C 1/2 C 1/ C .C.s C 1/ C D/s 2 D 1. 2B ACB CD 5A C 5B C 2C C D ACB D 0
D D D D
1 1 1 0
.set s D 0/I .set s D 1/I .set s D 1/I .equate coefficients of s 3 /:
Solving this system yields A D 1=2, B D 1=2, C D 1=2, D D 0; therefore 1 1 1 sC1 1 H.s/ D $ .1 t e 2 2 2 s s .s C 1/ C 1/ 2 Since
t
cos t/:
(G)
2.s C 1/ C 1 2s C 3 D $ e t .2 cos t C sin t/; s 2 C 2s C 2 .s C 1/2 C 1
(B) and (F)) imply that
y1 .t/ D
1 t 1 e .3 cos t C sin t/ C : 2 2
(H)
From (C), (F)), and (G), y2 .t/ D so y2 .t
1C e
2 / D
t C . 2
.t 2/
.
1/e
t
2
cos t
1/ cos t C
1 2
2
1 2
e
t
sin t;
sin t C 1
t : 2
(I)
From (D), (F)), and (G), y3 .t/ D
1 2
so y3 .t
3 / D
e t .3 cos t C .3 C 1/ sin t C t C 3 / ; 1 e 2
.t 3/
.3 cos t C .3 C 1/ sin t C t/ :
Now (E), (20), (I), and (J) 1 1 y D e t .3 cos t C sin t/ C imply that 2 2 2 1 .t 2/ u.t 2 / e . 1/ cos t C sin t C 1 2 1 .t 3/ u.t 3 / e .3 cos t C .3 C 1/ sin t/ C t . 2 8.5.22. (a) f .t/ D
1 X
nD0
u.t
1 1X n /; F .s/ D e s nD0
ns
t 2
(J)
1 X 1 ; Y .s/ D e s.s 2 C 1/ nD0
ns
;
1 D s.s 2 C 1/
s e ns $ 1 cos t; $ u.t n /.1 cos.t n // D u.t n /.1 . 1/n cos t/; s2 C 1 s.s 2 C 1/ 1 m X X y.t/ D u.t n /.1 . 1/n cos t/. If m t < .m C 1/ , y.t/ D .1 . 1/n cos t/. Therefore, nD0 nD0 ( 2m C 1 cos t; 2m t < .2m C 1/ .m D 0; 1; : : : / y.t/ D . 2m; .2m 1/ t < 2m .m D 1; 2; : : : / 1 s
150 Chapter 8 Laplace Transforms 1 X
(b) f .t/ D
2n /t; F .s/ D
u.t
nD0
1 X
1 X
2ns
e
nD0
L.t C 2n s/ D
1 2n 1 C D 2 e 2ns Yn .s/, where Yn .s/ D 2 2 2 s .s C 1/ s.s C 1/ s nD0 t
sin t C 2n
2n /yn .t/ D u.t
2n /.t
2n cos t/; therefore y.t/ D
sin t
m X
.t
1 X
1 2n C ; Y .s/ D s2 s
1 2n C C1 s
2n $ yn .t/ D s2 C 1
1 X
2ns
u.t
2n /.t
Yn .s/ $ u.t
sin t
2n cos t/.
nD0
2n cos t/ D .m C 1/.t
sin t
nD0
2ns
2n / D sin t, e
If 2m t < 2.m C 1/ , then y.t/ D
e
nD0
s2
2n / D cos t and sin.t
2n cos t. Since cos.t
1 X
sin t
m cos t/:
! 1 1 X X 1 1 n ns (c) f .t/ D 1C2 . 1/ u.t n /; F .s/ D 1C2 . 1/ e ; Y .s/ D 1 C 2 . 1/n e 2 C 1/ s s.s nD1 nD1 nD1 1 1 s e ns D $ 1 cos t; $ u.t n /.1 cos.t n // D u.t n /.1 s.s 2 C 1/ s s2 C 1 s.s 2 C 1/ 1 X . 1/n cos t/; y.t/ D 1 cos t C 2 . 1/n u.t n /.1 . 1/n cos t/. If m t < .m C 1/ , n
ns
nD1
y.t/ D 1 (d) f .t/ D
1 X
u.t
nD0
1 1 1 1 $ .e t Ce 2sC1 s 2 If m t < .m C 1/, y.t/
D D
cos t C 2
m X
. 1/n .1
nD1
1 1X n/; F .s/ D e s nD0 t
2/;
e s.s 2
m 1 X t e 2
nD0 mC1
ns
1/
n
.s 2
1 X
nD0
$
m
ns
; Y .s/ D
u.t
n/ 2
.t n/
Ce
1 e .e t 2.1 e/
(e) f .t/ D .sin tC2 cos t/
. 1/n cos t/ D . 1/m 1 s.s 2
et C e
1 2 D .e t 2
C e t/
u.t 2n /; F .s/ D
m
.2m C 1/ cos t:
1 X
1 1 1 D C s.s 2 1/ 2s 1 1 1X 2 ; y.t/ D u.t n/ e t n C e 2
1/ nD0 t
e
ns
;
nD0
m
C e t/
m X
en
m
1 1 C 2s X e s 2 C 1 nD0
2ns
; Y .s/ D
1 X 1 C 2s e .s 2 C 1/.s 2 C 2s C 2/ nD0
.As C B/..s C 1/2 C 1/ C .C.s C 1/ C D/.s 2 C 1/ D 1 C 2s: 1 1 3 0
1
1:
where D D D D
2 .
nD0
1 C 2s As C B C.s C 1/ C D D 2 C 2 C 1/.s C 2s C 2/ s C1 .s C 1/2 C 1
2B C C C D A C B C 2D 5A C 5B C 4C C 2D ACC
.t n/
.set s D 0/I .set s D 1/I .set s D 1/I .equate coefficients of s 3/:
2ns
;
!
;
Section 8.5 Constant Coeefficient Equations with Piecewise Continuous Forcing Functions
151
Solving this system yields A D 0, B D 1, C D 0, D D 1. Therefore, .s 2
Since sin.t
1 C 2s C 1/.s 2 C 2s C 2/
s2
$
2n / D sin t, e
2ns
1 C 2s $ u.t .s 2 C 1/.s 2 C 2s C 2/
so y.t/ D sin t
1 X
y.t/ D sin t 1 X
m X
1
nD0
n/; F .s/ D
u.t
nD0
s.s
.t 2n/
e
1 1/.s
2/
1 1X e s nD0
D
e ns s.s 1/.s y.t/ D If m t < m C 1, y.t/
D D
m X 1
2e t
nD0
mC1 2
et
1 2s 2/
ns
; Y .s/ D
s
1
1 u.t 2
1 1X u.t 2 nD0
n
C e 2.t
m1
e 1
n/
mC1
e
1
C
n/ 1
D
2/
1 1 1 $ 1 2s 2 2
n/ 1 2e t
mC1 2
1 C e 2.t 2
2e t n
m/ 1
e 1
n
m
sin t;
:
!
e
t
!
sin t:
;
2e t C e 2t I
C e 2.t
C e 2.t
et
e 2.mC1/ 1 e 2
1 1/.s
s.s
.t 2n/
e
.t 2n/
e
D mC1
1
$
2n / 1
2n / 1
u.t
nD0
If 2m t < 2.m C 1/ ,
(f) f .t/ D
1 1 C 1 .s C 1/2 C 1 1 e t sin t:
D
n/
m X
n/
I
:
1 e n C e 2.t 2 nD0
m/
e 2n
nD0
2mC2
e2
m X
:
by 0 .t/ cy.t/ on .˛; t0/ and .t0 ; ˇ/, y 00 .t0 C/ D a f .t0 C/ by 0 .t0 / cy.t0 / f .t0 / by 0 .t0 / cy.t0 / , and y 00 .t0 / D . This implies the conclusion. a00 a (b) Since y has a junp discontinouity at t0 , applying Exercise 8.4.23(c) to y 0 shows that y 0 is not differentiable at t0 . Therefore,y cannot satisfy (A) on .˛; ˇ/ if f has a jump discontinuity at some t0 in .˛; ˇ/. 8.5.24. (a) The assumptions imply that y 00 .t/ D
f .t/
8.5.26. If 0 t < t0 , then y.t/ D ´0 .t/. Therefore, y.0/ D ´0 .0/ D k0 and y 0 .0/ D ´0 .0/ D k1 , and ay 00 C by 0 C cy D a´000 C b´00 C c´0 D f0 .t/ D f .t/;
0 < t < t0 :
152 Chapter 8 Laplace Transforms Now P suppose that 1 m n. For convenience, define tnC1 D 1. If tm t < tmC1 , then y.t/ D m kD0 ´m .t/, so ay 00 C by C cy D
m X
.a´00k C b´0k C c´k / D f0 C
kD0
m X
.fk
fk
kD1
1/
D fm D f;
tm < t < tmC1 :
Thus, y satisfies ay 00 C by 0 C cy D f on any open interval that does not contain any of the points t1 , t2 ,. . . , tn . Since ´.tm / D ´0 .tm / for m D 1; 2; : : : , y and y 0 are continuous on Œ0; 1/. Since y 00 .t/ D .by 0 .t/ C cy.t//=a if t ¤ tm (m D 1; 2; : : : ), y 00 has limits from the left at t1 ; : : : ; tn . 8.6 CONVOLUTION a s as and cos bt $ 2 , so H.s/ D 2 . 2 2 2 Ca s Cb .s C a /.s 2 C b 2 / 1 a a (b) e t $ and sin at $ 2 , so H.s/ D . 2 s 1 s Ca .s 1/.s 2 C a2 / a 1 as (c) sinh at $ 2 and cosh at $ 2 , so H.s/ D 2 . 2 2 s a s a .s a2 /2 2!s s 2 !2 2!s.s 2 ! 2 / (d) t sin !t $ 2 and t cos !t $ , so H.s/ D . .s C ! 2 /2 .s 2 C ! 2 /2 .s 2 C ! 2 /4 Z t Z t (e) e t sin ! cos !.t / d D .e sin ! / e .t / cos !.t / d ; e t sin !t $
8.6.2. (a) sin at $
s2
0
0
s 1 .s 1/! and e cos !t $ , so H.s/ D . 2 2 .s 1/ C ! ..s 1/2 C ! 2 /2 Z t Z t (f) e t 2 .t /e d D 2 e 2 .t /e .t / d ; t 2 e 2t $
! 1/2 C ! 2
.s
t
0
H.s/ D (g) e
.sZ
2 2/3 .s
t
t
e 0
0
1/2
and te t $
2/3
cos !.t
/ d D
Z
t
e
2
e
.t /
0
2t
cos !.t / d ; te
0
/ d ; e 3t $
$
1 s
3
1 . 3/ ..s 1/2 1/ 1 2 2 (i) te 2t $ and sin 2t $ 2 , so H.s/ D . 2 2 .s 2/ s C4 .s 2/ .s 2 C 4/ 6 1 6 (j) t 3 $ 4 and e t $ , so H.s/ D 4 . s s 1 s .s 1/ 3 3 6Š 6Š (k) t 6 $ 7 and e t sin 3t $ , so H.s/ D 7 . 2 s .s C 1/ C 9 s Œ.s C 1/2 C 9 2 6 12 (l) t 2 $ 3 and t 3 $ 4 , so H.s/ D 7 . s s s 7Š 2 2 7Š 7 t (m) t $ 8 and e sin 2t $ , so H.s/ D 8 . s .s C 1/2 C 4 s Œ.s C 1/2 C 4 24 2 48 (n) t 4 $ 5 and sin 2t $ 2 , so H.s/ D 5 2 . s s C4 s .s C 4/
so H.s/ D
.s
1 .s
1/2
, so
.
sC1 sC1 , so H.s/ D . 2 2 2 .s C 1/ ZC ! .s C 2/ Œ.s C 1/2 C ! 2 Z t t (h) e t e 2 sinh.t / d D e 3 e .t / sinh.t 0
2 .s
1 and e .s C 2/2
and e t sinh t $
t
.s
cos !t $
1 1/2
1
,
Section 8.6 Convolution
153
1 Y .s/ 1 1 s2 C 1 1 1 8.6.4. (a) Y .s/ D 2 ; Y .s/ 1 C D ; Y .s/ D 2 ; Y .s/ D 2 , so y D sin t. 2 2 2 s s2 s s s s s C1 2 1 2sY .s/ 2s 1 .s C 1/ 1 (b) Y .s/ D 2 ; Y .s/ 1 C 2 D 2 ; Y .s/ 2 D 2 ; Y .s/ D 2 s C1 s C1 s C1 s C1 s C1 s C1 1 , so y D te t . .s C 1/2 2sY .s/ 2s 1 .s 1/2 1 .s 2 C 1/ A 1 ; Y .s/ 1 D ; Y .s/ 2 D ; Y .s/ D D C (c) Y .s/ D C 2 2 s s C1 s C1 s s C1 s s.s 1/2 s B C C , where A.s 1/2 C Bs.s 1/ C C s D s 2 C 1. Setting s D 0 and s D 1 shows that A D 1 s 2 .s 1/2 1 1 and C D 2; equating coefficients of s 2 yields A C B D 1, so B D 0. Therefore, Y .s/ D C , s .s 1/2 so y D 1 C 2te t . 1 Y .s/ 1 1 s 1 sC1 1 1 (d) Y .s/ D 2 C ; Y .s/ 1 D 2 ; Y .s/ D 2 ; Y .s/ D D 2 C 3, s sC1 s C1 s sC1 s s3 s s t2 so y D t C . 2 sY .s/ s 1 s3 4s 2 C 1 1 (e) sY .s/ 4 D 2 C 2 ; Y .s/ s D 4 C ; Y .s/ D ; Y .s/ D s s C1 s2 C 1 s2 s2 C 1 s2 2 2 4 2 .4s C 1/.s C 1/ 4s C 5s C 1 4 5 1 5 1 D D C 3 C 5 , so y D 4 C t 2 C t 4 . 5 5 s s s s 2 24 s Y .s/ 1 s 1 s2 s 1 s 1 s 1 (f) Y .s/ D 2 C 2 ; Y .s/ 1 D 2 ; Y .s/ 2 D 2 ; Y .s/ D 2 D 2 s C1 s C1 s C1 s C1 s C1 s C1 s 1 1 , so y D 1 t. s s2 Z t Z 0 Z t 8.6.6. Substituting x D t yields f .t /g. / d D f .x/g.t x/. dx/ D f .x/g.t 0 t 0 Z t x/ dx D f . /g.t / d . 0
k0 .as C b/ C k1 a F .s/ C . Since p.s/ p.s/ a.r1 C r2 /, (A) can be rewritten as
8.6.8. p.s/Y .s/ D F .s/ C a.k1 C k0 s/ C bk0 , so (A) Y .s/ D p.s/ D a.s
r1 /.s
r2 / and therefore b D
k0 .s r1 r2 / k1 C : .s r1 /.s r2 / .s r1 /.s r2 / 1 1 1 e r2 t e r1 t 1 D $ ; .s r1 /.s r2 / r2 r1 s r2 s r1 r2 r1 so the convolution theorem implies that Z F .s/ 1 t e r2 e r1 $ f .t / d : a.s r1 /.s r2 / a 0 r2 r1 Y .s/ D
s .s
a.s
F .s/ r1/.s
r2 /
C
r1 r2 r2 1 D r1 /.s r2 / r2 r1 s r1
r1 r2
1 r1 s
r2
$
r2 e r 1 t r2
r1 e r 2 t : r1
Therefore, y.t/ D k0
r2 e r 1 t r2
r1 e r 2 t e r2 t C k1 r1 r2
e r1 t 1 C a r1 a
Z
t 0
e r2 r2
e r1 f .t r1
/ d :
154 Chapter 8 Laplace Transforms F .s/ k0 .as C b/ C k1 a C . Since p.s/ p.s/ /2 C ! 2 and therefore b D 2a, (A) can be rewritten as
8.6.10. p.s/Y .s/ D F .s/ C a.k1 C k0 s/ C bk0 , so (A) Y .s/ D p.s/ D a.s
Y .s/ D .s
aŒ.s
k0 .s 2/ F .s/ C C /2 C ! 2 .s /2 C ! 2 .s
k1 : /2 C ! 2
1 1 $ e t sin !t, so the convolution theorem implies that 2 2 / C ! ! Z t F .s/ 1 $ e t f .t / sin ! d : aŒ.s /2 C ! 2 a! 0 s 2 .s / t D $e cos !t sin !t : .s /2 C ! 2 .s /2 C ! 2 !
Therefore,
y.t/ D e t k0 cos !t
sin !t !
C
Z t k1 1 sin !t C e t f .t ! a! 0
/ sin ! d :
8.6.12. (a) ay 00 C by 0 C cy D f0 .t/ C u.t p.s/Y .s/ D F0 .s/ C L.u.t Y .s/ D
f0 .t//; y.0/ D 0; y 0 .0/ D 0I
t1 /.f1 .t/ t1 /.f1 .t/
f0 .t/// D F0 .s/ C e
st1
L.g/I
st1
F0 .s/ C e G.s/ : p.s/
.B/
1 $ w.t/, the convolution theorem implies that p.s/ Z t Z t F0 .s/ G.s/ $ w.t /f0 . / d and $ w.t /g. / d : p.s/ p.s/ 0 0 Z t e st1 G.s/ Now Theorem 8.4.2 implies that $ u.t t1 / w.t t1 /g. / d , and (B) implies that p.s/ 0 Z t Z t t1 y.t/ D w.t /f0 . / d C u.t t1 / w.t t1 /g. / d : (b) Since F0 .s/ $ f0 .t/, G.s/ $ g.t/, and
0
0
Rt
Rt
(c) Let ´0 .t/ D 0 w.t /f0 . / d and ´1 .t/ D 0 w.t /g. / d . Then y.t/ D ´0 .t/ C u.t t1 /´1 .t t1 /. Using Leibniz’s rule as in the solution of Exercise 8.6.11(b) shows that Z t Z t 0 0 0 ´0 .t/ D w .t /f0 . / d ; ´1 .t/ D w 0 .t /g. / d ; t > 0; 0
´000 .t/ D
f0 .t/ C a
if t > 0, and that
Z
0
t
w 00.t 0
a´000 C b´00 C c´0 D f0 .t/
/f0 . / d ; ´001 .t/ D
and
g.t/ C a
Z
t
w 00 .t
/g. / d ;
t > 0;
a´001 C b´01 C c´1 D f1 .t C t1 /
f0 .t C t1 /;
t > 0:
0
Section 8.7 Constant Coefficient Equations with Impulses
155
This implies the stated conclusion for y 0 and y 00 on .0; t/ and .t; 1/, and that ay 00 C by 0 C cy D f .t/ on these intervals. (d) Since the functions ´0 .t/ and h.t/ D u.t t1 /´1 .t t1 / are both continuous on Œ0; 1/ and h.t/ D 0 if 0 t t1 , y is continuous on Œ0; 1/. From (c), y 0 is continuous on Œ0; t1/ and .t1 ; 1/, so we need only show that y 0 is continuous at t1 . For this it suffices to that h0 .t1 / D 0. Since h.t1 / D 0 if Z tshow t1 h.t/ h.t1 / D 0. If t > t1 , then h.t/ D w.t t1 /g. / d . Since h.t1 / D 0, t t1 , (B) lim t !t1 t t1 0 ˇ Z t t1 ˇ ˇ h.t/ h.t1 / ˇ ˇ ˇ jw.t t1 /g. /j d : .B/ ˇ ˇ t t 1 0 Since g is continuous from the right at 0, we can choose constants T > 0 and M > 0 so that jg. /j < M if 0 T . Then (B) implies that ˇ ˇ Z t t1 ˇ h.t/ h.t1 / ˇ ˇ ˇM jw.t t1 /j d ; t1 < t < t1 C T: .C/ ˇ t t ˇ 1 0
Now suppose > 0. Since w.0/ D 0, we can choose T1 such that 0 < T1 < T and jw.x/j < =M if 0 x < T1 . If t1 < t < t1 C T1 and 0 t t1 , then 0 t t1 < T1 , so (C) implies that ˇ ˇ ˇ h.t/ h.t1 / ˇ ˇ ˇ < ; t1 < t < t1 C T: ˇ t t ˇ 1 Therefore, lim
t !t1 C
h.t/ t
h.t1 / D 0. This and (B) imply that h0 .t1 / D 0. t1
8.7 CONSTANT COEFFICIENT EQUATIONS WITH IMPULSES 10 10 C .s C 1/.7s 2/ 2 5 C . 9 C 7s/ C 7; YO .s/ D D C ; sC1 .s 1/.s C 2/.s C 1/ s C 2 s C 1 1 1 1 1 1 1 e t e 2t yO D 2e 2t C 5e t ; D D ;w DL 1 D ; p.s/ .s C 2/.s 1/ 3 s 1 sC2 p.s/ 3 5 y D 2e 2t C 5e t C u.t 1/ e .t 1/ e 2.t 1/ . 3 8.7.2. .s 2 C s
2/YO .s/ D
8.7.4. .s 2 C 1/YO .s/ D
3 s2 C 9
1 C s;
3 s 1 3 1 1 s 1 1 8s 5 3 C D C D ; .s 2 C 1/.s 2 C 9/ s2 C 1 8 s2 C 1 s2 C 9 s2 C 1 8 s2 C 1 s2 C 9 1 1 1 1 1 yO D .8 cos t 5 sin t sin 3t/; D 2 ;w DL 1 D sin t; y D .8 cos t 5 sin t 8 p.s/ s C1 p.s/ 8 sin 3t/ 2u.t =2/ cos t. YO .s/ D
8.7.6. .s 2
1/YO .s/ D
8 C1 s
1 1 1 1 D D p.s/ .s 1/.s C 1/ 2s 1 3e t 8 C 2u.t 2/ sinh.t 2/; 8.7.8. .s 2 C 4/YO .s/ D
8 s
2
8 C s.1 s/ 4 3 8 D C ; yO D 4e t C 3e t 8; s.s 1/.s C 1/ s 1 s C 1 s 1 1 et C e t ;w D L 1 D D sinh t; y D 4e t C sC1 p.s/ 2
s; YO .s/ D
C 8s; (A) YO .s/ D
Bs C C where A.s 2 C 4/ C .Bs C C /.s s2 C 4
.s
8 8s C ; 2/.s 2 C 4/ s 2 C 4 .s
8 A D C 2/.s 2 C 4/ s 2
2/ D 8. Setting s D 2 yields A D 1; setting s D 0
156 Chapter 8 Laplace Transforms yields 4A 2C 8 .s 2/.s 2 C 4/ 1 L 1 D p.s/
D 8, so C D 2; A C B D 0 (coefficient of x 2 ), so B D A D 1; therefore 1 sC2 1 1 D , so (A) implies that yO D e 2t C7 cos 2t sin 2t; D 2 ;w D s 2 s2 C 4 p.s/ s C4 1 1 sin 2t. Since sin.2t / D sin 2t, y D e 2t C 7 cos 2t sin 2t u.t =2/ sin 2t. 2 2 1
1 s.s 1/ A B C D C C 2 s 1 .s 1/.s C 1/ s 1 s C 1 .s C 1/2 where A.s C 1/2 C .B.s C 1/ C C /.s 1/ D 1 s.s 1/. Setting s D 1 yields A D 1=4; setting s D 1 O yields C D 1=2; since A C B D 1 (coefficient of s 2), B D 1 A D 5=4. Therefore, Y .s/ D 1 1 5 1 1 1 1 t 1 t 1 1 1 C ; yO D e C e .2t 5/; D ;w D L 1 D te t ; 4 s 1 4 s C 1 2 .s C 1/2 4 4 p.s/ .s C 1/2 p.s/ 1 1 y D e t C e t .2t 5/ C 2u.t 2/.t 2/e .t 2/. 4 4 8.7.10. .s 2 C 2s C 1/YO .s/ D
C .2 s/ 2; YO .s/ D
1 .s C 1/ C 1 8.7.12. .s 2 C 2s C 2/YO .s/ D .2 s/ 2; Y .s/ D ; yO D e t .sin t cos t/; D 2C1 .s C 1/ p.s/ 1 1 ;w D L 1 D e t sin t. Since sin.t / D sin t and sin.t 2 / D sin t, .s C 1/2 C 1 p.s/ y D e t .sin t cos t/ e .t / u.t / sin t 3u.t 2 /e .t 2/ sin t. 1 C s.7 2s/ 7 1 6 1 1 1 C2.2 s/C3; YO .s/ D D ; s 2s.s C1=2/.s 2/ 10s 2 5 s C1=2 2s 1 1 1 1 1 7 2t 6 t =2 1 1 yO D e e ; D D ;wDL 1 D 10 5 2 p.s/ 2.s C 1=2/.s 2/ 5 s 2 s C 1=2 p.s/ 1 2t 7 2t 6 t =2 1 1 .e e t =2/; y D e e C u.t 2/ e 2.t 2/ e .t 2/=2 ; 5 10 5 2 5 s 1 1 s s s 2 O O 8.7.16. .s C 1/Y .s/ D 2 1; Y .s/ D 2 D s C4 .s C 1/.s 2 C 4/ s 2 C1 3 s2 C 1 s2 C 4 1 1 1 1 1 ; yO D .cos t cos 2t 3 sin t/; D 2 ;w DL 1 D sin t. Since sin.t s2 C 1 3 p.s/ s C1 p.s/ =2/ D cos t and sin.t / D sin t, 8.7.14. .2s 2 3s 2/YO .s/ D
yD
1 .cos t 3
8.7.18. .s 2 C 2s C 1/YO .s/ D
cos 2t 1 s
.s
1
3 sin t/
=2/ cos t C 3u.t
2u.t
1; (A) YO .s/ D
1 1/.s C 1/2
.s
/ sin t:
1 ; .s C 1/2
1 A B C D C C 2 1/.s C 1/ s 1 sC1 .s C 1/2
where A.s C 1/2 C .B.s C 1/ C C /.s 1/ D 1. Setting s D 1 yields A D 1=4; setting s D C D 1=2; since A C B D 0 (coefficient of s 2), B D A D 1=4. Therefore, .s
1 1 1 D 1/.s C 1/2 4s 1
1 1 4sC1
1 1 : 2 .s C 1/2
This and (A) imply that 1 1 YO .s/ D 4s 1
1 1 4sC1
3 1 I 2 .s C 1/2
1 yields
Section 8.7 Constant Coefficient Equations with Impulses 1 1 t 1 1 1 e e t .1 C 6t/ ; D ; w D L D te t ; 4 p.s/ .s C 1/2 p.s/ 1 t yD e e t .1 C 6t/ u.t 1/.t 1/e .t 1/ C 2u.t 2/.t 2/e 4
157
yO D
.t 2/
.
8.7.20. y 00 C 4y D 1 2u.t =2/ C ı.t / 3ı.t 3=2/; y.0/ D 1; y 0 .0/ D 1. .s 2 C 4/YO .s/ D 1 2e s=2 1 2e s=2 s 1 1 1 1 s O C s 1; Y .s/ D C 2 . Since D , YO .s/ D 2 s s.s s.s 2 C 4/ 4 s s2 C 4 C 4/ s C 4 1 3 s 1 1 s 3 1 1 1 C 2 2e s=2 . yO D cos 2t sin 2t C C u.t =2/.1Ccos 2t/. 2 2 4s 4 s C s s C4 4 2 4 4 4 s C 4 1 1 1 wDL D sin 2t. Since sin 2.t / D sin 2t and sin 2.t 3=2/ D sin 2t, p.s/ 2 3 1 1 1 1 3 y D cos 2t sin 2t C C u.t =2/.1 C cos 2t/ C u.t / sin 2t C u.t 3=2/ sin 2t. 4 2 4 4 2 2 8.7.26. w.t/ D e
t
st0
sin t; fh .t/ D
u.t
t0 /
u.t h
s.t0 Ch/
t0
h/
; .s 2 C 2s C 2/Yh.s/ D
1e h
st0
e s
s.t0 Ch/
;
1e e 1 .s C 1/ C 1 1 1 ; D C $ 1 e t .cos t C sin t/ ; 2 2 2 h s.s C 2s C 2/ s.s C 2s C 2/ 2 ..s C 1/ C 1/ 2s 2 8 0; 0 t < t0 ; ˆ ˆ ˆ ˆ h i < 1 1 e .t t0 / .cos.t t0 / C sin.t t0 // ; t0 t < t0 C h; yh .t/ D 2h ˆ ˆ i .t t0 / h ˆ ˆ : e e h .cos.t t0 h/ C sin.t t0 h// cos.t t0 / sin.t t0 / ; t t0 C h: 2h Yh .s/ D
8.7.28. w.t/ D e t
e
2t
1/2 2
t0 /
u.t h
8 ˆ ˆ ˆ ˆ ˆ
2 repeat this argument, starting from (B), with p1 replaced by p2 , and P2 replaced by P3 D p3 pn . 9.2.42. (a) .cos A C i sin A/.cos B C i sin B/
D .cos A cos B
sin A sin B/
C.cos A sin B C sin A cos B/ D cos.A C B/ C i sin.A C B/:
(b) Obvious for n D 0. If n D 1 write 1 cos C i sin
D D
1 cos i sin cos C i sin cos i sin cos i sin D cos i sin D cos. / C i sin. /: cos2 C sin2
(d) If n is a negative integer, then (B) .cos C i sin /n D
1 . From the hint, (C) .cos C i sin /jnj
1 D .cos i sin /jnj D .cos. / C i sin. //jnj . Replacing by and n by .cos C i sin /jnj jnj in (A) shows that (D) .cos. / C i sin. //jnj D cos. jnj/ C i sin. jnj/. Since jnj D n, (E) cos. jnj/ C i sin. jnj/ D cos n C i sin n. Now (B), (C), (D), and (E) imply (A). (e) From (A), ´nk D cos 2k C i sin 2k D 1 and kn D cos.2k C 1/ C i sin.2k C 1/ D cos.2k C 1/ D cos D 1. (f) From (e), 1=n ´0 ; : : : ; 1=n´n 1 are all zeros of ´n . Since they are distinct numbers, ´n has the stated factoriztion.
174 Chapter 9 Linear Higher Order Equations From (e), 1=n0 ; : : : ; 1=nn the stated factoriztion.
1
are all zeros of ´n C . Since they are distinct numbers, ´n C has
2k 2k 1 D .r ´0 /.r ´1 /.r ´2 / where ´k D cos C i sin , k D 0; 1; 2. 3 ! p p 3 1 3 1 3 1 2 3 Ci , and ´2 D i . Therefore, p.r / D .r 1/ rC C , Hence, ´0 D 1, ´1 D 2 2 2 2 2 4 ( p ! p !) 3 3 so e x ; e x=2 cos x ; e x=2 sin x is a fundamental set of solutions. 2 2 .2k C 1/ .2k C 1/ (b) p.r / D r 3 C1 D .r 0 /.r 1 /.r 2 / where k D cos Ci sin , k D 0; 1; 2. 3 ! p p 3 2 1 3 1 3 1 3 , 1 D 1 2 D i . Therefore, p.r / D .r C 1/ r C , so Hence, 0 D C i 2 2 2 2 2 4 ( p ! p !) 3 3 e x ; e x=2 cos x ; e x=2 sin x is a fundamental set of solutions. 2 2 p p p p .2k C 1/ (c) p.r / D r 4 C64 D .r 2 20 /.r 2 21 /.r 2 22 /.r 2 23 /, where k D cos C 4 1Ci 1Ci 1 i 1 i .2k C 1/ i sin , k D 0; 1; 2; 3. Therefore, 0 D p , 1 D p , 2 D p , and 3 D p , 4 2 2 2 2 so p.r / D ..r 2/2 C 4/..r C 2/2 C 4/ and fe 2x cos 2x; e 2x sin 2x; e 2x cos 2x; e 2x sin 2xg is a fundamental set of solutions. 2k 2k (d) p.r / D r 6 1 D .r ´0 /.r ´1 /.r ´2 /.r ´3 /.r ´4 /.r ´5 / where ´k D cos Ci sin , 6 6 p p p 1 3 1 3 1 3 k D 0; 1; 2; 3; 4; 5. Therefore, ´0 D 1, ´1 D C i , ´2 D Ci , ´3 D 1, ´4 D i , 2 2 2 2 2 2 ! ! p 2 2 1 3 1 3 1 3 and ´5 D i , so p.r / D .r 1/.r C 1/ r C rC C and 2 2 2 4 2 4 ( ! ! ! p p p p !) 3 3 3 3 x x x=2 x=2 x=2 x=2 e ; e ; e cos x ; e sin x ;e cos x ;e sin x is a funda2 2 2 2 mental set of solutions. (e) p.r / D r 6 C 64 D .r 20 /.r 21 /.r 22/.r 23/.rp 24/.r 25 / where p k D .2k C 1/ .2k C 1/ 3 i 3 i cos Ci sin , k D 0; 1; 2; 3; 4; 5. Therefore, 0 D C , 1 D i , 2 D C , 6 6 2 2 2 2 p p p p 3 i 3 i 3 D , 4 D i , and 5 D , so p.r / D .r 2 C 4/..r 3/2 C 1/..r C 3/2 C 1/ and 2 2 p 2 2p p p fcos 2x; sin 2x; e 3x cos x; e 3x sin x; e 3x cos x; e 3x sin xg is a fundamental set of solutions. (f) p.r / D .r 1/6 1 D .r 1 ´0 /.r 1 ´1 /.r 1 ´2 /.r 1 ´3 /.r 1p ´4 /.r 1 ´5 / where p ´k D 2k 2k 1 3 1 3 cos C i sin , k D 0; 1; 2; 3; 4; 5. Therefore, ´0 D 1, ´1 D C i , ´2 D Ci , ´3 D 6 6 2 ! 2 ! p p 2 22 2 1 3 1 3 3 3 1 3 1, ´4 D i , and ´5 D i , so p.r / D r .r 2/ r C r C 2 2 2 2 2 4 2 4 ( p ! p ! p ! p !) 3 3 3 3 and 1; e 2x ; e 3x=2 cos x ; e 3x=2 sin x ; e x=2 cos x ; e x=2 sin x is a funda2 2 2 2 mental set of solutions. r6 1 (g) p.r / D r 5 C r 4 C r 3 C r 2 C r C 1 D . Therefore, from the solution of (d) p.r / D r 1 9.2.43. (a) p.r / D r 3
Section 9.3 Undetermined Coefficients for Higher Order Equations ! ! 3 1 2 3 C rC C and 4 2 4 p ! p ! p ! 3 3 3 x x=2 x=2 x=2 e ;e cos x ;e sin x ;e cos x ;e 2 2 2 tal set of solutions. .r C 1/ r (
1 2
2
x=2
sin
p
3 x 2
!)
175
is a fundamen-
9.3 UNDETERMINED COEFFICIENTS FOR HIGHER ORDER EQUATIONS 9.3.2. If y D u 3x , then y 000 2y 00 5y 0 C 6y D e 3x Œ.u000 11u00 C 34u0 24u/ 2.u00 6u0 C 9u/ 5.u0 3u/ C 6u D e 3x .u000 11u00 C 34u0 24u/. Let up D A C Bx C Cx 2 , where . 24A C 34B 22C / C . 24B C 68C /x 24Cx 2 D 32 23x C 6x 2 . Then C D 1=4, B D 1=4, e 3x .3 x C x 2/. A D 3=4 and yp D 4 9.3.4. If y D ue 2x , then y 000 C 3y 00 y 0 3y D e 2x Œ.u000 6u00 C 12u0 8u/ C 3.u00 4u0 C 4u/ .u0 2u/ 3u D e 2x .u000 3u00 u0 C 3u/. Let up D A C Bx C Cx 2 , where .3A B 6C / C .3B 2C /x C 3Cx 2 D 2 17x C 3x 2 . Then C D 1, B D 5, A D 1, and yp D e 2x .1 5x C x 2 /. 9.3.6. If y D ue x , then y 000 Cy 00 2y D e x Œ.u000 C3u00 C3u0 Cu/C.u00 C2u0 Cu/ 2u D e x .u000 C4u00 C 5u0 /. Let up D x.ACBx CCx 2/, where .5AC8B C6C /C.10B C24C /x C15Cx 2 D 14C34x C15x 2. Then C D 1, B D 1, A D 0, and yp D x 2 e x .1 C x/. 9.3.8. If y D ue x , then y 000 y 00 y 0 C y D e x Œ.u000 C 3u00 C 3u0 C u/ .u00 C 2u0 C u/ .u0 C u/ C u D e x .u000 C 2u00 /. Let up D x 2.A C Bx/ where .4A C 6B/ C 12Bx D 7 C 6x. Then B D 1=2, A D 1, x2ex and yp D .2 C x/. 2 9.3.10. If y D ue 3x , then y 000 5y 00 C3y 0 C9y D e 3x Œ.u000 C9u00 C27u0 C27u/ 5.u00 C6u0 C9u/C3.u0 C 3u/C9u D e 3x .u000 C4u00 /. Let up D x 2 .ACBxCCx 2/, where .8AC6B/C.24B C24C /xC48Cx 2 D 22 48x 2 . Then C D 1, B D 1, A D 2, and yp D x 2 e 3x .2 C x x 2 /. 9.3.12. If y D ue x=2 , then 8y 000
12y 00 C 6y 0
y D e x=2Œ8.u000 C 3u00=2 C 3u0=4 C u=8/ 12.u00 C u0 C 1 C 4x u=4/ C 6.u0 C u=2/ u D 8e x=2u000 , so u000 D . Integrating three times and taking the constants 8 x3 x 3 e x=2 of integration to be zero yields up D .1 C x/. Therefore, yp D .1 C x/. 48 48 9.3.14. If y D ue 2x , then y .4/ C3y 000 Cy 00 3y 0 2y D e 2x Œ.u.4/ C8u000 C24u00 C32u0 C16u/C3.u000 C 6u00 C 12u0 C 8u/ C .u00 C 4u0 C 4u/ 3.u0 C 2u/ 2u D e 2x .u.4/ C 11u000 C 43u00 C 69u0 C 36u/. Let up D A C Bx where .36A C 69B/ C 36Bx D 33 36x. Then B D 1, A D 1, and yp D e 2x .1 x/. 9.3.16. If y D ue x , then 4y .4/ 11y 00 9y 0 2y D e x Œ4.u.4/ C 4u000 C 6u00 C 4u0 C u/ 11.u00 C 2u0 C u/ 9.u0 C u/ 2u D e x .4u.4/ C 16u000 C 13u00 15u0 18u/. Let up D A C Bx where ex .18A C 15B/ 18Bx D 1 C 6x. Then B D 1=3, A D 1=3, and yp D .1 x/. 3 9.3.18. If y D ue x , then y .4/ 4y 000 C6y 00 4y 0 C2y D e x Œ.u.4/ C4u000 C6u00 C4u0 Cu/ 4.u000 C3u00 C 3u0 C u/ C 6.u00 C 2u0 C u/ 4.u0 C u/ C 2u D e x .u.4/ C u/. Let up D A C Bx C Cx 2 C Dx 3 C Ex 4 where .A C 24E/ C Bx C Cx 2 C Dx 3 C Ex 4 D 24 C x C x 4. Then E D 1 D D 0, C D 0 B D 1, A D 0, and yp D xe x .1 C x 3 /. 9.3.20. If y D ue 2x , then y .4/ C y 000 2y 00 .u000 C 6u00 C 12u0 C 8u/ 2.u00 C 4u0 C 4u/
6y 0 4y D e 2x Œ.u.4/ C 8u000 C 24u00 C 32u0 C 16u/ C 6.u0 C 2u/ 4u D e 2x .u.4/ C 9u000 C 28u00 C 30u0 /. Let
176 Chapter 9 Linear Higher Order Equations up D x.AC Bx C Cx 2 / where .30AC 56B C 54C /C .60B C 168C /x C 90Cx 2 D .4 C 28x C 15x 2 /. xe 2x .1 x 2 /. Then C D 1=6, B D 0, A D 1=6, and yp D 6 9.3.22. If y D ue x , then y .4/ 5y 00 C 4y D e x Œ.u.4/ C 4u000 C 6u00 C 4u0 C u/ 5.u00 C 2u0 C u/ C 4u D e x .u.4/ C 4u000 C u00 6u0 /. Let up D x.A C Bx C Cx 2 / where . 6A C 2B C 24C / C . 12B C xe x .1 C x 2 /. 6C /x 18Cx 2 D 3 C x 3x 2 , so C D 1=6, B D 0, A D 1=6. Then yp D 6 9.3.24. If y D ue 2x , then y .4/ 3y 000 C 4y 0 D e 2x Œ.u.4/ C 8u000 C 24u00 C 32u0 C 16u/ 3.u000 C 6u00 C 12u0 C 8u/ C 4.u0 C 2u/ D e 2x .u.4/ C 5u000 C 6u00 /. Let up D x 2 .A C Bx C Cx 2 / where .12A C 30B C 24C / C .36B C 120C /x C 72Cx 3 D 15 C 26x C 12x 2. Then C D 1=6, B D 1=6, x 2 e 2x .3 C x C x 2 /. A D 1=2, and yp D 6 9.3.26. If y D ue x , then 2y .4/ 5y 000 C 3y 00 C y 0 y D e x Œ2.u.4/ C 4u000 C 6u00 C 4u0 C u/ 5.u000 C 3u00 C 3u0 C u/ C 3.u00 C 2u0 C u/ C .u0 C u/ u D e x .2u.4/ C 3u000 /. Let up D x 3 .A C Bx/, where x3ex .18A C 48B/ C 72Bx D 11 C 12x. Then B D 1=6, A D 1=6, and yp D .1 C x/. 6 9.3.28. If y D ue 2x , then y .4/ 7y 000 C 18y 00 20y 0 C 8y D e 2x Œ.u.4/ C 8u000 C 24u00 C 32u0 C 16u/ 7.u000 C 6u00 C 12u0 C 8u/ C 18.u00 C 4u0 C 4u/ 20.u0 C 2u/ C 8u D e 2x .u.4/ C u000 /. Let up D x 3 .A C Bx C Cx 2 / where .6A C 24B/ C .24B C 120C /x C 60Cx 2 D 3 8x 5x 2 . Then so x 3 e 2x C D 1=12 B D 1=12, A D 1=6, and yp D .2 C x x 2 /. 12 9.3.30. If y D ue x , then y 000 C y 00 4y 0 4y D e x Œ.u000 3u00 C 3u0 u/ C .u00 2u0 C u/ u/ 4u D e x .u000 2u00 3u0 /. Let up D .A0 C A1 x/ cos 2x C .B0 C B1 x/ sin 2x, where 8A1 14B1 14A1 C 8B1 8A0 14B0 15A1 8B1 14A0 C 8B0 C 8A1 15B1 Then A1 D 1, B1 D 1, A0 D 1, B0 D 1, and yp D e
x
Œ.1
D D D D
4.u0
22 6 1 1:
x/ cos 2x C .1 C x/ sin 2x.
9.3.32. If y D ue x , then y 000 2y 00 Cy 0 2y D e x Œ.u000 C3u00 C3u0 Cu/ 2.u00 C2u0 Cu/C.u0 Cu/ 2u D e x .u000 C u00 2u/. Let up D .A0 C A1 x C A2 x 2 / cos 2x C .B0 C B1 x C B2 x 2 / sin 2x where
6A0 8B0 8A0 6B0
6A2 8B2 8A2 6B2 6A1 8B1 24A2 C 8B2 8A1 6B1 8A2 24B2 12A1 C 4B1 C 2A2 C 12B2 4A1 12B1 12A2 C 2B2
D D D D D D
4 3 5 5 9 6:
Then A2 D 0, B2 D 1=2; A1 D 1=2, B1 D 1=2; A0 D 1=1, B0 D 1=2; and yp D .1
x C x 2 / sin 2x.
ex Œ.1 C x/ cos 2x C 2
9.3.34. If y D ue x , then y 000 y 00 C 2y D e x Œ.u000 C 3u00 C 3u0 C u/ .u00 C 2u0 C u/ C 2u D e x .u000 C 2u00 C u0 C 2u/. Since cos x and sin x satisfy u000 C 2u00 C u0 C 2u D 0, let up D xŒ.A0 C
Section 9.3 Undetermined Coefficients for Higher Order Equations
177
A1 x/ cos x C .B0 C B1 x/ sin x where 4A1 C 8B1 8A1 4B1 2A0 C 4B0 C 4A1 C 6B1 4A0 2B0 6A1 C 4B1
D 4 D 12 D 20 D 12:
Then A1 D 1, B1 D 1; A0 D 1, B0 D 3; and yp D xe x Œ.1 C x/ cos x C .3 C x/ sin x. 9.3.36. If y D ue 3x , then D e 3x Œ.u000 C 9u00 C 27u0 C 27u/ 6.u00 C 6u0 C 9u/ C 18.u0 C 3u/ D e 3x .u000 C 3u00 C 9u0 C 27u/. Since cos 3x and sin 3x satisfy u000 C 3u00 C 9u0 C 27u D 0, let up D xŒ.A0 C A1 x/ cos 3x C .B0 C B1 x/ sin 3x where 36A1 C 36B1 36A1 36B1 18A0 C 18B0 C 6A1 C 18B1 18A0 18B0 18A1 C 6B1 Then A1 D 1=12, B1 D 0; A0 D, B0 D 1=12; and yp D
D D D D
3 3 2 3:
xe 3x .x cos 3x C sin 3x/. 12
9.3.38. If y D ue x , then y .4/ 3y 000 C 2y 00 C 2y 0 4y D e x Œ.u.4/ C 4u000 C 6u00 C 4u0 C u/ 3.u000 C 3u00 C 3u0 C u/ C 2.u00 C 2u0 C u/ C 4.u0 C u/ C u D e x .u.4/ C u000 u00 C u0 2u/. Let up D A cos 2x C B sin 2x where 18A 6B D 2 and 6A C 18B D 1. Then A D 1=12, B D 1=12, ex and yp D .cos 2x sin 2x/. 12 9.3.40. If y D ue x , then y .4/ C 6y 000 C 13y 00 C 12y 0 C 4y D e x Œ.u.4/ 4u000 C 6u00 4u0 C u/ C 6.u000 3u00 C 3u0 u/ C 13.u00 2u0 C u/ C 12u0 u/ C 4u D e x .u.4/ C 2u000 C u00 /. Let up D .A0 C A1 x/ cos x C .B0 C B1x/ sin x where
2B0 6A1 2A0 C 2A1
2B1 2A1 2B1 6B1
D D D D
Then A1 D 1=2, B1 D 1=2, A0 D 1=2, B0 D 1, and yp D
1 1 4 5: x
e 2
Œ.1 C x/ cos x C .2
x/ sin x.
9.3.42. If y D ue x , then y .4/ 5y 000 C 13y 00 19y 0 C 10y D e x Œ.u.4/ C 4u000 C 6u00 C 4u0 C u/ 5.u000 C 3u00 C 3u0 C u/ C 13.u00 C 2u0 C u/ 19.u0 C u/ C 10u D e x .u.4/ u000 C 4u00 4u0 /. Since cos 2x and sin 2x satisfy u.4/ u000 C 4u00 4u0 D 0, let up D x.A cos 2x C B sin 2x/ where 8A 16B D 1 xe x and 16A C 8B D 1. Then A D 3=40, B0 D 1=40, and yp D .3 cos 2x sin 2x/. 40 9.3.44. If y D ue x , then y .4/ 5y 000 C 13y 00 19y 0 C 10y D e x Œ.u.4/ C 4u000 C 6u00 C 4u0 C u/ 5.u000 C 3u00 C 3u0 C u/ C 13.u00 C 2u0 C u/ 19.u0 C u/ C 10u D e x .u.4/ u000 C 4u00 4u0 /. Since cos 2x and sin 2x satisfy u.4/ u000 C 4u00 4u0 D 0, let up D xŒ.A0 C A1 x/ cos 2x C .B0 C B1 x/ sin 2x/ where 16A1 32B1 D 8 32A1 C 16B1 D 4 8A0 16B0 40A1 12B1 D 7 16A0 C 8B0 C 12A1 40B1 D 8:
178 Chapter 9 Linear Higher Order Equations Then A1 D 0, B1 D 1=4; A0 D 0, B0 D 1=4, and yp D
xe x .1 C x/ sin 2x. 4
9.3.46. If y D ue 2x , then y .4/ 8y 000 C 32y 00 64y 0 C 64y C 4y D e 2x Œ.u.4/ C 8u000 C 24u00 C 32u0 C 16u/ 8.u000 C6u00 C12u0 C8u/C32.u00 C4u0 C4u/ 64.u0 C2u/C64u D e 2x .u.4/ C8u00 C16u/. Since cos 2x, sin 2x, x cos 2x, and x sin 2x satisfy u.4/ C 8u00 C 16u D 0, let up D x 2 .A cos 2x C B sin 2x/ x 2e 2x .cos 2x sin 2x/. where 32A D 1 and 32B D 1. Then A D 1=32, B D 1=32, and yp D 32 9.3.48. Find particular solutions of (a) y 000 4y 00 C 5y 0 2y D 4e x , (b) y 000 4y 00 C 5y 0 2y D e 2x , and (c) y 000 4y 00 C 5y 0 2y D 2 cos x C 4 sin x. (a) If y D ue x , then y 000 4y 00 C5y 0 2y D e x Œ.u000 C3u00 C3u0 Cu/ 4.u00 C2u0 Cu/C5.u0 Cu/ 2u D x 000 e .u u00 /. Let u1p D Ax 2 where 2A D 4. Then A D 2, and y1p D 2x 2 e x . (b) If y D ue 2x , then y 000 4y 00 C 5y 0 2y D e 2x Œ.u000 C 6u00 C 12u0 C 8u/ 4.u00 C 4u0 C 4u/ C 5.u0 C 2u/ 2u D e 2x .u000 C 2u00 C u0 /. Let u2p D x. Then y2p D xe 2x . 000 00 0 (c) If y3p D A cos xCB sin x, then y3p 4y3p C5y3p 2y3p D .2AC4B/ cos xC. 4AC2B/ sin x D 2 cos x C 4 sin x if A D 1 and B D 0, so y3p D cos x. From the principle of superposition, yp D 2x 2 e x C xe 2x cos x. 9.3.50. Find particular solutions of (a) y 000 y 0 D 2.1 C x/, (b) y 000 y 0 D 4e x , (c) y 000 y 0 D 6e x , and (d) y 000 y 0 D 96e 3x 000 0 (a) Let y1p D x.A C Bx/. Then y1p y1p D A 2Bx D 2.1 C x/ if A D 2 and B D 1; therefore 2 y1p D 2x C 2x . (b) If y D ue x , then y 000 y 0 D e x Œ.u000 C 3u00 C 3u0 C u/ .u0 C u/ D e x .u000 C 3u00 C 2u0 /. Let u2p D 4x. Then y2p D 4xe x . (c) If y D ue x , then y 000 y 0 D e x Œ.u000 3u00 C 3u0 u/ .u0 u/ D e x .u000 3u00 C 2u0 /. Let u2p D 3x. Then y2p D 6xe x . 000 0 (d) Since e 3x does not satisfy the complementary equation, let y4p D Ae 3x . Then y4p y4p D 3x 4x 24Ae . Let A D 4; then y4p D 4e . From the principle of superposition, yp D 2x C x 2 C 2xe x 3xe x C 4e 3x 9.3.52. Find particular solutions of (a) y 000 C 3y 00 C 3y 0 C y D 12e x and (b) y 000 C 3y 00 C 3y 0 C y D 9 cos 2x 13 sin 2x. (a) If y D ue 2x , then y 000 C 3y 00 C 3y 0 C y D e 2x Œ.u000 3u00 C 3u0 u/ C 3.u00 2u0 C u/ C 3.u0 u/ C u D e x u000 . Let u000 1p D 12. Integrating three times and taking the constants of integration to be zero yields u1p D 2x 3 . Therefore,y1p D 2x 3. (b) Let y2p D A cos 2x C B sin 2x where 11A 2B D 9 and 2A 11B D 13. Then A D 1, B D 1, and y2p D cos 2x C sin 2x. From the principle of superposition, yp D 2x 3 e 2x cos 2x C sin 2x. 9.3.54. Find particular solutions of (a) y .4/ 5y 00 C 4y D 12e x , (b) y .4/ 5y 00 C 4y D 6e x , and (c) y .4/ 5y 00 C 4y D 10 cos x. (a) If y D ue x , then y .4/ 5y 00 C 4y D e x Œ.u.4/ C 4u000 C 6u00 C 4u0 C u/ 5.u00 C 2u0 C u/ C 4u D x .4/ e .u C 4u000 C u00 6u0 /. Let u1p D 2x. Then y1p D 2xe x . (b) If y D ue x , then y .4/ 5y 00 C 4y D e x Œ.u.4/ 4u000 C 6u00 4u0 C u/ 5.u00 2u0 C u/C 4u D e x .u.4/ 4u000 C u00 C 6u0 /. Let u2p D x. Then y2p D xe x . (c) Let y3p D A cos xCB sin x where 10A D 10 and 10B D 0. Then A D 1, B D 0, and y3p D cos x. From the principle of superposition, yp D 2xe x C xe x C cos x. 9.3.56. Find particular solutions of (a) y .4/ C 2y 000 3y 00 4y 0 C 4y D e 2x .
3y 00
4y 0 C 4y D 2e x .1 C x/ and (b) y .4/ C 2y 000
Section 9.3 Undetermined Coefficients for Higher Order Equations
179
(a) If y D ue x , then y .4/ C 2y 000 3y 00 4y 0 C 4y D e x Œ.u.4/ C 4u000 C 6u00 C 4u0 C u/ C 2.u000 C 3u00 C 3u0 C u/ 3.u00 C 2u0 C u/ 4.u0 C u/ C 4u D e x .u.4/ C 6u000 C 9u00 /. Let u1p D x 2 .A C Bx/ x2 where .18A C 36B/ C 54Bx D 2 C 2x. Then B D 1=27, A D 1=27, and y1p D .1 C x/e x . 27 (b) If y D ue 2x , then y .4/ C 2y 000 3y 00 4y 0 C 4y D e 2x Œ.u.4/ 8u000 C 24u00 32u0 C 16u/ C 2.u000 6u00 C 12u0 8u/ 3.u00 4u0 C 4u/ 4.u0 2u/ C 4u D e 2x .u.4/ 6u000 C 9u00 /. Let x 2 2x e . u2p D Ax 2 where 18A D 1. Then A D 1=18 and yp D 18 2 x Œ.2 C 2x/e x C 3e 2x . From the principle of superposition, yp D 54 9.3.58. Find particular solutions of (a) y .4/ C 5y 000 C 9y 00 C 7y 0 C 2y D e x .30 C 24x/ and (b) y .4/ C 5y 000 C 9y 00 C 7y 0 C 2y D e 2x . (a) If y D ue x , then y .4/ C 5y 000 C 9y 00 C 7y 0 C 2y D e x Œ.u.4/ 4u000 C 6u00 4u0 C u/ C 5.u000 3u00 C 3u0 u/ C 9.u00 2u0 C u/ C 7.u0 u/ C 2u D e x .u.4/ C u000 /. Let u1p D x 3 .A C Bx/ where .6A C 24B/ C 24Bx D 30 C 24x. The B D 1, A D 1, and y1p D x 3 .1 C x/e x . (b) If y D ue 2x , then y .4/ C 5y 000 C 9y 00 C 7y 0 C 2y D e 2x Œ.u.4/ 8u000 C 24u00 32u0 C 16u/ C 5.u000 6u00 C 12u0 8u/ C 9.u00 4u0 C 4u/ C 7.u0 2u/ C 2u D e 2x .u.4/ 3u000 C 3u00 u0 /. Let u2p D x. Then y2p D xe 2x . From the principle of superposition, yp D x 3 .1 C x/e x C xe 2x . 9.3.60. If y D ue 2x , then y 000 y 00 y 0 Cy D e 2x Œ.u000 C6u00 C12u0 C8u/ .u00 C4u0 C4u/ .u0 C2u/C u D e 2x .u000 C 5u00 C 7u0 C 3u/. Let up D A C Bx, where .3A C 7B/ C 3x D 10 C 3x. Then B D 1, A D 1 and yp D e 2x .1 C x/. Since p.r / D .r C 1/.r 1/2 , y D e 2x .1 C x/ C c1 e x C e x .c2 C c3x/ 9.3.62. If y D ue 2x , then y 000 6y 00 C11y 0 6y D e 2x Œ.u000 C6u00 C12u0 C8u/ 6.u00 C4u0 C4u/C11.u0 C 2u/ 6u D e 2x .u000 u0 /. Let up D x.ACBx CCx 2 / where . AC6C / 2Bx 3Cx 2 D 5 4x 3x 2. Then C D 1, B D 2, A D 1, and yp D xe 2x .1 C x/2 . Since p.r / D .r 1/.r 2/.r 3/, y D xe 2x .1 C x/2 C c1 e x C c2 e 2x C c3 e 3x . 9.3.64. If y D ue x , then y 000 3y 00 C 3y 0 y D e x Œ.u000 C 3u00 C 3u0 C u/ 3.u00 C 2u0 C u/ C 3.u0 C u/ u D e x u000 . Let u000 D 1 C x. Integrating three times and taking the constants of integration x3 x3ex to be zero yields u D .4 C x/. Therefore, yp D .4 C x/. Since p.r / D .r 1/3 , y D 24 24 x3ex .4 C x/ C e x .c1 C c2 x C c3 x 2 /. 24 9.3.66. If y D ue 2x , then y 000 C 2y 00 y 0 2y D e 2x Œ.u000 6u00 C 12u0 8u/ C 2.u00 4u0 C 4u/ .u0 2u/ 2u D e 2x .u000 4u00 C 3u0 /. Let up D .A0 C A1 x/ cos x C .B0 C B1 x/ sin x where 4A1 C 2B1 2A1 C 4B1 4A0 C 2B0 8B1 2A0 C 4B0 C 8A1
D D D D
2 9 23 8: x 3 2x Then A1 D 1=2, B1 D 2; A0 D 1, B0 D 3=2, and yp D e 1C cos x C 2x sin x . 2 2 x 3 Since p.r / D .r 1/.r C 1/.r C 2/, y D e 2x 1 C cos x C 2x sin x C c1 e x C c2 e x C 2 2 c3 e 2x .
180 Chapter 9 Linear Higher Order Equations 9.3.68. If y D ue x , then y .4/ 4y 000 C 14y 00 20y 0 C 25y D e x Œ.u.4/ C 4u000 C 6u00 C 4u0 C u/ 4.u000 C 3u00 C 3u0 C u/ C 14.u00 C 2u0 C u/ 20.u0 C u/ C 25u D e x .u.4/ C 8u00 C 16u/. Since cos 2x, sin 2x, x cos 2x, and x sin 2x satisfy u.4/ C8u00 C16u D 0, let up D x 2Œ.A0 CA1x/ cos 2x C.B0 CB1 x/ sin 2x where 96A1 D 6 96B1 D 0 32A0 C 48B1 D 2 48A1 32B0 D 3: Then A1 D Œ.r
1=16, B1 D 0; A0 D
1/2 C 12 , y D
1=16, B0 D 0, and yΠD
x2ex .1 C x/ cos 2x. Since p.r / D 16
x2ex .1 C x/ cos 2x C e x Œ.c1 C c2 x/ cos 2x C .c3 C c4 x/ sin 2x. 16
9.3.70. If y D ue x , then y 000 y 00 y 0 C y D e x Œ.u000 3u00 C 3u0 u/ .u00 2u0 C u/ .u0 u/ C u D e x .u000 4u00 C 4u0 /. Let up D x.A C Bx/, where .4A 8B/ C 8Bx D 4 C 8x. Then B D 1, A D 1, and yp D x.1 C x/e x . Since p.r / D .r C 1/.r 1/2 the general solution is y D x.1 C x/e x C c1 e x C c2 e x C c3 xe x . Therefore, 2 3 2 3 2 32 3 y x.1 C x/e x e x ex xe x c1 4 y 0 5 D 4 e x .x 2 x 1/ 5 C 4 e x e x e x .x C 1/ 5 4 c2 5 : y 00 e x .x 2 3x/ e x e x e x .x C 2/ c3 Setting x D 0 and imposing the initial conditions yields 32 3 3 2 2 3 2 c1 2 0 1 1 0 4 0 5 D 4 1 5 C 4 1 1 1 5 4 c2 5 ; c3 0 0 1 1 2
so c1 D 1, c2 D 1, c3 D 1, and y D e
x
.1 C x C x 2 / C .1
x/e x .
9.3.72. If y D ue x , then y 000 2y 00 5y 0 C 6y D e x Œ.u.4/ 4u000 C 6u00 4u0 C u/ C 2.u000 3u00 C 3u0 u/ C 2.u00 2u0 C u/ C 2.u0 u/ C u D e x .u.4/ 2u000 C 2u00 /. Let up D x 2 .A C Bx/, where .4A 12B/ C 12Bx D 20 12x. Then B D 1, A D 2, and yp D x 2.2 x/e x . Since p.r / D .r C 1/2 .r 2 C 1/, the general solution is y D x 2 .2 x/e x C e x .c1 C c2x/C c3 cos x C c4 sin x. Therefore, 2 3 2 3 2 32 3 y x 2 .2 x/e x e x xe x cos x sin x c1 6 y0 7 6 7 6 e x .1 x/e x 6 7 x.x 2 5x C 4/e x sin x cos x 7 6 00 7 D 6 7C6 7 6 c2 7 x x 3 2 x 4 y 5 4 .x 5 4 e 8x C 14x 4/e .x 2/e cos x sin x 5 4 c3 5 y 000 .x 3 11x 2 C 30x 18/e x e x .3 x/e x sin x cos x c4
Setting x D 0 and imposing the initial conditions yields 2 3 2 3 2 3 0 1 6 7 6 7 6 1 4 0 6 7D6 7C6 4 7 5 4 4 5 4 1 22 18 1
so c1 D 2, c2 D 1, c3 D 1, c4 D 1, and y D .2
0 1 2 3
1 0 1 0
x/.x 2 C 1/e
x
32 0 c1 6 c2 1 7 76 0 5 4 c3 1 c4
C cos x
3
7 7; 5
sin x.
9.3.74. If y D ue x , then y .4/ 3y 000 C 5y 00 2y 0 D e x Œ.u.4/ C 4u000 C 6u00 C 4u0 C u/ 3.u000 C 3u00 C 3u0 C u/ C 4.u00 C 2u0 C u/ 2.u0 C u/ D e x .u.4/ C u000 C u00 C u0 /. Since cos x and sin x satisfy
Section 9.4 Variation of Parameters for Higher Order Equations
181
u.4/ C u000 C u00 C u0 D 0, let up D x.A cos x C B sin x/ where 2A 2B D 2 and 2A 2B D 2. Then A D 1, B D 0, and yp D e x cos x. Since p.r / D r .r 1/Œ.r 1/2 C 1 the general solution is y D e x cos x C c1 C e x .c2 C c3 cos x C c4 sin x/. Therefore, 2 3 2 3 2 y xe x cos x 1 ex e x cos x e x sin x x x 6 y 0 7 6 e x ..x C 1/ cos x x sin x/ 7 6 0 e x e .cos x sin x/ e .cos x C sin x/ 6 00 7 D 6 7 6 4 y 5 4 e x .2 cos x 2.x C 1/ sin x/ 5C4 0 e x 2e x sin x 2e x cos x 000 x x x x y e .2x cos x C 2.x C 3/ sin x/ 0 e e .2 cos x C 2 sin x/ e .2 cos x 2 sin x/
Setting x D 0 and imposing the initial conditions yields 2 3 2 3 2 2 0 1 1 6 0 7 6 1 7 6 0 1 6 7 6 7 6 4 1 5 D4 2 5C4 0 1 5 0 0 1
1 1 0 2
32 0 c1 6 c2 1 7 76 2 5 4 c3 2 c4
so c1 D 2, c2 D 1, c3 D 1, c4 D 1, and 2 C e x Œ.1 C x/ cos x
sin x
3
7 7; 5
1.
9.4 VARIATION OF PARAMETERS FOR HIGHER ORDER EQUATIONS ˇ ˇ 2 2 2 ˇ ˇ e x xe x x2e x ˇ ˇ 2 ˇ ˇ 2 2 2 9.4.2. W D ˇ ˇ D 2e 3x ; 2xe x e x .1 2x 2 / 2xe x .1 x 2 / ˇ ˇ ˇ e x2 .4x 2 2/ 2xe x2 .2x 2 3/ 2e x2 .2x 4 5x 2 C 1/ ˇ ˇ ˇ ˇ ˇ 2 2 ˇ ˇ ˇ ˇ x2 2 x2 xe x x2e x 2 e x e ˇ ˇ ˇ ˇ 2 2x W1 D ˇ D x e ; W D ˇ ˇ ˇD 2 2 2 2 2 ˇ e x .1 2x 2 / 2xe x .1 x 2 / ˇ ˇ 2xe x 2xe x .1 x 2 / ˇ ˇ ˇ 2 2 ˇ ˇ 1 F W2 F W1 2 xe x 2 ˇ e x ˇ 2xe 2x ; W3 D ˇ D x 5=2 ; u02 D D ˇ D e 2x ; u01 D 2 2 x x 2 ˇ 2xe P0 W 2 P0 W e .1 2x / ˇ p F W2 2 5=2 x 3=2 ; u03 D D x=2; u1 D x 7=2=7; u2 D x ; u3 D x 3=2=3; yp D u1 y1 Cu2 y2 Cu3 y3 D P0 W 5 8 2 e x x 7=2. 105 ˇ ˇ ex e x ˇ ˇ ˇ ˇ ˇ 1 ˇ e x ex ˇ ˇ x x ˇ x x ˇ ˇ e .x 1/ e .x C 1/ x ˇ D 2=x 2 ; W1 D ˇˇ x x 9.4.4. W D ˇˇ 0 ˇ 2 2 e x .x C 1/ ˇ e .x 1/ x x ˇ ˇ ˇ ˇ e x .x 2 2x C 2/ e x .x 2 C 2x C 2/ ˇ x2 x2 ˇ 0 ˇ 3 3 x x ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ e x ex ˇ ˇ ˇ ˇ x 1 1 2 e .x C 1/ e x .x 1/ 0 ˇ ˇ ˇ ˇ x x ; W D D ; W D D ; u1 D ˇ ˇ ˇ ˇ x x 2 3 e .x C 1/ ˇ e .x 1/ ˇ ˇ ˇ x2 x2 x2 ˇ 0 ˇ ˇ 0 ˇ 2 2 x x F W1 F W2 F W2 x x 0 0 D 2; u2 D D e .x C 1/; u3 D D e .x 1/; u1 D 2x; u2 D e x .x C 2/; P0 W P0 W P0 W u3 D e x .x 2/; yp D u1 y1 C u2 y2 C u3 y3 D 2.x 2 C 2/=x. ˇ ˇ ˇ x ˇ ˇ ˇ 1 x ˇ e ˇ ˇ 1 ˇ e x ˇ ˇ ˇ ˇ 2 x x 2.x 2/ ˇ ˇ ˇ e x ˇˇ D e .x 1/ ; W2 D 9.4.6. W D ˇ e x e x 1=x 2 ˇ D ; W D ˇ 1 1 ˇ ˇ ˇ ˇ e x x3 x2 2 ˇ x ˇ ˇ ˇ x 2 e e ˇ ˇ x x3
ˇ ˇ ˇ ˇ ˇD ˇ ˇ
32
3 c1 7 6 c2 7 76 7 5 4 c3 5 c4
182 Chapter 9 Linear Higher Order Equations ˇ ˇ x ˇ e ˇ ˇ ˇ ex ˇ
1 x 1 x2
ˇ ˇ ˇ ˇ ˇ D ˇ ˇ
ˇ x ˇ e e x .x C 1/ ˇ x ; W D 3 ˇ e x2
ˇ e x ˇˇ D e x ˇ
2; u01 D
F W1 D e P0 W
x
.x
1/; u02 D
F W2 F W2 x2 D e x .x C 1/; u03 D D 2x 2 ; yp D u1 y1 C u2 y2 C u3 y3 D 2 . P0 W P0 W 3 ˇ ˇ 1 ˇ p ˇ ˇ ˇ ˇ x p x2 ˇ ˇ p 1 ˇ ˇ ˇ x ˇ 2 ˇˇ p x x2 ˇ ˇ ˇ p x ˇ 5 15 x 1 1 ˇ ˇ ˇ ˇ x ˇ ; W1 D ˇ ; W2 D ˇ 1 9.4.8. W D ˇ p 2x ˇ D ˇD 1 ˇ ˇ ˇ 2 x ˇ 4x 2 2x 3=2 ˇ 2px 2x ˇ 2x ˇ ˇ ˇ 3=2 1 3 ˇ ˇ 2x 2 ˇ ˇ 4x 5=2 ˇ ˇ 4x 3=2 ˇ p ˇ 1 ˇ ˇ x p 3=2 ˇ ˇ p 3x x ˇ ˇ D 1 ; u0 D F W1 D 5 x; u0 D F W2 D 3x 3=2; u0 D ; W3 D ˇ 1 1 2 3 ˇ 1 2 x P W P0 W 0 ˇ p ˇ ˇ 2 x ˇ 3=2 2x F W2 2 10 3=2 6 D ; u1 D x ; u2 D x 5=2 ; u3 D 2 ln jxj; yp D u1 y1 C u2 y2 C u3 y3 D 2x 2 ln jxj P0 W x 3 5 32 2 32 2 x . Since x satisfies the complementary equation we take yp D ln jxj. 15 15 ˇ ˇ ex ˇ ˇ ˇ ˇ ˇ ˇ x 1=x ˇ 1 ˇ ex ˇ ˇ x ˇ ˇ x x ˇ ˇ ex 2e .1 x/ e .x 1/ 1 ˇ ˇ ˇ D x x D 9.4.10. W D ˇˇ 1 ; W D ; ˇ ˇ x 1 ˇ 3 1 e .x 1/ ˇ ˇ x x2 x2 x2 ˇ ˇ x 2 ˇ ˇ ˇ 2 e .x 2x C 2/ ˇ x2 x2 ˇ ˇ 0 3 3 xˇ x ˇ ˇ ˇ ˇ ˇ ˇ ex 1 ˇ ˇ ˇ x ˇ ˇ x e .x 2/ ˇ x ˇ ˇ x x ˇˇ D 2 ; u0 D F W1 D 1; u0 D F W2 D W2 D ˇ D ; W D ˇ ˇ x 3 2 1 ˇ e .x 1/ ˇ ˇ ˇ 1 x x 1 P0 W P0 W ˇ 1 ˇ ˇ ˇ 2 2 x x F W2 x 2 .3 x/ x.2 x/; u03 D D 2xe x ; u1 D x; u2 D ; u3 D 2e x .x C 1/; yp D u1 y1 C P0 W 3 2x 3 C 3x 2 C 6x C 6 2 u2 y2 C u3 y3 D . Since x C satisfies the complementary equation we take 3x x 2 2x C 6 yp D . 3 ˇ ˇ ˇ x ex e x ˇ ˇ x ˇ ˇ ˇ ˇ ˇ ˇ e ˇ x e x ˇ e x ˇˇ x x ˇ ˇ ˇ ˇ ˇ D e x .x C 9.4.12. W D ˇ 1 e e D 2; W2 D ˇ ˇ D 2x; W1 D ˇ e x e x ˇ 1 e x ˇ ˇ 0 ex e x ˇ ˇ ˇ ˇ x ex ˇ ˇ ˇ D e x .x 1/; u0 D F W1 D 1; u0 D F W2 D e x .x C 1/=2; u0 D F W2 D 1/; W3 D ˇ 1 2 3 1 ex ˇ P0 W P0 W P0 W x x x e .x 1/=2; u1 D x; u2 D e .x C2/=2; u3 D e .x 2/=2; yp D u1 y1 Cu2 y2 Cu3 y3 D x 2 2.
ˇ ˇ ˇ ˇD ˇ ˇ
Section 9.4 Variation of Parameters for Higher Order Equations p x
183
p 1= x
ˇ ˇ 1 ˇ ˇ ˇ x 3=2 1 1 ˇ ˇ ˇ 3=2 x p ˇ p ˇ ˇ x 3=2 3=2 1 1 3 x 3 ˇ ˇ ˇ x x p ˇ ˇ ˇ p 3=2 5=2 12 ˇ 1 ˇ ˇ 3 x 3 2 2 x 2x 2x D 6 ; W1 D ˇ ˇD ˇ 1 3 3 15 5=2 ˇ 2x 3=2 ˇ ˇ x 2 2x p ˇ ˇ ˇ 3 15 3 ˇ ˇ 4x 3=2 4x 5=2 4 x 4x 7=2 ˇ p ˇ ˇ ˇ 5=2 7=2 3 15 3 105 ˇ 4x 4 x 4x ˇ 5=2 8x 7=2 8x 3=2 ˇ 8x 9=2 ˇ 8x p ˇ ˇ 1 1 1 ˇ ˇ ˇ p ˇ 3=2 ˇ ˇ ˇ x x ˇ x p 3=2 3=2 ˇ ˇ ˇ ˇ x x x p ˇ ˇ ˇ ˇ 1 3 x 3 6 6 1 1 3 ˇ ˇ ˇ ˇ p ; W D D ; W D ˇ ˇ ˇ ˇ D p 2 3 5=2 ˇ 7=2 5=2 3=2 5=2 ˇ ˇ 2 ˇ 2 x 2x x x 2 x 2x 2x ˇ ˇ ˇ ˇ 3 15 1 1 3 15 ˇ ˇ ˇ ˇ p ˇ ˇ ˇ ˇ 3=2 7=2 3=2 5=2 7=2 4x 4 x 4x 4x 4x 4x ˇ ˇ ˇ p ˇ 1 ˇ x p x 3=2 ˇˇ ˇ x ˇ p ˇ ˇ 2 F W2 F W1 2 1 1 3 x ˇˇ 0 0 ˇ ; W4 D ˇ p ˇ D x 3=2 ; u1 D P W D 3x; u2 D P W D 3=2 x 9=2 2 2 x 2x 0 0 ˇ ˇ ˇ 3 3 ˇˇ 1 ˇ p ˇ 4x 3=2 4x 5=2 4 x ˇ F W2 F W4 3x 2 x4 3x 2 ; u03 D D 1; u04 D D x 3 ; u1 D ; u2 D x 3 ; u3 D x; u4 D ; P0 W P0 W 2 4 x 5=2 yp D u1 y1 C u2 y2 C u3 y3 D . 4 ˇ ˇ ˇ 2 ˇ ˇ ˇ x x2 x3 x 4 ˇˇ ˇ x ˇ x x3 ˇ x3 x 4 ˇˇ x4 ˇ ˇ ˇ 1 2x 3x 2 4x 3 ˇ 2 3 2 4 6 ˇ ˇ 4x ˇˇ D 2x ; W2 D ˇˇ 1 3x 4x 3 9.4.16. W D ˇˇ 2 ˇ D 12x ; W1 D ˇ 2x 3x 0 2 6x 12x 2 ˇ ˇ ˇ 2 ˇ 0 6x 12x 2 6x 12x ˇ ˇ 0 0 ˇ 6 24x ˇ ˇ ˇ ˇ ˇ x x2 ˇ x x2 x3 ˇ x 4 ˇˇ ˇ ˇ ˇ F W1 x2 6x 5 ; W3 D ˇˇ 1 2x 4x 3 ˇˇ D 6x 4 ; W4 D ˇˇ 1 2x 3x 2 ˇˇ D 2x 3; u01 D D ; P0 W 6 ˇ 0 2 12x 2 ˇ ˇ 0 2 6x ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ 9.4.14. W D ˇ ˇ ˇ ˇ ˇ ˇ ˇ
x F W2 1 0 F W4 1 x3 x2 x F W2 D ; u03 D D ; u4 D D ; u1 D ; u2 D ; u3 D ; P0 W 2 P0 W 2 P0 W 6x 18 4 2 4 4 4 x ln jxj 11x 11x ln jxj u4 D ; yp D u1 y1 C u2 y2 C u3 y3 D . Since satisfies the complementary 6 6 36 36 4 x ln jxj equation we take yp D . 6 ˇ ˇ ˇ x x2 ˇ 2 ˇ 1=x 1=x 2 ˇˇ ˇ ˇ x 1=x 1=x 2 ˇˇ 2 3 ˇ ˇ ˇ 1 2x 1=x 2=x ˇ D 72=x 6; W1 D ˇ 2x 1=x 2 2=x 3 ˇˇ D 12=x 4; 9.4.18. W D ˇˇ ˇ 2=x 3 6=x 4 ˇˇ 3 ˇ 0 2 ˇ 2 2=x 6=x 4 ˇ ˇ 0 0 6=xˇ 4 24=x 5 ˇ ˇ ˇ ˇ ˇ ˇ x ˇ x x 2 1=x 2 ˇ ˇ x x2 1=x 1=x 2 ˇˇ 1=x ˇ ˇ ˇ ˇ 2 3 5 3 2 W2 D ˇˇ 1 1=x 2=x ˇˇ D 6=x ; W3 D ˇˇ 1 2x 2=x ˇˇ D 12=x ; W4 D ˇˇ 1 2x 1=x 2 3 4 4 ˇ 0 2=x ˇ 0 2 ˇ 0 2 6=x ˇ 6=x ˇ 2=x 3 F W1 F W2 F W2 F W4 6=x; u01 D D 2; u02 D D 1=x; u03 D D 2x 2 ; u04 D D x 3 ; u1 D 2x; P0 W P0 W P0 W P0 W u2 D ln jxj; u3 D 2x 3 =3; u4 D x 4 =4; yp D u1 y1 C u2 y2 C u3 y3 D x 2 ln jxj 19x 2=12. Since 19x 2 =12 satisfies the complementary equation we take yp D x 2 ln jxj.
ˇ ˇ ˇ ˇD ˇ ˇ
u02 D
ˇ ˇ ˇ ˇD ˇ ˇ
184 Chapter 9 Linear Higher Order Equations ˇ ˇ ˇ ˇ e 2x x ˇ e x e 2x ˇ e =x ˇ ˇ x ˇ ˇ x 2x ˇ x ˇ e .x 1/ e .2x 1/ 2x ˇ e ˇ 2e e 6x ˇ ˇ 2 2 x x 9.4.20. W D ˇ ; ˇD x 2 2x 2 e .x 2x C 2/ 2e .2x 2x C 1/ ˇ x ˇ x4 ˇ e ˇ 4e 2x ˇ ˇ x3 x3 ˇ x 3 3x 2 C 6x 6/ 2e 2x .4x 3 6x 2 C 6x 3/ ˇˇ ˇ e x 8e 2x e .x ˇ ˇ x4 x4 ˇ ˇ x 2x ˇ ˇ 2x e e ˇ ˇ e ˇ ˇ x x ˇ ˇ x 2x e 5x ˇ ˇ e .x 1/ e .2x 1/ W1 D ˇ 2e 2x ˇD 3 ; ˇ ˇ x x2 x2 ˇ ˇ x 2 2x 2 e .x 2x C 2/ 2e .2x 2x C 1/ ˇ ˇ ˇ ˇ 4e 2x x3 x3 ˇ ˇ x 2x ˇ ˇ e e ˇ ex ˇ ˇ ˇ x x ˇ ˇ x 2x e 4x ˇ ˇ x e .x 1/ e .2x 1/ W2 D ˇ e ; ˇD 2 2 ˇ ˇ x3 x x ˇ x 2 2x C 2/ 2e 2x .2x 2 2x C 1/ ˇˇ ˇ x e .x ˇ ˇ e 3 x xˇ 3 ˇ 2x ˇ ˇ e ˇ e x e 2x ˇ ˇ ˇ x ˇ ˇ 2x e 5x .2 x/ ˇ x e .2x 1/ ˇ 2x W3 D ˇ e ; ˇD 2e ˇ ˇ x3 x2 ˇ ˇ 2x 2 2e .2x 2x C 1/ ˇ ˇ x 4e 2x ˇ e ˇ 3 x ˇ ˇ x e ˇ ˇ x e 2x ˇ ˇ e ˇ ˇ x 4x x ˇ ˇ e .x 1/ ˇ D e .x C 2/ ; W4 D ˇˇ e x 2e 2x ˇ 2 x3 x ˇ ˇ x 2 ˇ ˇ x e .x 2x C 2/ ˇ ˇ e 4e 2x x3 F W2 F W2 F W4 F W1 u01 D D 3; u02 D D 3e x ; u03 D D 3.2 x/; u04 D D 3e x .x C 2/; P0 W P0 W P0 W P0 W u1 D 3x; u2 D 3e x ; u3 D 3x.4 x/=2; u4 D 3e x .x C 3/; yp D u1 y1 C u2 y2 C u3 y3 D 3e x .x 2 C 4x C 6/ 3e x .2x C 3/ . Since is a solution of the complementary equation we take yp D 2x x x 3xe . 2 ˇ ˇ ˇ x x3 ˇ 3 ˇ x ln x ˇˇ ˇ ˇ x x ln x ˇˇ 3 9.4.22. W D ˇˇ 1 3x 2 1 C ln x ˇˇ D 4x 2 ; W1 D ˇˇ 2x 3 ln x; W2 D 2 ˇ D x 3x 1 C ln x ˇ ˇ ˇ ˇ 0 6x ˇ1=x ˇ ˇ x ˇ ˇ x x3 ˇ F W1 1 0 F W2 1 x ln x 3 0 ˇ ˇ D x; W3 D ˇ ˇ ˇ 1 1 C ln x ˇ ˇ 1 3x 2 ˇ D 2x ; u1 D P W D 2 ln x=x x ; u2 D P W D x 3 ; 0 0 F W2 2 1 2 0 u3 D D ; u1 D .ln x/ ln x; u2 D ; u3 D 2 ln x; yp D u1 y1 C u2 y2 C u3 y3 D P0 W x 2x 2 x x x.ln x/2 x ln x . Since x ln x satisfies the complementary equation we take yp D x.ln x/2 2 2
Section 9.4 Variation of Parameters for Higher Order Equations
185
The general solution is y D x.ln x/2 C c1 x C c2 x 3 C c3 x ln x, so 3 2 32 3 2 3 2 x x3 x ln x x.ln x/2 y c1 2 2 1 C ln x 7 4 c 5 .ln x/ 2 ln x 7 C 6 1 3x 4 y0 5 D 6 5 4 5 4 2 2 ln x 2 1 00 y c3 0 6x x x x Setting x D 1 and imposing the initial conditions yields 2 3 2 3 2 32 3 4 0 1 1 0 c1 4 4 5 D 4 0 5 C 4 1 3 1 5 4 c2 5 : 2 2 0 6 1 c3
Solving this system yields c1 D 3, c2 D 1, c3 D 2. Therefore,y D x.ln x/2 C 3x C x 3 2x ln x. ˇ x ˇ 2x ˇ e ˇ ˇ 2x ˇ xe x ˇ x e 2x ˇ ˇ e ˇ xe x x 2x ˇ ˇ ˇ ˇ D e x .1 3x/; 9.4.24. W D ˇ e 2e e .1 x/ ˇ D e .6x 5/; W1 D ˇ 2x x 2e e .1 x/ ˇ ˇ e x 4e 2x e x .x 2/ ˇ ˇ x ˇ ˇ x ˇ ˇ e ˇ ˇ e F W1 xe x e 2x ˇˇ ˇ ˇ ˇ W2 D ˇ x D 1 2x; W3 D ˇ x D e 3x ; u01 D D 1 3x; u02 D e e x .1 x/ ˇ e 2e 2x ˇ P0 W F W2 x.2 3x/ e 2x F W2 D e x .2x 1/; u03 D D e 2x ; u1 D ; u2 D e x .2x C 1/; u3 D ; P0 W P0 W 2 2 x e yp D u1 y1 C u2 y2 C u3 y3 D e x .3x 2 C x C 2/=2. Since is a solution of the complementary 2 x e .3x C 1/x equation we take yp D . 2 x e .3x C 1/x The general solution is y D C c1 e x C c2 e 2x C c3 xe x , so 2 2 3 e x .3x C 1/x ˇ2 2 3 6 3 7 ˇˇ x ˇ c1 2 e 2x xe x y 6 7 ˇ e x 2 ˇ e .3x C 7x C 1/ 7 6 4 y0 5 D 6 7 C ˇˇ e x 2e 2x e x .1 x/ ˇˇ 4 c2 5 : 7 6 2 00 y 4 e x .3x 2 C 13x C 8/ 5 ˇ e x 4e 2x e x .x 2/ ˇ c3 2 Setting x D 0 and imposing the initial conditions yields 2 3 2 3 2 32 3 4 0 1 1 0 c1 3 5 4 1 5 4 c2 5 : D 4 12 5 C 4 1 2 2 1 4 2 c3 19 4 Solving this system yields c1 D
4xe
x
.
ˇ ˇ x x2 ex ˇ 9.4.26. W D ˇˇ 1 2x e x ˇ 0 2 ex ˇ ˇ x x2 e x .x 1/; W3 D ˇˇ 1 2x
3, c2 D
1, c3 D 4. Therefore, y D
e x .3x C 1/x 2
3e x
e 2x C
ˇ ˇ ˇ 2 ˇ ˇ ˇ ˇ ˇ ˇ x ex ˇ e x ˇˇ x 2 ˇ D e x .x 2 2xC2/; W1 D ˇ x ˇ ˇ ˇ ˇ 2x e x ˇ D e .x 2x/; W2 D ˇ 1 e x ˇ D ˇ ˇ ˇ ˇ D x 2 ; u0 D F W1 D x.x 2/; u0 D F W2 D 1 x; u0 D F W2 D 1 2 3 ˇ P0 W P0 W P0 W 4 2 2 x C 6x C 12x C 12 6x C 12x x 2 e x ; yp D u1 y1 C u2 y2 C u3 y3 D . Since is a solution of the 6 6 4 x C 12 complementary equations we take yp D . 6
186 Chapter 9 Linear Higher Order Equations x 4 C 12 C c1x C c2 x 2 C c3 e x , so 6 3 2 4 3 ˇ ˇ x x2 ex .x C 12/=6 ˇ 3 5D 4 5 C ˇ 1 2x e x 2x =3 ˇ ˇ 0 2 ex 2x 2
The general solution is y D 2
y 4 y0 y 00
Setting x D 0 and imposing the initial conditions yields 2 3 2 3 2 0 2 0 0 4 5 5 D 4 0 5C4 1 0 0 0 0 2
ˇ2 3 ˇ c1 ˇ ˇ 4 c2 5 : ˇ ˇ c3
32 3 1 c1 1 5 4 c2 5 : 1 c3
Solving this system yields c1 D 3, c2 D 1, c3 D 2. Therefore, y D
x 4 C 12 C 3x 6
x 2 C 2e x .
ˇ ˇ x ˇ ˇ e x e 3x ˇˇ ˇ e ˇ x C 1 e 3x e 3x ˇˇ 4x ˇ e x 3e 3x ˇˇ D 2e 4x .3x 1/; W1 D ˇˇ x D 2e ; W D 2 3x ˇ ˇ 1 e 3e 3e 3x x 3x ˇ e 9e ˇ ˇ ˇ x C 1 ex ˇ F W2 F W1 x 0 ˇ e 3x .3x C 2/; W3 D ˇˇ D 2e x ; u02 D D 3x 2; x ˇ D xe ; u1 D 1 e P0 W P0 W x.3x C 4/ e 2x .2x C 1/ F W2 D xe 2x ; u1 D 2e x ; u2 D ; u3 D ; yp D u1 y1 C u03 D P0 W 2 4 x 2 x e .6x C 2x 7/ 7e u2 y2 C u3 y3 D . Since is a solution of the complementary equation we take 4 4 x xe .3x C 1/ yp D 2 xe x .3x C 1/ The general solution is y D C c1 .x C 1/ C c2 e x C c3e 2x , so 2 ˇ 2 3 2 3 ˇ ˇ x C 1 e x e 3x ˇ y xe x .3x C 1/=2 ˇ ˇ 4 y 0 5 D 4 e x .3x 2 C 7x C 1/=2 5 C ˇ 1 e x 3e 3x ˇˇ : ˇ ˇ 0 y 00 e x .3x 2 C 13x C 8/=2 e x 9e 3x ˇ ˇ ˇ xC1 ˇ 9.4.28. W D ˇˇ 1 ˇ 0
Setting x D 0 and imposing the initial conditions yields 2 3 3 2 3 2 32 3 0 1 1 1 c1 4 4 5 5 D 4 1 5 C 4 1 1 3 5 4 c2 5 : 4 2 1 0 1 9 c3 4 4 Solving this system yields c1 D e 2x . 2
ˇ ˇ ˇ ˇ ˇ ˇ ˇ 9.4.30. W D ˇˇ ˇ ˇ ˇ ˇ ˇ
x
x
2
1
2x
0
2
0
0
1 x 1 x2 2 x3 6 x2
1 , c2 D 2
1 1 , c3 D . Therefore, y D 4 2
ˇ ˇ x ln x ˇˇ ˇ ln x C 1 ˇˇ ˇ D ˇ 1 ˇ ˇ x ˇ 1 ˇ ˇ x2
ˇ ˇ 2 ˇ x ˇ ˇ 12 ˇ 2x ; W D 1 ˇ x3 ˇ ˇ ˇ 2
xe x .3x C 1/ x C 1 C 2 2
1 x 1 x2 2 x3
ˇ ˇ ˇ ˇ ˇ ln x ln x C 1 ˇˇ D 6 x ˇ 1 ˇ ˇ x
ex C 4
x ln x
3 ; x
ˇ ˇ ˇD ˇ
Section 9.4 Variation of Parameters for Higher Order Equations
187
ˇ ˇ ˇ 1 ˇ 1 ˇ ˇ 2 ˇ ˇ ˇ ˇ x ln x ˇ 2 x x ˇ x x x ln x ˇˇ ˇ ˇ x x ˇˇ ˇ ˇ ˇ 4 1 1 ˇ ˇ ˇ ; W3 D ˇ 1 2x ln x C 1 ˇ D x; W4 D ˇˇ 1 2x D ln x C 1 ˇˇ D 2 2 ˇ 1 ˇ ˇ x x2 x ˇ ˇ ˇ ˇ ˇ 0 2 2 1 2 ˇ ˇ ˇ x ˇ ˇ 0 2 ˇ x3 x x3 F W1 9 0 F W2 3 F W2 3x 2 0 F W4 9 6 0 ; u1 D D 9 ln x=2 ; u2 D D ; u03 D D ; u4 D D ; x P0 W 4 P0 W x P0 W 4 P0 W 2 9x ln x 27x x3 9x u1 D ; u2 D 3 ln x; u3 D ; u4 D ; yp D u1 y1 Cu2 y2 Cu3 y3 D 3x 2 ln x 7x 2 . 2 4 4 2 Since 7x 2 satisfies the complementary equation we take yp D 3x 2 ln x. c3 C c4 x ln x, so The general solution is y D 3x 2 ln x C c1x C c2 x 2 C x ˇ ˇ ˇ ˇ 1 2 ˇ x ln x ˇˇ 2 3 ˇ x x 2 3 2 3x 2 ln x 3 x ˇ ˇ c1 y 1 ˇ ˇ 6 7 ln x C 1 ˇ 6 c 7 6 y 0 7 6 6x ln x C 3x 7 ˇ 1 2x x2 6 00 7 D 6 6 ln x C 9 7 C ˇ ˇ6 2 7: ˇ ˇ 4 c3 5 4 y 5 4 2 1 5 ˇ 0 2 ˇ 6 000 ˇ 3 ˇ c4 y x x ˇ ˇ 1 6 x ˇ 0 0 ˇ ˇ ˇ 2 2 x x ˇ ˇ ˇ x ˇ ˇ W2 D ˇˇ 1 ˇ ˇ ˇ 0
Setting x D 1 and imposing the initial conditions yields 2 3 2 3 ˇ ˇ 1 1 7 0 ˇ 6 11 7 6 3 7 ˇ 1 2 6 7D 6 7Cˇ 4 5 5 4 9 5 ˇ 0 2 ˇ ˇ 0 0 6 6
1 1 2 6
0 1 1 1
ˇ2 ˇ ˇ ˇ6 ˇ6 ˇ4 ˇ ˇ
3 c1 c2 7 7: c3 5 c4
Solving this system yields c1 D 0, c2 D 7, c3 D 0, c4 D 0. Therefore, y D 3x 2 ln x 7x 2. ˇ p p ˇ ˇ x ˇ ˇ p p x 1=x 1= x ˇ ˇ ˇ p x 1=x 1= x 2 3=2 ˇ ˇ 1 ˇ p 1=2 x 1=x 1=2x ˇ D 9=8x 6; W1 D ˇ 1=2 x 1=x 2 1=2x 3=2 9.4.32. W D ˇˇ 3=2 3 5=2 ˇ ˇ 0 1=4x 2=x 3=4x 3=2 3 ˇ ˇ ˇ 1=4x 2=x 3=4x 5=2 ˇ 0 3=8x 5=2 6=x 4 15=8x 7=2 ˇ ˇ ˇ ˇ ˇ p p p ˇ x ˇ x 1=x 1= x ˇˇ x 1= x ˇˇ ˇ ˇ p 1=2x 3=2 ˇˇ D 3=4x 2; 3=4x 4; W2 D ˇˇ 1 1=x 2 1=2x 3=2 ˇˇ D 3=2x 7=2; W3 D ˇˇ 1 1=2 x ˇ 0 2=x 3 ˇ 0 3=4xˇ 5=2 ˇ 1=4x 3=2 3=4x 5=2 ˇ ˇ p ˇ x ˇ x 1=x ˇ ˇ p p F W2 F W1 1=2 x 1=x 2 ˇˇ D 3=2x 5=2; u01 D W4 D ˇˇ 1 D 1=x; u02 D D 2= x; P0 W P0 W ˇ 0 1=4x 3=2 2=x 3 ˇ p p F W2 F W4 u03 D D x; u04 D D 2 x; u1 D ln x; u2 D 4 x; u3 D x 2 =2; u4 D 4x 3=2=3; P0 W P0 W yp D u1 y1 C u2 y2 C u3 y3 D x ln x 19x=6. since 19x=6 satisfies the complementary equation we take yp D x ln x. p p The general solution is y D x ln x C c1x C c2 x C c3 =x C c4= x, so ˇ2 p p 2 3 2 3 ˇ 3 ˇ x ˇ c1 x 1=x 1= x y x ln x ˇ ˇ p 6 y 0 7 6 ln x C 1 7 ˇ 1 7 1=2 x 1=x 2 1=2x 3=2 ˇˇ 6 6 00 7 D 6 7Cˇ 6 c2 7 3=2 3 5=2 ˇ 4 c3 5 : 4 y 5 4 1=x 5 ˇ 0 1=4x 2=x 3=4x ˇ ˇ ˇ 0 3=8x 5=2 y 000 1=x 2 6=x 4 15=8x 7=2 ˇ c4
ˇ ˇ ˇ ˇD ˇ ˇ
188 Chapter 9 Linear Higher Order Equations Setting x D 1 and imposing the initial conditions yields 2 3 2 3 ˇ ˇ 1 2 0 1 ˇ 6 0 7 6 1 7 ˇ 1 1=2 6 7 6 7 ˇ 4 4 5 D4 1 5Cˇ 0 1=4 ˇ 37 ˇ 0 3=8 1 4
Solving this system yields c1 D 1, c2 D
1 1 2 6
1 1=2 3=4 15=8
ˇ2 ˇ ˇ ˇ6 ˇ6 ˇ4 ˇ ˇ
3 c1 c2 7 7: c3 5 c4
1, c3 D 1, c4 D 1. Therefore, y D x ln x Cx
p 1 1 xC C p . x x
F Wj (1 j n), the argument used in the derivation of the method of P0 W variation of parameters implies that yp is a solution of (A). (b) Follows immediately from (a), since uj .x0 / D 0, j D 1; 2; : : : ; n. (c) Expand the determinant in cofactors of its nth row. (d) Just differentiate the determinant n 1ˇ times. @j G.x; t/ ˇˇ has two identical rows, and is therefore zero, while (e) If 0 j n 2, then @x j ˇxDt ˇ @n 1 G.x; t/ ˇˇ D W .t/ ˇ @x j xDt Z x Z x @G.x; t/ (f) Since yp .x/ D G.x; t/F .t/ dt, yp0 .x/ D G.x; x/F .x/C F .t/ dt. But G.x; x/ D @x x0 Z x0 x @G.x; t/ 0 from (e), so yp0 .x/ D F .t/ dt. Repeating this argument for j D 1; : : : ; n and invoking @x x0 (e) each time yields the conclusion. 9.4.34. (a) Since u0j D . 1/n
j
9.4.36. ˇ ˇ y1 .t/ ˇ 0 ˇ y .t/ ˇ 1 ˇ y1 .x/
y1 .t/ y10 .t/ y1 .x/
y2 .t/ y20 .t/ y2 .x/
ˇ ˇ ˇ ˇ t t2 1=t ˇ ˇ ˇ D ˇ 1 2t 1=t 2 ˇ ˇ 2 ˇ ˇ x x 1=x ˇ 2 ˇ ˇ t 1=t ˇˇ ˇ D xˇ 2t 1=t 2 ˇ D
ˇ ˇ t ˇ 3 Since P0 .t/ D t and W .t/ D ˇˇ 1 ˇ x Z x 2 .x t/ .2x C t/ F .t/ dt . 6xt 3 x0 9.4.38. ˇ ˇ y1 .t/ ˇ 0 ˇ y .t/ ˇ 1 ˇ y1 .x/
y1 .t/ y10 .t/ y1 .x/
y2 .t/ y20 .t/ y2 .x/
ˇ ˇ ˇ ˇ D ˇ ˇ
D
D
ˇ ˇ t x ˇˇ 1 2
ˇ ˇ t 1=t e t =t ˇ 2 t ˇ 1 1=t e .1=t 1=t 2 / ˇ ˇ x 1=x e x =x ˇ ˇ ˇ 1=t ˇ e t =t ˇ x ˇˇ 2 t 2 ˇ 1=t e .1=t 1=t / xe t t2
ˇ 1=t ˇˇ 1 C 1=t 2 ˇ x t/2 .2x C t/ : xt
x2 t2 .x C D t x ˇ ˇ 1=t ˇ 6 .x 1=t 2 ˇˇ D , G.x; t/ D t 1=x ˇ
3x C 2
t2 2t x2
ˇ ˇ ˇ ˇ ˇ ˇ
e t .t 2/ xt
ˇ ˇ ˇ ˇ ˇ ˇ
ˇ ˇ t t2 ˇ ˇ 1 2t
ˇ ˇ ˇ ˇ
t/2 .2x C t/ , so yp D 6xt 3
ˇ 1 ˇˇ t e t =t t ˇ x 1 e .1=t 1=t 2 / 2e x x 2 e t e t t.t 2/ 2te x D xt xt 2
ˇ ˇ ˇ ex ˇ t ˇC ˇ ˇ x ˇ 1
ˇ 1=t ˇˇ 1=t 2 ˇ
Section 9.4 Variation of Parameters for Higher Order Equations ˇ ˇ t 1=t e t =t ˇ 2 t ˇ Since P0 .t/ D t.1 t/ and W .t/ D ˇ 1 1=t e .1=t 1=t 2 / ˇ 0 2=t 3 e t .1=t 2=t 2 C 2=t 3/ Z x 2 x t.t 2/ 2te .x t / x 2 t.t 2/ 2te .x t / , so y D F .t/ dt . p 2x.t 1/2 2x.t 1/2 x0
189
ˇ ˇ t ˇ ˇ D 2e .1 t/ , G.x; t/ D ˇ t3 ˇ
9.4.40. ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ
y1 .t/ y10 .t/ y100 .t/ y1 .x/
y1 .t/ y10 .t/ y100 .t/ y1 .x/
y2 .t/ y20 .t/ y200 .t/ y2 .x/
y3 .t/ y30 .t/ y300 .t/ y3 .x/
ˇ ˇ ˇ ˇ 1 ˇ ˇ ˇ ˇ ˇ D ˇ 0 ˇ ˇ 0 ˇ ˇ ˇ ˇ 1 ˇ ˇ ˇ ˇ D ˇ ˇ
D ˇ ˇ ˇ ˇ Since P0 .t/ D t and W .t/ D ˇˇ ˇ ˇ 9.4.42. ˇ ˇ y1 .t/ ˇ 0 ˇ y .t/ ˇ 100 ˇ y .t/ ˇ 1 ˇ y1 .x/
y1 .t/ y10 .t/ y100 .t/ y1 .x/
y2 .t/ y20 .t/ y200 .t/ y2 .x/
1 0 0 0
y3 .t/ y30 .t/ y300 .t/ y3 .x/
t t2 1 2t 0 2 0 0
x0
1 0 0 0 t/ .1 C 2t/ C e
6 C t ˇ ˇ ˇ ˇ ˇD ˇ ˇ ˇ
6x t2
t2 2t 2 0
.x t/3 12 , G.x; t/ D , so yp D 4 t 6x
8/ C 16x 2 e 2.x ˇ e 2t ˇˇ 2e 2t ˇˇ D 128t, 4e 2t ˇˇ 2t ˇ 8e 2t/ 4x 2 C 4t 2 2 , so 4x 2 C 4t 2
ˇ ˇ ˇ ˇ ˇ ˇ
2x 2 2 2.t x/3 C D : t3 x xt 3
.16t 2
e 2t 2e 2t 4e 2t 8e 2t 2.x t / .1 32t 2 t/ .1 C 2t/ C e 2.x t /.1 2t/ 32t 2
ˇ ˇ ˇ ˇ Since P0 .t/ D t and W .t/ D ˇˇ ˇ ˇ
ˇ 1=t ˇˇ 1=t 2 ˇˇ 2=t 3 ˇˇ 1=x ˇ ˇ ˇ ˇ 1 t2 t2 1=t ˇˇ 1=t ˇ 2t 1=t 2 ˇˇ C x ˇˇ 0 2t 1=t 2 ˇ 0 2 2 2=t 3 ˇ 2=t 3 ˇ ˇ ˇ ˇ 1 t t2 ˇ 1 t 1=t ˇˇ ˇ ˇ 1 0 1 1=t 2 ˇˇ C ˇˇ 0 1 2t ˇˇ 0 0 2=t 3 ˇ x ˇ 0 0 2 ˇ
ˇ ˇ ˇ ˇ ˇ 1 t 2 e 2t e 2t ˇˇ ˇ ˇ 2t ˇ ˇ 2e 2t ˇˇ ˇ D ˇ 0 2t 2e2t ˇ ˇ 0 2 4e 4e 2t ˇˇ ˇ ˇ ˇ ˇ 1 x 2 e 2x e 2x ˇ ˇ 2 ˇ ˇ ˇ t ˇ 1 e 2t e 2t ˇˇ ˇ ˇ 2ˇ 2t 2t ˇ ˇ 2e Cx ˇ 0 D ˇ 2t 2e 2t ˇ ˇ 2 4e ˇ 0 4e 2t ˇ ˇ ˇ ˇ ˇ 1 t2 ˇ e 2t ˇˇ ˇ ˇ 2x ˇ 2x ˇ 2t ˇ e ˇ 0 2t 2e Ce ˇ ˇ ˇ 0 2 ˇ 4e 2t ˇ D
e 2.x G.x; t/ D Z x 2.x e yp D
1=t 1=t 2 2=t 3 6=t 4
t 1 0 ˇ ˇ ˇ x 2 ˇˇ ˇ
t2 2t 2 x2
t 1 0 x
2
t /t
e 2t 2e 2t 4e 2t 1 0 0
t2 2t 2
.8t C 4/ C e
F .t/ dt.
Z
x
.x
x0
e 2t 2e 2t 4e 2t ˇ e 2t ˇˇ 2e 2t ˇˇ 4e 2t ˇ 2.x t /
t/3 F .t/ dt. 6x
ˇ ˇ ˇ ˇ ˇ ˇ
.8t
4/:
CHAPTER 10 Linear Systems of Differential Equations
10.1 INTRODUCTION TO SYSTEMS OF DIFFERENTIAL EQUATIONS 10.1.2. Q10 D .rate in/1 .rate out/1 and Q20 D .rate in/2 .rate out/2 . The volumes of the solutions in T1 and T2 are V1 .t/ D 100 C 2t and V2 .t/ D 100 C 3t, respectively. T1 receives salt from the external source at the rate of (2 lb/gal) (6 gal/min) D 12 lb/min, and from 1 T2 at the rate of (lb/gal in T2 / (1 gal/min) D Q2 lb/min. Therefore, (A) (rate in)1 D 12 C 100 C 3t 1 Q2 . Solution leaves T1 at 5 gal/min, since 3 gal/min are drained and 2 gal/min are pumped to 100 C 3t 5 1 Q1 5 D Q1 . Now (A) T2 ; hence (B) .rate out/1 D . lb/gal in T1 / (5 gal/min) D 100 C 2t 100 C 2t 5 1 and (B) imply that (C) Q10 D 12 Q1 C Q2 . 100 C 2t 100 C 3t T2 receives salt from the external source at the rate of (1 lb/gal) (5 gal/min) D 5 lb/min, and from 1 1 T1 at the rate of (lb/gal in T1 / (2 gal/min) D Q1 2 D Q1 lb/min. Therefore, (D) 100 C 2t 50 C t 1 (rate in)2 D 5 C Q1 . Solution leaves T2 at 4 gal/min, since 3 gal/min are drained and 1 gal/min is 50 C t 1 4 pumped to T1 ; hence (E) .rate out/2 D . lb/gal in T2 / (4 gal/min) D Q2 4 D Q2 . 100 C 3t 100 C 3t 1 4 Now (D) and (E) imply that (F) Q20 D 5 C Q1 Q2 . Now (C) and (F) form the desired 50 C t 100 C 3t system. 10.1.4. mX00 D 10.1.8.
˛X0
mgR2
X ; see Example 10.1.3. kXk3
I1i J1i I2i
J2i y1;i C1 y2;i C1
D g1 .ti ; y1i ; y2i /; D g2 .ti ; y1i ; y2i /; D g1 .ti C h; y1i C hI1i ; y2i C hJ1i / ; D g2 .ti C h; y1i C hI1i ; y2i C hJ1i / ; h D y1i C .I1i C I2i /; 2 h D y2i C .J1i C J2i /: 2 191
192 Chapter 10 Linear Systems of Differential Equations 10.2 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS 10.2.6. Let yi D y .i 1/ , i D 1; 2; : : : ; n; then yi0 D yi C1 , i C Pn .t/y1 D F .t/, so 2 0 1 0 0 6 0 0 1 0 1 6 6 :: : : : :: :: : : ::: AD 6 P0 6 : 4 0 0 0 1 Pn Pn 1 Pn 2 P1
D 1; 2; : : : ; n 3 7 7 7 7 7 5
and
1 and P0 .t/yn0 C P1 .t/yn C
1 fD P0
2
0 0 :: :
6 6 6 6 6 4 0 F
3
7 7 7 7: 7 5
If P0 ; P1 ; : : : ; Pn and F are continuous and P0 has no zeros on .a; b/, then P1 =P0 ; : : : ; Pn =P0 and F=P0 are contnuous on .a; b/. 10.2.7. (a) .c1 P Cc2 Q/0ij D .c1 pij Cc2 qij /0 D c1 pij0 Cc2 qij0 D .c1 P 0 Cc2Q0 /ij ; hence .c1 P Cc2Q/0 D c1 P 0 C c2 Q 0 . r X (b) Let P be k r and Q be r s; then PQ is k s and .PQ/ij D pi ł qłj . Therefore, .PQ/0ij D
r X łD1
pi0 ł qłj C
r X łD1
łD1
pi ł qłj0 D .P 0 Q/ij C .PQ0 /ij . Therefore, .PQ/0 D P 0 Q C PQ0 .
10.2.10. (a) From Exercise 10.2.7(b) with P D Q D X, .X 2 /0 D .XX/0 D X 0 X C XX 0 . (b) By starting from Exercise 10.2.7(b) and using induction it can be shown if P1 ; P2 ; : : : ; Pn are square matrices of the same order, then .P1 P2 Pn /0 D P10 P2 Pn C P1 P20 Pn C C P1P2 Pn0 . Taking P1 D P2 D D Pn D X yields (A) .Y n /0 D Y 0 Y n 1 C Y Y 0 Y n 2 C Y 2 Y 0 Y n 3 C C n X1 Y n 1Y 0 D Y r Y 0Y n r 1. r D0
(c) If Y is a scalar function, then (A) reduces to the familiar result .Y n /0 D nY n
1
Y 0.
10.2.12. From Exercise 10.2.6, the initial value problem (A) P0 .x/y .n/ C P1 .x/y .n 1/ C C Pn .x/y D F .x/, y.x0 / D k0 ; y 0 .x0 / D k1 ; : : : ; y .n 1/ .x0 / D kn 1 is equivalent to the initial value problem (B) y0 D A.t/y C f.t/, with 2 3 2 3 2 3 0 1 0 0 0 k0 6 0 7 6 0 7 0 1 0 7 6 7 6 7 1 6 6 :: :: :: :: 7 ; f D 1 6 :: 7 ; and k D 6 k1 7 : :: AD 6 : 7 6 7 6 : 7 : : 7 : : P0 6 P0 6 : 7 4 :: 5 4 0 5 4 5 0 0 1 0 kn 1 Pn Pn 1 Pn 2 P1 F
Since Theorem 10.2.1 implies that (B) has a unique solution on .a; b/, it follows that (A) does also.
10.3 BASIC THEORY OF HOMOGENEOUS LINEAR SYSTEM 1 0 1 y 0 10.3.2. (a) The system equivalent of (A) is (B) y D y, where y D . y0 P0 .x/ P2 .x/ P1 .x/ y1 y2 Let y1 D and y1 D . Then the Wronskian of fy1; y2 g as defined in this section is y10 y20 y1 y2 D W. y10 y20
Section 10.3 Basic Theory of Homogeneous Linear System
193
(b) The trace of the matrix in (B) is P1 .x/=P0 .x/, so Eqn. 10.3.6 implies that W .x/ D W .x0 / exp 10.3.4.ˇ (a) See the ˇsolution of Exercise 9.1.18. ˇ 0 ˇ y0 ˇ ˇ a11 y11 C a12 y21 a11 y12 C a12 y22 y 11 12 ˇDˇ (c) ˇˇ y21 y22 ˇ ˇ y21 y 22 y11 y12 a11 W C a12 0 D a11 W . Similarly, D a22 W . 0 0 y21 y22
ˇ ˇ ˇ ˇ ˇ D a11 ˇ y11 ˇ ˇ y21
ˇ ˇ ˇ y21 y12 ˇˇ ˇ Ca 12 ˇ ˇ y22 y21
Z
x x0
ˇ y22 ˇˇ D y22 ˇ
10.3.6. (a) From the equivalence of Theorem 10.3.3(b)and (e), Y .t0 / is invertible. (b) From the equivalence of Theorem 10.3.3(a)and (b), the solution of the initial value problem is y D Y .t/c, where c is a constant vector. To satisfy y.t0 / D k, we must have Y .t0 /c D k, so c D Y 1 .t0 /k and y D Y 1 .t0 /Y .t/k. e 4t 2e 3t c 2c2 D 10 10.3.8. (b) y D C c where 1 ; so c1 D 6, c2 D 2, and 2 e 4t 5e 3t c1 C 5c2 D 4 6e 4t C 4e 3t yD . 6e 4t 10e 3t 1 e 4t 2e 3t 1 2 5 2 1 (c) Y .t/ D ; Y .0/ D ; Y .0/ D ; y D Y .t/Y 1 .0/k D e 4t 5e 3t 1 5 1 1 7 1 5e 4t C 2e 3t 2e 4t 2e 3t k. 7 5e 4t 5e 3t 2e 4t C 5e 3t 3t e et c C c2 D 2 10.3.10. (b) y1 D c1 C c , where 1 ; so c1 D 5, c2 D 3, and 2 e 3t et c1 c2 D 8 5e 3t 3e t yD . 5e 3t C 3e t 3t e et 1 1 (c) Y .t/ D ; Y .0/ D ; 3e 3t et 1 1 1 1 1 e 3t C e t e 3t e t 1 1 1 Y .0/ D ; y D Y .t/Y .0/k D k. 1 2 1 2 e 3t e t e 3t C e t 2 3 2 3 2 4t 3 e 2t e 2t e c1 c2 C c3 D 0 5 C c2 4 e 2t 5 C c3 4 e 4t 5, where 10.3.12. (b) y D c1 4 0 c2 C c3 D 9 ; so e 2t 0 e 4t c1 C c3 D 12 2 3 e 2t C e 4t 1 c1 D 11, c2 D 10, c3 D 1, and y D 4 10e 2t C e 4t 5. 3 11e 2t C e 4t 2 3 2 3 2 3 2t 2t 4t e e e 1 1 1 1 1 2 1 (c) Y .t/ D 4 0 e 2t e 4t 5; Y .0/ D 4 0 1 1 5; Y 1 .0/ D 4 1 2 1 5; 3 2t 4t 1 1 1 e 0 e 1 0 1 2 3 2e 2t C e 4t e 2t C e 4t e 2t C e 4t 1 e 2t C e 4t 5 k. y D Y .t/Y 1 .0/k D 4 e 2t C e 4t 2e 2t C e 4t 3 2t 4t 2t 4t e Ce e Ce 2e 2t C e 4t
10.3.14. If Y and Z are both fundamental matrices for y0 D A.t/y, then Z D C Y , where C is a constant invertible matrix. Therefore, Z Y 1 D C and Y Z 1 D C 1 .
10.3.16. (a) The Wronskian of fy1 ; y2; : : : ; yn g equals one when t D t0 . Apply Theorem 10.3.3. (b) Let Y be the matrix with columns fy1; y2 ; : : : ; yng. From (a), Y is a fundamental matrix for y0 D A.t/y on .a; b/. From Exercise 10.3.15(b), so is Z D Y C if C is any invertible constant matrix.
P1 .s/ ds . P0 .s/
194 Chapter 10 Linear Systems of Differential Equations 10.3.18. (a) 10 .t/ D Z 0 .t/Z.s/ D AZ.t/Z.s/ D A.t/ and 1 .0/ D Z.s/, since Z.0/ D I . 20 .t/ D Z 0 .t C s/ D AZ.t C s/ D A2 .t/ (since A is constant) and 2 .0/ D Z.s/. Applying Theorem 10.2.1 to the columns of 1 and 2 shows that 1 D 2 . (b) With s D t, (a) implies that Z.t/Z. t/ D Z.0/ D I ; therefore .Z.t// 1 D Z. t/. (c) e 0A D I is analogous to e oa D e 0 D 1 when a is a scalar, while e .t Cs/A D e tAe sA is analagous to e .t Cs/a D e t a e sa when a is a scalar. 10.4
CONSTANT COEFFICIENT HOMOGENEOUS SYSTEMS I
ˇ ˇ ˇ 1 ˇ 5 4 3 ˇ D . C 1=2/. C 2/. Eigenvectors associated with 1 D 1=2 satisfy 10.4.2. ˇˇ ˇ 4 3 5 4 3 3 x1 0 1 D , so x1 D x2 . Taking x2 D 1 yields y1 D e t =2. Eigenvectors 4 4 x2 0 1 3 3 x1 0 D , so x1 D x2 . Taking x2 D 1 yields associated with 2 D 2 satisfy 4 4 x2 0 1 1 1 1 1 y2 D e 2t . Hence y D c1 e t =2 C c2 e 2t . 1 1 1 ˇ ˇ ˇ 1 4 ˇˇ ˇ 10.4.4. ˇ D . 1/. C 3/. Eigenvectors associated with 1 D 3 satisfy 1 1 ˇ 2 4 x1 0 2 D , so x1 D 2x2. Taking x2 D 1 yields y1 D e 3t . Eigenvectors 1 2 x2 0 1 2 4 x1 0 associated with 2 D 1 satisfy D , so x1 D 2x2 . Taking x2 D 1 yields 1 2 x2 0 2 2 2 y2 D e t . Hence y D c1 e 3t C c2 et . 1 1 1 ˇ ˇ ˇ 4 3 ˇˇ 2 3 x1 0 ˇ 10.4.6. ˇ D . 2/. 1/. Eigenvectors associated with 1 D 2 satisfy D , 2 1 ˇ 2 3 x2 0 3 3 so x1 D x2 . Taking x2 D 2 yields y1 D e 2t . Eigenvectors associated with 2 D 1 sat2 2 3 3 x1 0 1 isfy D , so x1 D x2. Taking x2 D 1 yields y2 D e t . Hence y D 2 2 x2 0 1 3 1 c1 e 2t C c2 et . 2 1 ˇ ˇ ˇ 1 1 2 ˇˇ ˇ 10.4.8. ˇˇ 1 2 3 ˇˇ D . C 3/. C 1/. 2/. The eigenvectors associated with ˇ 4 1 1 ˇ 2 3 :: : 0 4 1 2 6 7 6 7 :: with 1 D 3 satisfy the system with augmented matrix 6 1 , which is row 1 3 : 0 7 4 5 :: 4 1 2 : 0 2 3 :: 1 : 0 7 6 1 0 6 7 : equivalent to 6 0 1 . Hence x1 D x3 and x2 D 2x3. Taking x3 D 1 yields y1 D 2 :: 0 7 4 5 :: 0 0 0 : 0
Section 10.4 Constant Coefficient Homogeneous Systems I 2
3 1 4 2 5e 1 2
3t
1
: 2 :: : 3 :: : 0 ::
3
1 satisfy the system with augmented ma2
3 :: 1 : 0 7 7 : . Hence x1 D 4 0:: 0 7 5 :: 0 : 0
0 7 6 1 0 7 6 , which is row equivalent to 6 0 1 x3 1 1 0 7 5 4 4 1 0 0 0 2 3 1 and x2 D 4x3. Taking x3 D 1 yields y2 D 4 4 5e t . The eigenvectors associated with with 1 2 3 :: 1 1 2 : 0 6 7 6 7 :: 3 D 2 satisfy the system with augmented matrix 6 1 , which is row equivalent 4 3 : 0 7 4 5 :: 4 1 3 : 0 2 3 :: 2 3 1 6 1 0 1 : 0 7 6 7 to 6 0 1 1 ::: 0 7. Hence x1 D x3 and x2 D x3 . Taking x3 D 1 yields y3 D 4 1 5e 2t . 4 5 1 : 0 0 0 :: 0 2 3 2 3 2 3 1 1 1 Hence y D c1 4 2 5 e 3t C c2 4 4 5 e t C c3 4 1 5 e 2t . 1 1 1 ˇ ˇ ˇ 3 ˇ 5 8 ˇ ˇ ˇ 1 2 ˇˇ D . 1/. C 2/. 2/. The eigenvectors associated with 10.4.10. ˇ 1 ˇ 1 1 1 ˇ 2 3 :: 2 5 8 : 0 6 7 6 7 :: with 1 D 1 satisfy the system with augmented matrix 6 1 , which is row equiv2 2 : 0 7 5 4 :: 1 1 2 : 0 2 3 : 1 0 23 :: 0 6 7 6 7 alent to 6 0 1 4 ::: 0 7. Hence x1 D 32 x3 and x2 D 43 x3 . Taking x3 D 3 yields y1 D 4 5 3 :: 0 0 0 : 0 2 3 2 4 4 5e t . The eigenvectors associated with with 2 D 2 satisfy the system with augmented ma3 2 3 2 3 :: :: 5 5 8 : 0 1 1 0 : 0 6 7 6 7 6 7 6 7 :: :: trix 6 1 , which is row equivalent to 7 6 7. Hence x1 D x2 1 2 : 0 5 4 4 0 0 1 : 0 5 :: :: 1 1 1 : 0 0 0 0 : 0 2 3 1 and x3 D 0. Taking x2 D 1 yields y2 D 4 1 5e 2t . The eigenvectors associated with with 0 6 6 trix 6 4
2
. The eigenvectors associated with with 2 D
195
196 Chapter 10 Linear Systems of Differential Equations 2
6 6 3 D 2 satisfy the system with augmented matrix 6 4
1
5
1
3
: 8 :: : 2 :: : 3 ::
3 0 7 7 , which is row equivalent to 0 7 5
1 1 0 3 : 2 3 1 0 74 :: 0 7 6 7 6 7 6 0 1 5 ::: 0 7. Hence x1 D 47 x3 and x2 D 54 x3 . Taking x3 D 4 yields y3 D 4 5 5e 2t . 4 5 4 4 :: 0 0 0 : 0 2 3 2 3 2 3 2 1 7 Hence y D c1 4 4 5 e t C c2 4 1 5 e 2t C c3 4 5 5 e 2t . 3 0 4 ˇ ˇ ˇ 4 1 4 ˇˇ ˇ 3 2 ˇˇ D . 3/. C 2/. C 1/. The eigenvectors associated with 10.4.12. ˇˇ 4 ˇ 1 1 1 ˇ 3 2 :: 1 1 4 : 0 7 6 7 6 :: 1 D 3 satisfy the system with augmented matrix 6 4 , which is row equiva6 2 : 0 7 5 4 :: 1 1 4 : 0 2 3 :: 11 : 0 7 6 1 0 6 7 : lent to 6 0 1 . Hence x1 D 11x3 and x2 D 7x3. Taking x3 D 1 yields y1 D 7 :: 0 7 4 5 : 0 0 0 :: 0 2 3 11 4 7 5e 3t . The eigenvectors associated with 2 D 2 satisfy the system with augmented matrix 1 2 3 2 3 :: :: 1 4 : 0 7 1 : 0 7 6 6 6 1 0 6 7 6 7 :: : , which is row equivalent to . Hence x1 D x3 and x2 D 6 4 6 0 1 1 2 : 0 7 2 :: 0 7 4 5 4 5 : : 1 1 1 :: 0 0 0 0 :: 0 3 2 1 2x3 . Taking x3 D 1 yields y2 D 4 2 5e 2t . The eigenvectors associated with 3 D 1 satisfy the sys1 3 3 2 2 :: :: 1 0 1 : 0 : 0 5 1 4 6 7 7 6 7 6 6 7 :: :: . , which is row equivalent to tem with augmented matrix 6 4 7 6 1 : 0 7 2 2 : 0 5 5 4 0 1 4 :: :: 0 0 0 : 0 1 1 0 : 0 2 3 2 3 2 3 1 11 1 Hence x1 D x3 and x2 D x3 . Taking x3 D 1 yields y3 D 4 1 5e t . Hence y D c1 4 7 5 e 3t C c2 4 2 5 e 1 1 1 ˇ ˇ ˇ 3 2 2 ˇˇ ˇ 10.4.14. ˇˇ 2 7 2 ˇˇ D . C 5/. 5/2 . The eigenvectors associated with 1 D ˇ 10 10 5 ˇ 2
2t
2
3 1 C c3 4 1 5 e t . 1
Section 10.4 Constant Coefficient Homogeneous Systems I 2
6 6 5 satisfy the system with augmented matrix 6 4
2
1 6 6 6 0 4
0
:: 1 : 5 :: 1 : 5 : 0 ::
0
3
8
2
2
2 12
2
10 10
0
197
3 :: : 0 7 7 :: , which is row equivalent to : 0 7 5 :: : 0
2 3 1 7 1 1 7 . Hence x1 D x3 and x2 D x3 . Taking x3 D 5 yields y1 D 4 1 5e 5t . The 1 0 7 5 5 5 5 0 0 0 2 3 :: 2 2 2 : 0 7 6 6 7 : eigenvectors associated with 2 D 5 satisfy the system with augmented matrix 6 , 2 2 2 :: 0 7 4 5 : 10 10 10 :: 0 2 3 : 1 1 :: 0 7 6 1 6 7 : which is row equivalent to 6 0 . Hence x1 D x2 x3 . Taking x2 D 0 and x3 D 0 0 :: 0 7 5 4 : 0 0 0 :: 0 2 3 2 3 1 1 1 yields y2 D 4 0 5e 5t . Taking x2 D 1 and x3 D 0 yields y3 D 4 1 5e 5t . Hence y D 1 0 3 2 3 2 2 3 1 1 1 c1 4 1 5 e 5t C c2 4 0 5 e 5t C c3 4 1 5 e 5t . 0 5 1 ˇ ˇ ˇ 7 4 ˇˇ 12 4 x1 0 10.4.16. ˇˇ D . 5/.C5/. Eigenvectors associated with D 5 satisfy D , 1 6 7 ˇ 6 2 x2 0 x2 1 so x1 D . Taking x2 D 3 yields y1 D e 5t . Eigenvectors associated with 2 D 5 satisfy 3 3 2 4 x1 0 2 D , so x1 D 2x2. Taking x2 D 1 yields y2 D e 5t . The general 6 12 x2 0 1 1 2 2 1 2 2 5t 5t solution is y D c1 e C c2 e . Now y.0/ D ) c1 C c2 D , 3 1 4 3 1 4 2 4 so c1 D 2 and c2 D 2. Therefore, y D e 5t C e 5t . 6 2 ˇ ˇ ˇ 21 12 ˇˇ ˇ 10.4.18. ˇ D . 9/. C 3/. Eigenvectors associated with 1 D 9 satisfy 24 15 ˇ 12 12 x1 0 1 D , so x1 D x2 . Taking x2 D 1 yields y1 D e 9t . Eigenvec24 24 x2 0 1 1 24 12 x1 0 tors associated with 2 D 3 D , so x1 D x2 . Taking x2 D 2 yields 24 12 x2 0 2 1 1 1 5 3t 9t y2 D e . The general solution is y D c1 e C c2 e 3t . Now y.0/ D ) 2 1 2 3 1 1 5 7 2 c1 C c2 D , so c1 D 7 and c2 D 2. Therefore, y D e 9t e 3t . 1 2 3 7 4
198 Chapter 10 Linear Systems of Differential Equations ˇ ˇ ˇ ˇ 10.4.20. ˇ ˇ ˇ 1 D
1 6
1 3
2 3
0
1 6
0
ˇ ˇ ˇ ˇ 0 ˇD ˇ ˇ 0
1 2
1=2/2 . The eigenvectors associated with with
. C 1=2/. 2
6 6 1=2 satisfy the system with augmented matrix 6 4
2 3
1 3
2 3
1 3
: 0 :: : 0 :: : 1 ::
0
3
7 7 , which is row equivalent to 0 7 5
0 0 0 3 : 2 3 1 21 0 :: 0 1 6 7 x 6 7 2 and x3 D 0. Taking x2 D 2 yields y1 D 4 2 5e t =2. 6 0 0 1 ::: 0 7. Hence x1 D 4 5 2 0 : 0 0 0 :: 0 The eigenvectors associated with with 2 D 3 D 1=2 satisfy the system with augmented matrix 3 2 3 2 :: :: 1 1 0 : 0 1 1 0 : 0 3 7 6 7 6 3 7 6 7 6 :: :: , which is row equivalent to . Hence x1 D x2 and 7 6 6 2 2 0 0 : 0 7 0 : 0 5 5 4 0 4 3 3 : : 0 0 0 :: 0 0 0 0 :: 0 2 3 1 x3 is arbitrary. Taking x2 D 1 and x3 D 0 yields y2 D 4 1 5e t =2. Taking x2 D 0 and x3 D 1 yields 0 2 3 2 3 2 3 2 3 0 1 1 0 y3 D 4 0 5e t =2. The general solution is y D c1 4 2 5 e t =2 C c2 4 1 5 e t =2 C c3 4 0 5 e t =2. 1 0 0 1 2 2 2 2 3 2 3 3 3 3 4 1 1 0 4 Now y.0/ D 4 7 5 ) c1 4 2 5 C c2 4 1 5 C c3 4 0 5 e t =2 D 4 7 5, so c1 D 1, c2 D 5, and 1 0 0 1 1 2 3 2 3 2 3 1 5 0 c3 D 1. Hence y D 4 2 5 e t =2 C 4 5 5 e t =2 C 4 0 5 e t =2. 0 0 1 ˇ ˇ ˇ 6 3 8 ˇˇ ˇ ˇ 1 2 ˇˇ D . 1/. C 2/. 3/. The eigenvectors associated with 10.4.22. ˇ 2 ˇ 3 3 5 ˇ 3 2 :: 3 8 : 0 7 6 5 6 7 : 1 D 1 satisfy the system with augmented matrix 6 2 , which is row equivalent to 0 2 :: 0 7 5 4 :: 3 3 6 : 0 2 3 :: 2 3 1 : 0 7 1 6 1 0 6 7 : . Hence x1 D x3 and x2 D x3 . Taking x3 D yields y1 D 4 1 5e t . The 6 0 1 1 :: 0 7 4 5 1 : 0 0 0 :: 0 2 3 :: 8 3 8 : 0 6 7 6 7 :: eigenvectors associated with 2 D 2 satisfy the system with augmented matrix 6 2 , 3 2 : 0 7 4 5 :: 3 3 3 : 0 2
Section 10.4 Constant Coefficient Homogeneous Systems I 2
6 1 0 6 which is row equivalent to 6 0 1 4 0 0
2
3 1 y2 D 4 0 5e 1 2 3 8 6 3 6 6 2 2 2 4 3 3 8
2t
1 0 0
199
3 :: : 0 7 7 :: . Hence x1 D x3 and x2 D 0. Taking x3 D 1 yields : 0 7 5 :: : 0
. The eigenvectors associated with 3 D 3 satisfy the system with augmented matrix
:: : :: : :: :
3
2
: 1 0 :: : 0 1 :: : 0 0 ::
3
0 7 0 7 6 1 6 7 7 , which is row equivalent to 6 0 . Hence x1 D x2 and x3 D 0 7 0 7 5 4 5 0 0 0 2 3 2 3 2 3 1 1 1 0. Taking x2 D 1 yields y3 D 4 1 5e 3t . The general solution is y D c1 4 1 5e t C c2 4 0 5e 2t C 0 1 1 2 3 2 3 2 3 2 3 2 3 2 3 1 0 1 1 1 0 c3 4 1 5e 3t . Now y.0/ D 4 1 5 ) c1 4 1 5 C c2 4 0 5 C c34 1 5 D 4 1 5, so c1 D 2, 0 1 1 1 0 1 2 3 2 3 2 3 2 3 1 c2 D 3, and c3 D 1. Therefore, y D 4 2 5 e t 4 0 5 e 2t C 4 1 5 e 3t . 2 3 0 ˇ ˇ ˇ 3 0 1 ˇˇ ˇ 2 7 ˇˇ D . 2/. C 2/. 4/. The eigenvectors associated with 10.4.24. ˇˇ 11 ˇ 1 0 3 ˇ 3 2 :: : 0 1 0 1 6 7 6 7 :: with 1 D 2 satisfy the system with augmented matrix 6 11 7, which is row equiv4 7 : 0 5 4 :: 1 0 1 : 0 2 3 :: 1 0 1 : 0 6 7 6 7 alent to 6 0 1 1 ::: 0 7. Hence x1 D x3 and x2 D x3. Taking x3 D 1 yields y1 D 4 5 : 0 0 0 :: 0 2 3 1 4 1 5e 2t . The eigenvectors associated with with 2 D 2 satisfy the system with augmented ma1 2 3 2 3 :: :: 5 0 1 : 0 1 0 0 : 0 6 7 6 7 6 7 6 7 trix 6 11 0 7 ::: 0 7, which is row equivalent to 6 0 0 1 ::: 0 7. Hence x1 D x3 D 0 4 5 4 5 :: :: 1 0 5 : 0 0 0 0 : 0 2 3 0 and x2 is arbitrary. Taking x3 D 1 yields y2 D 4 1 5e 2t . The eigenvectors associated with with 0
200 Chapter 10 Linear Systems of Differential Equations 3 :: : 0 6 7 6 7 :: 3 D 4 satisfy the system with augmented matrix 6 11 , which is row equivalent 6 7 : 0 7 4 5 : 1 0 1 :: 0 2 3 : 2 3 1 :: 0 7 1 6 1 0 6 7 : to 6 0 1 . Hence x1 D x3 and x2 D 3x3. Taking x3 D 1 yields y3 D 4 3 5e 4t . 3 :: 0 7 4 5 1 : 0 0 0 :: 0 2 3 2 3 2 3 2 3 1 0 1 2 The general solution is y D c14 1 5e 2t C c2 4 1 5e 2t C c3 4 3 5e 4t . Now y.0/ D 4 7 5 ) 1 0 1 6 2 3 2 3 2 3 2 3 1 0 1 2 c1 4 1 5 C c24 1 5 C c3 4 3 5 D 4 7 5, so c1 D 2, c2 D 3, and c3 D 4. Hence y D 1 0 1 6 2 3 2 3 2 3 2 0 4 4 2 5 e 2t 4 3 5 e 2t C 4 12 5 e 4t 2 0 4 ˇ ˇ ˇ 3 1 0 ˇˇ ˇ 2 0 ˇˇ D .C1/. 2/2 . The eigenvectors associated with 1 D 1 sat10.4.26. ˇˇ 4 ˇ 4 4 2 ˇ 2 3 2 :: 1 1 0 : 0 7 4 6 4 6 1 0 7 6 6 : : isfy the system with augmented matrix 6 4 , which is row equivalent to 6 0 1 1 0 : 0 7 1 4 5 4 :: 4 4 3 : 0 0 0 0 2 3 1 Hence x1 D x2 =4 and x2 D x3 . Taking x3 D 4 yields y1 D 4 4 5e t . The eigenvectors associated 4 2 3 :: 1 0 : 0 7 6 1 6 7 : , which is row with with 2 D 3 D 2 satisfy the system with augmented matrix 6 4 4 0 :: 0 7 4 5 :: 4 4 0 : 0 2 3 :: 1 0 : 0 7 6 1 6 7 : equivalent to 6 0 . Hence x1 D x2 and x3 is arbitrary. Taking x2 D 1 and x3 D 0 0 0 :: 0 7 4 5 :: 0 0 0 : 0 2 3 2 3 1 0 yields y2 D 4 1 5e 2t . Taking x2 D 0 and x3 D 1 yields y3 D 4 0 5e 2t . The general solution is 0 1 2 3 2 3 2 3 2 3 2 3 2 3 1 1 0 7 1 1 y D c1 4 4 5e t C c24 1 5e 2t C c3 4 0 5e 2t . Now y.0/ D 4 10 5 ) c1 4 4 5 C c2 4 1 5 C 4 0 1 2 4 0 2 3 2 3 2 3 2 3 0 7 1 6 c3 4 0 5 D 4 10 5, so c1 D 1, c2 D 6, and c3 D 2. Hence y D 4 4 5 e t C 4 6 5 e 2t . 1 2 4 2 2
1
0
1
:: : :: : :: :
3
0 7 7 0 7 5 0
Section 10.5 Constant Coefficient Homogeneous Systems II
201
10.4.28. (a) If y.t0 / D 0, then y is the solution of the initial value problem y0 D Ay; y.t0 / D 0. Since y 0 is a solution of this problem, Theorem 10.2.1 implies the conclusion. (b) It is given that y01 .t/ D Ay1 .t/ for all t. Replacing t by t shows that y01.t / D Ay1 .t / D Ay2 .t/ for all t. Since y02.t/ D y01 .t / by the chain rule, this implies that y02 .t/ D Ay2 .t/ for all t. (c) If z.t/ D y1 .t /, then z.t2 / D y1 .t1 / D y2 .t2 /; therefore z and y2 are both solutions of the initial value problem y0 D Ay; y.t2 / D k, where k D y2 .t2 /. 10.4.42. The characteristic polynomial of A is p./ D 2 .a C b/ C ab ˛ˇ, so the eigenvalues of p aCb aCb C b a C
A are 1 D and 1 D , where D .a b/2 C 4˛ˇ; x1 D and 2ˇ 2 2 b a x2 D are associated eigenvectors. Since > jb aj, if L1 and L2 are lines through 2ˇ the origin parallel to x1 and x2 , then L1 is in the first and third quadrants and L2 is in the second and 2ˇ > 0. If Q0 D P0 there are three possibilities: fourth quadrants. The slope of L1 is D b aC (i) if ˛ˇ D ab, then 1 D 0 and P .t/ D P0 , Q.t/ D Q0 for all t > 0; (ii) if ˛ˇ < ab, then 1 > 0 and limt !1 P .t/ D limt !1 Q.t/ D 1 (monotonically); (iii) if ˛ˇ > ab, then 1 < 0 and limt !1 P .t/ D limt !1 Q.t/ D 0 (monotonically). Now suppose Q0 ¤ P0 , so that the trajectory cannot intersect L1 , and assume for the moment that (A) makes sense for all t > 0; that is, even if one or the other of P and Q is negative. Since 2 > 0 it follows that either limt !1 P .t/ D 1 or limt !1 Q.t/ D 1 (or both), and the trajectory is asymptotically parallel to L2 . Therefore,the trajectory must cross into the third quadrant (so P .T / D 0 and Q.T / > 0 for some finite T ) if Q0 > P0 , or into the fourth quadrant (so Q.T / D 0 and P .T / > 0 for some finite T ) if Q0 < P0 . 10.5
CONSTANT COEFFICIENT HOMOGENEOUS SYSTEMS II
ˇ ˇ 10.5.2. ˇˇ 1
ˇ ˇ 1 x1 0 ˇ D .C1/2 . Hence 1 D 1. Eigenvectors satisfy 1 D , 2 ˇ 1 1 x2 0 1 so x1 D x2 . Taking x2 D 1 yields y1 D e t . For a second solution we need a vector u such that 1 1 1 u1 1 1 1 t D . Let u1 D 1 and u2 D 0. Then y2 D e C te t . The general 1 1 u2 1 0 1 1 1 1 t t t solution is y D c1 e C c2 e C te . 1 0 1 ˇ ˇ ˇ 3 1 ˇˇ 1 1 x1 0 2 10.5.4. y0 D ˇˇ D . 2/ . Hence D 2. Eigenvectors satisfy D , 1 1 1 ˇ 1 1 x2 0 1 so x1 D x2 . Taking x2 D 1 yields y1 D e 2t . For a second solution we need a vector u such that 1 1 1 u1 1 1 1 2t D . Let u1 D 1 and u2 D 0. Then y2 D e C te 2t . 1 1 u2 1 0 1 1 1 1 The general solution is y D c1 e 2t C c2 e 2t C te 2t . 1 0 1 ˇ ˇ ˇ 10 9 ˇˇ 6 9 x1 0 2 10.5.6. ˇˇ D .C4/ . Hence D 4. Eigenvectors satisfy D , 1 4 2 ˇ 4 6 x2 0 3 so x1 D 32 x2 . Taking x2 D 2 yields y1 D e 4t . For a second solution we need a vector u such that 2 6 9 u1 3 1 e 4t 3 1 D . Let u1 D 2 and u2 D 0. Then y2 D C te 4t . The 4 6 u2 2 0 2 2 1
202 Chapter 10 Linear Systems of Differential Equations general solution is y D c1 2
10.5.8. 4 4 0
2 6
3 2
1 1
4
2
e
4t
C c2
3
5D
.
1 0
e
4t
2
C
3 2
te
2
:: : :: 1 : 2 : 0 ::
.
4/2 . Hence 1 D 0 and 2 D 3 D 4. The eigenvectors
6 6 associated with 1 D 0 satisfy the system with augmented matrix 6 4 2
4t
0 2 4 6 0 4
: 1 :: : 1 :: : 2 ::
3 0 7 7 , which is 0 7 5 0
3
1 2
0 7 6 1 0 1 1 7 6 . Hence x1 D row equivalent to 6 0 1 x3 and x2 D x3. Taking x3 D 2 yields 0 7 5 4 2 2 0 0 0 2 3 1 y1 D 4 1 5. The eigenvectors associated with 2 D 4 satisfy the system with augmented matrix 2 3 2 3 2 :: :: 1 : 0 7 1 : 0 7 2 6 4 2 6 1 0 1 7 6 7 6 :: :: 1 . Hence x1 D x3 and , which is row equivalent to 6 0 1 6 4 2 : 0 7 1 : 0 7 2 5 2 4 5 4 : : 0 0 0 :: 0 0 4 2 :: 0 2 3 1 1 x2 D x3. Taking x3 D 2 yields y2 D 4 1 5e 4t . For a third solution we need a vector u such 2 2 32 3 2 3 2 4 2 1 u1 1 that 4 4 2 1 5 4 u2 5 D 4 1 5. The augmented matrix of this system is row equivalent to 0 4 2 2 u 2 3 3 : 2 3 2 3 1 :: 0 0 1 2 7 6 1 0 4t 1 e 6 :: 1 7. Let u D 0, u D 0, and u D . Then y D 4 1 5 4 1 1 5te 4t . The C 7 6 0 1 3 1 2 3 : 2 5 2 4 2 2 0 2 : 0 0 0 :: 0 general solution is 2 3 2 3 02 3 2 3 1 1 1 0 1 4t e y D c1 4 1 5 C c2 4 1 5 e 4t C c3 @4 1 5 C 4 1 5 te 4t A: 2 2 2 0 2 2
10.5.10. 4
1
2 1
1 3
1 2 1
3
5D
.
2/. C 2/2 . Hence 1 D 2 and 2 D 3 D 2
6 6 eigenvectors associated with 1 D 2 satisfy the system with augmented matrix 6 4
3
1
2
2
1
3
2. The
: 1 :: : 2 :: : 3 ::
3
0 7 7 , 0 7 5 0
Section 10.5 Constant Coefficient Homogeneous Systems II 2
6 1 0 6 which is row equivalent to 6 0 1 4
0 1
0 0
2
3 0 y1 D 4 1 5e 2t . The 1 2 : 1 :: 6 1 1 6 : trix 6 2 2 2 :: 4 : 1 3 1 ::
0
203
3 :: : 0 7 7 :: . Hence x1 D 0 and x2 D x3 . Taking x3 D 1 yields : 0 7 5 :: : 0
eigenvectors associated with 2 D 3
2
2 satisfy the system with augmented ma: 1 :: : 0 :: : 0 ::
3
0 7 0 7 6 1 0 6 7 7 , which is row equivalent to 6 0 1 . Hence x1 D x3 and 0 7 0 7 5 4 5 0 0 0 0 2 3 1 x2 D 0. Taking x3 D 1 yields y2 D 4 0 5e 2t . For a third solution we need a vector u such 1 2 32 3 2 3 1 1 1 u1 1 that 4 2 2 2 5 4 u2 5 D 4 0 5. The augmented matrix of this system is row equivalent to 1 3 1 u3 1 3 2 :: 1 2 3 2 3 1 : 2 7 1 1 6 1 0 2t 1 1 e 6 :: 1 7. Let u D 0, u D , and u D . Then y D 4 1 5 4 0 5te 2t . C 7 6 0 1 3 1 2 3 0 : 2 5 2 2 2 4 0 1 : 0 0 0 :: 0 The general solution is 2 3 1 2 3 2 3 02 3 0 1 1 1 2t e 2t 2t 2t y D c1 4 1 5 e C c2 4 0 5 e C c 3 @4 1 5 C 4 0 5 te A: 2 1 1 0 1 2
10.5.12. 4
6
2 2
5 1
3 3
1
1
3
5D
. C 2/.
4/2 . Hence 1 D
2 and 2 D 3 D 4. The 2
3 :: : 0 7 7 :: , : 0 7 5 :: : 0
5 3 6 8 6 eigenvectors associated with 1 D 2 satisfy the system with augmented matrix 6 2 1 3 4 2 1 3 2 3 :: 6 1 0 1 : 0 7 6 7 which is row equivalent to 6 0 1 1 ::: 0 7. Hence x1 D x3 and x2 D x3. Taking x3 D 1 4 5 :: 0 0 0 : 0 2 3 1 yields y1 D 4 1 5e 2t . The eigenvectors associated with 2 D 4 satisfy the system with augmented 1 2 3 2 3 :: :: 2 5 3 : 0 1 0 1 : 0 6 7 6 7 6 7 6 7 :: :: matrix 6 2 , which is row equivalent to . Hence x1 D x3 7 6 5 3 : 0 5 1 : 0 7 4 4 0 1 5 :: :: 2 1 3 : 0 0 0 0 : 0
204 Chapter 10 Linear Systems of Differential Equations 2
3 1 and x2 D x3 . Taking x3 D 1 yields y2 D 4 1 5e 4t . For a third solution we need a vector u such 1 2 32 3 2 3 2 5 3 u1 1 5 3 5 4 u2 5 D 4 1 5. The augmented matrix of this system is row equivalent to that 4 2 2 1 3 u3 1 2 3 :: 1 2 3 2 3 1 : 2 7 1 1 6 1 0 4t 1 e 6 7 : . Let u3 D 0, u1 D , and u2 D 0. Then y3 D 4 0 5 C 4 1 5te 4t . The 6 0 1 1 :: 0 7 4 5 2 2 0 1 : 0 0 0 :: 0 2 3 1 2 3 2 3 02 3 1 1 1 1 4t e C 4 1 5 te 4t A. general solution is y D c1 4 1 5 e 2t C c2 4 1 5 e 4t C c3 @4 0 5 2 1 1 1 0 ˇ ˇ ˇ 15 9 ˇˇ 12 9 x1 2 10.5.14. ˇˇ D . 3/ . Hence D 3. Eigenvectors satisfy D 1 16 9 ˇ 16 12 x2 3 3 so x1 D x2 . Taking x2 D 4 yields y1 D e 3t . For a second solution we need a vector u such 4 4 1 12 9 u1 3 1 e 3t 3 that D . Let u1 D and u2 D 0. Then y2 D C te 3t . 16 12 u2 4 0 4 4 4 3t 3 5 3 1 e C te 3t . Now y.0/ D ) The general solution is y D c1 e 3t C c2 4 8 4 0 4 1 3 5 5 12 c1 C c2 4 D , so c1 D 2 and c2 D 4. Therefore, y D e 3t te 3t . 4 0 8 8 16 ˇ ˇ ˇ 7 24 ˇˇ 12 24 x1 2 ˇ D . 5/ . Hence D 5. Eigenvectors satisfy D 10.5.16. ˇ 1 6 17 ˇ 6 12 x2 2 so x1 D 2x2. Taking x2 D 1 yields y1 D e 5t . For a second solution we need a vector u such 1 1 2 12 24 u1 1 e 5t 2 D . Let u1 D and u2 D 0. Then y2 D that C te 5t . 6 12 u2 1 0 1 6 6 5t 2 1 e 2 3 The general solution is y D c1 e 5t C c2 C te 5t . Now y.0/ D ) 1 0 1 1 6 1 2 3 3 12 c1 C c2 6 D , so c1 D 1 and c2 D 6. Therefore, y D e 5t te 5t . 1 1 1 6 0 2 3 1 1 0 10.5.18. 4 1 1 2 5 D . 1/. C 2/2 . Hence 1 D 1 and 2 D 3 D 2. The 1 1 1 2 3 :: 2 1 0 : 0 6 7 6 7 :: eigenvectors associated with 1 D 1 satisfy the system with augmented matrix 6 1 , 2 2 : 0 7 4 5 :: 1 1 2 : 0 2 3 : 2 : 6 1 0 3 : 0 7 2 4 6 7 which is row equivalent to 6 0 1 4 ::: 0 7. Hence x1 D x3 and x2 D x3 . Taking 3 4 5 3 3 :: 0 0 0 : 0
0 0
,
0 0
,
Section 10.5 Constant Coefficient Homogeneous Systems II
205
x 23 D 3 yields 1 6 1 6 6 1 1 4 1 1
The eigenvectors associated with 2 D 2 2 satisfy the system 3 3 with augmented matrix :: :: 0 : 0 7 6 1 1 0 : 0 7 7 6 7 :: , which is row equivalent to 6 0 0 1 ::: 0 7. Hence x1 D x2 and 2 : 0 7 5 4 5 :: :: 1 : 0 0 0 0 : 0 2 3 1 x3 D 0. Taking x2 D 1 yields y2 D 4 1 5e 2t . For a third solution we need a vector u such 0 3 2 32 3 2 1 1 0 u1 1 1 2 5 4 u2 5 D 4 1 5. The augmented matrix of this system is row equivathat 4 1 1 1 1 u3 0 2 3 :: 2 3 1 7 1 6 1 1 0 : 6 7 lent to 6 0 0 1 ::: . Let u2 D 0, u1 D 1, and u3 D 1. Then y3 D 4 0 5e 2t C 1 7 4 5 1 : 0 0 0 :: 0 3 2 3 02 3 2 3 3 2 2 1 1 1 1 2 4 1 5te 2t . The general solution is y D c14 4 5e t Cc24 1 5e 2t Cc3 @4 0 5e 2t C 4 1 5te 0 0 1 0 3 2 3 2 3 2 3 2 3 2 3 6 2 1 1 6 Now y.0/ D 4 5 5 ) c14 4 5 C c2 4 1 5 C c34 0 5 D 4 5 5, so c1 D 2, c2 D 3, 7 3 0 1 7 2 3 2 3 2 3 4 2 1 and c3 D 1. Therefore, y D 4 8 5 e t C 4 3 5 e 2t C 4 1 5 te 2t . 6 1 0 2 3 7 4 4 10.5.20. 4 1 0 1 5 D . C 3/. 1/2 . Hence 1 D 3 and 2 D 3 D 1. The 9 5 6 2 3 :: 4 4 : 0 7 6 4 6 7 : , eigenvectors associated with 1 D 3 satisfy the system with augmented matrix 6 1 3 1 :: 0 7 4 5 : 9 5 9 :: 0 2 3 : 1 :: 0 7 6 1 0 6 7 : which is row equivalent to 6 0 1 . Hence x1 D x3 and x2 D 0. Taking x3 D 1 yields 0 :: 0 7 4 5 : 0 0 0 :: 0 2 3 1 y1 D 4 0 5e 3t . The eigenvectors associated with 2 D 1 satisfy the system with augmented ma1 2 3 2 3 :: :: 8 4 4 : 0 1 0 0 : 0 6 6 7 7 6 7 6 7 :: :: trix 6 1 , which is row equivalent to 7 6 7. Hence x1 D 0 1 1 : 0 0 1 1 : 0 4 5 4 5 :: :: 9 5 5 : 0 0 0 0 : 0
1
2t A
.
206 Chapter 10 Linear Systems of Differential Equations 2
3 0 and x2 D x3 . Taking x3 D 1 yields y2 D 4 1 5e t . For a third solution we need a vector u such 1 2 32 3 2 3 8 4 4 u1 0 1 1 5 4 u2 5 D 4 1 5. The augmented matrix of this system is row equivalent to that 4 1 9 5 5 u 1 2 3 3 :: 2 3 2 3 0 : 1 7 1 0 6 1 0 6 7 :: 4 2 5e t C 4 1 5te t . . Let u D 0, u D 1, and u D 2. Then y D 6 0 1 7 3 1 2 3 1 : 2 5 4 0 1 :: 0 0 0 : 0 2 3 2 3 02 3 2 3 1 0 1 0 1 The general solution is y D c14 0 5e 3t C c2 4 1 5e t C c3 @4 2 5e t C 4 1 5te t A. Now 1 1 0 1 2 3 2 3 2 3 2 3 2 3 6 1 0 1 6 y.0/ D 4 9 5 ) c1 4 0 5 C c24 1 5 C c3 4 2 5 D 4 9 5, so c1 D 2, c2 D 1, and 1 1 1 0 1 2 3 2 3 2 3 4 0 2 c3 D 4. Therefore, y D 4 0 5 e 3t C 4 9 5 e t 4 4 5 te t . 1 4 2 2 3 4 8 4 3 1 3 5 D . C 4/. 8/2 . Hence 1 D 4 and 2 D 3 D 8. The 10.5.22. 4 1 1 9 2 3 :: 8 8 4 : 0 6 7 6 7 :: , eigenvectors associated with 1 D 4 satisfy the system with augmented matrix 6 3 3 3 : 0 7 4 5 :: 1 1 13 : 0 2 3 :: 1 0 : 0 7 6 1 6 7 : which is row equivalent to 6 0 . Hence x1 D x2 and x3 D 0. Taking x2 D 1 yields 0 1 :: 0 7 4 5 :: 0 0 0 : 0 2 3 1 y1 D 4 1 5e t . The eigenvectors associated with 2 D 8 satisfy the system with augmented matrix 0 2 3 2 3 :: :: 8 4 : 0 7 6 4 6 1 0 1 : 0 7 6 7 6 7 :: : , which is row equivalent to 6 3 6 0 1 0 :: 0 7. Hence x1 D x3 and 9 3 : 0 7 4 5 4 5 :: :: 1 1 1 : 0 0 0 0 : 0 2 3 1 x2 D 0. Taking x3 D 1 yields y2 D 4 0 5e 8t . For a third solution we need a vector u such that 1 2 32 3 2 3 4 8 4 u1 1 4 3 9 3 5 4 u2 5 D 4 0 5. The augmented matrix of this system is row equivalent to 1 1 1 u3 1
Section 10.5 Constant Coefficient Homogeneous Systems II 2
6 1 6 6 0 4 0
: 0 1 :: : 1 0 :: : 0 0 ::
3 4 1 4
0
3
3 2 3 3 1 8t 1 e . Then y3 D 4 1 5 C 4 0 5te 8t . 4 4 0 1 3 02 3 2 3 1 3 1 8t e 5e 8t C c3 @4 1 5 C 4 0 5te 8t A. Now y.0/ D 4 0 1 3 2 3 3 4 4 1 5 D 4 1 5, so c1 D 1, c2 D 3, and c3 D 8. 4 3 0 3 2 8 C 4 0 5 te 8t . 8
7 3 7 7. Let u3 D 0, u1 D , and u2 D 5 4
2 3 1 1 The general solution is c1 4 1 5e t C c24 0 0 1 2 3 2 3 2 3 2 4 1 1 4 1 5 ) c1 4 1 5 C c2 4 0 5 C c3 4 3 0 1 3 3 2 2 1 3 Therefore, y D 4 1 5 e 4t C 4 2 5 e 8t 0 3 2 3 5 1 1 1 9 3 5 D . 10.5.24. 4 2 2 4 2 : 1 1 :: 6 1 6 : with augmented matrix 6 1 3 3 :: 4 : 2 2 2 :: 2
207
2
6/3 . Hence 1 D 6. The eigenvectors satisfy the system 3
2
3 :: : 0 7 7 :: . : 0 7 5 :: : 0
0 7 0 6 1 0 7 6 , which is row equivalent to 6 0 1 1 0 7 5 4 0 0 0 0 2 3 0 Hence x1 D 0 and x2 D x3 . Taking x3 D 1 yields y1 D 4 1 5e 6t . For a second solution we need a 1 2 32 3 2 3 0 1 1 1 u1 vector u such that 4 1 3 3 5 4 u2 5 D 4 1 5. The augmented matrix of this system is row 1 2 2 2 u3 2 3 :: 2 3 1 0 : 1 4 7 6 1 0 1 1 e 6t 6 7 :: 1 7. Let u3 D 0, u1 D equivalent to 6 0 1 , and u2 D . Then y2 D 4 1 5 C 1 : 4 5 4 4 4 4 0 : 0 0 0 :: 0 2 3 2 32 3 2 1 3 4 0 1 1 1 v1 6 1 7 4 1 5te 6t . For a third solution we need a vector v such that 4 1 3 3 5 4 v2 5 D 4 4 5. 1 2 2 2 v3 0 2 3 :: 1 0 : 8 7 6 1 0 6 :: 1 7. Let v D 0, v D 1 , The augmented matrix of this system is row equivalent to 6 0 1 3 1 1 : 8 7 4 5 8 :: 0 0 0 : 0 2 3 2 3 2 3 1 1 0 6t 6t 2 6t 1 e te t e and v2 D . Then y3 D 4 1 5 C4 1 5 C4 1 5 . The general solution is y D 8 8 4 2 0 0 1
208 Chapter 10 Linear Systems of Differential Equations 2
3 02 0 c1 4 1 5 e 6t C c2 @4 1
2
6
2
2
3 2 3 1 1 0 e 6t 1 5 C 4 1 5 te 6t A 4 0 1
2 3 1 e 6t C4 Cc3 @4 1 5 8 0 02
3 2 3 1 1 0 te 6t t 2 e 6t A: 1 5 C4 1 5 4 2 0 1
3 4 2 1 1 5 D .C2/3 . Hence 1 D 2. The eigenvectors satisfy the sys10.5.26. 4 2 3 1 2 3 2 3 :: :: 4 4 4 : 0 1 0 0 : 0 6 7 6 7 6 7 6 7 :: :: tem with augmented matrix 6 2 , which is row equivalent to 7 6 7. 1 1 : 0 5 4 4 0 1 1 : 0 5 : : 2 3 3 :: 0 0 0 0 :: 0 2 3 0 Hence x1 D 0 and x2 D x3 . Taking x3 D 1 yields y1 D 4 1 5e 2t . For a second solution we need a 1 3 2 3 32 2 0 4 4 4 u1 vector u such that 4 2 1 1 5 4 u2 5 D 4 1 5. The augmented matrix of this system is row 1 2 3 3 u3 2 3 :: 2 3 1 7 1 6 1 0 0 : 6 7 :: equivalent to 6 0 1 1 : . Let u3 D 0, u1 D 1, and u2 D 1. Then y2 D 4 1 5 e 2t C 1 7 4 5 0 : 0 0 0 :: 0 3 2 32 3 2 v1 4 4 4 0 4 1 5 te 2t . For a third solution we need a vector v such that 4 2 1 1 5 4 v2 5 D v3 2 3 3 1 2 3 : 2 3 : 3 1 4 7 6 1 0 0 : 7 : 4 1 5. The augmented matrix of this system is row equivalent to 6 1 7. Let v3 D 0, 6 0 1 1 :: 2 5 4 0 : 0 0 0 :: 0 2 3 2 3 2 3 3 1 0 3 1 e 2t t 2 e 2t v1 D , and v2 D . Then y3 D 4 2 5 C 4 1 5 te 2t C 4 1 5 . The general 4 2 4 2 0 0 1 2 3 02 3 2 3 1 0 1 0 solution is y D c1 4 1 5 e 2t C c2 @4 1 5 e 2t C 4 1 5 te 2t A 1 0 1 02 3 2 3 2 3 1 3 1 0 2t 2 2t e t e A Cc3 @4 2 5 C 4 1 5 te 2t C 4 1 5 4 2 0 0 1 10.5.28. 4
2 2
4
12 24 24
10 11 8
3
5D
. C 6/3 . Hence 1 D
6. The eigenvectors satisfy the
Section 10.5 Constant Coefficient Homogeneous Systems II : 12 10 :: : 18 11 :: : 24 14 ::
2
6 4 6 system with augmented matrix 6 2 4
3 2 0 7 6 1 7 6 , which is row equivalent to 6 0 0 7 5 4
0
1
1
1 2
209 3 :: : 0 7 7 :: . : 0 7 5 :: : 0
0 0 0 3 2 D 2 yields y1 D 4 1 5e 6t . For a second solu2 32 3 2 3 10 u1 2 11 5 4 u2 5 D 4 1 5. The augmented matrix 14 u3 2 3 :: 1 : 1 7 7 : :: 1 7. Let u3 D 0, u1 D 1 1, and u2 D 2 6 5 :: 0 : 0
2
0
2
x3 . Taking x3 2 2 4 12 tion we need a vector u such that 4 2 18 2 24 2 6 1 0 6 of this system is row equivalent to 6 0 1 4 0 0 2 3 2 3 6 2 1 e 6t . Then y2 D 4 1 5 C 4 1 5 te 6t . For a third solution we need a vector v such that 6 6 0 2 3 2 2 32 3 4 12 10 v1 1 4 2 18 11 5 4 v2 5 D 4 16 5. The augmented matrix of this system is row equivalent to 2 24 14 v3 0 2 3 :: 2 3 1 1 : 12 3 7 6 1 0 1 1 e 6t 6 7 : 1 :: 1 7. Let v3 D 0, v1 D , and v2 D . Then y3 D 4 1 5 6 0 1 2 36 5 4 3 36 36 0 :: 0 0 0 : 0 2 3 2 3 2 3 0 2 3 2 6 2 2 6 6t te 6t t 2 e 6t e 4 1 5 C4 1 5 . The general solution is y D c1 4 1 5 e 6t C c2 @ 4 1 5 C4 6 2 6 0 2 2 0 Hence x1 D
x3 and x2 D
0 2
3 12 e 6t Cc3 @ 4 1 5 36 0
2
10.5.30. 4
4
1 1
0 3
1 1
0
2 2
0
3 2 6 te 6t 4 1 5 C4 6 0
3
5D
: 1 :: : 1 :: : 1 ::
3 1 2 t 2 e 6t A: 1 5 2 2
.C3/3 . Hence 1 D 3. The eigenvectors satisfy the sys3
2
3 :: : 0 7 7 :: . : 0 7 5 :: : 0
0 7 6 1 0 1 7 6 , which is row equivalent to 6 0 0 0 1 0 0 7 5 4 1 0 0 0 0 0 2 3 1 Hence x1 D x3 and x2 is arbitrary. Taking x2 D 0 and x3 D 1 yields y1 D 4 0 5 e 3t . Taking x2 D 1 2 3 0 1 and x3 D 0 yields y2 D 4 1 5 e 3t . For a third solution we need constants ˛ and ˇ and a vector u such 0
6 6 tem with augmented matrix 6 4
1
2
3 2 1 5 te 2
1
6t A
210 Chapter 10 Linear Systems of Differential Equations 2
32 3 2 3 2 3 1 u1 1 0 that 4 1 5 4 u2 5 D ˛ 4 0 5 C ˇ 4 1 5. The augmented matrix of this system is row 1 u3 1 0 3 :: ˛ 6 1 0 1 : 7 6 7 : : equivalent to 6 0 0 0 : ˛ C ˇ 7; hence the system has a solution if ˛ D ˇ D 1, which yields 4 5 :: 0 0 0 : 0 2 3 2 3 1 1 the eigenvector x3 D 4 1 5. Taking u1 D 1 and u2 D u3 D 0 yields the solution y3 D 4 0 5 e 3t C 1 0 2 3 2 3 2 3 02 3 2 1 0 1 1 4 1 5 te 3t . The general solution is y D c1 4 0 5 e 3t C c2 4 1 5 e 3t C c3 @4 0 5 e 3t C 4 1 1 0 0 2 3 3 1 0 5 D . C 2/3 . Hence 1 D 2. The eigenvectors satisfy the 1 1 0 10.5.32. 4 1 1 2 2 3 2 3 :: :: 1 0 : 0 7 6 1 6 1 1 0 : 0 7 6 7 6 7 :: : system with augmented matrix 6 1 , which is row equivalent to 6 0 0 0 :: 0 7. 1 0 : 0 7 4 5 4 5 :: :: 1 1 0 : 0 0 0 0 : 0 2 3 1 Hence x1 D x2 and x3 is arbitrary. Taking x2 D 1 and x3 D 0 yields y1 D 4 1 5 e 2t . Taking x2 D 0 2 3 0 0 and x3 D 1 yields y2 D 4 0 5 e 2t . For a third solution we need constants ˛ and ˇ and a vector u such 1 32 3 2 3 2 3 2 1 0 u1 1 1 0 5 4 5 4 5 4 4 1 u 1 1 0 D˛ C ˇ 0 5. The augmented matrix of this system is row that 2 0 1 u3 1 1 0 2 3 :: ˛ 6 1 1 0 : 7 6 7 : equivalent to 6 0 0 0 :: ˛ C ˇ 7; hence the system has a solution if ˛ D ˇ D 1, which yields 4 5 :: 0 0 0 : 0 2 3 2 3 1 1 the eigenvector x3 D 4 1 5. Taking u1 D 1 and u2 D u3 D 0 yields the solution y3 D 4 0 5 e 2t C 1 0 2 3 2 3 2 3 02 3 2 1 1 0 1 4 1 5 te 2t . The general solution is y D c1 4 1 5 e 2t C c2 4 0 5 e 2t C c3 @4 0 5 e 2t C 4 1 0 1 0 1 1 1
0 0 0 2
10.5.34.
y03
Ay3
D .1 I C.1 I D
ue 1 t
A/ve 1 t C .1 I
A/ute 1 t C ue 1 t
t 2 e 1 t C xte 1 t 2 xte 1 t C ue 1 t C 0 C xte 1 t D 0:
A/x
3 1 1 5 te 1
3 1 1 5 te 1
1
3t A
1
2t A
.
Section 10.6 Constant Coefficient Homogeneous Systems III
211
Now suppose that c1 y1 C c2 y2 C c3y3 D 0. Then t2 c1x C c2 .u C tx/ C c3 v C tu C x D 0: 2
.A/
Differentiating this twice yields c3 x D 0, so c3 D 0 since x ¤ 0. Therefore,(A) reduces to (B) c1x C c2 .u C tx/ D 0. Differentiating this yields c2 x D 0, so c2 D 0 since x ¤ 0. Therefore,(B) reduces to c3 x D 0, so c1 D 0 since x ¤ 0. Therefore,y1 , y2 , and y3 are linearly independendent. 10.6
CONSTANT COEFFICIENT HOMOGENEOUS SYSTEMS III
ˇ ˇ 10.6.2. ˇˇ 2 10 is 4
ˇ ˇ ˇ D . C 1/2 C 4. The augmented 9 ˇ 3 2 :: 2i 4 : 0 5 , which is row equivalent to 4 : 26 10 2i :: 0 5 i .5 i /x2 =13. Taking x2 D 13 yields the eigenvector x D 13 5 i t yields e .cos 2t C i sin 2t// 13 11 26
4
y D c1 e ˇ ˇ 5 10.6.4. ˇˇ 3
6
t
5 cos 2t C sin 2t 13 cos 2t
ˇ ˇ ˇ D . ˇ
C c2 e
t
matrix of .A 1
5Ci 13
:: : :: :
. 1 C 2i / I / x D 0 3
0 5 . Therefore,x1 D 0
0 0 . Taking real and imaginary parts of
5 sin 2t cos 2t 13 sin 2t
:
2/2 C 9. Hence, D 2 C 3i is an eigenvalue of A. The associated 2 3 3i 6 eigenvectors satisfy .A .2 C 3i / I / x D 0. The augmented matrix of this system is 4 3 3 3i 2 3 :: 1 1 i : 0 5 which is row equivalent to 4 . Therefore,x1 D .1 C i /x2 . Taking x2 D 1 yields :: 0 0 : 0 1Ci x1 D 1 C i , so x D is an eigenvector. Taking real and imaginary parts of e 2t .cos 3t C 1 sin 3t C cos 3t cos 3t sin 3t 1Ci C c2 e 2t . i sin 3t/ yields y D c1 e 2t 1 cos 3t sin 3t ˇ ˇ ˇ 3 3 1 ˇˇ ˇ 1 5 3 ˇˇ D . C 1/ . C 2/2 C 4 . The augmented matrix of .A C 10.6.6. ˇˇ ˇ 3 7 3 ˇ 3 2 3 2 :: :: 2 3 1 : 0 1 0 1 : 0 6 7 6 7 6 7 6 7 :: :: I /x D 0 is 6 1 , which is row equivalent to 7. Therefore,x1 D 7 6 4 3 : 0 5 4 4 0 1 1 : 0 5 :: :: 3 7 4 : 0 0 0 0 : 0 2 3 1 x2 D x3 . Taking x3 D 1 yields y1 D 4 1 5 e t . The augmented matrix of .A . 2 C 2i /I / x D 0 1 1
3 :: : 0 5 , :: : 0
212 Chapter 10 Linear Systems of Differential Equations 2
1
6 6 is 6 4
2i
3
1
3
3
1 2i
7
3 5
.1 C i / x3 and x2 D 2
.1
2i i/
2
:: : :: : :: :
3 2 0 7 6 1 0 7 6 , which is row equivalent to 6 0 1 0 7 5 4 0
0 0
x3 . Taking x3 D 2 yields the eigenvector x2 D 2
3 :: : 0 7 7 :: 1 i . Therefore,x1 D : 0 7 2 5 :: 0 : 0 2 3 1Ci 4 1 C i 5. The real 2 3 cos 2t sin 2t cos 2t sin 2t 5 and 2 cos 2t 1Ci 2
3 2 1Ci and imaginary parts of e 2t .cos 2t C i sin 2t/ 4 1 C i 5 are y2 D e 2t 4 2 3 2 sin 2t C cos 2t y3 D e 2t 4 sin 2t C cos 2t 5. Therefore, 2 sin 2t 2 3 2 3 2 3 1 cos 2t sin 2t sin 2t C cos 2t y D c1 4 1 5 e t C c2e 2t 4 cos 2t sin 2t 5 C c3 e 2t 4 sin 2t C cos 2t 5: 1 2 cos 2t 2 sin 2t ˇ ˇ ˇ 10.6.8. ˇˇ ˇ 2 6 4 6 is 6 4 4 4
ˇ 3 ˇˇ 4 1 2 ˇˇ D . 1/ . C 1/2 C 4 . The augmented matrix of .A I /x D 0 4 2 3 ˇ 3 2 3 :: :: 1 3 : 0 7 6 1 0 1 : 0 7 7 7 6 :: : , which is row equivalent to 6 0 1 1 :: 0 7. Therefore,x1 D x2 D 2 2 : 0 7 5 4 5 :: :: 2 2 : 0 0 0 0 : 0 2 3 1 x3 . Taking x3 D 1 yields y1 D 4 1 5 e t . The augmented matrix of .A . 1 C 2i /I / x D 0 is 1 2 3 3 2 :: :: 1 i 1 3 : 0 7 : 0 7 2 6 2 2i 6 1 0 6 7 7 6 :: : . Therefore,x1 D , which is row equivalent to 6 0 1 6 4 2i 2 1 :: 0 7 : 0 7 4 5 5 4 :: :: 4 2 4 2i : 0 0 0 0 : 0 2 3 1Ci 1 i 5. The real and 2 x3 and x2 D x3 . Taking x3 D 2 yields the eigenvector x2 D 4 2 2 2 3 2 3 1Ci sin 2t cos 2t 5 are y2 D e t 4 5 and y3 D imaginary parts of e t .cos 2t C i sin 2t/ 4 2 2 cos 2t 2 2 cos 2t 2 3 cos 2t sin 2t 5. Therefore, 2 sin 2t e t4 2 sin 2t 2 3 2 3 2 3 1 sin 2t cos 2t cos 2t sin 2t 5 C c3 e t 4 5: 2 cos 2t 2 sin 2t y D c1 4 1 5 e t C c2 e t 4 1 2 cos 2t 2 sin 2t 3
1
Section 10.6 Constant Coefficient Homogeneous Systems III
213
ˇ ˇ 1 ˇˇ 7 3 5 ˇˇ 10.6.10. D . 2/2 C 1. The augmented matrix of .A .2 C i /I /x D 0 is ˇ 2 5 3 ˇ 3 2 3 2 3 :: :: 1C3i 1 5 : 0 5 1 : 0 5 2 4 1 3i , which is row equivalent to 4 . Therefore,x1 D :: :: 3 2 1 3i : 0 0 0 : 0 1 C 3i 1 C 3i x2 . Taking x2 D 2 yields the eigenvector x D . Taking real and imaginary parts of 2 2 1 C 3i e 2t .cos t C i sin t/ yields 2 cos t 3 sin t sin t C 3 cos t y D c1 e 2t C c2 e 2t : 2 cos t 2 sin t ˇ ˇ ˇ 34 ˇ 52 ˇ ˇ D . 2/2 C 16. The augmented matrix of .A .2 C 4i / I / x D 0 10.6.12. ˇ 20 30 ˇ 2 3 2 3 :: :: 8Ci : 0 32 4i 52 : 0 1 5 5, which is row equivalent to 4 5. Therefore,x1 D is 4 :: :: 20 32 4i : 0 0 0 : 0 .8 C i / 8 i x2 . Taking x2 D 5 yields the eigenvector x D . Taking real and imaginary parts of 5 5 8 i sin 4t 8 cos 4t cos 4t 8 sin 4t 2t 2t 2t e .cos 4tCi sin 4t/ yields y D c1 e C c2 e . 5 5 cos 4t 5 sin 4t ˇ ˇ ˇ 3 4 2 ˇˇ ˇ 10.6.14. ˇˇ 5 7 8 ˇˇ D . C 2/ . 2/2 C 9 . The augmented matrix of .A C ˇ 10 13 8 ˇ 2 3 2 3 :: :: 5 4 2 : 0 7 2 : 0 7 6 6 1 0 6 7 6 7 :: : 2I /x D 0 is 6 , which is row equivalent to . Therefore,x1 D 6 0 1 5 9 8 : 0 7 2 :: 0 7 5 5 4 4 : : 10 13 6 :: 0 0 0 0 :: 0 2 3 2 x2 D 2x3. Taking x3 D 1 yields y1 D 4 2 5 e 2t . The augmented matrix of .A .2 C 3i /I / x D 0 is 1 2 3 2 3 :: :: 1 3i 4 2 : 0 1 0 1 i : 0 6 7 6 7 6 7 6 7 :: :: , which is row equivalent to 6 7 6 7. Therefore,x1 D 5 5 3i 8 : 0 0 1 i : 0 4 5 4 5 :: :: 10 13 10 3i : 0 0 0 0 : 0 2 3 1Ci 5. The real .1 i /x3 and x2 D ix3 . Taking x3 D 1 yields the eigenvector x2 D 4 i 1 3 2 3 2 cos 3t sin 3t 1Ci 5 are y2 D e 2t 4 5 and sin 3t i and imaginary parts of e 2t .cos 3t C i sin 3t/ 4 1 cos 3t
214 Chapter 10 Linear Systems of Differential Equations 2
3 sin 3t C cos 3t 5. Therefore, y3 D c3 e 2t 4 cos 3t sin 3t 2 3 2 3 2 2 cos 3t sin 3t 2t 2t 2t 5 C c3 e 4 y D c1 4 2 5 e C c2 e 4 sin 3t 1 cos 3t ˇ ˇ 1 ˇ 10.6.16. ˇˇ 0 ˇ 1 2 2 6 0 2 6 is 6 0 1 1 4 1 0 1
2 2
2 1
0
:: : :: : :: :
ˇ ˇ ˇ ˇD ˇ ˇ
.
2/ .
1/2 C 1 . The augmented matrix of .A
3 2 0 7 6 1 7 6 , which is row equivalent to 6 0 0 7 5 4 0
0
2
3
0 1 0
1 Taking x3 D 1 yields y1 D 4 1 5 e t . The augmented matrix of .A 1 2
6 1 0 6 which is row equivalent to 6 0 1 4 0 0
1
i
:: : :: : :: :
3 sin 3t C cos 3t 5: cos 3t sin 3t
3
: 1 :: : 1 :: : 0 ::
I /x D 0
3 0 7 7 . Therefore,x1 D x2 D 1. 0 7 5 0
2
i
6 6 .1 C i /I / x D 0 is 6 0 4 1
2
1
2
i 0
1 1
i
0 7 .1 C i / 7 1Ci . Therefore,x1 D .1 C i /x3 and x2 D x3 . 0 7 2 5 2 0 0 2 3 2 C 2i Taking x3 D 2 yields the eigenvector x2 D 4 1 C i 5. The real and imaginary parts of e 4t .cos t C 2 2 3 2 3 2 3 2 C 2i 2 cos t 2 sin t 2 sin t C 2 cos t i sin t/ 4 1 C i 5 are y2 D e t 4 cos t sin t 5 and y3 D c3 e t 4 cos t C sin t 5. Therefore, 2 2 cos t 2 sin t 2 3 2 3 2 3 1 2 cos t 2 sin t 2 sin t C 2 cos t y D c1 4 1 5 e t C c2e t 4 cos t sin t 5 C c3e t 4 cos t C sin t 5: 1 2 cos t 2 sin t ˇ ˇ 7 15 10.6.18. ˇˇ 3 1 2 :: 3 6i 15 : 4 is :: 3 3 6i :
ˇ ˇ ˇ D . 4/2 C 36. The augmented matrix of .A .4 C 6i / I / x D 0 ˇ 3 3 2 :: 1 1 C 2i : 0 0 5 5. Therefore,x1 D , which is row equivalent to 4 :: 0 0 0 : 0 1 2i .1C2i /x2 . Taking x2 D 1 yields the eigenvector x D . Taking real and imaginary parts of 1 1 2i 2 sin 6t cos 6t 2 cos 6t sin 6t e 4t .cos 6tCi sin 6t/ yields y D c1e 4t C c2 e 4t . 1 cos 6t sin 6t 5 1 2 c1 5 5 cos 6t C 5 sin 6t Now y.0/ D ) D , so c1 D 1, c2 D 3, and y D e 4t . 1 1 0 c2 1 cos 6t 3 sin 6t
:: : :: : :: :
3 0 7 7 , 0 7 5 0
Section 10.6 Constant Coefficient Homogeneous Systems III
215
ˇ ˇ 1 ˇˇ 4 6 1 2 1 1Ci 2 ˇˇ 10.6.20. D C . The augmented matrix of A I x D 0 ˇ 5 2 6 ˇ 2 4 2 6 3 2 3 2 :: :: 1C3i 1 1 3i 2 : 0 5 1 : 0 5 5 4 4 , which is row equivalent to . Therefore,x1 D is :: :: 6 5 1 3i : 0 0 0 : 0 1 C 3i 1 C 3i x2 . Taking x2 D 5 yields the eigenvector x D . Taking real and imaginary parts of 5 5 1 C 3i cos t=2 3 sin t=2 sin t=2 C 3 cos t=2 t =2 t =2 t =2 e .cos t=2Ci sin t=2/ yields y D c1 e C c2 e . 5 5 cos t=2 5 sin t=2 1 2 cos.t=2/ C sin.t=2/ 1 1 3 c1 1 , c2 D , and y D e t =2 . Now y.0/ D ) D , so c1 D cos.t=2/ C 2 sin.t=2/ 1 5 0 c2 1 5 5 ˇ ˇ ˇ 4 ˇ 4 0 ˇ ˇ ˇ 10.6.22. ˇ 8 10 20 ˇˇ D . 8/ . 2/2 C 4 . The augmented matrix of .A ˇ 2 3 2 ˇ 2 3 2 3 :: :: 0 : 0 7 2 : 0 7 6 1 0 6 0 4 7 6 7 6 : : : . Therefore,x1 D 8I /x D 0 is 6 8 6 , which is row equivalent to 6 0 1 2 :: 0 7 20 : 0 7 4 5 4 5 :: :: 2 3 6 : 0 0 0 0 : 0 2 3 2 x2 D 2x3 . Taking x3 D 2 yields y1 D 4 2 5 e 8t . The augmented matrix of .A .2 C 2i /I / x D 0 is 1 2 3 2 3 :: :: 2 2i 4 0 : 0 1 0 2 C 2i : 0 6 7 6 7 6 7 6 7 :: :: , which is row equivalent to . Therefore,x1 D 6 7 6 8 8 2i 20 : 0 5 2i : 0 7 4 4 0 1 5 :: :: 2 3 4 2i : 0 0 0 0 : 0 2 3 2 2i .2 2i /x3 and x2 D 2ix3 . Taking x3 D 1 yields the eigenvector x2 D 4 2i 5. The real and 1 2 3 2 3 2 cos 2t C 2 sin 2t 2 2i 5 and y3 D 2 sin 2t imaginary parts of e 2t .cos 2t C i sin 2t/ 4 2i 5 are y2 D e 2t 4 2 cos 2t 1 2 3 2 3 2 3 2 sin 2t C 2 cos 2t 2 2 cos 2t C 2 sin 2t 5, so the general solution is y D c1 4 2 5 e 8t Cc2 e 2t 4 5C e 2t 4 2 cos 2t 2 sin 2t sin 2t 1 2 cos 2t 2 3 2 3 2 32 3 2 3 2 sin 2t C 2 cos 2t 8 2 2 2 c1 8 5. Now y.0/ D 4 6 5 ) 4 2 0 c3 e 2t 4 2 cos 2t 2 5 4 c2 5 D 4 6 5, so c1 D 2, sin 2t 5 1 1 0 c3 5 2 3 2 3 4 4 cos 2t C 8 sin 2t c2 D 3, c3 D 1, and y D 4 4 5 e 8t C e 2t 4 6 sin 2t C 2 cos 2t 5. 2 3 cos 2t C sin 2t ˇ ˇ ˇ 4 4 4 ˇˇ ˇ 10 3 15 ˇˇ D . 8/.2 C 16/. The augmented matrix of .A 8I /x D 0 is 10.6.24. ˇˇ ˇ 2 3 1 ˇ
216 Chapter 10 Linear Systems of Differential Equations 3 2 3 :: 0 1 0 2 : 0 6 7 6 7 6 7 6 7 :: , which is row equivalent to . Therefore,x1 D 2x3 6 10 7 6 5 0 5 1 : 0 7 4 4 0 1 5 : 2 3 0 0 0 0 :: 0 2 3 2 and x2 D x3 . Taking x3 D 1 yields y1 D 4 1 5 e 8t . The augmented matrix of .A 4iI / x D 0 is 1 2 3 2 3 :: :: 4 4 : 0 7 1Ci : 0 7 6 4 4i 6 1 0 6 7 6 7 :: : , which is row equivalent to 6 0 1 . Therefore,x1 D 6 10 3 4i 15 : 0 7 1 C 2i :: 0 7 4 5 4 5 :: :: 2 3 1 4i : 0 0 0 0 : 0 2 3 1 i .1 i /x3 and x2 D .1 2i /x3 . Taking x3 D 1 yields the eigenvector x2 D 4 1 2i 5. The 1 2 3 2 3 1 i cos 4t C sin 4t real and imaginary parts of .cos 4t C i sin 4t/ 4 1 2i 5 are y2 D 4 cos 4t C 2 sin 4t 5 and y3 D 1 cos 4t 3 2 2 3 2 3 2 cos 4t C sin 4t sin 4t cos 4t 4 sin 4t 2 cos 4t 5, so the general solution is y D c1 4 1 5 e 8t C c2 4 cos 4t C 2 sin 4t 5 C 1 cos 4t sin 4t 2 3 2 3 2 32 3 2 3 sin 4t cos 4t 16 2 1 1 c1 16 c3 4 sin 4t 2 cos 4t 5. Now y.0/ D 4 14 5 ) 4 1 1 2 5 4 c2 5 D 4 14 5, so c1 D 3, sin 4t 6 1 1 0 c3 6 2 3 2 3 6 10 cos 4t 4 sin 4t c2 D 3, c3 D 7, and y D 4 3 5 e 8t C 4 17 cos 4t sin 4t 5. 3 3 cos 4t 7 sin 4t 2
4
4
: 4 :: : 15 :: : 7 ::
10.6.28. (a) From the quadratic formula the roots are p kuk2 kv2k C .kuk2 kv2 k/2 C 4.u; v/2 k1 D 2.u; v/ p 2 2 kuk kv k .kuk2 kv2k/2 C 4.u; v/2 k2 D : 2.u; v/
Clearly k1 > 0 and k2 < 0. Moreover, k1 k2 D (b) Since k2 D
.kuk2
kv2k/2
.kuk2 kv2k/2 C 4.u; v/2 D 1: 4.u; v/2
1= k1, .2/
u1
D u
k2 v D u C
v.2/ 1
D v C k2 u D v
1 1 1 .1/ vD .v C k1 u/ D v k1 k1 k1 1 1 1 1 .1/ uD .u k1 v/ D u : k1 k1 k1 1
Section 10.6 Constant Coefficient Homogeneous Systems III
217
2 3 ˇ :: 10 ˇˇ 15 5i 10 : 0 5, D 2 C25. The augmented matrix of .A 5iI /x D 0 is 4 :: 15 ˇ 25 15 5i : 0 2 3 : 3Ci :: 0 1 5 5. Therefore,x1 D .3 i / x2 . Taking x2 D 5 yields the which is row equivalent to 4 :: 5 0 0 : 0 3 i 3 1 eigenvector x D , so u D and v D . The quadratic equation is 3k 2 33kC3 D 5 5 0 :5257 :8507 0, with positive root k :0902. Routine calculations yield U ,V . :8507 :5257 2 3 ˇ ˇ :: ˇ 3 ˇ 15 3 6i 15 : 0 ˇ D 2 C36. The augmented matrix of .A 6iI /x D 0 is 4 5, 10.6.32. ˇˇ :: 3 3 ˇ 3 3 6i : 0 2 3 :: 1 1 2i : 0 5 which is row equivalent to 4 . Therefore,x1 D .1 2i /x2 . Taking x2 D 1 :: 0 0 : 0 1 C 2i 1 2 yields the eigenvector x D , so u D and v D . The quadratic equation is 1 1 0 :9732 2k 2 C 2k C 2 D 0, with positive root k 1:6180. Routine calculations yield U , :2298 :2298 V . :9732 ˇ ˇ ˇ 5 12 ˇˇ ˇ 10.6.34. ˇ D . C 1/2 C 36. The augmented matrix of .A . 1 C 6i /I /x D 0 6 7 ˇ 2 3 2 3 :: :: 6 6i 12 : 0 1 .1 C i / : 0 5, which is row equivalent to 4 5. Therefore,x1 D is 4 :: :: 6 6 6i : 0 0 0 : 0 1Ci 1 1 .1 C i /x2 . Taking x2 D 1 yields the eigenvector x D , so u D and v D . 1 1 0 2 The quadratic equation is k 1 D 0, with positive root k 1:6180. Routine calculations yield k :5257 :8507 U ,V . :8507 :5257 ˇ ˇ ˇ 4 9 ˇˇ ˇ 10.6.36. ˇ D . C 1/2 C 36. The augmented matrix of .A . 1 C 6i /I /x D 0 5 2 ˇ 2 3 2 3 :: :: 3 6i 3 6i 9 : 0 1 : 0 5 5, which is row equivalent to 4 5. Therefore,x1 D is 4 :: :: 5 3 6i : 0 0 0 : 0 3 6i 3 6i 3 6 x2 . Taking x2 D 5 yields the eigenvector x D , so u D and v D . The 5 5 0 5 2 quadratic equation is 18k C 2kC 18 D 0, with positive root k 1:0571. Routine calculations yield :8817 :4719 U ,V . :4719 :8817 ˇ ˇ ˇ 1 5 ˇˇ 10.6.38. ˇˇ D . C 1/2 C 100. The augmented matrix of .A . 1 C 10i /I /x D 0 20 1 ˇ ˇ ˇ 10.6.30. ˇˇ
15 25
218 Chapter 10 Linear Systems of Differential Equations 3 3 2 :: i i : 0 0 1 2 5. Therefore,x1 D x2 . 5, which is row equivalent to 4 is 4 :: 2 0 0 : 0 20 10i 0 i 0 1 Taking x2 D 2 yields the eigenvector x D , so u D and v D . Since .u; v/ D 0 we 2 2 0 0 1 just normalize u and v to obtain U D ,VD . 1 0 ˇ ˇ ˇ 7 6 ˇˇ ˇ D . C 1/2 C 36. The augmented matrix of .A . 1 C 6i /I /x D 0 10.6.40. ˇ 12 5 ˇ 2 3 2 3 :: :: 1 i 6 6i 6 : 0 5 1 : 0 5 2 is 4 , which is row equivalent to 4 . Therefore,x1 D :: :: 12 6 6i : 0 0 0 : 0 1 i 1 i 1 1 x2 . Taking x2 D 2 yields the eigenvector x D , so u D and v D . The 2 2 0 2 2 quadratic 4k C equation is k 1 D 0, with positive root k :2361. Routine calculations yield :5257 :8507 U ,V . :8507 :5257 2
10.7
10i
5
:: : :: :
VARIATION OF PARAMETERS FOR NONHOMOGENEOUS LINEAR SYSTEMS
1 3e t e 2t 2e t et 0 1 ; u D Y f D e t 2e 2t e 2t 3e 2t 5 4t 3t 3t 5e 13e C 3e ; yp D Y u D . 6e t e 3t 11e 3t e 2t e t e 2t e 2t 0 1 10.7.4. Y D ; u D Y f D 2e 2t e t 2e t et t 2t t 2e e 5 3e uD ; yp D Y u D . e 2t 4e t 5e t 6 sin t cos t sin t cos t 0 1 10.7.6. Y D ; u D Y f D cos t sin t cos t sin t t sin t t uD ; yp D Y u D . t cos t 0 2 3 2 e 3t e 2t e t 3e 3t 1 3t t 5 0 1 4 4 10.7.8. Y D e 0 3e ;u D Y f D 4e 2t 6 3t 2t t e e 7e et 2 3 2 3 3t 2t 2t 3t 3e 9e 9e 2e 14 1 4 8e t C 4e 2t 5; u D 16e t 4e 2t 5; yp D Y u D 6 12 e 2t e t e 2t 2e t 10.7.2. Y D e 2t uD 2e 5t
50e 37 10e 3t
2 2e t
1 t
20e 4t 2e 10e 5t C 6e
D
2e 2t 2e 2t
D
t cos t C sin t t sin t cos t
2e 4e t
t
2t t
;
;
;
32 3 3e 3t 1 2t 2e 2t 5 4 e t 5 D 0 et et 2 3 t 3e C 4 14 6e t 4 5. 6 10 2 3 2 32 t 3 et e t te t 9e t 0 9e t e 1 4 e t 3e t te t 5; u0 D Y 1 f D 3e t .3 2t/ 6te t 9e t 5 4 e t 5 D 10.7.10. Y D 4 e t 18 et e t te t 6e t 6e t 0 et 2 3 2 3 2 3 0 0 2e t 1 1 1 4 e 2t .3 2t/ 5; u D 4 e 2t .2 t/ 5; yp D Y u D 4 e t 5. 3 3 3 2e 2t e 2t 2e t 6e 6e
3t
D
Section 10.7 Variation of Parameters for Nonhomogeneous Linear Systems
219
1 1 e t .t C 1/ 1 e t e t t e t .t C 2/ 10.7.12. u0 D Y 1 f D D ; u D t t 2 t e t e .1 t/ e t .2 t/ 2t e 2 t 22 t t t e e e .t C 2/ t D . yp D Y u D e t e t .2 t/ 2t 2 et 2t 1 1 2e 2t e 3t 1 2 e t e 3e 2t C e 3t 10.7.14. u0 D Y 1 f D D ; u D t 2t 2t 3t e 2 e 2e e e 3t 3e 2t 3 9 3 2t 2t 3t 3t 1 2 e t 1 3e C e 5e e D . yp D Y u D 2 e 3t 3e 2t 9 et 9 e 3t 5e 2t 2 1 1 2e t C t 2 t e t t 1 t e t 0 1 D ;u D 10.7.16. u D Y f D 2 et t t2 1 t et t 2 2e t t 1 2 t t 2 t 3 1 1 te .t C 2/ C t t e 2e C t 2 yp D Y u D D . t 2 t t 3 e t t 2e te .t 2/ C t C 2 2 2 2 32 t 3 2 3 2 e 5t e 4t e 3t e e 4t e 4t 1 1 1 0 1 2t t t 5 4 5 4 5 4 4 10.7.18. u D Y f D 2e e 1 0 D 2e ;uD 8e t 3 3 12 2t t t e 2e 1 0 e 4e t 2 5t 32 3 2 3 2t 4t t e e 0 e 3e 1 4 4t 1 e 0 e t 5 4 8e t 5 D 4 1 5. yp D Y u D 12 4 e 3t 1 1 4e t e t 10.7.20. u0 D Y
1
f D
2
1 4 2t
2 t e 1 4 et yp D Y u D Y D 4t et
;
;
;
3
5;
e 2t 1 2
e t e t e t
32 t 3 2 3 2 3 te t te t C e 2t e 1 2t 1 te t te t 1 5 4 0 5 D 4 e 2t 5; u D 4 e 2t 5; 4 t 0 2 e 0 0 32 3 2 3 t 2t 2t C 1 et 4 2t 1 5. e t 5 4 e 2t 5 D 4t 0 0 2t C 1
10.7.22. (c) If yp D Y u, then y0p D Y 0 u C Y u0 D AY u C Y u0 , so (E) y0p D Ayp C Y u0 . However, from the derivation of the method of variation of parameters in Section 9.4, Y u0 D f as defined in the solution of (a). This and (E) imply the conclusion. (d) Since Y u0 D f with f as defined in the solution of (a), u1 ; u2 ; : : : ; un satisfy the conditions required in the derivation of the method of variation of parameters in Section 9.4; hence, yp D c1y1 C c2 y2 C C cn yn is a particular solution of (A).
CHAPTER 11 Boundary Value Problems and Fourier Expansions 11.1
EIGENVALUE PROBLEMS FOR y 00 C y D 0
11.1.2. From Theorem 11.1.2 with L D , n D n2 , yn D sin nx, n D 1; 2; 3; : : : 1/2
.2n
11.1.4. From Theorem 11.1.4 with L D , n D
4
, yn D sin
.2n
1/x 2
, n D 1; 2; 3; : : : ;
11.1.6. From Theorem 11.1.6 with L D , 0 D 0, y0 D 1, n D n2 , y1n D cos nx, y2n D sin nx, n D 1; 2; 3; : : : 11.1.8. From Theorem 11.1.5 with L D 1, n D
.2n
.2n 1/ x 1/2 2 , yn D cos , n D 1; 2; 3; : : : 4 2
11.1.10. From Theorem 11.1.6 with L D 1, 0 D 0, y0 D 1, n D n2 2 , y1n D cos n x, y2n D sin n x, n D 1; 2; 3; : : : 11.1.12. From Theorem 11.1.6 with L D 2, 0 D 0, y0 D 1, n D n x sin , n D 1; 2; 3; : : : 2 11.1.14. From Theorem 11.1.5 with L D 3, n D 11.1.16. From Theorem 11.1.3 with L D 5, n D
.2n
n x n2 2 , y1n D cos , y2n D 4 2
1/2 2 .2n 1/ x , yn D cos , n D 1; 2; 3; : : : 36 6
n2 2 n x , yn D cos , n D 1; 2; 3; : : : 25 5
11.1.18. From Theorem 11.1.1, any eigenvalues of Problemp11.1.4 must beppositive. If > 0, then every solution of y 00 C y D cos x C c2 sin x where c1 and c2 are p 0 is of thepform y D c1p 0 constants. Therefore, y D . c sin x C c cos x/. Sincep y 0 .0/ D 0, c2 D 0. Therefore, 1 2 p p y D c1 cos x. Since y.L/ D 0, c1 cos L D 0. To make c1 cos L D 0 with c1 ¤ 0 we must p .2n 1/ .2n 1/2 2 choose D , where n is a positive integer. Therefore,n D is an eigenvalue 2L 4L2 .2n 1/ x and yn D cos is an associated eigenfunction. 2L
221
222 Chapter 11 Boundary Value Problems and Fourier Expansions ˇ rx L r x ˇˇL dx D sin D 0, so y0 D 1 is orthogonal 11.1.20. If r is a positive integer, then cos L r L ˇ L L Z L m x n x to all the other eigenfunctions. If m and n are distinct positive integers, then cos cos dx D L L 0 Z L 1 m x n x cos cos dx D 0, from Example 11.1.4. 2 L L L Z
L
11.1.22. Let m and n be distinct positive integers. From the identity cos A cos B D
cos.A C B/ with A D .2m Z
L
cos 0
.2m
1/ x=2L and B D .2n
1/ x .2n 1/ x 1 cos dx D 2L 2L 2
Z
L 0
1/ x=2L,
.m cos
1 Œcos.A 2
n/ x .m C n 1/ x C cos L L
11.1.24. If y D c1 C c2 x, then y 0 .0/ D 0 implies that c2 D 0, so y D c1 . Now Z L c1 dx D c1 L D 0 only if c1 D 0. Therefore,zero is not an eigenvalue.
Z
B/ C
dx D 0:
L 0
y.x/ dx D
0
If y D c1 cosh kx C c2 sinh kx, then y 0 .0/ D 0 implies that c2 D 0, so y D c1 cosh kx. Now Z L Z L sinh kL y.x/ dx D c1 cosh kx dx D c1 D 0 with k > 0 only if c1 D 0. Therefore, there are no k 0 0 negative eigenvalues. If y D c1 cos kx C c2 sin kx, then y 0 .0/ D 0 implies that c2 D 0, so y D c1 cos kx. Now Z L Z L sin kL n y.x/ dx D c1 cos kx dx D c2 D 0 if k D , where n is a positive integer. Therefore, k L 0 0 n x n2 2 n D and yn D cos , n D 1; 2; 3; : : : . 2 L L Z L 11.1.26. If y D c1 C c2 .x L/, then y 0 .L/ D 0 implies that c2 D 0, so y D c1 . Now y.x/ dx D 0 Z L c1 dx D c1 L D 0 only if c1 D 0. Therefore,zero is not an eigenvalue. 0
If y D c1 cosh k.x L/ C c2 sinh k.x L/, then y 0 .L/ D 0 implies that c2 D 0, so y D c1 cosh k.x Z L Z L sinh kL L/. Now y.x/ dx D c1 cosh k.x L/ dx D c1 D 0 with k > 0 only if c1 D 0. k 0 0 Therefore,there are no negative eigenvalues. If y D c1 cos k.x L/ C c2 sin k.x L/, then y 0 .L/ D 0 implies that c2 D 0, so y D c1 cos k.x L/. Z L Z L sin kL n Now y.x/ dx D c1 cos k.x L/ dx D c2 D 0 if k D , where n is a positive integer. k L 0 0 n2 2 n.x L/ n x Therefore, n D and yn D cos , or, equivalently, yn D cos , n D 1; 2; 3; : : : . 2 L L L
Section 11.2 Fourier Expansions I 11.2 FOURIER EXPANSIONS I 11.2.2. 1 2 Z
D
a0
D
an
Z
D
bn
1
.2 1
x/ dx D
1
.2 1 1
.2 1 "
Z
1
2 dx D 2I
0
D
x/ cos n x dx D 4
0
Z
ˇ1 ˇ 4 sin n x ˇˇ D 0I cos n x dx D n 0
1 0
x sin n x dx #
1 2 X . 1/n sin n x. From Theorem 11.2.4, nD1 n
8
0, if and only if X.0/ D X 0 .L/ D 0. Since we are interested in nontrivial solutions, X must be a nontrivial solution of (B) X 00 C X D 0; X.0/ D 0; X 0 .L/ D 0. From Theorem 11.1.4, n D .2n 1/2 2=4L2 is an .2n 1/ x eigenvalue of (B) with associated eigenfunction Xn D sin , n D 1; 2; 3; : : : . Substituting 2L 2 2 2 0 2 2 2 2 D .2n 1/ =4L into (A) yields T D ..2n 1/ a =4L /T , which has the solution Tn D 2 2 2 2 e .2n 1/ a t =4L . .2n 1/ x 2 2 2 2 , n D 1; 2; 3; : : : We have now shown that the functions un .x; t/ D e .2n 1/ a t =4L sin 2L 2 satisfy ut D a uxx and the boundary conditions u.0; t/ D ux .L; t/ D 0; t > 0. Any finite sum m X .2n 1/ x 2 2 2 2 dn e .2n 1/ a t =4L sin also has these properties. Therefore,it is plausible to expect 2L nD1 1 X .2n 1/ x 2 2 2 2 that that this is also true of the infinite series (C) u.x; t/ D dn e .2n 1/ a t =4L sin 2L nD1 1 X .2n 1/ x under suitable conditions on the coefficients fdn g. Since u.x; 0/ D dn sin , if fdn g are 2L nD1 the mixed Fourier sine coefficients of f on Œ0; L, then u.x; 0/ D f .x/ at all points x in Œ0; L where the mixed Fourier sine series converges to f .x/. In this case (C) is a formal solution of the initial-boundary value problem of Definition 12.1.3. 12.1.8. Since f .0/ D f .1/ D 0 and f 00 .x/ D 2, Theorem 11.3.5(b) implies that ˛n
D D
S.x/ D
ˇ1 Z 1 ˇ 4 4 ˇ D sin n x dx D cos n x ˇ n2 2 0 n3 3 0 8 8 < ; if n D 2m 1; .2m 1/3 3 : 0; if n D 2mI
4 .cos n n3 2
1 8 X 1 .2n 1/ x sin . From Definition 12.1.1, 3 3 nD1 .2n 1/ L
u.x; t/ D
1 8 X 1 e 3 nD1 .2n 1/3
239
.2n 1/2 2 t
sin.2n
1/ x:
1/
240 Chapter 12 Fourier Solutions of Partial Differential 12.1.10. ˛1
2 2
D D
if n 2, then ˛n
D D D D
S.x/ D
ˇ Z Z 1 x 2 ˇˇ 1 x sin x dx D x.1 cos 2x/ dx D x cos 2x dx 0 2 ˇ0 0 0 ˇ ˇ Z ˇ 1 sin 2x ˇˇ x sin 2x ˇˇ cos 2x dx D C D I ˇ 2 2 4 0 2 0 0
Z
2
Z Z 2 1 x sin x sin nx dx D xŒcos.n 1/x cos.n C 1/x dx 0 0 ˇ Z 1 sin.n 1/x sin.n C 1/x ˇˇ sin.n 1/x sin.n C 1/x x dx ˇ n 1 nC1 n 1 nC1 0 0 ˇ 1 cos.n 1/x cos.n C 1/x ˇˇ 1 1 1 D . 1/nC1 1 ˇ 2 2 2 2 .n 1/ .n C 1/ .n 1/ .n C 1/ 8 0 0 if n D 2m 1; < 4n nC1 16m . 1/ 1 D if n D 2mI : .n2 1/2 .4m2 1/
sin x 2
1 16 X n sin 2nx. From Definition 12.1.1, nD1 .4n2 1/2
u.x; t/ D
e 2
3t
sin x
1 16 X n e nD1 .4n2 1/2
12n2 t
sin 2nx:
12.1.12. Since f .0/ D f .L/ D 0 and f 00 .x/ D 6x, Theorem 11.3.5(b) implies that " # ˇ Z 3 Z 3 36 n x 108 n x ˇˇ3 n x ˛n D x sin dx D x cos cos dx n2 2 0 3 n3 3 3 ˇ0 3 0 ˇ 108 n x ˇˇ3 324 108 D . 1/nC1 3 3 C 4 4 sin D . 1/nC1 3 3 I n n 3 ˇ0 n S.x/ D
1 324 X . 1/n n x sin . From Definition 12.1.1, u.x; t/ D 3 3 nD1 n 3
1 324 X . 1/n e 3 nD1 n3
4n2 2 t =9
sin
n x . 3
12.1.14. Since f .0/ D f .1/ D f 00 .0/ D f 00 .L/ D 0 and f .4/ D 360x, Theorem 11.3.5(b) and Exercise 35(b) of Section 11.3 imply that " # ˇ1 Z 1 Z 1 ˇ 720 720 ˛n D x sin n x dx D x cos n x ˇˇ cos n x dx n4 4 0 n5 5 0 0 ˇ 720 n x ˇˇ1 720 nC1 720 D . 1/ C 6 6 sin D . 1/nC1 5 5 I n5 5 n L ˇ0 n S.x/ D
1 720 X . 1/n sin n x. From Definition 12.1.1, u.x; t/ D 5 n5 nD1
1 720 X . 1/n e 5 n5 nD1
7n2 2 t
sin n x.
Section 12.1 The Heat Equation
241
12.1.16. Since f .0/ D f .1/ D f 00 .0/ D f 00 .L/ D 0 and f .4/ D 120.3x 1/, Theorem 11.3.5(b) and Exercise 35(b) of Section 11.3 imply that " # ˇ1 Z 1 Z 1 ˇ 240 240 ˛n D .3x 1/ sin n x dx D .3x 1/ cos n x ˇˇ 3 cos n x dx n4 4 0 n5 5 0 0 ˇ1 ˇ 240 720 240 n D Œ. 1/ 2 C 1 C 6 6 sin n x ˇˇ D Œ1 C . 1/n 2 I 5 5 n5 5 n n 0 1 240 X 1 C . 1/n 2 sin n x. From Definition 12.1.1, 5 nD1 n5
S.x/ D
1 240 X 1 C . 1/n 2 e 5 nD1 n5
u.x; t/ D
12.1.18. ˛0 D ˛n
Z
2
.x 2 0
4x/ dx D
1 L
x3 3
2x 2
ˇ2 ˇ ˇ D ˇ 0
"
sin n x:
8 ; if n 1, 3
ˇ n x 2 n x ˇˇ2 2 2 .x 4x/ cos dx D .x 4x/ sin 2 n 2 ˇ0 0 " # ˇ Z 2 n x ˇˇ2 n x 16 8 .x 2/ cos cos dx D 2 2 ˇ 2 2 n 2 0 2 n 0
Z
D D
C.x/ D
1 2
2n2 2 t
2
1 8 16 X 1 n x C 2 cos . From Definition 12.1.3, u.x; t/ D 2 3 nD1 n 2
# n x 2 .x 2/ sin dx 2 0 ˇ 32 n x ˇˇ2 16 sin D 2 2I ˇ 3 3 n 2 0 n Z
2
1 8 16 X 1 C 2 e 3 nD1 n2
n2 2 t
cos
n x . 2
1 1 .2n 1/ x 384 X 12.1.20. From Example 11.3.5, C.x/ D 4 cos . From Definition 12.1.3, 4 nD1 .2n 1/4 2 1 1 .2n 1/ x 384 X 2 2 u.x; t/ D 4 e 3.2n 1/ t =4 cos . 4 nD1 .2n 1/4 2
ˇ1 Z ˇ 1 1 1 3x 5 2 4 3 4 ˇ 12.1.22. ˛0 D .3x 4Lx / dx D x D . Since f 0 .0/ D f 0 .1/ D 0 and ˇ L 0 L 5 5 0 f 000 .x/ D 24.3x 1/, Theorem 11.3.5(a) implies that " # ˇ1 Z 1 Z 1 ˇ 48 48 ˇ ˛n D .3x 1/ sin n x dx D .3x 1/ cos n x ˇ 3 cos n x dx n3 3 0 n4 4 0 0 ˇ1 ˇ 48 144 48 n ˇ D D Œ. 1/ 2 C 1 C sin n x Œ1 C . 1/n 2 ; n 1I ˇ 4 4 5 5 4 4 n n n 0
C.x/ D
2 5
1 48 X 1 C . 1/n 2 cos n x. From Definition 12.1.3, 4 nD1 n4
u.x; t/ D
2 5
1 48 X 1 C . 1/n 2 e 4 n4 nD1
3n2 2 t
cos n x:
242 Chapter 12 Fourier Solutions of Partial Differential Z 1 4 1 x5 x4 2x3 3 2 2 12.1.24. ˛0 D .x 2 x C x / dx D C 0 5 2 3 0 000 f . / D 0 and f .x/ D 12.2x /, Theorem 11.3.5(a) implies that ˇ Z ˇ 24 24 ˇ .2x / sin nx dx D .2x / cos nx ˛n D ˇ 3 4 n 0 n 0 ˇ ˇ 24 48 24 n n D Œ. 1/ C C 5 sin nx ˇˇ D Œ1 C . 1/ n4 n n4 0 ( 0 if n D 2m 1; 3 D n 1I if n D 2m; m4 C.x/ D
4 30
3
1 X 1 4 cos 2nx. From Definition 12.1.3, u.x; t/ D n4 30 nD1
3
ˇ 4 ˇ ˇ D . Since f 0 .0/ D ˇ 30 0 2
Z
cos nx dx
0
1 X 1 e n4 nD1
4n2 t
cos 2nx.
12.1.26. 2
D D D D SM .x/ D 8
.2n 1/x x 2/ sin dx 2 0 ˇ Z .2n 1/x ˇˇ .2n 1/x 4 2 . x x / cos . 2x/ cos dx ˇ .2n 1/ 2 2 0 0 ˇ Z 8 .2n 1/x ˇˇ .2n 1/x . 2x/ sin C 2 sin dx ˇ .2n 1/2 2 2 0 0 ˇ ˇ 8 32 .2n 1/x ˇ . 1/n cos ˇ .2n 1/2 .2n 1/3 2 0 8 32 n . 1/ C I .2n 1/2 .2n 1/3
D
˛n
1 X
nD1
Z
. x
1 .2n
1/2
u.x; t/ D 8
1 X
nD1
4
n
. 1/ C
.2n
1 .2n
1/2
1/
sin
.2n
12.1.28. Since f .0/ D f 0 .1/ D 0, and f 00 .x/ D 6.1 ˛n
D D D D
1/x 2
4
n
. 1/ C
.2n
1/
e
. From Definition 12.1.4,
3.2n 1/2 t =4
sin
.2n
1/x 2
:
2x/, Theorem 11.3.5(d) implies that
Z 1 48 .2n 1/ x .1 2x/ sin dx 2 2 .2n 1/ 0 2 " # ˇ Z 1 96 .2n 1/ x ˇˇ1 .2n 1/ x .1 2x/ cos cos dx ˇ C2 .2n 1/3 3 2 2 0 0 " ˇ # 96 4 .2n 1/ x ˇˇ1 1C sin ˇ .2n 1/3 3 .2n 1/ 2 0 96 4 1 C . 1/n I .2n 1/3 3 .2n 1/
Section 12.1 The Heat Equation
SM .x/ D
243
1 96 X 1 4 .2n 1/ x n 1 C . 1/ sin . From Definition 12.1.4, 3 3 .2n 1/ .2n 1/ 2 nD1
u.x; t/ D
1 96 X 1 4 n 1 C . 1/ e 3 nD1 .2n 1/3 .2n 1/
.2n 1/2 2 t
sin
.2n
1/ x : 2
12.1.30. Since f .0/ D f 0 .L/ D f 00 .0/ D 0 and f 000 .x/ D 6, Theorem 11.3.5(d) and Exercise 11.3.50(b) imply that Z 1 96 .2n 1/ x cos dx ˛n D .2n 1/3 3 0 2 ˇ1 192 .2n 1/ x ˇˇ 192 D sin D . 1/n I ˇ 4 4 .2n 1/ 2 .2n 1/4 4 0
SM .x/ D
1 X
. 1/n .2n 1/ x 192 sin . From Definition 12.1.4, 4 nD1 .2n 1/4 2 u.x; t/ D
1
192 X . 1/n e 4 nD1 .2n 1/4
.2n 1/2 2 t
sin
.2n
1/ x : 2
12.1.32. Since f .0/ D f 0 .1/ D f 00 .0/ D 0 and f 000 .x/ D 12.2x 1/, Theorem 11.3.5(d) and Exercise 11.3.50(b) imply that Z 1 192 .2n 1/ x ˛n D .2x 1/ cos dx 3 3 .2n 1/ 0 2 " # ˇ Z 1 384 .2n 1/ x ˇˇ1 .2n 1/ x D .2x 1/ sin 2 sin dx ˇ .2n 1/4 4 2 2 0 0 " ˇ # 4 .2n 1/ x ˇˇ1 384 nC1 D . 1/ 1C cos ˇ .2n 1/4 4 .2n 1/ 2 0 384 384 4 4 nC1 n D . 1/ 1 D . 1/ C I .2n 1/4 4 .2n 1/ .2n 1/4 4 .2n 1/ 1 384 X 1 4 .2n 1/ x n SM .x/ D 4 . 1/ C sin . From Definition 12.1.4, 4 .2n 1/ .2n 1/ 2 nD1
1 384 X 1 4 n u.x; t/ D 4 . 1/ C e .2n 1/4 .2n 1/
.2n 1/2 2 t
sin
.2n
nD1
12.1.36. From Example 11.3.3, CM .x/ D u.x; t/ D
1 8 X 1 .2n 1/ x cos . From Definition 12.1.5, 2 2 .2n 1/ 2 nD1
1 8 X 1 e 2 .2n 1/2 nD1
1/ x : 2
3.2n 1/2 2 t =4
cos
.2n
1/ x : 2
244 Chapter 12 Fourier Solutions of Partial Differential 12.1.38. Since f 0 .0/ D f . / D 0 and f 00 .x/ D 2, Theorem 11.3.5(c) implies that ˇ Z 16 .2n 1/x 32 .2n 1/x ˇˇ 32 cos dx D sin D . 1/nC1 I ˛n D ˇ 2 3 .2n 1/ 0 2 .2n 1/ 2 .2n 1/3 0 CM .x/ D
1 32 X . 1/n .2n 1/x cos . From Definition 12.1.5, nD1 .2n 1/3 2
u.x; t/ D
1 32 X . 1/n e nD1 .2n 1/3
7.2n 1/2 t =4
cos
.2n
1/x 2
:
12.1.40. Since f 0 .0/ D f .1/ D 0 and f 00 .x/ D 6.2x C 1/, Theorem 11.3.5(c) implies that Z 1 48L .2n 1/ x ˛n D .2x C 1/ cos dx .2n 1/2 2 0 2 " # ˇ Z 1 .2n 1/ x ˇˇ1 .2n 1/ x 96 D .2x C 1/ sin 2 sin dx ˇ .2n 1/3 3 2 2 0 0 " ˇ # 4 .2n 1/ x ˇˇ1 96 nC1 D . 1/ 3 cos ˇ .2n 1/3 3 .2n 1/ 2 0 96 4 D . 1/n 3 C I .2n 1/3 3 .2n 1/ 1 1 4 .2n 1/ x 96 X n CM .x/ D 3 . 1/ 3 C cos . From Definition 12.1.5, .2n 1/3 .2n 1/ 2 nD1
1 96 X 1 4 n u.x; t/ D 3 . 1/ 3 C e .2n 1/3 .2n 1/
.2n 1/2 2 t =4
cos
.2n
nD1
1/ x : 2
12.1.42. Theorem 11.3.5(c) and Exercise 11.3.42(b) imply that Since f 0 .0/ D f .1/ D f 00 .1/ D 0 and f 000 .x/ D 12.2x 1/, Z 1 192 .2n 1/ x ˛n D .2x 1/ sin dx 3 3 .2n 1/ 0 2 " # ˇ Z 1 384 .2n 1/ x ˇˇ1 .2n 1/ x D .2x 1/ cos 2 cos dx ˇ .2n 1/4 4 2 2 0 0 " ˇ # 384 4 .2n 1/ x ˇˇ1 384 . 1/n 4 D 1 sin ˇ D .2n 1/4 4 1 C .2n 1/ I .2n 1/4 4 .2n 1/ 2 0 CM .x/ D
1 384 X 1 . 1/n 4 .2n 1/ x 1C cos . From Definition 12.1.5, 4 4 .2n 1/ .2n 1/ 2 nD1
u.x; t/ D
1 384 X 1 . 1/n 4 1 C e 4 .2n 1/4 .2n 1/ nD1
.2n 1/2 2 t =4
cos
.2n
1/ x : 2
Section 12.1 The Heat Equation ˇ Z 2 L=2 n x 2 n x ˇˇL=2 2 h sin dx D cos D 1 12.1.44. ˛n D L 0 L n L ˇ0 n 1 2 X1h n i n x S.x/ D 1 cos sin . From Definition 12.1.1, nD1 n 2 L 1 2 X1h 1 nD1 n
u.x; t/ D
cos
n i e 2
n2 2 t 2 =L2
cos
sin
245
n i ; 2
n x : L
12.1.46. ˛n
ˇ .2n 1/ x 4 .2n 1/ x ˇˇL=2 sin dx D cos ˇ 2L .2n 1/ 2L 0 0 4 .2n 1/ / 1 cos I .2n 1/ 4
D SM .x/ D
Z
2 L
D
1 4 X 1 1 nD1 2n 1
u.x; t/ D
L=2
cos
.2n
1/ / 4
1 4 X 1 1 2n 1
cos
nD1
sin
.2n
.2n
1/ / 4
1/ x . From Definition 12.1.4, 2L
e
.2n 1/2 2 a 2 t =4L2
sin
.2n
1/ x : 2L
12.1.48. Let u.x; t/ D v.x; t/ C q.x/; then ut D vt and uxx D vxx C q 00 , so vt D 9vxx C 9q 00 54x; 0 < x < 4; t > 0; v.0; t/ D 1 q.0/; v.4; t/ D 61 q.4/; t > 0; v.x; 0/ D 2 x C x 3 q.x/; 0 x 4:
.A/
We want q 00 .x/ D 6x; q.0/ D 1; q.4/ D 61; q.x/ D x 3 C a1 C a2 x; q.0/ D 1 ) a1 D 1; q.x/ D x 3 C 1 C a2 x; q.4/ D 61 ) a2 D 1; q.x/ D x 3 C 1 x. Now (A) reduces to vt D 9vxx ; 0 < x < 4; t > 0; v.0; t/ D 0; v.4; t/ D 0; t > 0; v.x; 0/ D 1; 0 x 4; which we solve by separation of variables. ˛n
D D
S.x/ D
ˇ n x 2 n x ˇˇ4 sin dx D cos 4 n 4 ˇ0 0 8 4 < 2 if n D 2m 1; Œ1 . 1/n D .2m 1/ : n 0 if n D 2mI
1 2
Z
L
1 4 X 1 .2n 1/ x sin . From Definition 12.1.1, nD1 .2n 1/ 4
v.x; t/ D
1 4 X 1 e .2n 1/ nD1
9 2 n2 t =16
sin
.2n
1/ x : 4
246 Chapter 12 Fourier Solutions of Partial Differential Therefore, x C x3 C
u.x; t/ D 1
2 2 1 4 X e 9 n t =16 .2n 1/ x sin : nD1 .2n 1/ 4
12.1.50. Let u.x; t/ D v.x; t/ C q.x/; then ut D vt and uxx D vxx C q 00 , so
vt D 3uxx C 3q 00 18x; 0 < x < 1; t > 0; vx .0; t/ D 1 q 0 .0/; v.1; t/ D 1 q.1/; t > 0; v.x; 0/ D x 3 2x q.x/; 0 x 1:
.A/
We want q 00 .x/ D 6x; q 0 .0/ D 1; q.1/ D 1; q 0 .x/ D 3x 2 C a2 ; q 0 .0/ D 1 ) a2 D 1; q 0 .x/ D 3x 2 1; q.x/ D x 3 x C a1 x; q.1/ D 1 ) a1 D 1; q.x/ D x 3 x 1. Now (A) reduces to vt D 3uxx ; 0 < x < 1; t > 0; vx .0; t/ D 0; v.1; t/ D 0; t > 0; v.x; 0/ D 1 x; 0 x 1: From Example 11.3.3, CM .x/ D 1 1 8 X e 2 nD1 .2n 1/2
nD1
1/2 2 t =4
3.2n
u.x; t/ D
1 8 X 1 .2n 1/ x cos . From Definition 12.1.5, v.x; t/ D 2 2 .2n 1/ 2
1
cos
.2n
x C x3 C
1/ x . Therefore, 2
1 8 X 1 e 2 nD1 .2n 1/2
3.2n 1/2 2 t =4
.2n
cos
1/ x : 2
12.1.52. Let u.x; t/ D v.x; t/ C q.x/; then ut D vt and uxx D vxx C q 00 , so
vt D vxx C q 00 C 2 sin x; 0 < x < 1; t > 0; v.0; t/ D q.0/; vx .1; t/ D q 0 .1/; t > 0; v.x; 0/ D 2 sin x q.x/; 0 x 1:
.A/
We want q 00 .x/ D 2 sin x; q.0/ D 0; q 0 .1/ D ; q 0 .x/ D cos x Ca2; q 0 .1/ D ) a2 D 0; q 0 .x/ D cos x; q.x/ D sin x C a1 ; q.0/ D 0 ) a1 D 0; q.x/ D sin x. Now (A) reduces to vt D vxx ; 0 < x < 1; t > 0; v.0; t/ D 0; vx .1; t/ D 0; t > 0; v.x; 0/ D sin x; 0 x 1: ˛n
D 2
Z
1
sin x sin
.2n
0
1/ x dx D 2
Z
1 0
2
ˇ sin.2n 3/ x=2 sin.2n C 1/ x=2 ˇˇ1 ˇ .2n 3/ .2n C 1/ 0 1 1 1 n 2 n8 D . 1/ D . 1/ 2n 3 2n C 1 .2n C 1/.2n D
SM .x/ D
2
1 8 X . 1/n nD1 .2n C 1/.2n
v.x; t/ D
3/
sin
.2n
3/
I
1/ x . From Definition 12.1.4, 2
1 8 X . 1/n .2n C 1/.2n nD1
cos.2n C 1/ x 2
3/ x
cos.2n
3/
e
.2n 1/2 2 t =4
sin
.2n
1/ x : 2
dx
Section 12.2 The Wave Equation 1 8 X . 1/n .2n C 1/.2n
Therefore, u.x; t/ D sin x C
nD1
3/
e
.2n 1/2 2 t =4
sin
.2n
247
1/ x . 2
12.1.54. (a) Since f is piecewise smooth of Œ0; ˇL, there is a constant K such that jf .x/j K, 0 x ˇZ Z ˇ 2 ˇˇ L n x 2 L 2 2 2 ˇ L. Therefore, j˛n j D ˇ f .x/ sin dx ˇ jf .x/j dx D 2K. Hence, j˛n e n a t =L j ˇ L 0 Lˇ 0 L 2
2 2
2Ke n a t =L , so u.x; t/ converges for all x if t > 0, by the comparison test. n x 2 2 2 (b) Let t be a fixed positive number. Apply Theorem 12.1.2 with ´ D x and wn .x/ D ˛n e n t =L sin . L n x 2K 2 2 2 2 2 2 , so jwn0 .x/j ne n t =L , 1 < x < 1. Since Then wn0 .x/ D n˛n e n t =L cos L L L 1 X n2 2 a 2 t =L2 ne converges if t > 0, Theorem 12.1.1 (with ´1 D x1 and ´2 D x2 arbitrary) implies
nD1
the concclusion. 1 X (c) Since n2 e
n2 2 a 2 t =L2
nD1 n2 2 t =L2
also converges if t > 0, an argument like that in (b) with wn .x/ D
n x yields the conclusion. L (d) Let x be arbitrary, but fixed. Apply Theorem 12.1.2 with ´ D t and n x 2a2 2 n x 2 2 2 2 2 2 2 2 wn .t/ D ˛n e n a t =L sin . Then wn0 .t/ D n ˛n e n a t =L sin , so jwn0 .t/j L L2 L 1 X 2K 2 a2 2 n2 2 a 2t0 =L2 2 2 2 2 n e if t > t0 . Since n2 e n a t0 =L converges, Theorem 12.1.1 (with ´1 D L nD1 t0 > 0 and ´2 D t1 arbitrary implies the conclusion for t t0 . However, since t0 is an arbitrary positive number, this holds for t > 0.
n˛n e
cos
12.2 THE WAVE EQUATION hR R1 1=2 12.2.1. ˇn D 2 0 x sin n x C 1=2 .1 Z
1=2
x sin n x dx 0
D D
Z
1=2
.1 0
x/ sin n x dx
i x/ sin n x dx ;
" ˇ1=2 ˇ 1 x cos n x ˇˇ n 0
Z
1=2
cos n x dx 0
ˇ1=2 ˇ n 1 1 cos C 2 2 sin n x ˇˇ D 2n 2 n 0 D D
1 n
"
.1
1 n cos 2n 2
ˇ1 ˇ x/ cos n x ˇˇ
1=2
C
Z
#
1 n 1 n cos C 2 2 sin I 2n 2 n 2
1=2
cos n x dx 0
#
ˇ1 ˇ 1 1 n 1 n ˇ sin n x D cos C 2 2 sin I ˇ 2 2 n 2n 2 n 2 1=2
8 4 < 4 n . 1/mC1 bn D 2 2 sin D .2m 1/2 2 : n 2 0
if n D 2m
if n D 2mI
1
248 Chapter 12 Fourier Solutions of Partial Differential Equations
Sg .x/ D
1 4 X . 1/nC1 sin.2n 2 .2n 1/2
1/ x. From Definition 12.1.1,
nD1
1 4 X . 1/nC1 u.x; t/ D sin 3.2n 3 3 nD1 .2n 1/3
1/ t sin.2n
1/ x:
12.2.2. Since f .0/ D f .1/ D 0 and f 00 .x/ D 2, Theorem 11.3.5(b) implies that ˇ1 Z 1 ˇ 4 4 4 sin n x dx D cos n x ˇˇ D .cos n ˛n D 3 2 n2 2 0 n3 3 n 0 8 8 < ; if n D 2m 1; D .2m 1/3 3 : 0; if n D 2mI
Sf .x/ D
1 1 8 X sin.2n 3 .2n 1/3
1/ x. From Definition 12.1.1,
nD1
u.x; t/ D
1 8 X 1 cos 3.2n 3 nD1 .2n 1/3
1/ t sin.2n
1/ x:
12.2.4. Since g.0/ D g.1/ D 0 and g00 .x/ D 2, Theorem 11.3.5(b) implies that ˇ1 Z 1 ˇ 4 4 4 ˇn D sin n x dx D cos n x ˇˇ D .cos n 2 2 3 3 3 n 0 n n 2 0 8 8 < ; if n D 2m 1; D .2m 1/3 3 : 0; if n D 2mI
Sg .x/ D
1/
1 8 X 1 sin.2n 3 .2n 1/3
1/
1/ x. From Definition 12.1.1,
nD1
1 1 8 X u.x; t/ D sin 3.2n 3 4 nD1 .2n 1/4
12.2.6. From Example 11.2.6, Sf .x/ D u.x; t/ D
12.2.8. From Example 11.2.6 Sg .x/ D u.x; t/ D
1/ t sin.2n
1/ x:
1 324 X . 1/n n x sin . From Definition 12.1.1, 3 3 n 3 nD1
1 324 X . 1/n 8 nt n x cos sin : 3 3 nD1 n 3 3 1 324 X . 1/n n x sin . From Definition 12.1.1, 3 n3 3 nD1
1 81 X . 1/n 8 nt n x sin sin : 4 4 2 n 3 3 nD1
Section 12.2 The Wave Equation 12.2.10. ˛1
2 2
D D
if n 2, then D
˛n
D D D Sf .x/ D
ˇ Z Z 1 x 2 ˇˇ 1 x sin x dx D x.1 cos 2x/ dx D x cos 2x dx 0 2 ˇ0 0 0 ˇ Z ˇ ˇ 1 sin 2x ˇˇ x sin 2x ˇˇ cos 2x dx D C D I ˇ 2 2 4 0 2 0 0
Z
2
Z Z 2 1 x sin x sin nx dx D xŒcos.n 1/x cos.n C 1/x dx 0 0 ˇ Z 1 sin.n 1/x sin.n C 1/x ˇˇ sin.n 1/x sin.n C 1/x x dx ˇ n 1 nC1 n 1 nC1 0 0 ˇ ˇ 1 cos.n 1/x cos.n C 1/x ˇ 1 1 1 D . 1/nC1 1 ˇ 2 2 2 2 .n 1/ .n C 1/ .n 1/ .n C 1/ 8 0 0 if n D 2m 1; < 4n nC1 16m . 1/ 1 D if n D 2mI : .n2 1/2 .4m2 1/ 1 16 X n sin 2nx. From Definition 12.1.1, 2 nD1 .4n 1/2
sin x 2
u.x; t/ D
p cos 5 t sin x 2
1 p 16 X n cos 2n 5 t sin 2nx: 2 2 nD1 .4n 1/
12.2.12. ˇ1
2 2
D D
if n 2 then ˇn
D D D D
Sg .x/ D
Z
ˇ Z Z 1 x 2 ˇˇ 1 x sin x dx D x.1 cos 2x/ dx D x cos 2x dx 0 2 ˇ0 0 0 ˇ ˇ Z ˇ 1 sin 2x ˇˇ x sin 2x ˇˇ cos 2x dx D C D I ˇ 2 2 4 0 2 0 0
2
Z Z 1 2 x sin x sin nx dx D xŒcos.n 1/x cos.n C 1/x dx 0 0 ˇ Z 1 sin.n 1/x sin.n C 1/x ˇˇ sin.n 1/x sin.n C 1/x x dx ˇ n 1 nC1 n 1 nC1 0 0 ˇ 1 cos.n 1/x cos.n C 1/x ˇˇ 1 1 1 D . 1/nC1 1 ˇ 2 2 2 2 .n 1/ .n C 1/ .n 1/ .n C 1/ 8 0 0 if n D 2m 1; < 4n nC1 16m . 1/ 1 D if n D 2mI : .n2 1/2 .4m2 1/
sin x 2
1 16 X n sin 2nx. From Definition 12.1.1, nD1 .4n2 1/2
p u.x; t/ D p sin 5 t sin x 2 5
1 p 8 X 1 p sin 2n 5 t sin 2nx: 2 2 1/ 5 nD1 .4n
249
250 Chapter 12 Fourier Solutions of Partial Differential Equations 12.2.14. Since f .0/ D f .1/ D f 00 .0/ D f 00 .L/ D 0 and f .4/ D 360x, Theorem 11.3.5(b) and Exercise 35(b) of Section 11.3 imply that " # ˇ1 Z 1 Z 1 ˇ 720 720 ˛n D x sin n x dx D x cos n x ˇˇ cos n x dx n4 4 0 n5 5 0 0 ˇ 720 n x ˇˇ1 720 nC1 720 C 6 6 sin D . 1/nC1 5 5 I D . 1/ n5 5 n L ˇ0 n Sf .x/ D
1 720 X . 1/n sin n x. From Definition 12.1.1, u.x; t/ D 5 nD1 n5
1
720 X . 1/n cos 3n t sin n x. 5 nD1 n5
mL 1 L 12.2.16. (a) t must be in some interval of the form ŒmL=a; .m C 1/L=a. If t mC , a 2 a 1 L .m C 1/L then (i) holds with 0 L=2a. If m C t , then (ii) holds with 0 L=2a. 2 a a (b) Suppose that (i) holds. Since mL .2n 1/ a .2n 1/ a .2n 1/ a cos C D cos cos.2n 1/m D . 1/m cos ; L a L L (A) implies that u.x; t/ D . 1/m u.x; /. Suppose that (ii) holds. Since .2n 1/ a .m C 1/L cos C L a
1/ a cos.2n 1/.m C 1/ L .2n 1/ a D . 1/mC1 cos ; L D cos
.2n
(B) implies that that u.x; t/ D . 1/mC1 u.x; /. 12.2.18. Since f 0 .0/ D f .2/ D 0 and f 00 .x/ D 2, Theorem 11.3.5(c) implies that Z 2 32 .2n 1/ x ˛n D cos dx 2 2 .2n 1/ 0 4 ˇ .2n 1/ x ˇˇ4 128 128 nC1 D sin I ˇ D . 1/ .2n 1/3 3 4 .2n 1/3 3 0 CMf .x/ D
1 128 X . 1/n .2n 1/ x cos . From Exercise 12.2.17, 3 .2n 1/3 4 nD1
u.x; t/ D
1
128 X . 1/n 3.2n 1/ t .2n 1/ x cos cos : 3 .2n 1/3 4 4 nD1
12.2.20. Since g0 .0/ D g.2/ D 0 and g00 .x/ D 2, Theorem 11.3.5(c) implies that Z 2 32 .2n 1/ x ˇn D cos dx .2n 1/2 2 0 4 ˇ 128 .2n 1/ x ˇˇ4 128 D sin D . 1/nC1 I ˇ 3 3 .2n 1/ 4 .2n 1/3 3 0
Section 12.2 The Wave Equation 1 128 X . 1/n .2n 1/ x cos . From Exercise 12.2.17, 3 3 .2n 1/ 4
CMf .x/ D
nD1
u.x; t/ D
1 512 X . 1/n 3.2n 1/ t .2n 1/ x sin cos : 3 4 nD1 .2n 1/4 4 4
12.2.22. Since f 0 .0/ D f .1/ D 0 and f 00 .x/ D 6.2x C 1/, Theorem 11.3.5(c) implies that ˛n
D D D D
48L .2n 1/2 2
Z
1
1/ x dx 2 0 " # ˇ Z 1 96 .2n 1/ x ˇˇ1 .2n 1/ x .2x C 1/ sin 2 sin dx ˇ .2n 1/3 3 2 2 0 0 " ˇ1 # ˇ 4 .2n 1/ x 96 nC1 ˇ . 1/ 3 cos ˇ .2n 1/3 3 .2n 1/ 2 0 4 96 . 1/n 3 C I .2n 1/3 3 .2n 1/ .2x C 1/ cos
.2n
1 96 X 1 4 .2n 1/ x n CMf .x/ D 3 . 1/ 3 C cos : nD1 .2n 1/3 .2n 1/ 2
From Exercise 12.2.17,
1 1 4 .2n 96 X n u.x; t/ D 3 . 1/ 3 C cos nD1 .2n 1/3 .2n 1/
p 1/ 5 t .2n 1/ x cos : 2 2
12.2.24. Since g0 .0/ D g.1/ D 0 and g00 .x/ D 6.2x C 1/, Theorem 11.3.5(c) implies that ˇn
D D D D
CMg .x/ D
48L .2n 1/2 2
1
1/ x dx 2 "0 # ˇ Z 1 .2n 1/ x ˇˇ1 .2n 1/ x 96 .2x C 1/ sin 2 sin dx ˇ .2n 1/3 3 2 2 0 0 " ˇ # 96 4 .2n 1/ x ˇˇ1 nC1 . 1/ 3 cos ˇ .2n 1/3 3 .2n 1/ 2 0 96 4 . 1/n 3 C I .2n 1/3 3 .2n 1/ Z
.2x C 1/ cos
.2n
1 96 X 1 4 .2n 1/ x n . 1/ 3 C cos . From Exercise 12.2.17, 3 3 nD1 .2n 1/ .2n 1/ 2
1 192 X 1 4 .2n n u.x; t/ D p . 1/ 3 C sin .2n 1/ 4 5 nD1 .2n 1/4
p 1/ 5 t .2n 1/ x cos : 2 2
251
252 Chapter 12 Fourier Solutions of Partial Differential Equations 12.2.26. Since f 0 .0/ D f .1/ D f 00 .1/ D 0 and f 000 .x/ D 24.x cise 42(b) of Section 11.3 imply that ˛n
D D D D
CMf .x/ D
384 .2n 1/3 3
1
Z
1/ x dx 2 0 " # ˇ Z 1 768 .2n 1/ x ˇˇ1 .2n 1/ x .x 1/ cos cos dx ˇ .2n 1/4 4 2 2 0 0 " ˇ # 768 2 .2n 1/ x ˇˇ1 1 sin ˇ .2n 1/4 4 .2n 1/ 2 0 768 . 1/n 2 1C I .2n 1/4 4 .2n 1/ .x
1/ sin
.2n
1 1 . 1/n 4 .2n 1/ x 384 X 1C cos . From Exercise 12.2.17, 4 nD1 .2n 1/4 .2n 1/ 2
u.x; t/ D
1 384 X 3.2n 1/ t .2n 1/ x 1 . 1/n 4 1C cos cos : 4 4 nD1 .2n 1/ .2n 1/ 2 2
12.2.28. Since g0 .0/ D g.1/ D g00 .1/ D 0 and g000 .x/ D 24.x cise 11.2.42(b) imply that ˇn
1/, Theorem 11.3.5(c) and Exer-
D D D D
384 .2n 1/3 3
1
.x 0
1/ x dx 2 ˇ Z .2n 1/ x ˇˇ1 1/ cos ˇ 2 1/ sin
.2n
" # 1 .2n 1/ x 768 .x cos dx .2n 1/4 4 2 0 0 " ˇ # 2 .2n 1/ x ˇˇ1 768 1 sin ˇ .2n 1/4 4 .2n 1/ 2 0 768 . 1/n 2 1C I .2n 1/4 4 .2n 1/
CMg .x/ D From Exercise 12.2.17, u.x; t/ D
Z
1/, Theorem 11.3.5(c) and Exer-
1 384 X 1 . 1/n 4 .2n 1/ x 1 C cos : 4 4 nD1 .2n 1/ .2n 1/ 2
1 768 X 1 . 1/n 4 3.2n 1/ t .2n 1/ x 1 C sin cos : 5 5 3 nD1 .2n 1/ .2n 1/ 2 2
12.2.30. Since f 0 .0/ D f .1/ D f 00 .1/ D 0 and f 000 .x/ D 24.x
1/, Theorem 11.3.5(c) and Exer-
Section 12.2 The Wave Equation
253
cise 42(b) of Section 11.3 imply that Z 1 384 .2n 1/ x .x 1/ sin dx ˛n D .2n 1/3 3 0 2 " # ˇ Z 1 768 .2n 1/ x ˇˇ1 .2n 1/ x D .x 1/ cos cos dx ˇ .2n 1/4 4 2 2 0 0 " ˇ # 768 2 .2n 1/ x ˇˇ1 D 1 sin ˇ .2n 1/4 4 .2n 1/ 2 0 768 . 1/n 2 1C I D .2n 1/4 4 .2n 1/ CMf .x/ D
1 768 X 1 . 1/n 2 .2n 1/ x 1 C cos . From Exercise 12.2.17, 4 nD1 .2n 1/4 .2n 1/ 2
u.x; t/ D
1 768 X 1 . 1/n 2 .2n 1/ t .2n 1/ x 1C cos cos : 4 .2n 1/4 .2n 1/ 2 2 nD1
12.2.32. Setting A D .2n 1/ x=2L and B D .2n 1/ at=2L in the identities 1 1 cos A cos B D Œcos.A C B/ C cos.A B/ and cos A sin B D Œsin.A C B/ sin.A B/ yields 2 2 .2n 1/ at .2n 1/ x 1 .2n 1/.x C at/ .2n 1/.x at/ cos cos D cos C cos .A/ 2L 2L 2 2L 2L and sin
.2n
.2n 1/ x 1/ at cos 2L 2L
D D
Since CMf .x/ D 1 X
nD1
1 X
˛n cos
nD1
˛n cos
.2n
.2n
1 .2n 1/.x C at/ .2n 1/.x sin sin 2 2L 2L Z .2n 1/ xCat .2n 1/ cos d : 4L 2L x at
at/
.B/
1/ x , (A) implies that 2L
.2n 1/ x 1 1/ at cos D ŒCMf .x C at/ C CMf .x 2L 2L 2
at/:
.C/
Since it can be shown that a mixed Fourier cosine series can be integrated term by term between any two limits, (B) implies that 1 X
nD1
2Lˇn .2n 1/ at .2n 1/ x sin cos .2n 1/ a 2L 2L
D D D
Z xCat 1 1 X .2n 1/ ˇn cos d 2a 2L x at nD1 ! Z xCat X 1 1 .2n 1/ ˇn cos d 2a x at 2L nD1 Z xCat 1 CMg . / d : 2a x at
254 Chapter 12 Fourier Solutions of Partial Differential Equations This and (C) imply that u.x; t/ D
1 ŒCMf .x C at/ C CMf .x 2
at/ C
1 2a
Z
xCat
CMg . / d : x at
12.2.34. We begin by looking for functions of the form v.x; t/ D X.x/T .t/ that are not identically zero and satisfy vt t D a2 vxx , v.0; t/ D 0, vx .L; t/ D 0 for all .x; t/. As shown in the text, X and T must satisfy X 00 C X D 0 and (B) T 00 C a2 T D 0 for the same value of . Since v.0; t/ D X.0/T .t/ and vx .L; t/ D X 0 .L/T .t/ and we don’t want T to be identically zero, X.0/ D 0 and X 0 .L/ D 0. Therefore, must be an eigenvalue of (C) X 00 C X D 0, X.0/ D 0, X 0 .L/ D 0, and X must be .2n 1/2 , integer), with a -eigenfunction. From Theorem 11.1.4, the eigenvalues of (C) are n D 2=4L2 .2n 1/ x .2n 1/2 2 , n D 1, 2, 3,. . . . Substituting D into associated eigenfunctions Xn D sin 2L 4L2 (B) yields T 00 C ..2n 1/2 2a2 =4L2/T D 0, which has the general solution Tn D ˛n cos
.2n
1/ at 2ˇn L .2n 1/ at C sin ; 2L .2n 1/ a 2L
where ˛n and ˇn are constants. Now let .2n vn .x; t/ D Xn .x/Tn .t/ D ˛n cos
1/ at 2ˇn L .2n 1/ at C sin 2L .2n 1/ a 2L
sin
.2n
1/ x : 2L
Then @vn .x; t/ D @t
.2n
1/ a .2n 1/ at .2n 1/ at ˛n sin C ˇn cos 2L 2L 2L
so
sin
.2n
1/ x ; 2L
@vn .2n 1/ x .x; 0/ D ˇn sin : @t 2L .2n 1/ x .2n 1/ x Therefore,vn satisfies (A) with f .x/ D ˛n sin and g.x/ D ˇn sin . More gen2L 2L erally, if ˛1; ˛2 ; : : : ; ˛m and ˇ1 ; ˇ2 ; : : : ; ˇm are constants and vn .x; 0/ D ˛n sin
um .x; t/ D
.2n
m X .2n ˛n cos
nD1
1/ x 2L
and
1/ at 2ˇn L .2n 1/ at C sin 2L .2n 1/ a 2L
sin
.2n
1/ x ; 2L
then um satisfies (A) with f .x/ D
m X
˛n sin
nD1
This motivates the definition.
.2n
1/ x 2L
and
g.x/ D
m X
nD1
ˇn sin
.2n
1/ x : 2L
Section 12.2 The Wave Equation
255
12.2.36. Since f .0/ D f 0 .1/ D 0, and f 00 .x/ D 6.1 2x/, Theorem 11.3.5(d) implies that Z 1 48 .2n 1/ x ˛n D .1 2x/ sin dx .2n 1/2 2 0 2 " # ˇ Z 1 96 .2n 1/ x ˇˇ1 .2n 1/ x .1 2x/ cos cos dx D ˇ C2 .2n 1/3 3 2 2 0 0 " ˇ # 96 4 .2n 1/ x ˇˇ1 D 1C sin ˇ .2n 1/3 3 .2n 1/ 2 0 96 4 1 C . 1/n I D .2n 1/3 3 .2n 1/ 1 96 X 1 4 .2n 1/ x n SMf .x/ D 1 C . 1/ sin . From Exercise 12.2.34, 3 3 nD1 .2n 1/ .2n 1/ 2 1 1 4 3.2n 1/ t .2n 1/ x 96 X n 1 C . 1/ cos sin : 3 .2n 1/3 .2n 1/ 2 2
u.x; t/ D
nD1
12.2.38. Since g.0/ D g0 .1/ D 0, and g00 .x/ D 6.1 2x/, Theorem 11.3.5(d) implies that Z 1 48 .2n 1/ x .1 2x/ sin dx ˇn D .2n 1/2 2 0 2 " # ˇ Z 1 96 .2n 1/ x ˇˇ1 .2n 1/ x D .1 2x/ cos cos dx ˇ C2 .2n 1/3 3 2 2 0 0 " ˇ # 96 4 .2n 1/ x ˇˇ1 D 1C sin ˇ .2n 1/3 3 .2n 1/ 2 0 4 96 D 1 C . 1/n I .2n 1/3 3 .2n 1/ 1 96 X .2n 1/ x 1 4 n SMg .x/ D 1 C . 1/ sin . From Exercise 12.2.34, 3 3 nD1 .2n 1/ .2n 1/ 2 1 64 X 1 4 3.2n 1/ t .2n 1/ x n 1 C . 1/ sin sin : 4 nD1 .2n 1/4 .2n 1/ 2 2
u.x; t/ D
12.2.40. Since f .0/ D f 0 . / D f 00 .0/ D 0 and f 000 .x/ D 6, Theorem 11.3.5(d) and Exercise 11.3.50(b) imply that ˇ Z 96 .2n 1/x 192 .2n 1/x ˇˇ 192 ˛n D cos dx D sin D . 1/n I ˇ 3 4 .2n 1/ 0 2 .2n 1/ 2 .2n 1/4 0 1
192 X . 1/n .2n 1/x SMf .x/ D sin . From Exercise 12.2.34, nD1 .2n 1/4 2 1 192 X . 1/n .2n u.x; t/ D cos 4 .2n 1/ nD1
p 1/ 3 t .2n 1/x sin : 2 2
256 Chapter 12 Fourier Solutions of Partial Differential Equations 12.2.42. Since g.0/ D g0 . / D g00 .0/ D 0 and g000 .x/ D 6, Theorem 11.3.5(d) and Exercise 50(b) imply that ˇ Z 96 .2n 1/x 192 .2n 1/x ˇˇ 192 n ˇn D cos dx D sin ˇ D . 1/ .2n 1/4 I .2n 1/3 0 2 .2n 1/4 2 0 SMg .x/ D
1
.2n 1/x 192 X . 1/n sin . From Exercise 12.2.34, 4 nD1 .2n 1/ 2 p 1 384 X . 1/n .2n 1/ 3 t .2n 1/x u.x; t/ D p sin sin : 5 .2n 1/ 2 2 3 nD1
12.2.44. Since f .0/ D f 0 .1/ D f 00 .0/ D 0 and f 000 .x/ D 12.2x 1/, Theorem 11.3.5(d) and Exercise 11.3.50(b) imply that Z 1 192 .2n 1/ x .2x 1/ cos dx ˛n D .2n 1/3 3 0 2 " # ˇ Z 1 384 .2n 1/ x ˇˇ1 .2n 1/ x D .2x 1/ sin 2 sin dx ˇ .2n 1/4 4 2 2 0 0 " ˇ1 # ˇ 4 .2n 1/ x 384 nC1 ˇ D . 1/ C cos ˇ .2n 1/4 4 .2n 1/ 2 0 4 384 4 384 nC1 n D . 1/ D . 1/ C I .2n 1/4 4 .2n 1/ .2n 1/4 4 .2n 1/ 1 384 X 1 4 .2n 1/ x n SMf .x/ D 4 . 1/ C sin . From Exercise 12.2.34, .2n 1/4 .2n 1/ 2 nD1
1 384 X 1 4 n u.x; t/ D 4 . 1/ C cos.2n nD1 .2n 1/4 .2n 1/
1/ t sin
.2n
1/ x : 2
12.2.46. Since g.0/ D g0 .1/ D g00 .0/ D 0 and g000 .x/ D 12.2x 1/, Theorem 11.3.5(d) and Exercise 11.3.50(b) imply that Z 1 .2n 1/ x 192 ˇn D .2x 1/ cos dx 3 3 .2n 1/ 0 2 " # ˇ Z 1 384 .2n 1/ x ˇˇ1 .2n 1/ x D .2x 1/ sin 2 sin dx ˇ .2n 1/4 4 2 2 0 0 " ˇ # 384 4 .2n 1/ x ˇˇ1 nC1 D . 1/ C cos ˇ .2n 1/4 4 .2n 1/ 2 0 384 4 384 4 nC1 n D . 1/ D . 1/ C I .2n 1/4 4 .2n 1/ .2n 1/4 4 .2n 1/ 1 384 X 1 4 .2n 1/ x n SMg .x/ D 4 . 1/ C sin . From Exercise 12.2.34, 4 nD1 .2n 1/ .2n 1/ 2 u.x; t/ D
1 384 X 1 4 n . 1/ C sin.2n 5 .2n 1/5 .2n 1/ nD1
1/ t sin
.2n
1/ x : 2
Section 12.2 The Wave Equation
257
12.2.48. Since f is continuous on Œ0; L and f 0 .L/ D 0, Theorem 11.3.4 implies that SMf .x/ D f .x/, 0 x L. From Exercise 11.3.58, SMf is the odd periodic extension (with period 2L) of the function f .x/; 0 x L; r .x/ D which is continuous on Œ0; 2L. Since r .0/ D r .2L/ D f .0/ D f .2L x/; L < x 2L; f 0 .x/; 0 < x < L; 0, SMf is continuous on . 1; 1/. Moreover, r 0 .x/ D r 0 .0/ D 0 f .2L x/; L < x < 2L; C fC0 .0/, r 0 .2L/ D fC0 .0/, and, since f 0 .L/ D 0, r 0 .L/ D 0. Hence, r is differentiable on Œ0; 2L. Since r .0/ D r .2L/ D f .0/ D 0, Theorem 12.2.3(a) with h D r , p D SMf , and L replaced by 2L implies that SMf is differentiable on 00. 1; 1/. Similarly, SMg is differentiable on . 1; 1/. f .x/; 0 < x < L; 00 00 Now we note that r .x/ D r 00 .L/ D f 00 .L/, and rC .0/ D r 00 .2L/ D f 00 .2L x/; L < x < 2L; 0 0 fC00 .0/ D 0. Since SMf is the even periodic extension of r 0 , Theorem 12.2.3(b) with h D r 0 , q D SMf , 0 and L replaced by 2L implies that SMf is differentiable on . 1; 1/. Now follow the argument used to complete the proof of Theorem 12.2.4. 12.2.50. From Example 11.3.5, Cf .x/ D 4 u.x; t/ D 4
1
1 .2n 1/ x 768 X cos . From Exercise 12.2.49, 4 4 .2n 1/ 2 nD1
1
768 X 1 cos 4 nD1 .2n 1/4
p 5.2n 1/ t .2n 1/ x cos : 2 2
ˇ ˇ 3x 5 2 4 x 4 ˇˇ D . Since f 0 .0/ D f 0 . / D 0 and 5 5 0 0 f 000 .x/ D 24.3x /, Theorem 11.3.5(a) implies that ˇ Z Z ˇ 48 48 ˇ ˛n D .3x / sin nx dx D .3x / cos nx 3 cos nx dx ˇ n3 0 n4 0 0 ˇ ˇ 48 144 48 D Œ. 1/n 2 C C 5 sin nx ˇˇ D Œ1 C . 1/n 2 ; n 1I 4 n4 n n 0 12.2.52. ˛0 D
Cf .x/ D
Z
.3x 4
2 4 5
48
4Lx 3 / dx D
1 X 1 C . 1/n 2 cos nx. From Exercise 12.2.49, n4
nD1
u.x; t/ D
2 4 5
48
1 X 1 C . 1/n 2 cos 2nt cos nx: n4 nD1
ˇ ˇ 1 3x 5 2 4 4 ˇ x D . Since g0 .0/ D g0 . / D 0 and 12.2.54. ˇ0 D .3x 4Lx / dx D ˇ 5 5 0 0 g000 .x/ D 24.3x /, Theorem 11.3.5(a) implies that ˇ Z Z ˇ 48 48 ˇ ˇn D .3x / sin nx dx D .3x / cos nx ˇ 3 cos nx dx n3 0 n4 0 0 ˇ ˇ 48 144 48 D Œ. 1/n 2 C C 5 sin nx ˇˇ D Œ1 C . 1/n 2 ; n 1I 4 n n n4 0 Z
4
3
258 Chapter 12 Fourier Solutions of Partial Differential Equations
Cg .x/ D
2 4 5
48
1 X 1 C . 1/n 2 cos nx. From Exercise 12.2.49, n4
nD1
u.x; t/ D
2 4t 5
24
1 X 1 C . 1/n 2 sin 2nt cos nx: n5
nD1
ˇ 4 ˇ ˇ D . Since f 0 .0/ D ˇ 30 0
Z 1 4 1 x5 x4 2x3 .x 2 x 3 C 2x 2 / dx D C 0 5 2 3 f 0 . / D 0 and f 000 .x/ D 12.2x /, Theorem 11.3.5(a) implies that ˇ Z ˇ 24 24 ˇ ˛n D .2x / sin nx dx D .2x / cos nx ˇ 3 4 n 0 n 0 ˇ ˇ 24 48 24 D Œ. 1/n C C 5 sin nx ˇˇ D Œ1 C . 1/n n4 n n4 0 ( 0 if n D 2m 1; 3 D n 1I if n D 2m; m4 12.2.56. ˛0 D
Cf .x/ D
4 30
3
1 X 4 1 cos 2nx. From Exercise 12.2.49, u.x; t/ D n4 30 nD1
3
2
4 30
3
1 X 1 4t cos 2nx. From Exercise 12.2.49, u.x; t/ D n4 30 nD1
cos nx dx
0
1 X 1 cos 8nt cos 2nx. n4 nD1
Z 1 x5 x4 2x 3 1 4 12.2.58. ˇ0 D .x 2 x 3 C 2 x 2/ dx D C 0 5 2 3 g0 . / D 0 and g000 .x/ D 12.2x /, Theorem 11.3.5(a) implies that ˇ Z ˇ 24 24 ˇ ˇn D .2x / sin nx dx D .2x / cos nx ˇ 3 4 n 0 n 0 ˇ ˇ 24 48 24 n n D Œ. 1/ C C 5 sin nx ˇˇ D Œ1 C . 1/ n4 n n4 0 ( 0 if n D 2m 1; 3 1I D if n D 2m; m4 Cg .x/ D
Z
ˇ 4 ˇ ˇ D . Since g0 .0/ D ˇ 30 0 2
Z
cos nx dx 0
1 3X 1 sin 8nt cos 2nx. 8 nD1 n5
1 12.2.60. Setting A D n x=L and B D n at=L in the identities cos A cos B D Œcos.AC B/C cos.A 2 1 B/ and cos A sin B D Œsin.A C B/ sin.A B/ yields 2 n at n x 1 n.x C at/ n.x at/ cos D cos C cos .A/ cos L L 2 L L and
n at n x sin cos L L
D D
1 n.x C at/ n.x at/ sin sin 2 Z L L xCat n n cos d : 2L x at L
.B/
Section 12.2 The Wave Equation
Since Cf .x/ D ˛0 C
1 X
˛n sin
nD1
˛0 C
1 X
n x , (A) implies that L
˛n cos
nD1
259
n at n x 1 cos D ŒCf .x C at/ C Cf .x L L 2
at/:
.C/
Since it can be shown that a Fourier sine series can be integrated term by term between any two limits, (B) implies that ˇ0 t C
1 X ˇn L n at n x sin cos n a L L nD1
Z xCat 1 1 X n ˇn cos d 2a nD1 L x at ! Z xCat 1 X 1 n ˇ0 C ˇn cos d 2a x at L nD1 Z xCat 1 Cg . / d : 2a x at
D ˇ0 t C D D
This and (C) imply that 1 u.x; t/ D ŒCf .x C at/ C Cf .x 2
at/ C
1 2a
Z
xCat
Cg . / d : x at
12.2.62.(a). Since jpn .x/j 1 and jqn .t/j 1 for all t, jkn pn .x/qn .t/j jkn j for all .x; t/, and the comparison test implies the conclusion. (b) If t is fixed but arbitrary, then jkn pn0 .x/qn .t/j jjnjkn j, so Theorem 12.1.2 with ´ D x and wn .x/ D kn pn .x/qn .t/ justifies term by term differentiation with respect to x on . 1; 1/. If x is fixed but arbitrary, then jkn pn .x/qn0 .t/j jjnjkn j, so Theorem 12.1.2 with ´ D t and wn .t/ D kn pn .x/qn .t/ justifies term by term differentiation with respect to t on . 1; 1/. (c) The argument is similar to argument use in (b). 1 1 X X n x n at n x n at ˇn L (d) Apply (b) and (c) to the series ˛n cos sin and sin sin , recallL L n a L L nD1 nD1 ing that the individual terms in the series satisfy ut t D a2 uxx for all .x; t/. Z xCat f .x C ct/ C f .x ct/ 1 (d) u.x; t/ D C g.u/ du. 2 2a x at
12.2.64. u.x; t/
D
.x C at/ C .x 2 2
D x C .x C at/
at/ .x
C
1 2a 2
Z
xCat x at
4au du D x C 2
at/ D x.1 C 4at/:
Z
xCat x at
ˇxCat ˇ u du D x C u2 ˇˇ x at
12.2.66. u.x; t/
D D
sin.x C at/ C sin.x 2 sin.x C at/ C sin.x 2
Z xCat 1 C a cos u du 2a x at at/ sin.x C at/ sin.x at/ C D sin.x C at/: 2 at/
260 Chapter 12 Fourier Solutions of Partial Differential Equations 12.2.68. u.x; t/
D x sin x cos at C at cos x sin at C
Z xCat .x C at/ sin.x C at/ C .x at/ sin.x at/ 1 C sin u du D 2 2a x at xŒsin.x C at/ C sin.x at/ atŒsin.x C at/ sin.x at/ D C 2 2 cos.x at/ cos.x C at/ C 2a
sin x sin at : a
12.3 LAPLACE’S EQUATION IN RECTANGULAR COORDINATES 12.3.2. Since f .0/ D f .1/ D 0 and f 00 .x/ D 2 6x, Theorem 11.3.5(b) implies that " # ˇ Z 1 Z 2 n x 8 n x ˇˇ2 n x 4 .4 6x/ sin dx D 3 3 .4 6x/ cos C6 cos dx ˛n D n2 2 0 2 n 2 ˇ0 2 0 ˇ 32 96 n x ˇˇ2 32 n D .1 C . 1/ 2/ C 4 4 sin D Œ1 C . 1/n 2 I ˇ 3 3 3 n n 2 0 n 3
S.x/ D
1 32 X Œ1 C . 1/n 2 n x sin . From Example 12.3.1, 3 n3 2 nD1
u.x; y/ D
1 32 X Œ1 C . 1/n 2 sinh n.3 3 nD1 n3 sinh 3n=2
y/=2
sin
n x : 2
12.3.4. ˛1
2 2
D D
if n 2, then ˛n
D D D D
S.x/ D
ˇ Z Z 1 x 2 ˇˇ 1 x sin x dx D x.1 cos 2x/ dx D x cos 2x dx 0 2 ˇ0 0 0 ˇ Z ˇ ˇ 1 sin 2x ˇˇ x sin 2x ˇˇ cos 2x dx D C D I ˇ 2 2 4 0 2 0 0
Z
2
Z Z 2 1 x sin x sin nx dx D xŒcos.n 1/x cos.n C 1/x dx 0 0 ˇ Z 1 sin.n 1/x sin.n C 1/x ˇˇ sin.n 1/x sin.n C 1/x x dx ˇ n 1 nC1 n 1 nC1 0 0 ˇ ˇ 1 cos.n 1/x cos.n C 1/x ˇ 1 1 1 nC1 D . 1/ 1 ˇ .n 1/2 .n C 1/2 .n 1/2 .n C 1/2 8 0 0 if n D 2m 1; < 4n nC1 16m . 1/ 1 D if n D 2mI : .n2 1/2 .4m2 1/
sin x 2
1 16 X n sin 2nx. From Example 12.3.1, 2 nD1 .4n 1/2
u.x; y/ D
sinh.1 y/ sin x 2 sinh 1
1 16 X n sinh 2n.1 y/ sin 2nx: .4n2 1/2 sinh 2n nD1
Section 12.3 Laplace’s Equation in Rectangular Coordinates
261
ˇ x/2 ˇˇ1 1 12.3.6. ˛0 D .1 x/ dx D D ; if n 1, ˇ 2 2 0 0 " # ˇ1 Z 1 Z 1 ˇ 2 .1 x/ sin n x ˇˇ C sin n x dx ˛n D 2 .1 x/ cos n x dx D n 0 0 0 8 ˇ1 4 < ˇ 2 if n D 2m 1; ˇ D 2 Œ1 . 1/n D cos n x D .2m 1/2 2 ˇ 2 2 : n2 2 n 0 0 if n D 2mI Z
C.x/ D
1
.1
1 1 4 X 1 C 2 cos.2n 2 .2n 1/2
1/ x. From Example 12.3.3,
nD1
u.x; y/ D
1 y 4 X sinh.2n 1/y C 3 cos.2n 2 .2n 1/3 cosh 2.2n 1/
1/ x:
nD1
ˇ 1/3 ˇˇ1 1 12.3.8. ˛0 D .x 1/ dx D D ; if n 1, then ˇ 3 3 0 0 " # ˇ1 Z 1 Z 1 ˇ 2 2 2 ˛n D 2 .x 1/ cos n x dx D .x 1/ sin n x ˇˇ 2 .x 1/ sin n x dx n 0 0 0 " # ˇ1 Z 1 ˇ1 ˇ ˇ 4 4 4 4 ˇ D .x 1/ cos n x ˇ cos n x dx D 2 2 sin n x ˇˇ D 2 2 I 2 2 3 3 n n n n 0 0 0 Z
C.x/ D
1
2
.x
1 1 4 X 1 y 4 X sinh ny 1 C 2 cos n x. From Example 12.3.3, u.x; y/ D C cos n x. 3 nD1 n2 3 3 nD1 n3 cosh n
12.3.10. Since g.0/ D g0 .1/ D 0, and g00 .y/ D 6.1 2y/, Theorem 11.3.5(d) implies that Z 1 48 .2n 1/y ˛n D .1 2y/ sin dy 2 2 .2n 1/ 0 2 " # ˇ Z 1 96 .2n 1/y ˇˇ1 .2n 1/y D .1 2y/ cos cos dy ˇ C2 .2n 1/3 3 2 2 0 0 " ˇ # 4 .2n 1/y ˇˇ1 96 D 1C sin ˇ .2n 1/3 3 .2n 1/ 2 0 96 4 D 1 C . 1/n I .2n 1/3 3 .2n 1/ SM .y/ D
1 96 X 1 4 .2n 1/y n 1 C . 1/ sin . From Example 12.3.5, 3 3 nD1 .2n 1/ .2n 1/ 2
u.x; y/ D
1 96 X 4 cosh.2n 1/.x 2/=2 .2n 1/y n 1 C . 1/ sin : 3 3 .2n 1/ .2n 1/ cosh 2.2n 1/=2 2 nD1
262 Chapter 12 Fourier Solutions of Partial Differential Equations 12.3.12. From Example 11.3.8.3, SM .y/ D
1 96 X 1 4 .2n 1/y n 3 C . 1/ sin : 3 3 .2n 1/ .2n 1/ 2 nD1
From Example 12.3.5, 1 96 X 4 cosh.2n 1/.x 3/=2 .2n 1/y n 3 C . 1/ sin : u.x; y/ D 3 3 nD1 .2n 1/ .2n 1/ cosh 3.2n 1/=2 2 12.3.14. cn
D D
2 3
Z
3
.3y 0
4 .2n
1/
y 2 / cos " .3y
" 24 .3 .2n 1/2 2
.2n
1/y 6
y 2 / sin
.2n
dy ˇ 1/y ˇˇ3 ˇ 6 0
Z
0
3
.3
2y/ sin
.2n
1/y 6
dy
#
# ˇ Z 3 1/y ˇˇ3 .2n 1/y D 2y/ cos cos dy ˇ C2 6 6 0 0 ˇ 72 288 .2n 1/y ˇˇ3 288 .2n 1/y D C sin ˇ D .2n 1/3 3 sin .2n 1/2 2 .2n 1/3 3 6 2 0 288 72 D C . 1/n 1 I .2n 1/2 2 .2n 1/3 3 1 72 X .2n 1/y 1 4. 1/n CM .y/ D 1C cos . From Example 12.3.7, 2 2 nD1 .2n 1/ .2n 1/ 6 u.x; y/ D
.2n
1 4. 1/n cosh.2n 1/ x=6 .2n 1/y 432 X 1 C cos : 3 .2n 1/ .2n 1/3 sinh.2n 1/=3 6 nD1
12.3.16. Since g0 .0/ D g.1/ D 0 and g00 .y/ D 6y, Theorem 11.3.5(c) implies that Z 1 48 .2n 1/y ˛n D y cos dy 2 2 .2n 1/ 0 2 " # ˇ Z 1 96 .2n 1/y ˇˇ1 .2n 1/y D y sin sin dy ˇ .2n 1/3 3 2 2 0 0 " ˇ # 96 2 .2n 1/y ˇˇ1 nC1 D . 1/ C cos ˇ .2n 1/3 3 .2n 1/ 2 0 96 2 D . 1/n C dyI .2n 1/3 3 .2n 1/ 1 96 X 1 2 .2n 1/y n CM .y/ D . 1/ C cos . From Example 12.3.7, 3 nD1 .2n 1/3 .2n 1/ 2 u.x; y/ D
1 192 X cosh.2n 1/ x=2 2 .2n 1/y n . 1/ C cos : 4 4 .2n 1/ sinh.2n 1/=2 .2n 1/ 2 nD1
Section 12.3 Laplace’s Equation in Rectangular Coordinates
263
12.3.18. The boundary conditions require products v.x; y/ D X.x/Y .y/ such that (A) X 00 C X D 0, X 0 .0/ D 0, X 0 .a/ D 0, and (B) Y 00 Y D 0, Y .0/ D 1, Y .b/ D 0. From Theorem 11.1.3, the eigenvaln2 2 ues of (A) are D 0, with associated eigenfunction X0 D 1, and n D , with associated eigenfunca2 n x , n D 1, 2, 3,. . . . Substituting D 0 into (B) yields Y000 D 0, Y0 .0/ D 1, Y0 .b/ D 0, tions Yn D cos a y n2 2 . Substituting D into (B) yields Yn00 .n2 2=a2 /Yn D 0, Yn .0/ D 1, Yn .b/ D so Y0 .y/ D 1 b a2 sinh n.b y/=a sinh n.b y/=a n x 0, so Yn D . Then vn .x; y/ D Xn .x/Yn .y/ D cos , so sinh n b=a sinh n b=a a n x n x vn .x; 0/ D cos . Therefore,vn is solution of the given problem with f .x/ D cos . More genera a m X sinh n.b y/=a y n x ally, if ˛0 ; : : : ; ˛m are arbitrary constants, then um .x; y/ D ˛0 1 C ˛n cos b sinh n b=a a nD1 m X n x . Therefore, if f is an arbiis a solution of the given problem with f .x/ D ˛0 C ˛n cos a nD1 trary piecewise smooth function on Œ0; a we define the formal solution of the given problem to be 1 1 X y X sinh n.b y/=a n x n x u.x; y/ D ˛0 1 C ˛n cos , where C.x/ D ˛0 C ˛n cos is b sinh n b=a a a nD1 nD1 Z Z 2 a n x 1 a the Fourier cosine series of f on Œ0; a; that is, ˛0 D f .x/ dx and ˛n D f .x/ cos dx, a 0 a 0 a n 1. ˇ Z 1 x5 4x 3 ˇˇ2 6 1 2 4 3 2 4 Now consider the special case. ˛0 D .x 4x C 4x / dx D x C D . ˇ 2 0 2 5 3 5 0 Since f 0 .0/ D f 0 .2/ D 0 and f 000 .x/ D 12.2x 2/, Theorem 11.3.5(a) implies that " # ˇ Z 2 Z 2 96 n x 192 n x ˇˇ2 n x ˛n D .2x 2/ sin dx D .2x 2/ cos 2 cos dx n3 3 0 2 n4 4 2 ˇ0 2 0 ˇ 768 n x ˇˇ2 384 192 n D Œ. 1/ 2 C 2 C sin D Œ1 C . 1/n n4 4 n5 5 2 ˇ0 n4 4 ( 0 if n D 2m 1; 48 D n 1: if n D 2m; m4 4 C.x/ D
8 15
1 48 X 1 8.1 y/ cos n x; u.x; y/ D 4 4 n 15 nD1
1 48 X 1 sinh n.1 y/ cos n x. 4 n4 sinh n nD1
12.3.20. The boundary conditions require products v.x; y/ D X.x/Y .y/ such that (A) X 00 C X D 0, X.0/ D 0, X 0 .a/ D 0, and (B) Y 00 Y D 0, Y .0/ D 1, Y .b/ D 0. From Theorem 11.1.4, the .2n 1/2 2 .2n 1/ x , with associated eigenfunctions Yn D sin , eigenvalues of (A) are n D 2 4a 2a 2 2 .2n 1/ n D 1, 2, 3,. . . . Substituting D into (B) yields Yn00 ..2n 1/2 2=4a2 /Yn D 0, 4a2 sinh.2n 1/.b y/=2a Yn .0/ D 1, Yn .b/ D 0, so Yn D . Then vn .x; y/ D Xn .x/Yn .y/ D sinh.2n 1/ b=2a sinh.2n 1/.b y/=2a .2n 1/ x .2n 1/ x sin , so vn .x; 0/ D sin . Therefore,vn is solution of sinh.2n 1/ b=2a 2a 2a
264 Chapter 12 Fourier Solutions of Partial Differential Equations 1/ x . More generally, if ˛1 ; : : : ; ˛m are arbitrary constants, 2a m X sinh.2n 1/.b y/=2a .2n 1/ x cos is a solution of the given problem then um .x; y/ D ˛n sinh.2n 1/ b=2a 2a nD1 m X .2n 1/ x with f .x/ D ˛n sin . Therefore, if f is an arbitrary piecewise smooth function on Œ0; a 2a nD1 1 X sinh.2n 1/.b y/=2a .2n 1/ x sin , we define the formal solution of the given problem to be u.x; y/ D ˛n sinh.2n 1/ b=2a 2a nD1 1 X .2n 1/ x is the mixed Fourier sine series of f on Œ0; a; that is, ˛n D where Sm .x/ D ˛n sin 2a nD1 Z 2 a .2n 1/ x f .x/ sin . a 0 2a Now consider the special case. Since f .0/ D f 0 .L/ D 0 and f 00 .x/ D 2, Theorem 11.3.5(d) implies that ˇ Z 3 .2n 1/ x 288 .2n 1/ x ˇˇ3 288 48 sin dx D cos D I ˛n D ˇ 2 2 3 3 .2n 1/ 0 6 .2n 1/ 6 .2n 1/3 3 0 the given problem with f .x/ D sin
.2n
1 288 X .2n 1/ x 1 SM .x/ D 3 sin I nD1 .2n 1/3 6
u.x; y/ D
1 288 X sinh.2n 1/.2 3 nD1 .2n 1/3 sinh.2n
y/=6 .2n 1/ x sin : 1/=3 6
12.3.22. The boundary conditions require products v.x; y/ D X.x/Y .y/ such that (A) X 00 C X D 0, X 0 .0/ D 0, X 0 .a/ D 0, and (B) Y 00 Y D 0, Y 0 .0/ D 0, Y .b/ D 1. From Theorem 11.1.3, the n2 2 eigenvalues of (A) are D 0, with associated eigenfunction X0 D 1, and n D , with associated a2 n x eigenfunctions Yn D cos , n D 1, 2, 3,. . . . Substituting D 0 into (B) yields Y000 D 0, Y00 .0/ D 0, a n2 2 Y0 .b/ D 1, so Y0 D 1. Substituting D into (B) yields Yn00 .n2 2 =a2 /Yn D 0, Yn0 .0/ D a2 cosh ny=a sinh ny=a n x 0, Yn .b/ D 1, so Yn D . Then vn .x; y/ D Xn .x/Yn .y/ D cos , so cosh n b=a cosh n b=a a n x n x vn .x; b/ D cos . Therefore,vn is solution of the given problem with f .x/ D cos . More a a m X cosh ny=a n x generally, if ˛0 ; : : : ; ˛m are arbitrary constants, then um .x; y/ D ˛0 C ˛n cos is cosh n b=a a nD1 m X n x a solution of the given problem with f .x/ D ˛0 C ˛n cos . Therefore, if f is an arbitrary a nD1 piecewise smooth function on Œ0; a we define the formal solution of the given problem to be u.x; y/ D 1 1 X X cosh ny=a n x n x ˛0 C ˛n cos , where C.x/ D ˛0 C ˛n cos is the Fourier cosine series of cosh n b=a a a nD1 Z a Z a nD1 1 2 n x f on Œ0; a; that is, ˛0 D f .x/ dx and ˛n D f .x/ cos dx, n 1. a 0 a 0 a
Section 12.3 Laplace’s Equation in Rectangular Coordinates Now consider the special case. Z 1 4 1 x5 ˛0 D .x 2 x 3 C 2x 2 / dx D 0 5
x4 2x 3 C 2 3
265
ˇ 4 ˇ ˇ D : ˇ 30 0
Since f 0 .0/ D f 0 . / D 0 and f 000 .x/ D 12.2x /, Theorem 11.3.5(a) implies that ˇ Z Z ˇ 24 24 ˇ .2x / sin nx dx D .2x / cos nx ˇ 2 cos nx dx ˛n D n3 0 n4 0 0 ˇ ˇ 24 48 24 D Œ. 1/n C C 5 sin nx ˇˇ D Œ1 C . 1/n 4 4 n n n 0 ( 0 if n D 2m 1; 3 n 1I D if n D 2m; 4 m C.x/ D
4 30
3
1 X 1 4 cos 2nx; u.x; y/ D 4 n 30
nD1
3
1 X 1 cosh 2ny cos 2nx n4 cos 2n
nD1
12.3.24. The boundary conditions require products v.x; y/ D X.x/Y .y/ such that (A) X 00 X D 0, X 0 .0/ D 0, X.a/ D 1, and (B) Y 00 C Y D 0, Y .0/ D 0, Y .b/ D 0. From Theorem 11.1.2, n2 2 ny the eigenvalues of (B) are n D , with associated eigenfunctions Yn D sin , n D 1, 2, b2 b 2 2 n 3,. . . . Substituting D into (A) yields Xn00 .n2 2=b 2 /Xn D 0, Xn0 .0/ D 0, Xn .a/ D 1, so b2 cosh n x=b cosh n x=b ny ny Xn D . Then vn .x; y/ D Xn .x/Yn .y/ D sin , so vn .a; y/ D sin . cosh n a=b cosh n a=b b b ny Therefore,vn is solution of the given problem with g.y/ D sin . More generally, if ˛1 ; : : : ; ˛m b m X ny cosh n x=b are arbitrary constants, then um .x; y/ D ˛n sin is a solution of the given problem cosh n a=b b nD1 m X ny with g.y/ D ˛n sin . Therefore, if g is an arbitrary piecewise smooth function on Œ0; b we b nD1 1 X cosh n x=b ny define the formal solution of the given problem to be u.x; y/ D ˛n sin , where cosh n a=b b nD1 Z 1 X 2 b ny ny S.y/ D ˛n sin is the Fourier sine series of g on Œ0; b; that is, ˛n D g.y/ sin dy. b b 0 b nD1
Now consider the special case. Since g.0/ D g.1/ D g00 .0/ D g00 .L/ D 0 and f .4/ .y/ D 24, Theorem 11.3.5(b) and Exercise 35(b) of Section 11.3 imply that ˇ1 Z 1 ˇ 48 48 ˛n D sin ny dy D cos ny ˇˇ 4 4 5 5 n 0 n 0 8 96 < 48 if n D 2m 1 D Œ. 1/n 1 D .2m 1/5 5 5 5 : n 0 if n D 2mI
S.y/ D
1 96 X 1 sin.2n 5 .2n 1/5 nD1
1/y; u.x; y/ D
1 96 X cosh.2n 1/ x sin.2n 5 .2n 1/5 cosh.2n 1/ nD1
1/y.
266 Chapter 12 Fourier Solutions of Partial Differential Equations 12.3.26. The boundary conditions require products v.x; y/ D X.x/Y .y/ such that (A) X 00 X D 0, X 0 .0/ D 0, X 0 .a/ D 1, and (B) Y 00 C Y D 0, Y .0/ D 0, Y .b/ D 0. From Theorem 11.1.2, the n2 2 ny eigenvalues of (B) are n D , with associated eigenfunctions Yn D sin , n D 1, 2, 3,. . . . 2 b b 2 2 n into (A) yields Xn00 .n2 2 =b 2/Xn D 0, Xn0 .0/ D 0, Xn0 .a/ D 1, so Xn D Substituting D b2 b cosh n x=b b cosh n x=b ny @vn ny . Then vn .x; y/ D Xn .x/Yn .y/ D sin , so .a; y/ D sin . n sinh n a=b n sinh n a=b b @x b ny Therefore,vn is solution of the given problem with g.y/ D sin . More generally, if ˛1; : : : ; ˛m are b m b X cosh n x=b ny arbitrary constants, then um .x; y/ D ˛n sin is a solution of the given problem nD1 n sinh n a=b b m X ny with g.y/ D ˛n sin . Therefore, if g is an arbitrary piecewise smooth function on Œ0; b we b nD1 1 cosh n x=b ny b X ˛n sin , where define the formal solution of the given problem to be u.x; y/ D nD1 n sinh n a=b b Z 1 X ny 2 b ny S.y/ D ˛n sin is the Fourier sine series of g on Œ0; b; that is, ˛n D g.y/ sin dy. b b 0 b nD1 Z 2 Z 4 ny ny 1 y sin C .4 y/ sin dy ; Now consider the special case. ˛n D 2 0 4 4 2 " # ˇ Z 2 Z 2 ny 4 ny ˇˇ2 ny y sin dy D y cos cos dy 4 n 4 ˇ0 4 0 0 ˇ n 4 ny ˇˇ2 2 n 4 n 2 cos C 2 2 sin D cos C 2 2 sin I D n 2 n 4 ˇ0 n 2 n 2 Z
2
.4 0
# ˇ Z 2 ny ˇˇ4 ny D y/ cos C cos dy 4 ˇ2 4 0 ˇ 2 n 4 ny ˇˇ4 2 n 4 n D cos sin D cos C 2 2 sin I n 2 n2 2 4 ˇ2 n 2 n 2 8 16 < n 16 . 1/mC1 if n D 2m 1 ˛n D 2 2 sin D .2m 1/2 2 : n 2 0 if n D 2mI
ny y/ sin dy 4
" 2 .4 n
S.y/ D
u.x; y/ D
1 16 X . 1/nC1 .2n 1/y sin I 2 2 nD1 .2n 1/ 4
1 64 X cosh.2n 1/ x=4 .2n 1/y . 1/nC1 sin : 3 3 nD1 .2n 1/ sinh.2n 1/=4 4
12.3.28. The boundary conditions require products v.x; y/ D X.x/Y .y/ such that (A) X 00 X D 0, X 0 .0/ D 1, X.a/ D 0, and (B) Y 00 C Y D 0, Y 0 .0/ D 0, Y 0 .b/ D 0. From Theorem 11.1.3, the
Section 12.3 Laplace’s Equation in Rectangular Coordinates
267
n2 2 , with associeigenvalues of (B) are 0 D 0, with associated eigenfunction Y0 D 1, and n D b2 ny ated eigenfunctions Yn D cos , n D 1, 2, 3,. . . . Substituting 0 D 0 into (A) yields X000 D 0, b n2 2 into (A) yields Xn00 .n2 2=b 2 /Xn D X00 .0/ D 1 X.a/ D 0, so X0 D x a. Substituting D b2 b sinh n.x a/=b 0, Xn0 .0/ D 1, Xn .a/ D 0, so Xn D . Then vn .x; y/ D Xn .x/Yn .y/ D n cosh n a=b ny @vn ny b sinh n.x a/=b cos , so .0; y/ D cos . Therefore,vn is solution of the given probn cosh n a=b b @x b ny . More generally, if ˛0 ; : : : ; ˛m are arbitrary constants, then um .x; y/ D lem with g.y/ D cos b m b X sinh n.x a/=b ny ˛0 .x a/ C ˛n cos is a solution of the given problem with g.y/ D nD1 n cosh n a=b b m X ny ˛n cos . Therefore, if g is an arbitrary piecewise smooth function on Œ0; b we define the formal b nD1 1 b X sinh n.x a/=b ny solution of the given problem to be u.x; y/ D ˛0 .x a/ C ˛n cos where nD1 n cosh n a=b b Z 1 X 1 b ny ny C.y/ D ˛0 C ˛n cos is the Fourier cosine series of g on Œ0; b; that is, ˛0 D g.y/ cos dy, b b 0 b nD1 Z 2 b ny ˛n D g.y/ cos dy, n 1. b 0 b Now consider the special case. From Example 11.3.1, C.y/ D u.x; y/ D
.x
2/ 2
2
1 4 X 1 cos.2n nD1 .2n 1/2
1/yI
1 4 X sinh.2n 1/.x 2/ cos.2n nD1 .2n 1/3 cosh 2.2n 1/
1/y:
12.3.30. The boundary conditions require products v.x; y/ D X.x/Y .y/ such that (A) X 00 C X D 0, X 0 .0/ D 0, X.a/ D 0, and (B) Y 00 Y D 0, Y .0/ D 1, and Y is bounded. From Theorem 11.1.5, the .2n 1/2 2 .2n 1/ x eigenvalues of (A) are n D , with associated eigenfunctions Yn D cos , n D 1, 2 4a 2a 2 2 .2n 1/ 2, 3,. . . . Substituting D into (B) yields Yn00 ..2n 1/2 2=4a2/Yn D 0, Yn .0/ D 1, so 4a2 .2n 1/ x Yn D e .2n 1/y=2a . Then vn .x; y/ D Xn .x/Yn .y/ D e .2n 1/y=2a cos , so vn .x; 0/ D 2a .2n 1/ x .2n 1/ x cos . Therefore,vn is solution of the given problem with f .x/ D cos . More 2a 2a m X .2n 1/ x generally, if ˛1 ; : : : ; ˛m are arbitrary constants, then um .x; y/ D ˛n e .2n 1/y=2a cos 2a nD1 m X .2n 1/ x is a solution of the given problem with f .x/ D ˛n cos . Therefore, if f is an ar2a nD1 bitrary piecewise smooth function on Œ0; a we define the formal solution of the given problem to be
268 Chapter 12 Fourier Solutions of Partial Differential Equations 1 X
1 X 1/ x .2n 1/ x , where Cm .x/ D ˛n cos is the mixed 2a 2a nD1 nD1 Z 2 a .2n 1/ x f .x/ cos . Fourier cosine series of f on Œ0; a; that is, ˛n D a 0 2a Now consider the special case. Since f 0 .0/ D f .L/ D 0 and f 00 .x/ D 2, Theorem 11.3.5(d) implies that Z 3 16L .2n 1/ x ˛n D cos dx 2 2 .2n 1/ 0 6 ˇ 288 .2n 1/ x ˇˇ3 288 nC1 D sin I ˇ D . 1/ .2n 1/3 3 6 .2n 1/3 3
u.x; y/ D
˛n e
.2n 1/y=2a
cos
.2n
0
CM .x/ D
u.x; y/ D
1 X
288 . 1/n .2n 1/ x cos I 3 nD1 .2n 1/3 6
1 288 X . 1/n e 3 .2n 1/3
.2n 1/y=6
cos
nD1
.2n
1/ x : 6
12.3.32. The boundary conditions require products v.x; y/ D X.x/Y .y/ such that (A) X 00 C X D 0, X.0/ D 0, X.a/ D 0, and (B) Y 00 Y D 0, Y 0 .0/ D 1, and Y is bounded. From Theon2 2 n x rem 11.1.2, the eigenvalues of (A) are n D , with associated eigenfunctions Yn D sin , 2 a a 2 2 n n D 1, 2, 3,. . . . Substituting D into (B) yields Yn00 .n2 2=a2 /Yn D 0, Yn0 .0/ D 1, so a2 a n x @vn n x a Yn D e ny=a . Then vn .x; y/ D Xn .x/Yn .y/ D e ny=a sin , so .x; 0/ D sin . n n a @y a n x Therefore,vn is solution of the given problem with f .x/ D sin . More generally, if ˛1 ; : : : ; ˛m are a m a X ˛n ny=a n x arbitrary constants, then um .x; y/ D e sin is a solution of the given problem n a nD1 m X n x with f .x/ D ˛n sin . Therefore, if f is an arbitrary piecewise smooth function on Œ0; a we a nD1 1 n x a X ˛n ny=a define the formal solution of the given problem to be u.x; y/ D e sin , where nD1 n a Z 1 X n x 2 a n x C.x/ D ˛n sin is the Fourier sine series of f on Œ0; a; that is, ˛n D f .x/ sin dx. a a a 0 nD1 Now consider the special case. Since f .0/ D f . / D 0 and f 00 .x/ D 2 6x, Theorem 11.3.5(b) implies that ˇ Z Z ˇ 2 2 ˇ ˛n D .2 6x/ sin nx dx D 3 .2 6x/ cos nx ˇ C 6 cos nx dx n2 0 n 0 0 ˇ ˇ 4 12 4 D Œ1 C . 1/n 2 C 4 sin nx ˇˇ D Œ1 C . 1/n 2 I 3 n n n3 0 S.x/ D 4
1 1 X X Œ1 C . 1/n 2 Œ1 C . 1/n 2 sin nx; u.x/ D 4 e n3 n4
nD1
nD1
ny
sin nx;
Section 12.3 Laplace’s Equation in Rectangular Coordinates
269
12.3.34. The boundary conditions require products v.x; y/ D X.x/Y .y/ such that (A) X 00 C X D 0, X.0/ D 0, X 0 .a/ D 0, and (B) Y 00 Y D 0, Y 0 .0/ D 1, and Y is bounded. From Theorem 11.1.4, the .2n 1/2 2 .2n 1/ x eigenvalues of (A) are n D , with associated eigenfunctions Yn D sin , n D 1, 2 4a 2a 2 2 .2n 1/ into (B) yields Yn00 ..2n 1/2 2=4a2/Yn D 0, Yn0 .0/ D 1, so 2, 3,. . . . Substituting D 4a2 2a 2a .2n 1/ x e .2n 1/y=2a . Then vn .x; y/ D Xn .x/Yn .y/ D e .2n 1/y=2a sin , Yn D .2n 1/ .2n 1/ 2a @vn .2n 1/ x .2n 1/ x so .x; 0/ D sin . Therefore,vn is solution of the given problem with f .x/ D sin . @y 2a 2a m 2a X ˛n .2n 1/ x More generally, if ˛1 ; : : : ; ˛m are arbitrary constants, then um .x; y/ D e .2n 1/y=2a sin nD1 2n 1 2a m X .2n 1/ x . Therefore, if f is an aris a solution of the given problem with f .x/ D ˛n sin 2a nD1 bitrary piecewise smooth function on Œ0; a we define the formal solution of the given problem to be 1 1 X 2a X ˛n .2n 1/ x .2n 1/ x u.x; y/ D e .2n 1/y=2a sin ,where Sm .x/ D ˛n sin is nD1 2n 1 2a 2a nD1 Z .2n 1/ x 2 a the mixed Fourier sine series of f on Œ0; a; that is, ˛n D f .x/ sin . a 0 2a Now consider the special case. ˛n
D D D D D
2 5
Z
5
x 2/ sin "
.5x 0
.2n
1/ x dx 10
# ˇ Z 5 1/ x ˇˇ5 .2n 1/ x .5x x / cos .5 2x/ cos dx ˇ .2n 1/ 10 10 0 0 " # ˇ Z 5 40 .2n 1/ x ˇˇ5 .2n 1/ x .5 2x/ sin sin dx ˇ C2 .2n 1/2 2 10 10 0 0 ˇ 200 800 .2n 1/ x ˇˇ5 n . 1/ cos ˇ .2n 1/2 2 .2n 1/3 3 10 0 200 800 . 1/n C I .2n 1/2 2 .2n 1/3 3 4
SM .x/ D u.x; y/ D
2
.2n
1 200 X 1 4 .2n 1/ x n . 1/ C sin I 2 2 nD1 .2n 1/ .2n 1/ 10
1 2000 X 1 4 n . 1/ C e 3 nD1 .2n 1/3 .2n 1/
.2n 1/y=10
sin
.2n
1/ x : 10
12.3.36. Solving BVP.1; 1; 1; 1/.f0; 0; 0; 0/ requires products X.x/Y .y/ such that X 00 C X D 0; Hence, Xn D cos
X 0 .0/ D 0;
n x , Yn D a
X 0 .a/ D 0I
Y 00
Y D 0;
a cosh n.y b/=a , and c1 n sinh n b=a
Y 0 .b/ D 0;
Y 0 .0/ D 1:
1 a X cosh n.y b/=a n x An cos is n sinh n b=a a nD1
270 Chapter 12 Fourier Solutions of Partial Differential Equations 1 X n x a formal solution of BVP.1; 1; 1; 1/.f0; 0; 0; 0/ if c1 is any constant and An cos is the Fourier a nD1 Z a cosine expansion of f0 on Œ0; a, which is possible if and only if f0 .x/ dx D 0. 0
1 a X cosh ny=a n x Bn cos is a formal solution of BVP.1; 1; 1; 1/.0; f1; 0; 0/ if Similarly, c2 C n sinh n b=a a nD1 1 X n x c2 is any constant and Bn cos is the Fourier cosine expansion of f1 on Œ0; a, which is possible a nD1 Z a if and only if f1 .x/ dx D 0. 0
1 b X cosh n.x a/=b ny Cn cos is a formal so nD1 n sinh n a=b b 1 X ny is the Fourier cosine expanlution of BVP.1; 1; 1; 1/.0; 0; g0; 0/ if c3 is any constant and Cn cos b nD1 Z b 1 b X cosh n x=b ny sion of g0 on Œ0; b, which is possible if and only if g0 .x/ dx D 0, and c4 C Dn cos nD1 n sinh n a=b b 0 1 X ny is a formal solution of BVP.1; 1; 1; 1/.0; 0; 0; g1/ if c4 is any constant and Dn cos is the Fourier b nD1 Z b cosine expansion of g1 on Œ0; b, which is possible if and only if g1 .x/ dx D 0.
Interchanging x and y and a and b shows that c3
0
Adding the four solutions yields u.x; y/
D
CC C
1 a X Bn cosh ny=a An cosh n.y nD1 n sinh n b=a
1 b X Dn cosh n x=b Cn cosh n.x nD1 n sinh n a=b
b/=a a/=b
cos
n x a
cos
ny ; b
where C is an arbitrary constant. 12.4 LAPLACE’S EQUATION IN POLAR COORDINATES 12.4.2. v.r; / D R.r /‚./ where (A) r 2R00 C rR0 R D 0 and ‚00 C ‚ D 0, ‚.0/ D 0, n2 2 n n2 2 , ‚ D sin , n D 1, 2, 3,. . . . Substituting D ‚. / D 0. From Theorem 11.1.2, n D n
2
2 2 2 n into (A) yields the Euler equation r 2 Rn00 C rRn0 Rn D 0 for Rn . The indicial polynomial is
2 n n sC , so Rn D c1 r n= C c2 r n= , by Theorem 7.4.3. We want Rn ./ D 1 and s
n= n= n= r 0 r n= Rn .0 / D 0, so Rn .r / D 0 n= ; 0 n= 0n= n= vn .r; / D
0
n= n=
0
0
n= n=
0n= n=
r
n=
r
n=
sin
n I
Section 12.4 Laplace’s Equation in Polar Coordinates
1 X n n , where S./ D
n sin is the Fourier n= n= n=
0 0 n= nD1 nD1 Z 1 n sine series if f on Œ0; ; that is, ˛n D f ./ sin d, n D 1, 2, 3,. . . .
0
u.r; / D
1 X
271
˛n
0
n= n=
r
n=
0
r
n=
sin
12.4.4. v.r; / D R.r /‚./ where (A) r 2R00 C rR0 R D 0 and ‚00 C ‚ D 0, ‚0 .0/ D 0, ‚. / D 0. .2n 1/2 2 .2n 1/ From Theorem 11.1.5, n D , ‚n D cos , n D 1, 2, 3,. . . . Substituting 4 2 2 2 2 .2n 1/ .2n 1/2 2 2 00 0 into (A) yields the Euler equation r R C rR Rn D 0 for Rn . The D n n 4 2 4 2 .2n 1/ .2n 1/ indicial polynomial is s sC , so Rn D c1r .2n 1/=2 C c2 r .2n 1/=2 , 2 2 by Theorem 7.4.3. We want Rn to be bounded as r ! 0C and Rn ./ D 1, so we take Rn .r / D 1 X r .2n 1/=2 r .2n 1/=2 .2n 1/ r .2n 1/=2 .2n 1/ ; v .r; / D cos ; u.r; / D ˛ cos , n n .2n 1/=2 2 2 .2n 1/=2 .2n 1/=2 nD1 1 X .2n 1/ where CM ./ D ˛n cos is the mixed Fourier cosine series of f on Œ0; ; that is, ˛n D 2 nD1 Z 2 .2n 1/ f ./ cos d, n D 1, 2, 3,. . . .
0 2 12.4.6. v.r; / D R.r /‚./ where (A) r 2R00 C rR0 R D 0 and ‚00 C ‚ D 0, ‚0 .0/ D 0, ‚0 . / D 0. n n2 2 , ‚n D cos , n D 1, 2, 3,. . . . Substituting From Theorem 11.1.3, 0 D 0, ‚0 D 1; n D 2
D 0 into (A) yields the equation r 2R000 C rR00 D 0 for R0 ; R0 D c1 C c2 ln r . Since we want R0 to be bounded as r ! 0C and R0 ./ D 1, R0 .r / D 1; therefore v0 .r; / D 1. n2 2 n2 2 2 00 0 Substituting D into (A) yields the Euler equation r R C rR Rn D 0 for Rn . The n n
2
2 n n indicial polynomial is s sC , so Rn D c1 r n= C c2r n= , by Theorem 7.4.3. Since
r n= r n= n we want Rn to be bounded as r ! 0C and Rn ./ D 1, Rn .r / D n= ; vn .r; / D n= cos ,
1 1 X r n= X n n n D 1, 2, 3,. . . ; u.r; / D ˛0 C ˛n n= cos , where F ./ D ˛0 C ˛n cos is the
nD1 nD1 Z Z 2 n 1 Fourier cosine series of f on Œ0; ; that is, ˛0 D f ./ d and ˛n D f ./ cos d,
0
0
n D 1, 2, 3,. . . .
CHAPTER 13 Boundary Value Problems for Second Order Ordinary Differential Equations
13.1 BOUNDARY VALUE PROBLEMS x x; y D x C c1e xC c2 e ; y.0/ D 2 H) c1 C c2 D 2; 2e 2 1 1 c1 2 y.1/ D 1 H) 1 C c1 e C c2 =e D 1; D ; c1 D ; c2 D ; e 1=e c2 2 e 1 1 e 2 e x e .x 1/ yD xC e 1
13.1.2. By inspection, yp D
13.1.4. By inspection, yp D 3 H)
x; y D
1 C 2c1 D 3; c1 D 2; y.1/
x C c1 e x C c2 e
x ; y 0 D 1 C c1 e x c2 e x ; y.0/ C y 0 .0/ D 2c 2 y 0 .1/ D 2 H) D 2; c2 D e; y D x C 2e x C e .x 1/ e
13.1.6. yp D Ax 2e x ; yp0 D A.x 2 C 2xe x /; yp00 D A.x 2 e x C 4xe x C 2e x /; yp00 2yp0 C yp D 2Ae x D e x if A D 1; yp D x 2 e x ; y D .x 2 C c1 C c2 x/e x ; y 0 D .x 2 C 2x C c1 C c2 C c2 x/e x ; B1 .y/ D 3 and B2 .y/ D 6e H) c1 2c2 D 3, 2c1 C 3c2 D 24; c1 D 13; c2 D 8; y D .x 2 8x C 13/e x . 13.1.8. B1.y/ D y.0/; B2 .y/ D y.1/ y 0 .1/. Let y1 D x, y2 D 1; B1 .y1 / D B2.y1 / D 0. By variation of parameters, if yp D u1 x C u2 where u01 x C u02 D 0 and u01 D F , then yp00 D F .x/. Let Z 1 Z x Z 1 Z x u01 D F , u02 D xF ; u1 D F .t/ dt, u2 D tF .t/ dt; yp D x F .t/ dt tF .t/ dt; x 0 x 0 Z 1 yp0 D F .t/ dt; y D yp C c1x C c2 . Since B1 .yp / D 0, B1 .x/ D 0 and B1 .1/ D 1, B1 .y/ D x Z 1 0 H) c2 D 0; hence, y D yp C c1x. Since B2 .yp / D tF .t/ dt and B2 .x/ D 0, B2.y/ D 0 H) 0 Z 1 tF .t/ dt D 0: There is no solution if this conditions does not hold. If it does hold, then the solutions 0
are y D yp C c1x, with c1 arbitrary.
13.1.10. (a) The condition is b a ¤ .kC1=2/ (k D integer). Let y1 D sin.x a/ and y2 D cos.x b/. Then y1 .a/ D y20 .b/ D 0 and fy1 ; y2 g is linearly independent if b a ¤ .k C 1=2/ (k D integer), since ˇ ˇ ˇ sin.x a/ cos.x b/ ˇˇ ˇ D cos.b a/ ¤ 0: ˇ cos.x a/ sin.x b/ ˇ 273
274 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations Now Theorem 13.1.2 implies that (A) has a unique solution for any continuous F and constants k1 and k2 . If y D u1 sin.x a/ C u2 cos.x b/ where u01 sin.x a/ C u02 cos.x u01 cos.x a/ u02 sin.x then y 00 C y D F .
u1 D
u01 D F .x/
1 cos.b a/
yD
Z
u02 D
b
F .t/ cos.t Z
sin.x cos.b
a/ ; a/ Z x F .t/ sin.t
1 cos.b a/ a Z cos.x b/ x F .t/ sin.t cos.b a/ a
b
F .t/ cos.t
F .x/
u2 D
b/ dt;
x
a/ a/
sin.x cos.b
b/ ; a/
cos.x cos.b
D 0 D F;
b/ b/
b/ dt
x
b/ dtI
a/ dt:
(b) If b a D .k C 1=2/ (k D integer), then y1 D sin.x a/ satisfies both boundary conditions y.a/ D 0 and y 0 .b/ D 0. Let y2 D cos.x a/. If yp D u1 sin.x a/ C u2 cos.x/ where u01 sin.x u01
cos.x
then yp00 C yp D F ; u01 D F cos.x Z
u1 D yp D
sin.x
yp0 D
cos.x
a/ C u02 cos.x a/
a/; u02 D
u02
sin.x
a/
F sin.x
a/;
b
F .t/ cos.t
a/ dt;
x
a/ a/
Z
u2 D
b
F .t/ cos.t
a/ dt
D 0;
a/
Z
D F; x
F .t/ sin.t
cos.x
a/
x
Z
b
F .t/ cos.t x
a/ dtI
a
a/ dt C sin.x
a/
Z
x
F .t/ sin.t
a/ dtI
F .t/ sin.t
a/ dt:
a
Z
x a
The general solution of y 00 Cy D F is y D yp Cc1 sin.x a/Cc2 cos.x a/. Since b a D .k C1=2/ , Z b Z b y 0 .b/ D 0 H) yp0 .b/ D . 1/k F .t/ sin.t a/ dt D 0; therefore, F must satisfy F .t/ sin.t a
a
a/ dt D 0. In this case, the solutions of the boundary value problem are y D yp C c1 sin.x arbitrary. a/ and y2 D sinh.x b/. Then y1 .a/ D 0, y2 .b/ D 0, and ˇ ˇ ˇ sinh.x a/ sinh.x b/ ˇ ˇ ˇ D sinh.b a/ ¤ 0: W .x/ D ˇ cosh.x a/ cosh.x b/ ˇ
a/, with c1
13.1.12. Let y1 D sinh.x
(Since W is constant (Theorem 5.1.4), evaluate it by setting x D b.) From Theorem 13.1.2, (A) has a unique solution for any continuous F and constants k1 and k2 . If y D u1 sinh.x a/ C u2 sinh.x b/ where u01 sinh.x u01 cosh.x then y 00
y D F.
u01 D F .x/
a/ C u02 sinh.x a/ C u02 cosh.x
sinh.x sinh.b
b/ ; a/
b/ D 0 b/ D F;
u02 D F .x/
sinh.x sinh.b
a/ ; a/
Section 13.1 Boundary Value Problems 1 u1 D sinh.b
a/
Z
b
F .t/ sinh.t x
Z
a/ a/
sinh.x yD sinh.b
b/ dt;
1 u2 D sinh.b
b
b/ dt C
F .t/ sinh.t x
sinh.x sinh.b
b/ a/
a/ Z x
Z
275
x
F .t/ sinh.t
a/ dtI
a
F .t/ sinh.t
a/ dt:
a
a/ and y2 D cosh.x b/. Then y10 .a/ D y20 .b/ D 0 and ˇ ˇ ˇ cosh.x a/ cosh.x b/ ˇ ˇ ˇ D sinh.b a/ ¤ 0: W .x/ D ˇ sinh.x a/ sinh.x b/ ˇ
13.1.14. Let y1 D cosh.x
(Since W is constant (Theorem 5.1.40, evaluate it by setting x D b.) If y D u1 cosh.x a/ C u2 cosh.x b/ where u01 cosh.x u01 then y 00
y D F.
u1 D
1 sinh.b
yD
sinh.x
u01 D F .x/ a/
cosh.x sinh.b
Z
a/ C u02 cosh.x a/ C u02 sinh.x b/ ; a/
cosh.x sinh.b
u02 D
b
F .t/ cosh.t
b/ dt;
x
a/ a/
Z
u2 D
b
F .t/ cosh.t x
b/ dt
b/ D 0
b/ D F; F .x/
1 sinh.b
cosh.x sinh.b
a/ ; a/
cosh.x sinh.b
b/ a/
a/ Z x
Z
x
F .t/ cosh.t
F .t/ cosh.t
a/ dt:
a
13.1.16. Let y1 D sin !x, y2 sin !.x /; then y1 .0/ D 0, y2 . / D 0, ˇ ˇ ˇ sin !x sin !.x / ˇˇ ˇ W .x/ D ˇ D ! sin ! ¤ 0 ! cos !x ! cos !.x / ˇ
if and only if ! is not an integer. If this is so, then y D u1 sin !x C u2 sin !.x u01 sin !x C u02 sin !.x / !.u01 cos !x C u02 cos !.x //
a/ dtI
a
/ if
D 0 D FI
sin !.x / sin !x u01 D F ; u02 D F I ! sin ! ! sin ! Z Z x 1 1 u1 D F .t/ sin !.t / dtI u2 D F .t/ sin !t dtI ! sin ! x ! sin ! 0 Z Z 1 yD sin !x F .t/ sin !.t / dt C sin !.x / F .t/ sin !t dt : ! sin ! x 0
If ! D n (positive integer), then y1ˇ D sin nx is a nontrivialˇsolution of y 00 C y D 0, y.0/ D 0, y. / D ˇ sin nx cos nx ˇˇ 0. Let y2 D cos nx; then W .x/ D ˇˇ D n, and yp D u1 sin nx C u2 cos nx n cos nx n sin nx ˇ
satisfies yp00 C n2 yp D 0 if
u01 sin nx C u02 cos nx nu01 cos nx nu02 sin nx
D 0 D FI
276 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations u01 D
Z Z 1 1 1 x F sin nx; u1 D F .t/ cos nt dt; u2 D F .t/ sin nt dt; n n x n 0 Z Z x 1 sin nx F .t/ cos nt dt C cos nx F .t/ sin nt dt I n x 0
1 F cos nx, u02 D n yp D
y D yp CZc1 sin nx C c2 cos nx. Since yp D 0, y.0/ D 0, so c2 D 0; y D yp C c1 sin nx. Since y. / D 0, F .t/ sin nt dt D 0 is necessary for existence of a solution. If this hold, then the solutions 0
are y D yp C c1 sin nx, with c1 arbitrary.
13.1.18. Let y1 D cos !x; y2 D sin !.x /; then y10 .0/ D y2 . / D 0, and ˇ ˇ ˇ cos !x sin !.x / ˇˇ ˇ W .x/ D ˇ D ! cos ! ¤ 0 ! sin !x ! cos !.x / ˇ
if and only if ! ¤ n C 1=2 (n D integer). If this is so, then y D u1 cos !x C u2 sin !.x y 00 C ! 2 y D F .x/ if !. then
u01
u01 cos !x C u02 sin !.x sin !x
C u02
cos !.x
/ satisfies
/ D 0
//!
D FI
F cos !x ; ! cos ! Z Z x 1 1 u1 D F .t/ sin !.t /; dt; u2 D F .t/ cos !t dt; ! cos ! x ! cos ! 0 Z Z x 1 yD sin !x F .t/ sin !.t /; dt C sin !.x / F .t/ cos !t dt : ! cos ! x 0 F sin !.x / ; ! cos !
u01 D
u02 D
If ! D n C 1=2 (n D integer), then y1 D cos.n C 1=2/x is a nontrivial solution y 00 C y D 0, y .0/ D y. / D 0. Let y2 D sin.n C 1=2/x; then ˇ ˇ ˇ ˇ cos.n C 1=2/x sin.n C 1=2/x ˇ ˇ D n C 1=2; W .x/ D ˇ .n C 1=2/ sin.n C 1=2/x .n C 1=2/ cos.n C 1=2/x ˇ 0
so yp D u1 cos.n C 1=2/x C u2 sin.n C 1=2/x satisfies yp00 C .n C 1=2/2 yp D F if u01 cos.n C 1=2/x C u02 sin.n C 1=2/x .n C 1=2/u01 sin.n C 1=2/x C .n C 1=2/u02 cos.n C 1=2/x u01 D
yp D
1 n C 1=2
yp0 D
u02 D
F cos.n C 1=2/x I n C 1=2
Z x F .t/ sin.n C 1=2/t F .t/ cos.n C 1=2/t u1 D dtI u2 D dtI n C 1=2 n C 1=2 x 0 Z Z x cos.n C 1=2/x F .t/ sin.n C 1=2/t dt C sin.n C 1=2/x F .t/ cos.n C 1=2/t dt I Z
F sin.n C 1=2/x I n C 1=2
D 0 D F:
sin.n C 1=2/x
x
Z
x
F .t/ sin.n C 1=2/t dt C cos.n C 1=2/x
0
Z
x
0
F .t/ cos.n C 1=2/t dtI
Section 13.1 Boundary Value Problems
0
yp0 .0/
y D
0
y D yp C c1 cos.n C 1=2/x C c2 sin.n C 1=2/xI
yp0
C .n C 1=2/. c1 sin.n C 1=2/x C c2 cos.n C 1=2//x:
D 0, y .0/ D 0 H) c2 D 0; y D yp C c1 cos.n C 1=2/x; Z 0 0 y D yp C .n C 1=2/c1 .n C 1=2/x; y. / D 0 H) yp . / D 0. Hence, Since
277
0
F .y/ cos.n C 1=2/t dt D 0
is a necessary condition for existence of a solution. If this holds, then the solutions are y D yp C c1 cos.n C 1=2/x with c1 arbitrary.
13.1.20. Suppose y D c1 ´1 C c2 ´2 is a nontrivial solution of the homogeneous boundary value problem. Then B1 .y/ D c1 B1 .´1 / C c2 B1 .´2 / D 0. From Theorem 13.1.1 we may assume without B1 .´1 / loss of generality that B1 .´2 / ¤ 0. Then c2 D c1 . Therefore, y is constant multiple of B1 .´2 / y0 D B1 .´2 /´1 B1.´1 /´2 ¤ 0. To that check y satisfies the boundary conditions, note that B1 .y0 / D B1 .´2 /B1 .´1 / B1 .´1 /B1 .´2 / D 0 B2 .y0 / D B1 .´2 /B2 .´1 / B1 .´1 /B2 .´2 / D 0, by Theorem 13.1.2. 13.1.22. y1 D a1 C a2 x; y1 .0/ 2y10 .0/ D a1 2a2 D 0 if a1 D 2, a2 D 1; y1 D 2 C x. y2 D b1 C b2x; y2 .1/ D 2y10 .1/ D b1 C 3b2 D 0 if b1 D 3, b2 D 1; y2 D 3 x. 8 ˆ ˇ ˇ ˆ .2 C t/.3 x/ ; 0 t x; < ˇ 2Cx 3 x ˇ 5 ˇ D 5I G.x; t/ D W .x/ D ˇˇ 1 1 ˇ ˆ .2 C x/.3 t/ ˆ : ; x t 1: 5 Z 1 Z x 1 yD .2 C x/ .3 t/F .t/ dt C .3 x/ .2 C t/F .t/ dt : (B) 5 x 0 (a) With F .x/ D 1, (B) becomes Z 1 Z x 1 y D .2 C x/ .3 t/ dt C .3 x/ .2 C t/ dt 5 x 0 2 2 x 6x C 5 x C 4x 1 D .2 C x/ C .3 x/ 5 2 2 2 x x 2 D : 2 (b) With F .x/ D x, (B) becomes Z 1 Z x 1 y D .2 C x/ .3t t 2 / dt C .3 x/ .2t C t 2 / dt 5 x 0 3 3 2 1 2x 9x C 7 x C 3x 2 D .2 C x/ C .3 x/ 5 6 3 3 5x 7x 14 D : 30 (b) With F .x/ D x 2 , (B) becomes Z 1 Z x 1 .2 C x/ .3t 2 t 3 / dt C .3 x/ .2t 2 C t 3 / dt y D 5 x 0 4 4 2 1 x 4x C 3 3x C 8x 3 D .2 C x/ C .3 x/ 5 4 12 4 5x 9x 18 D : 60
278 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations ˇ 2 ˇ x x, y2 D x 2 2x; then y1 .1/ D 0, y2 .2/ D 0; W .x/ D ˇˇ 2x 8 .t 1/x.x 2/ ˆ ˆ ; 1 t x; < t3 2 : Since P0 .x/ D x , G.x; t/ D ˆ x.x 1/.t 2/ ˆ : ; x t 2: t3 Z 2 Z x t 2 y D x.x 1/ F .t/ dt C x.x 2/ F .t/ dt: t3 x 1 13.1.24. y1 D x 2
(a) With F .x/ D 2x 3 , (B) becomes Z 2 y D 2x.x 1/ .t x
D
x.x
1/.x
2/ dt C 2x.x 2/2 C x.x
(b) With F .x/ D 6x 4, (B) becomes Z 2 y D 6x.x 1/ .t 2/t dt C 6x.x x
D
2x.x
1/.x C 1/.x
2/2 C x.x
2/
Z
2/
2/.x
Z
x 1
ˇ x 2 2x ˇˇ D x2. 2x 2 ˇ
(B)
x
1 2
.t
1/ dt
1/ D x.x
1/.x
2/:
x
.t
1/t dt
1
2/.x
1/2 .2x C 1/ D x.x
1/.x
2/.x C 3/:
13.1.26. y1 D a1 C a2 x; y10 D a2 ; B1 .y1 / D ˛a1 C ˇa2 D 0 if a1 D ˇ, a2 D ˛; y1 D ˇ ˛x. 0 y2 D b1 C b2 x; y2 D b2 ; B2.y2 / D b1 C . C ı/b2 D 0 if b1 D C ı, b2 D ; y2 D C ı x; ˇ ˛x C ı x W .x/ D D ˛. C ı/ ˇ. From Theorem 13.1.2, (A) has a unique solution ˛ if and only if ˛. C ı/ ˇ ¤ 0. Then 8 .ˇ ˛t/. C ı x/ ˆ ˆ ; 0 t x; < ˛. C ı/ ˇ G.x; t/ D .ˇ ˛x/. C ı t/ ˆ ˆ : ; x t 1: ˛. C ı/ ˇ
13.1.28. y1 D a1 cos x Ca2 sin x; y10 D a1 sin x Ca2 cos x; B1 .y1 / D ˛a1 Cˇa2 D 0 if a1 D ˇ, a2 D ˛. y1 D ˇ cos x ˛ sin x. y2 D b1 cos x C b2 sin x; y20 D b1 sin x C b2 cos x; B2 .y2 / D b2 ıb1 D ˇ cos x ˛ sin x cos x C ı sin x 0 if b1 D , b2 D ı; y2 D cos x C ı sin x; W .x/ D Since ˇ sin x ˛ cos x sin x C ı cos x ˇ W is constant, we can evaluate it with x D 0: W D D ˛ C ˇı. From Theorem 13.1.2, (A) ˛ ı has a unique solution if and only if ˛ C ˇı ¤ 0. Then 8 .ˇ cos t ˛ sin t/. cos x C ı sin x/ ˆ ˆ ; 0 t x; < ˛ C ˇı G.x; t/ D .ˇ cos x ˛ sin x/. cos t C ı sin t// ˆ ˆ : ; x t : ˛ C ˇı 13.1.30. y1 D e x .a1 cos x C a2 sin x/; y10 D e x Œa1 .cos x sin x/ C a2 .sin x C cos x/; B1 .y1 / D .˛ C ˇ/a1 C ˇa2 D 0 if a1 D ˇ, a2 D .˛ C ˇ/.
Section 13.2 Sturm-Liouville Problems y1 D e x .ˇ cos x .˛ C ˇ/ sin x. y2 D e x .b1 cos x C b2 sin x/; y20 D e x Œ.b1 .cos x sin x// C b2 .sin x C cos x/; B2 .y2 / D e =2Œ. C ı/b2 ıb1 D 0 if b1 D ı, b2 D . C ı/; y2 D e x Œ. C ı/ cos x C ı sin x/; To evaluate W .x/, we write y1 D e x v1 and y2 D e x v2 , where v1 D ˇ cos x .˛ C ˇ/ sin x and v2 D . C ı/ cos x C ı sin x. Then y10 D y1 C e x v10 and y2 D y2 C e x y20 , ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ y1 ˇ y1 y2 y2 ˇˇ v2 2x ˇ v1 ˇ ˇ W .x/ D ˇˇ D D e x 0 x 0 ˇ x 0 0 ˇ ˇ ˇ v 0 xv 0 y1 C e v1 y2 C e v2 e v1 xv2 1 2 ˇ ˇ ˇ .ˇ cos x .˛ C ˇ/ sin x ˇ . C ı/ cos x C ı sin x ˇ D ˇˇ : . ˇ sin x .˛ C ˇ/ cos x . . C ı/ sin x C ı cos x ˇ
279
ˇ ˇ ˇ ˇ
Since vi00 C vi D 0, i D 1, 2, Theorem 5.1.4 implies that W .x/ D Ke 2x , where is a constant that can be determined by setting x D 0 in the determinant: ˇ ˇ ˇ ˇ C ı ˇˇ W .x/e 2x ˇˇ D Œˇı C .˛ C ˇ/. C ı/: ˛ ˇ ı ˇ
From Theorem 13.1.2, the boundary value problem has a unique solution if and only if ˇı C .˛ C ˇ/. C ı/ ¤ 0. In this case the Green’s function is 8 Œˇ cos t .˛ C ˇ/ sin tŒ C ı/ cos x C ı sin x ˆ ˆ ; atx 0: r D 1 ˙ k; y D e x .c1 cosh kx C c2 sinh kx/; y 0 D c1e x . cosh kx C k sinh kx/ C c2 e x . sinh kx C k cosh kx/. The boundary conditions require that c1 C c2 k D 0 and . cosh k C k sinh k/c1 C . sinh k C k cosh k/c2 D 0. This system has a nontrivial solution if and only if .1 k 2 / sinh k D 0. Let k D 1 and c1 D c2 D 1; then D 1 is the only negative eigenvalue, with associated eigenfunction y D 1. D k 2 , k > 0: r D 1 ˙ i k; y D e x .c1 cos kx C c2 sin kx/; y 0 D c1 e x . cos kx k sin kx/ C c2 e x . sin x C k cos kx/. The boundary conditions require that c1 C c2 k D 0 and . cos k k sin k/c1 C . sin k C k cos k/c2 D 0. This system has a nontrivial solution if and only if .1 C k 2 / sin k D 0. Let k D n (k a positive integer)and c1 D n , c2 D 1; then n D n2 2 is an eigenvalue, with associated eigenfunction yn D e x .n cos n x C sin n x/. 13.2.12. Characteristic equation: r 2 C D 0. D 0 W y D c1 C c2x. y.0/ D 0 H) c1 D 0, so y D c2x. Now y.1/ 2y 0 .1/ D 0 H) c2 D 0. Therefore D 0 is not an eigenvalue. D k 2 , k > 0: y D c1 cosh kx C c2 sinh kx; y 0 D k.c1 sinh kx C c2 cosh kx/. y 0 .0/ H) c2 D 0, so y D c1 cosh kx. Now y.1/ 2y 0 .1/ D 0 H) c1.cosh k 2k sinh k/ D 0, which is possible with 1 . Graphing both sides of this equation on the same axes show that it c1 ¤ 0 if and only if tanh k D 2k has one positive solution k0 ; y0 D cosh k0 x is a k02 -eigefunction. D k 2 , k > 0: y D c1 cos kx C c2 sin kx; y 0 D k. c1 sin kx C c2 cos kx/. y 0 .0/ H) c2 D 0, so y D c1 cos kx. Now y.1/ 2y 0 .1/ D 0 H) c1.cos k C 2k sin k/ D 0, which is possible with c1 ¤ 0 1 if and only if tan k D . Graphing both sides of this equation on the same axes shows that it has a 2k solution kn in ..2n 1/=2; n /, n D 1, 2, 3, . . . ; yn D cos kn x is a kn2 -eigenfunction. 13.2.14. Characteristic equation: r 2 C D 0. D 0 W y D c1 C c2x. The boundary conditions require that c1 C 2c2 D 0 and c1 C c2 D 0, which imply that c1 D c2 D 0, so D 0 is not an eigenvalue. D k 2 , k > 0: y D c1 cosh kx C c2 sinh kx; y 0 D k.c1 sinh kx C c2 cosh kx/. The boundary conditions require that c1 C 2kc2 D 0 and c1 cosh k C c2 sinh 2k D 0. This system has a nontrivial solution if and only if tanh k D 2k. Graphing both sides of this equation
Section 13.2 Sturm-Liouville Problems
281
on the same axes shows that it has a solution k0 in .0; /; y0 D 2k0 cosh k0 x sinh k0 x is a k02 eigenfunction. D k 2 , k > 0: y D c1 cos kx C c2 sin kx; y 0 D k. c1 sin kx C c2 cos kx/. The boundary conditions require that c1 C 2kc2 D 0 and c1 cos k C c2 sin k D 0. This system has a nontrivial solution if and only if tan k D 2k. Graphing both sides of this equation on the same axes shows that it has a solution kn in .n; n C 1=2/, n D 1, 2,3, . . . ; yn D 2kn cos kn x sin kn x is a kn2 -eigenfunction. 13.2.16. Characteristic equation: r 2 C D 0. D 0 W y D c1 C c2x. The boundary conditions require that c1 C c2 D 0 and c1 C 4c2 D 0, so c1 D c2 D 0. Therefore D 0 is not an eigenvalue. D k 2 , k > 0: y D c1 cosh kx C c2 sinh kx; y 0 D k.c1 sinh kx C c2 cosh kx/. The boundary conditions require that c1 C kc2 D 0 and .cosh 2k C 2k sinh 2k/c1 C .sinh 2k C 2k cosh 2k/c2 D 0. k This system has a nontrivial solution if and only if tanh 2k D . Graphing both sides of this 1 p 2k 2 equation on the same axes shows that it has a solution k0 in .1= 2/; y0 D k0 cosh k0 x sinh k0 x is a k02 - eigenfunction. D k 2 , k > 0: y D c1 cos kx C c2 sin kx; y 0 D k. c1 sin kx C c2 cos kx/. The boundary conditions require that c1 C kc2 D 0 and .cos 2k 2k sin 2k/c1 C .sin 2k C 2k cos 2k/c2 D 0. k This system has a nontrivial solution if and only if tan 2k D . Graphing both sides of this 1 C 2k 2 equation on the same axes shows that it has a solution kn in ..2n 1/=4; n=2/, n D 1, 2,3, . . . ; yn D kn cos kn x sin kn x is a kn2 -eigenfunction. 13.2.18. Characteristic equation: r 2 C D 0. D 0 W y D c1 C c2 x. The boundary conditions require that 3c1 C 2c2 D 0 and 3c1 C 4c2 D 0, so c1 D c2 D 0. Therefore D 0 is not an eigenvalue. D k 2 , k > 0: y D c1 cosh kx C c2 sinh kx; y 0 D k.c1 sinh kx C c2 cosh kx/. The boundary conditions require that 3c1 C kc2 D 0 and .3 cosh 2k 2k sinh 2k/c1 C .3 sinh 3k 2k cosh 2k/c2 D 0. 9k This system has a nontrivial solution if and only if tanh 2k D . Graphing both sides of this equa9 C 2k 2 tion on the same axes shows that it has solutions y1 in .1; 2/ and y2 in .5=2; 7=2/; yn D kn cosh kn x 3 sinh kn x is a kn2 -eigenfunction, k D 1, 2. D k 2 , k > 0: y D c1 cos kx C c2 sin kx; y 0 D k. c1 sin kx C c2 cos kx/. The boundary conditions require that 3c1 C kc2 D 0 and .3 cos 2k C 2k sin 2k/c1 C .3 sin 2k 2k cos 2k/c2 D 0. 9k This system has a nontrivial solution if and only if tan 2k D . Graphing both sides of this 9 p 2k 2 equation on the same axes shows that it has solutions k0 in .3= 2; / and kn in ..2n C 3/=4; .n C 2/=3/, n D 1, 2,3, . . . ; yn D kn cos kn x 3 sin kn x is a kn2 -eigenfunction. 13.2.20. Characteristic equation: r 2 C D 0. D 0 W y D c1 C c2 x. The boundary conditions require that 5c1 C 2c2 D 0 and 5c1 C 3c2, so c1 D c2 D 0. Therefore D 0 is not an eigenvalue. D k 2 , k > 0: y D c1 cosh kx C c2 sinh kx; y 0 D k.c1 sinh kx C c2 cosh kx/. The boundary conditions require that 5c1 C 2kc2 D 0 and .5 cosh k 2k sinh k/c1 C .5 sinh k 2k cosh k/c2 D 0.
282 Chapter 13 Boundary Value Problems for Second Order Ordinary Differential Equations 20k . Graphing both sides of this equa25 C 4k 2 tion on the same axes shows that it has solutions k1 in .1; 2/ and k2 in .5=2; 7=2/; yn D 2kn cosh kn x sinh kn x is kn -eigenfunction, n D 1, 2. D k 2 , k > 0: y D c1 cos kx C c2 sin kx; y 0 D k. c1 sin kx C c2 cos kx/. The boundary conditions require that 5c1 C 2kc2 D 0 and .5 cos k C 2k sin k/c1 C .5 sin k 2k cos k/c2 D 0. 20k . Graphing both sides of this This system has a nontrivial solution if and only if tan k D 25 4k 2 equation on the same axes shows that it has a solution kn in ..2n C 1/=2; .n C 1/ /, n D 1, 2,3, . . . ; yn D 2kn cos kn x 3 sin kn x is a kn2 -eigenfunction. This system has a nontrivial solution if and only if tanh k D
13.2.22. D 0: x 2 y 00 2xy 0 C 2y D 0 is an Euler equation with indicial equation r .r 1/ 2r C 2 D .r 1/.r 2/ D 0. y D x.c1 Cc2x/; y.1/ D y.2/ D 0 H) c1 Cc2 D c1 C2c2 D 0 H) c1 D c2 D 0, so D 0 is not an eigenvalue. D k 2 ; k > 0: y D x.c1 cosh k.x 1/ C c2 sinh k.x 1//; y.1/ D 0 H) c1 D 0; y D c2 x sinh k.x 1/; y.2/ D 0 H) 2c2 sinh k D 0 H) c2 D 0; is not an eigenvalue. D k 2 ; k > 0: y D x.c1 cos k.x 1/Cc2 sin k.x 1//; y.1/ D 0 H) c1 D 0; y D c2 x sin k.x 1/; y.2/ D 0 with c2 ¤ 0 if k D n (n a positive integer); n D n2 2 ; yn D x sin n.x 1/ is a kn2 eigenfunction. 13.2.24. D 0: x 2 y 00 2xy 0 C 2y D 0 is an Euler equation with indicial equation r .r 1/ 2r C 2 D .r 1/.r 2/ D 0. y D x.c1 C c2 x/; y 0 D c1 C 2c2x; y.1/ D y 0 .2/ D 0 H) c1 C c2 D c1 C 4c2 D 0 H) c1 D c2 D 0, so D 0 is not an eigenvalue. D k 2 ; k > 0: y D x.c1 cosh k.x 1/ C c2 sinh k.x 1//; y.1/ D 0 H) c1 D 0; y D c2 x sinh k.x 1/; y 0 D c2 .sinh k.x 1/Ckx cosh k.x 1//; y 0 .2/ D 0 H) c2 .sinh k Ck cosh k/ H) c2 D 0; is not an eigenvalue. D k 2 ; k > 0: y D x.c1 cos k.x 1/Cc2 sin k.x 1//; y.1/ D 0 H) c1 D 0; y D c2 x sin k.x 1/; 0 y D c2.sin k.x 1/ C kx cos k.x 1//; y 0 .2/ D 0 with c2 ¤ 0 if and only if sin k C 2k cos k D 0 or, equivalently, tan k D 2k. Graphing both sides of this equation on the same axes shows that it has a solution kn in ..2n 1/=2; n /, n D 1, 2,3, . . . ; yn D x sin kn .x 1/ is a kn2 -eigenfunction. 13.2.26. D 0: y D c1 C c2x. The boundary conditions require that c1 C ˛c2 D 0 and c1 C . C ˛/c2 D 0, so c1 D c2 D 0. Therefore D 0 is not an eigenvalue of (A). D k 2 , k > 0: y D c1 cosh kx C c2 sinh kx; y 0 D k.c1 sinh kx C c2 cosh kx/. The boundary conditions require that c1 C ˛kc2 D 0 .cosh k C ˛k sinh k /c1 C .sinh k C ˛k cosh k /c2 D 0:
(D)
This system has a nontrivial solution if and only if .1 k 2 ˛ 2/ sinh k D 0, which holds with k > 0 if and only if k 2 D ˙1=˛. Therefore D 1=˛ 2 is the only negative eigenvalue. We can choose k D ˙1=˛. Either way, the first equation in (D) implies that e x=˛ is an associated eigenfunction. D k 2 , k > 0: y D c1 cos kx C c2 sin kx; y 0 D k. c1 sin kx C c2 cos kx/. The boundary conditions require that c1 C ˛kc2 D 0 (E) .cos k ˛k sin k /c1 C .sin k C ˛k cos k /c2 D 0: This system has a nontrivial solution if and only if .1 C k 2 ˛ 2 / sin k D 0. Choosing k D n produces eigenvalues n D n2 2. Setting k D n in the first equation in (E) yields c1 C ˛ nc2 D 0, so yn D n˛ cos nx sin nx.
Section 13.2 Sturm-Liouville Problems
283
13.2.28. y D c1 C c2 x. The boundary conditions require that ˛c1 C ˇc2 D 0 and c1 C .L C ı/c2 D 0. This system has a nontrivial solution if and only if ˛.L C ı/ ˇ D 0. 13.2.30. (a) y D c1 cos kx C c2 sin kx; y 0 D k. c1 sin kx C c2 cos kx/. The boundary conditions require that ˛c1 C ˇkc2 D 0 and . cos kL ık sin kL/c1 C . sin kL C ık cos kL/c2 D 0. This system has a nontrivial solution if and only if its determinant is zero. This implies the conclusion. (b) If ˛ı ˇ D 0, (A) reduces to .˛ C k 2 ˇı/ sin kL D 0:
(B)
From the solution of Exercise 13.2.29(b), ˛ C k 2 ˇı > 0 for all k > 0. Therefore the positive zeros of (B) are kn D n=L, n D 1, 2, 3, . . . , so the positive eigenvalues (SL) are n D n2 2=L2 , n D 1, 2, 3, .... 13.2.32. Suppose is an eigenvalue and y is an associated eigenfunction. From the solution of Exercise 13.2.31,
Z
b 2
a
0
r .x/y .x/ dx D p.a/y.a/y .a/
0
p.b/y.b/y .b/ C
Z
b
p.x/.y 0 .x//2 dx:
If ˛ˇ D 0 then either y.a/ D 0 or y 0 .a/ D 0, so y.a/y 0 .a/ D 0. If ˛ˇ < 0 then y.a/ D y.a/y 0 .a/ D
(A)
a
ˇ 0 .y .a//2 : ˛
ˇ 0 y .a/, so ˛ (B)
Moreover, y 0 .a/ ¤ 0 because if y 0 .a/ D 0 then y.a/ D 0, from (B), and y 0, a contradiction. Since ˇ > 0 if ˛ˇ < 0, we conclude that if ˛ˇ 0, then ˛ p.a/y.a/y 0 .a/ 0;
(C)
with equality if and only if ı D 0. A similar argument shows that if ı 0, then p.b/y.b/y 0 .b/ 0;
(D)
with equality if and only if ˛ˇ D 0. Since .˛ˇ/2 C .ı/2 > 0, the inequality must hold in at least one of (C) and (D). Now (A) implies that > 0.
