Y

PART

A

Ordinary Differential Equations (ODEs)

Part A p1

Advanced Engineering Mathematics, 10/e by Erwin Kreyszig Copyright 2011 by John Wiley & Sons. All rights reserved.

CHAPTER

2

Second-Order Linear ODEs

Chapter 2 p2


2.1 2.1 Homogeneous Homogeneous Linear Linear ODEs ODEs of of Second Second Order Order

Section 2.1 p3


2.1 Homogeneous Linear ODEs of Second Order

A second‐order ODE is called linear if it can be written A second‐order ODE is called linear if it can be written (1) y” (1) y”+ p(x)y’ + p(x)y’+ q(x)y = + q(x)y =r(x) r(x) and nonlinear if it cannot be written in this form. and nonlinear if it cannot be written in this form.

Section 2.1 p4



If r(x) ≣ If r(x) ≣0 (that is, r(x) = 0 for all x considered; read “r(x) is 0 (that is, r(x) = 0 for all x considered; read “r(x) is identically zero”), then (1) reduces to identically zero”), then (1) reduces to (2) y” + p(x)y’ + q(x)y = 0 (2) y” + p(x)y’ + q(x)y = 0 and is called homogeneous. If r(x) ≣ and is called homogeneous. If r(x) ≣0, then (1) is called 0, then (1) is called nonhomogeneous. nonhomogeneous.

Section 2.1 p5



An example of a nonhomogeneous linear ODE is An example of a nonhomogeneous linear ODE is −x cos x, y” + 25y = e y” + 25y = e−x cos x,

and a homogeneous linear ODE is and a homogeneous linear ODE is xy” xy”+ y’ + y’+ xy = + xy =0, 0, written in standard form written in standard form 1 ′′ y + y′ + y = 0. x

Section 2.1 p6



An example of a nonlinear ODE is An example of a nonlinear ODE is 2 = 0. y”y + y’ y”y + y’2 = 0.

The functions p and q in (1) and (2) are called the The functions p and q in (1) and (2) are called the coefficients coefficients of the ODEs. of the ODEs.

Section 2.1 p7



Solutions Solutions are defined similarly as for first‐order ODEs in are defined similarly as for first‐order ODEs in Chap. 1. Chap. 1. A function A function

y = y =h(x) h(x) is called a solution of a (linear or nonlinear) second‐order is called a solution of a (linear or nonlinear) second‐order ODE on some open interval I if h is defined and twice ODE on some open interval I if h is defined and twice differentiable throughout that interval and is such that the differentiable throughout that interval and is such that the ODE becomes an identity if we replace the unknown y by h, ODE becomes an identity if we replace the unknown y by h, the derivative y’ the derivative y’by h’, and the second derivative y” by h’, and the second derivative y”by h”. by h”.

Section 2.1 p8



Homogeneous Linear ODEs: Superposition Principle Linear ODEs have a rich solution structure. For the Linear ODEs have a rich solution structure. For the homogeneous equation the backbone of this structure is the homogeneous equation the backbone of this structure is the superposition principle or linearity principle, which says that superposition principle or linearity principle, which says that we can obtain further solutions from given ones by adding we can obtain further solutions from given ones by adding them or by multiplying them with any constants. them or by multiplying them with any constants. Of course, this is a great advantage of homogeneous linear Of course, this is a great advantage of homogeneous linear ODEs. ODEs.

Section 2.1 p9



Theorem 1 Fundamental Theorem Fundamental Theorem for the Homogeneous Linear ODE (2) for the Homogeneous Linear ODE (2)

For a homogeneous linear ODE (2), any linear combination of For a homogeneous linear ODE (2), any linear combination of two solutions on an open interval I is again a solution of (2) on I. two solutions on an open interval I is again a solution of (2) on I. In particular, for such an equation, sums and constant multiples In particular, for such an equation, sums and constant multiples of solutions are again solutions. of solutions are again solutions.

Section 2.1 p10



Initial Value Problem. Basis. General Solution For a second‐order homogeneous linear ODE (2) an For a second‐order homogeneous linear ODE (2) an initial initial value problem value problem consists of (2) and two consists of (2) and two initial conditions initial conditions y(x y’(x (4) (4) y(x00) = K ) = K00, , y’(x00) = K ) = K11. . These conditions prescribe given values K These conditions prescribe given values K00and K and K11of the of the solution and its first derivative (the slope of its curve) at the solution and its first derivative (the slope of its curve) at the same given x = same given x =xx0 in the open interval considered. in the open interval considered. 0

Section 2.1 p11



The conditions (4) are used to determine the two arbitrary The conditions (4) are used to determine the two arbitrary constants c constants c1 and c and c2 in a in a general solution general solution 1

2

y = (5) (5) y =cc1y1y11+ c + c22yy22 of the ODE; of the ODE; here, y here, y11and y and y22are suitable solutions of the ODE. are suitable solutions of the ODE. This results in a unique solution, passing through the This results in a unique solution, passing through the point (x point (x00, K , K00) with K ) with K11as the tangent direction (the slope) at as the tangent direction (the slope) at that point. that point. That solution is called a That solution is called a particular solution particular solution of the ODE (2). of the ODE (2).

Section 2.1 p12



Definition General Solution, Basis, Particular Solution General Solution, Basis, Particular Solution A A general solution general solution of an ODE (2) on an open interval I is a of an ODE (2) on an open interval I is a solution (5) in which y solution (5) in which y1 1 and y and y22are solutions of (2) on I that are are solutions of (2) on I that are not proportional, and c not proportional, and c11and c and c22are arbitrary constants. are arbitrary constants. These y , y22are called a are called a basis basis (or a (or a fundamental system fundamental system) of ) of These y11, y solutions of (2) on I. solutions of (2) on I. A A particular solution particular solution of (2) on I is obtained if we assign of (2) on I is obtained if we assign specific values to c specific values to c1 and c and c2 in (5). in (5). 1

Section 2.1 p13

2



Furthermore, Furthermore,as asusual, usual,yy11and andyy22are arecalled called proportional proportionalon onIIififfor forall allxxon onI,I, (b) (6) (a) ky22 or or (b)yy22==lyly11 (6) (a)yy11==ky where wherekkand andl lare arenumbers, numbers,zero zeroor ornot. not.(Note (Notethat that(a) (a)implies implies(b) (b)ifif and andonly onlyififkk≠ ≠0). 0).

Section 2.1 p14



Two functions y Two functions y11and y and y22are called are called linearly independent linearly independent on an on an interval I where they are defined if interval I where they are defined if (7) 1 = 0 and k 2 = 0. (7)kk1y1y11(x) + k (x) + k22yy2(x) = 0 everywhere on I implies k (x) = 0 everywhere on I implies k = 0 and k 2 1 2 = 0. And y And y11and y and y22are called are called linearly dependent linearly dependent on I if (7) also on I if (7) also holds for some constants k holds for some constants k11, k , k2 2 not both zero. not both zero. Then, if k Then, if k11≠ ≠0 or k 0 or k22≠ ≠0, we can divide and see that y 0, we can divide and see that y11and y and y22 are are k2 k1 proportional, proportional,y1 = − k y2 or y2 = − k y1 . 1 2 In contrast, in the case of linear independence these functions In contrast, in the case of linear independence these functions are not proportional because then we cannot divide in (7). This are not proportional because then we cannot divide in (7). This gives the following definition. ( .) gives the following definition. (See next slide See next slide.) Section 2.1 p15



Definition Basis (Reformulated) Basis (Reformulated) A A basis basis of solutions of (2) on an open interval I is a pair of of solutions of (2) on an open interval I is a pair of linearly independent solutions of (2) on I. linearly independent solutions of (2) on I.

Section 2.1 p16



Find a Basis If One Solution Is Known. Reduction of Order It happens quite often that one solution can be found by inspection or in some other way. Then a second linearly independent solution can be obtained by solving a first-order ODE. This is called the method of reduction of order. We first show how this method works in an example and then in general.

Section 2.1 p17



EXAMPLE 7 Reduction of Order If a Solution Is Known. Basis

Find a basis of solutions of the ODE (x2 − x)y” − xy’ + y = 0. Solution. (See next slide.)

Section 2.1 p18



EXAMPLE 7 (continued)

Solution. Inspection shows that y1 = x is a solution because y’1 = 1 and y”1 = 0, so that the first term vanishes identically and the second and third terms cancel. The idea of the method is to substitute

y = uy1 = ux,

y’ = u’x + u, y” = u”x + 2u’

into the ODE. This gives (x2 − x)(u”x + 2u’) − x(u’x + u) + ux = 0. ux and –xu cancel and we are left with the following ODE, which we divide by x, order, and simplify, (x2 − x)(u”x + 2u’) − x2u’ = 0, 2)u’ = 0. Section 2.1 p19

(x2 − x)u” + (x –



EXAMPLE 7 (continued)

Solution. (continued 1) This ODE is of first order in v = u’, namely, (x2 − x)v’ + (x – 2)v = 0. Separation of variables and integration gives

2⎞ dv x−2 ⎛ 1 =− 2 − ⎟ dx , dx = ⎜ v x −x ⎝ x −1 x⎠ x −1 ln v = ln x − 1 − 2 ln x = ln 2 . x (continued)

Section 2.1 p20


EXAMPLE 7

2.1 Homogeneous Linear ODEs of Second Order (continued)

Solution. (continued 2) We need no constant of integration because we want to obtain a particular solution; similarly in the next integration. Taking exponents and integrating again, we obtain x −1 1 1 1 v = 2 = − 2 , u = ∫ v dx = ln x + x x x x hence y2 = ux = x ln x + 1. Since y1 = x and y2 = x ln|x| + 1 are linearly independent (their quotient is not constant), we have obtained a basis of solutions, valid for all positive x.

Section 2.1 p21


2.2 2.2 Homogeneous Homogeneous Linear Linear ODEs ODEs with with Constant Constant Coefficients Coefficients

Section 2.2 p22


2.2 Homogeneous Linear ODEs with Constant Coefficients

We shall now consider second‐order homogeneous linear We shall now consider second‐order homogeneous linear ODEs whose coefficients a and b are constant, ODEs whose coefficients a and b are constant, (1) (1)

y” y”+ ay’ + ay’+ by = + by =0.0.

These equations have important applications in mechanical These equations have important applications in mechanical and electrical vibrations. and electrical vibrations.

Section 2.2 p23



To solve (1), we recall from Sec. 1.5 that the solution of the To solve (1), we recall from Sec. 1.5 that the solution of the first‐order linear ODE with a constant coefficient k first‐order linear ODE with a constant coefficient k y’y’+ ky = + ky =00 −kx. is an exponential function y = ce is an exponential function y = ce−kx. This gives us the idea to try as a solution of (1) the function This gives us the idea to try as a solution of (1) the function

(2) (2)

Section 2.2 p24

λx. y = e y = eλx.



Substituting (2) and its derivatives Substituting (2) and its derivatives λx and y’’ = λ2eλ x y’y’= λe = λeλx and y’’ = λ2eλ x

into our equation (1), we obtain into our equation (1), we obtain 2 + aλ + b)eλx = 0. (λ (λ2 + aλ + b)eλx = 0. Hence if λ Hence if λis a solution of the important is a solution of the important characteristic characteristic equation equation (or auxiliary equation) (or auxiliary equation) (3) (3)

λλ22+ aλ + aλ+ b + b= 0 = 0

then the exponential function (2) is a solution of the ODE then the exponential function (2) is a solution of the ODE (1). (1). Section 2.2 p25



Now from algebra we recall that the roots of this Now from algebra we recall that the roots of this quadratic equation (3) are quadratic equation (3) are 1 1 2 2 (4) λ = − + − λ = − − − 4b ). ( a a 4 b ), ( a a 2 (4) 1 2 2 (3) and (4) will be basic because our derivation shows that (3) and (4) will be basic because our derivation shows that the functions the functions (5) (5)

y1 = e λ1x and y2 = e λ2 x

are solutions of (1). are solutions of (1).

Section 2.2 p26



From algebra we further know that the quadratic From algebra we further know that the quadratic equation (3) may have three kinds of roots, depending on equation (3) may have three kinds of roots, depending on 2 − 4b, namely, the sign of the discriminant a the sign of the discriminant a2 − 4b, namely, (Case I) (Case I)

2 − 4b > 0, Two real roots if a Two real roots if a2 − 4b > 0,

(Case II) (Case II)

2 − 4b = 0, A real double root if a A real double root if a2 − 4b = 0,

(Case III) (Case III)

2 − 4b 0) where ω where ω2= b = b −−(¼)a (¼)a2. .

Section 2.2 p35



1 1 Case III. Complex Root − a + iω and − a − iω 2 2 (continued) It can be verified by substitution that these are solutions It can be verified by substitution that these are solutions in the present case. in the present case. They form a basis on any interval since their quotient They form a basis on any interval since their quotient cot ωx is not constant. cot ωx is not constant. Hence a real general solution in Case III is Hence a real general solution in Case III is (9) (9)

−ax/2 (A cos ωx + B sin ωx) y = e y = e−ax/2 (A cos ωx + B sin ωx)

Section 2.2 p36

(A, B arbitrary) (A, B arbitrary)



Summary of Cases I−III Case I

II

III

Roots of (2) Basis of (1) Distinct real λ1, λ2

e λ1x , e λ2 x

Real double root e−ax/2, xe−ax/2 λ = (−½)a Complex conjugate e−ax/2 cos ωx λ1 = (−½)a + iω e−ax/2 sin ωx λ2 = (−½)a − iω

Section 2.2 p37

General Solution of (1) y = c1e λ1x + c2 e λ2 x .

y = (c1 + c2x)e−ax/2.

y = e−ax/2 (A cos ωx + B sin ωx)



Derivation in Case III. Complex Exponential Function it = cos t + i sin t, (11) e (11) eit = cos t + i sin t,

called the Euler formula. . called the Euler formula

Section 2.2 p38


2.3 2.3 Differential Differential Operators. Operators. Optional Optional

Section 2.3 p39


2.3 Differential Operators. Optional

Operational calculus Operational calculus means the technique and application means the technique and application of operators. of operators. Here, an Here, an operator operator is a transformation that transforms a is a transformation that transforms a function into another function. function into another function. Hence differential calculus involves an operator, the Hence differential calculus involves an operator, the differential operator differential operator D, which transforms a (differentiable) D, which transforms a (differentiable) function into its derivative. function into its derivative. In operator notation we write d/dx In operator notation we write DD==d/dx and and (1) Dy = (1) Dy =y’y’= dy/dx. = dy/dx.

Section 2.3 p40



For a homogeneous linear ODE y” For a homogeneous linear ODE y”+ ay’ + ay’+ by = + by =0 with 0 with constant coefficients we can now introduce the constant coefficients we can now introduce the second‐ second‐ order differential operator order differential operator 2 + aD + bI, L = P(D) = D L = P(D) = D2 + aD + bI, where I is the identity operator defined by Iy = defined by Iy =y. y. where I is the identity operator Then we can write that ODE as Then we can write that ODE as

(2) (2)

2 + aD + bI)y = 0. Ly = P(D)y = (D Ly = P(D)y = (D2 + aD + bI)y = 0.

Section 2.3 p41



P suggests “polynomial.” P suggests “polynomial.”L is a L is a linear operator linear operator. . By definition this means that if Ly and By definition this means that if Ly and Lw Lwexist (this is the exist (this is the case if y and w are twice differentiable), then case if y and w are twice differentiable), then L(cy L(cy++kw) kw) exists for any constants c and k, and exists for any constants c and k, and L(cy L(cy++kw) kw)==cLy cLy++kLw. kLw. The point of this operational calculus is that P(D) can be treated The point of this operational calculus is that P(D) can be treated just like an algebraic quantity. just like an algebraic quantity.

Section 2.3 p42


2.4 2.4 Modeling Modeling of of Free Free Oscillations Oscillations of of aa Mass—Spring Mass—Spring System System

Section 2.4 p43


2.4 Modeling of Free Oscillations of a Mass—Spring System

Linear ODEs with constant coefficients have important Linear ODEs with constant coefficients have important applications in mechanics, as we show in this section as applications in mechanics, as we show in this section as well as in Sec. 2.8, and in electrical circuits as we show in well as in Sec. 2.8, and in electrical circuits as we show in Sec. 2.9. Sec. 2.9. In this section we model and solve a basic mechanical In this section we model and solve a basic mechanical system consisting of a mass on an elastic spring (a so‐ system consisting of a mass on an elastic spring (a so‐ called “mass–spring system,” called “mass–spring system,”Fig. 33), which moves up Fig. 33), which moves up and down. and down.

Section 2.4 p44



Setting Up the Model We take an ordinary coil spring that resists extension as We take an ordinary coil spring that resists extension as well as compression. well as compression. We suspend it vertically from a fixed support and attach a We suspend it vertically from a fixed support and attach a body at its lower end, for instance, an iron ball, as shown body at its lower end, for instance, an iron ball, as shown in Fig. 33. in Fig. 33. We let y = We let y =0 denote the position of the ball when the 0 denote the position of the ball when the system is at rest (Fig. 33b). system is at rest (Fig. 33b). Furthermore, we choose Furthermore, we choose the downward direction as the downward direction as positive positive, thus regarding downward forces as positive and , thus regarding downward forces as positive and upward forces as negative. upward forces as negative. Section 2.4 p45



Setting Up the Model (continued 1)

Section 2.4 p46



Setting Up the Model (continued 2) We now let the ball move, as follows. We pull it down by We now let the ball move, as follows. We pull it down by an amount y > an amount y >0 (Fig. 33c). 0 (Fig. 33c). This causes a spring force This causes a spring force (1) FF1 = = −−ky (Hookes’s law) (1) ky (Hookes’s law) 1 proportional to the stretch y, with k (> 0) called the proportional to the stretch y, with k (> 0) called the spring spring constant constant. . The minus sign indicates that F The minus sign indicates that F11points upward, against points upward, against the displacement. the displacement. It is a restoring force: It wants to restore the system, that is, It is a restoring force: It wants to restore the system, that is, to pull it back to y = to pull it back to y =0. 0. Stiff springs have large k. Stiff springs have large k. Section 2.4 p47



Setting Up the Model (continued 3) Note that an additional force Note that an additional force −−FF00is present in the spring, is present in the spring, caused by stretching it in fastening the ball, but F caused by stretching it in fastening the ball, but F00has no has no effect on the motion because it is in equilibrium with the effect on the motion because it is in equilibrium with the weight W of the ball, weight W of the ball, −−FF0 = W = mg, = W = mg, 2 0 2 = 32.17 ft/sec2 is where g = 980 cm/sec = 9.8 m/sec 2 where g = 980 cm/sec = 9.8 m/sec2 = 32.17 ft/sec2 is the the constant of gravity at the Earth’s surface constant of gravity at the Earth’s surface (not to be (not to be confused with the universal gravitational constant confused with the universal gravitational constant 2/M = 6.67 ∙ 10−11 nt m2/kg2, which we shall not need; G = gR −11 nt m2/kg2, which we shall not need; 10 G = gR2/M = 6.67 ∙ 6 m and M = 5.98 ∙ 1024 kg here R = 6.37 ∙ 10 here R = 6.37 ∙ 106 m and M = 5.98 ∙ 1024 kg are the Earth’s radius and mass, respectively). are the Earth’s radius and mass, respectively).

Section 2.4 p48



Setting Up the Model (continued 4) The motion of our mass–spring system is determined by The motion of our mass–spring system is determined by Newton’s second law Newton’s second law (2) (2) Mass Mass ××Acceleration = my” Acceleration = my”= Force = Force 2y/dt2 and “Force” is the resultant of all the where y” = d where y” = d2y/dt2 and “Force” is the resultant of all the forces acting on the ball. forces acting on the ball.

Section 2.4 p49



ODE of the Undamped System Every system has damping. Otherwise it would keep Every system has damping. Otherwise it would keep moving forever. moving forever. But if the damping is small and the motion of the system But if the damping is small and the motion of the system is considered over a relatively short time, we may is considered over a relatively short time, we may disregard damping. disregard damping. Then Newton’s law with F = Then Newton’s law with F =−−FF11gives the model gives the model my” my”= = −−FF1 = = −−ky; thus ky; thus 1

(3) (3)

Section 2.4 p50

my” my”+ ky = + ky =0.0.



ODE of the Undamped System (continued 1) This is a homogeneous linear ODE with constant This is a homogeneous linear ODE with constant coefficients. A general solution is obtained, namely coefficients. A general solution is obtained, namely k ω0 = . m This motion is called a harmonic oscillation (Fig. 34, see (Fig. 34, see This motion is called a harmonic oscillation next slide). next slide).

(4) (4)

y(t) = A cos ω y(t) = A cos ω00t + t +B sin ω B sin ω00tt

Section 2.4 p51



ODE of the Undamped System (continued 2)

Section 2.4 p52



ODE of the Undamped System (continued 3) Its frequency is f = Its frequency is f =ωω00/2/2ππHertz (= cycles/sec) because cos Hertz (= cycles/sec) because cos and sin in (4) have the period 2 and sin in (4) have the period 2ππ/ω /ω00. . The frequency f is called the The frequency f is called the natural frequency natural frequency of the of the system. system.

Section 2.4 p53



ODE of the Undamped System (continued 4) An alternative representation of (4), which shows the An alternative representation of (4), which shows the physical characteristics of amplitude and phase shift of (4), physical characteristics of amplitude and phase shift of (4), isis (4*) y(t) = C cos (ω (4*) y(t) = C cos (ω0t t −−δ)δ) 0

2 2 with C = A + B with and phase angle δ, where tan δ and phase angle δ, where tan δ= B/A. = B/A.

Section 2.4 p54



ODE of the Damped System To our model my” To our model my”= = −−ky we now add a damping force ky we now add a damping force FF2 = = −−cy’, cy’, 2 obtaining my” ky −−cy’; cy’; obtaining my”= = −−ky thus the ODE of the damped mass–spring system is thus the ODE of the damped mass–spring system is (5) my” (Fig. 36) (5) my”+ cy’ + cy’+ ky = + ky =0.0. (Fig. 36) Physically this can be done by connecting the ball to a Physically this can be done by connecting the ball to a dashpot; see Fig. 36 (next slide). dashpot; see Fig. 36 (next slide). We assume this damping force to be proportional to the We assume this damping force to be proportional to the velocity y’ velocity y’= dy/dt. This is generally a good approximation = dy/dt. This is generally a good approximation for small velocities. for small velocities.

Section 2.4 p55



ODE of the Damped System (continued 1)

Section 2.4 p56



ODE of the Damped System (continued 2) The constant c is called the damping constant. Let us show The constant c is called the damping constant. Let us show that c is positive. that c is positive. Indeed, the damping force F cy’acts against the motion; acts against the motion; Indeed, the damping force F22= = −−cy’ hence for a downward motion we have y’ hence for a downward motion we have y’> 0, which for > 0, which for positive c makes F negative (an upward force), as it positive c makes F negative (an upward force), as it should be. should be. Similarly, for an upward motion we have y’ Similarly, for an upward motion we have y’0 makes F 0 makes F2 positive (a downward force). positive (a downward force). 2

Section 2.4 p57



ODE of the Damped System (continued 3) The ODE (5) is homogeneous linear and has constant The ODE (5) is homogeneous linear and has constant coefficients. Hence we can solve it by the method in coefficients. Hence we can solve it by the method in Sec. 2.2. The characteristic equation is (divide (5) by m) Sec. 2.2. The characteristic equation is (divide (5) by m) c k 2 λ + λ + = 0. m m By the usual formula for the roots of a quadratic equation By the usual formula for the roots of a quadratic equation we obtain, as in Sec. 2.2, we obtain, as in Sec. 2.2, (6) (6) λλ11==−α −α++β,β,λλ11==−α −α−−ββ c 1 where α = and β = c 2 − 4 mk . 2m 2m

Section 2.4 p58



ODE of the Damped System (continued 4) It is now interesting that depending on the amount of It is now interesting that depending on the amount of damping present—whether a lot of damping, a medium damping present—whether a lot of damping, a medium amount of damping, or little damping—three types of amount of damping, or little damping—three types of motions occur, respectively: motions occur, respectively: 2 > 4mk. Distinct real roots λ , λ . (Overdamping) Case I. c Case I. c2 > 4mk. Distinct real roots λ11, λ22. (Overdamping) 2 = 4mk. A real double root. (Critical damping) c Case II. Case II. c2 = 4mk. A real double root. (Critical damping) 2 4mk, then λλ1 and λ 2 are distinct real roots. and λ 1 2 are distinct real roots. In this case the corresponding general solution of (5) is In this case the corresponding general solution of (5) is −(α−β)t + c e−(α+β)t. (7) y(t) = c e 1 (7) y(t) = c1e−(α−β)t + c22e−(α+β)t. We see that in this case, damping takes out energy so We see that in this case, damping takes out energy so quickly that the body does not oscillate. quickly that the body does not oscillate. > 0, For t > 0 both exponents in (7) are negative because α > 0, For t >0 both exponents in (7) are negative because α 2 2 2 ββ> 0, and β > 0, and β2= α = α2−−k/m < k/m 0)is, is,the thelarger largerisisω* ω*and andthe themore morerapid rapid the theoscillations oscillationsbecome. become.IfIfccapproaches approaches0,0,then thenω* ω*approaches approaches ω0 = k / m , giving givingthe theharmonic harmonicoscillation oscillation(4), (4),whose whosefrequency frequencyωω00/(2π) /(2π) isisthe thenatural naturalfrequency frequencyof ofthe thesystem. system.

Section 2.4 p66



Discussion of the Three Cases (continued 7) Case III. Case III. Underdamping Underdamping (continued 3) (continued 3)

Section 2.4 p67


2.5 2.5 Euler—Cauchy Euler—Cauchy Equations Equations

Section 2.5 p68


2.5 Euler—Cauchy Equations

Euler–Cauchy equations are ODEs of the form Euler–Cauchy equations are ODEs of the form 2 (1) xx2y” (1) y”+ axy’ + axy’+ by = + by =00 with given constants a and b and unknown function with given constants a and b and unknown functiony(x). y(x). We substitute We substitute m m−1, m−2 y = x , y’ = mx y” = m(m − 1)x m m−1 y = x , y’ = mx , y” = m(m − 1)xm−2 into (1). This gives into (1). This gives m−2 + axmxm−1 + bxm = 0 xx22m(m − 1)x m(m − 1)xm−2m+ axmxm−1 + bxm = 0 and we now see that y = and we now see that y =xxmwas a rather natural choice was a rather natural choice m. Dropping it, because we have obtained a common factor x because we have obtained a common factor xm. Dropping it, we have the auxiliary equation m(m −−1) + am + b = 0 or we have the auxiliary equation m(m 1) + am + b = 0 or 2 (2) (Note: a (2) mm2+ (a + (a −−1)m + 1)m +b = b =0. 0. (Note: a −−1, not a.) 1, not a.)

Section 2.5 p69



Euler–Cauchy equations Euler–Cauchy equations (continued) (continued) m is a solution of (1) if and only if m is a root of Hence y = x Hence y = xm is a solution of (1) if and only if m is a root of (2). The roots of (2) are (2). The roots of (2) are

1 1 1 1 2 2 (3) = − + − − = − − − m (1 a ) (1 a ) b , m (1 a ) (1 a ) − b. 2 (3) 1 2 4 2 4

Section 2.5 p70



Case I. Real different roots Case I. Real different roots mm11and m and m22give two real give two real solutions solutions y1 ( x) = x m1 and y2 ( x) = x m2 . These are linearly independent since their quotient is not These are linearly independent since their quotient is not constant. Hence they constitute a basis of solutions of (1) for constant. Hence they constitute a basis of solutions of (1) for all x for which they are real. The corresponding general all x for which they are real. The corresponding general solution for all these x is solution for all these x is (4) (4)

Section 2.5 p71

y1 ( x) = c1 x m1 + c2 x m2

(c(c1, c 2 arbitrary). , c 1 2 arbitrary).



Case II. A real double root Case II. A real double root 1 1 m1 = (1 − a) occurs if and only if b = ( a − 1)2 because then 2 4 1 (2) becomes [m + ( a − 1)]2 , as can be readily verified. Then 2 (1−a)/2, and (1) is of the form a solution is y = x a solution is y1 = x(1−a)/2, and (1) is of the form 1

2 1 a (1 − a ) 2 2 (5) ′′ ′ (1 ) x y + axy + − a y = 0 or y′′ + y′ + y = 0. 2 (5) 4 x 4x

Section 2.5 p72



Case II. Case II. (continued) (continued)

Thus, y Thus, y22= uy = uy11= y = y11ln x, and y ln x, and y11and y and y2 2 are linearly are linearly independent since their quotient is not constant. The independent since their quotient is not constant. The general solution corresponding to this basis is general solution corresponding to this basis is (6) (6)

Section 2.5 p73

m, y = (c + c ln x) x y = (c11 + c22 ln x) xm,

1 m = (1 − a). 2



Case III. Complex conjugate roots Case III. Complex conjugate roots are of minor practical are of minor practical importance, and we discuss the derivation of real importance, and we discuss the derivation of real solutions from complex ones just in terms of a typical solutions from complex ones just in terms of a typical example. example.

Section 2.5 p74



EXAMPLE 3 Real General Solution in the Case of Complex Roots Real General Solution in the Case of Complex Roots 2y” + 0.6xy’ + 16.04y = 0 The Euler–Cauchy equation x The Euler–Cauchy equation x2y” + 0.6xy’ + 16.04y = 0 2 − 0.4m + 16.04 = 0. has the auxiliary equation m has the auxiliary equation m2 − 0.4m + 16.04 = 0. The roots are complex conjugate The roots are complex conjugate i = −1. mm1 = 0.2 + 4i and m = 0.2 + 4i and m2 = 0.2 = 0.2 −−4i, where 4i, where 1

2

The corresponding real general solution for all positive x is The corresponding real general solution for all positive x is (8) (8)

0.2[A cos (4 ln x) + B sin (4 ln x)]. y = x y = x0.2[A cos (4 ln x) + B sin (4 ln x)].

Section 2.5 p75



Figure 48 shows typical solution curves in the three cases Figure 48 shows typical solution curves in the three cases discussed, in particular the real basis functions in Examples discussed, in particular the real basis functions in Examples 1 and 3. 1 and 3.

Section 2.5 p76


2.6 2.6 Existence Existence and and Uniqueness Uniqueness of of Solutions. Solutions. Wronskian Wronskian

Section 2.6 p77


2.6 Existence and Uniqueness of Solutions. Wronskian

In this section we shall discuss the general theory of In this section we shall discuss the general theory of homogeneous linear ODEs homogeneous linear ODEs (1) y” (1) y”+ p(x)y’ + p(x)y’+ q(x)y = + q(x)y =00 with continuous, but otherwise arbitrary, with continuous, but otherwise arbitrary, variable variable coefficients coefficients p and q. p and q. This will concern the existence and form of a general This will concern the existence and form of a general solution of (1) as well as the uniqueness of the solution of solution of (1) as well as the uniqueness of the solution of initial value problems consisting of such an ODE and two initial value problems consisting of such an ODE and two initial conditions initial conditions (2) y(x (2) y(x00) = K ) = K00, , y’(x y’(x00) = K ) = K11 with given x with given x0, K , K0, and K , and K1. . 0

Section 2.6 p78

0

1



The two main results will be Theorem 1, stating that such The two main results will be Theorem 1, stating that such an initial value problem always has a solution which is an initial value problem always has a solution which is unique, and Theorem 4, stating that a general solution unique, and Theorem 4, stating that a general solution (c(c1, c y = (3) 2 arbitrary) + c22yy22 , c (3) y =cc1y1y11+ c 1 2 arbitrary) includes all solutions. Hence linear ODEs with continuous includes all solutions. Hence linear ODEs with continuous coefficients have no “singular solutions” coefficients have no “singular solutions”(solutions not (solutions not obtainable from a general solution). obtainable from a general solution).

Section 2.6 p79



Theorem 1 Existence Existenceand andUniqueness UniquenessTheorem Theorem for forInitial InitialValue ValueProblems Problems

If p(x) and q(x) are continuous functions on some open interval I If p(x) and q(x) are continuous functions on some open interval I (see Sec. 1.1) and x is on I, then the initial value problem (see Sec. 1.1) and x00is on I, then the initial value problem consisting of (1) and (2) has a unique solution y(x) on the interval I. consisting of (1) and (2) has a unique solution y(x) on the interval I.

Section 2.6 p80



Linear Independence of Solutions A general solution on an open interval I is made up from a A general solution on an open interval I is made up from a basis basis yy11, y , y22on I, that is, from a pair of linearly independent on I, that is, from a pair of linearly independent solutions on I. Here we call y , y22linearly independent linearly independent on I on I solutions on I. Here we call y11, y if the equation if the equation (4) kk1yy1(x) + k 2yy2(x) = 0 on I implies k 1 = 0, k 2 = 0. (x) + k (x) = 0 on I implies k = 0, k (4) 1 1 2 2 1 2 = 0. We call y , y22linearly dependent linearly dependent on I if this equation also on I if this equation also We call y11, y holds for constants k 2 not both 0. In this case, and only in , k holds for constants k1, k 1 2 not both 0. In this case, and only in this case, y and y2 2 are proportional on I, that is (see Sec. 2.1), are proportional on I, that is (see Sec. 2.1), this case, y11and y (5) (a) y or (b) y for all on I. = ky2 or (b) y2 = ly = ly1 for all on I. (5) (a) y1 = ky 1

Section 2.6 p81

2

2

1



Theorem 2 Linear LinearDependence Dependenceand andIndependence Independenceof ofSolutions Solutions

Let the ODE (1) have continuous coefficients p(x) and q(x) on an Let the ODE (1) have continuous coefficients p(x) and q(x) on an open interval I. Then two solutions y and y22of (1) on I are linearly of (1) on I are linearly open interval I. Then two solutions y11and y dependent on I if and only if their “ Wronskian”” dependent on I if and only if their “Wronskian (6) W(y (6) W(y11, y , y22) = y ) = y11yy2’2’−−yy22yy1’1’ is 0 at some x is 0 at some x00in I. Furthermore, if W = in I. Furthermore, if W =0 at an x = 0 at an x =xx00in I, then in I, then W = W =0 on I; hence, if there is an x 0 on I; hence, if there is an x11in I at which W is not 0, in I at which W is not 0, then y , y2 are linearly independent on I. are linearly independent on I. then y1, y 1

2

Section 2.6 p82



Remark. Determinants. Determinants. Students familiar with second‐order Students familiar with second‐order determinants may have noticed that determinants may have noticed that

y1 y2 W ( y1 , y2 ) = = y1 y′2 − y2 y1′ . y1′ y′2

This determinant is called the Wronski determinant or, This determinant is called the Wronski determinantor, briefly, the Wronskian, of two solutions y , of two solutions y11and y and y22of (1), as of (1), as briefly, the Wronskian has already been mentioned in (6). Note that its four entries has already been mentioned in (6). Note that its four entries occupy the same positions as in the linear system (7). occupy the same positions as in the linear system (7).

Section 2.6 p83



A General Solution of (1) Includes All Solutions

Theorem 3 Existence Existenceof ofaaGeneral GeneralSolution Solution IfIfp(x) p(x)and andq(x) q(x)are arecontinuous continuouson onan anopen openinterval intervalI,I,then then(1) (1)has hasaa general generalsolution solutionon onI.I.

Section 2.6 p84



Theorem 4 AAGeneral GeneralSolution SolutionIncludes IncludesAll AllSolutions Solutions

If the ODE (1) has continuous coefficients p(x) and q(x) on some If the ODE (1) has continuous coefficients p(x) and q(x) on some open interval I, then every solution y = open interval I, then every solution y =Y(x) of (1) on I is of the form Y(x) of (1) on I is of the form (8) Y(x) = C (8) Y(x) = C11yy11(x) + C (x) + C22yy22(x) (x) where y where y11, y , y22is any basis of solutions of (1) on I and C is any basis of solutions of (1) on I and C11, C , C22are are suitable constants. suitable constants. Hence (1) does not have singular solutions (that is, solutions (that is, solutions Hence (1) does not have singular solutions not obtainable from a general solution). not obtainable from a general solution).

Section 2.6 p85


2.7 2.7 Nonhomogeneous Nonhomogeneous ODEs ODEs

Section 2.7 p86


2.7 Nonhomogeneous ODEs

We now advance from homogeneous to nonhomogeneous We now advance from homogeneous to nonhomogeneous linear ODEs. linear ODEs.

Consider the second‐order nonhomogeneous linear ODE Consider the second‐order nonhomogeneous linear ODE (1) y” (1) y”+ p(x)y’ + p(x)y’+ q(x)y = + q(x)y =r(x) r(x) where r(x) where r(x) ≠ ≠0.We shall see that a “general solution” 0.We shall see that a “general solution”of (1) of (1) is the sum of a general solution of the corresponding is the sum of a general solution of the corresponding homogeneous ODE homogeneous ODE (2) y” (2) y”+ p(x)y’ + p(x)y’+ q(x)y = + q(x)y =00 and a “particular solution” and a “particular solution”of (1). These two new terms of (1). These two new terms “general solution of (1)” “general solution of (1)”and “particular solution of (1)” and “particular solution of (1)”are are defined as follows. defined as follows. Section 2.7 p87



DEFINITION General GeneralSolution, Solution,Particular ParticularSolution Solution A A general solution general solution of the nonhomogeneous ODE (1) on an of the nonhomogeneous ODE (1) on an open interval I is a solution of the form open interval I is a solution of the form (3) y(x) = y (3) y(x) = yhh(x) + y (x) + ypp(x); (x); here, y here, yhh= = cc1y1y11++cc22yy22is a general solution of the homogeneous is a general solution of the homogeneous ODE (2) on I and y ODE (2) on I and yp p is any solution of (1) on I containing no is any solution of (1) on I containing no arbitrary constants. arbitrary constants. A A particular solution particular solution of (1) on I is a solution obtained of (1) on I is a solution obtained from (3) by assigning specific values to the arbitrary constants from (3) by assigning specific values to the arbitrary constants cc1 and c and c2 in y in yh. . 1

2

Section 2.7 p88

h



THEOREM 1 Relations Relationsof ofSolutions Solutionsof of(1) (1)to toThose Thoseof of(2) (2) (a) (a) The sum of a solution y of (1) on some open interval I and a The sum of a solution y of (1) on some open interval I and a solution solution ỹỹof (2) on I is a solution of (1) on I. In particular, (3) is a of (2) on I is a solution of (1) on I. In particular, (3) is a solution of (1) on I. solution of (1) on I. (b) (b) The difference of two solutions of (1) on I is a solution of (2) on I. The difference of two solutions of (1) on I is a solution of (2) on I.

Section 2.7 p89



THEOREM 2 AAGeneral GeneralSolution Solutionof ofaaNonhomogeneous NonhomogeneousODE ODE Includes IncludesAll AllSolutions Solutions

If the coefficients p(x), q(x), and the function r(x) in (1) are If the coefficients p(x), q(x), and the function r(x) in (1) are continuous on some open interval I, then every solution of (1) on I is continuous on some open interval I, then every solution of (1) on I is obtained by assigning suitable values to the arbitrary constants obtained by assigning suitable values to the arbitrary constants cc1 and c and c2 in a general solution (3) of (1) on I. in a general solution (3) of (1) on I. 1

2

Section 2.7 p90



Method of Undetermined Coefficients To solve the nonhomogeneous ODE (1) or an initial value To solve the nonhomogeneous ODE (1) or an initial value problem for (1), we have to solve the homogeneous ODE (2) and problem for (1), we have to solve the homogeneous ODE (2) and find any solution y of (1), so that we obtain a general find any solution yp p of (1), so that we obtain a general solution (3) of (1). solution (3) of (1). How can we find a solution y How can we find a solution yppof (1)? of (1)? One method is the so‐called One method is the so‐called method of undetermined method of undetermined coefficients coefficients. It is much simpler than another, more general, . It is much simpler than another, more general, method (given in Sec. 2.10). Since it applies to models of method (given in Sec. 2.10). Since it applies to models of vibrational systems and electric circuits to be shown in the vibrational systems and electric circuits to be shown in the next two sections, it is frequently used in engineering. next two sections, it is frequently used in engineering.

Section 2.7 p91



Method of Undetermined Coefficients (continued) More precisely, the method of undetermined coefficients is More precisely, the method of undetermined coefficients is suitable for linear ODEs with suitable for linear ODEs with constant coefficients a and b constant coefficients a and b (4) y” (4) y”+ ay’ + ay’+ by = + by =r(x) r(x) when whenr(x) is an exponential function, a power of x, a cosine r(x) is an exponential function, a power of x, a cosine or sine, or sums or products of such functions. These or sine, or sums or products of such functions. These functions have derivatives similar to r(x) itself. This gives functions have derivatives similar to r(x) itself. This gives the idea. the idea. We choose a form for y We choose a form for yppsimilar to r(x), but with unknown similar to r(x), but with unknown coefficients to be determined by substituting that y and its coefficients to be determined by substituting that yppand its derivatives into the ODE. Table 2.1 on p. 82 shows the derivatives into the ODE. Table 2.1 on p. 82 shows the choice of y choice of yppfor practically important forms of r(x). for practically important forms of r(x). Corresponding rules are as follows. Corresponding rules are as follows. Section 2.7 p92



Choice Rules for the Method of Undetermined Coefficients Choice Rules for the Method of Undetermined Coefficients (a) Basic Rule. (a) Basic Rule. If r(x) in (4) is one of the functions in the first If r(x) in (4) is one of the functions in the first column in Table 2.1, choose y column in Table 2.1, choose yppin the same line and determine its in the same line and determine its undetermined coefficients by substituting y undetermined coefficients by substituting yppand its derivatives into and its derivatives into (4). (4). (b) Modification Rule. (b) Modification Rule. If a term in your choice for y If a term in your choice for ypphappens to happens to be a solution of the homogeneous ODE corresponding to (4), be a solution of the homogeneous ODE corresponding to (4), 2 if this solution corresponds to a multiply this term by x (or by x multiply this term by x (or by x2 if this solution corresponds to a double root of the characteristic equation of the homogeneous ODE). double root of the characteristic equation of the homogeneous ODE). (c) Sum Rule. (c) Sum Rule. If r(x) is a sum of functions in the first column of If r(x) is a sum of functions in the first column of Table 2.1, choose for y Table 2.1, choose for yppthe sum of the functions in the the sum of the functions in the corresponding lines of the second column. corresponding lines of the second column. Section 2.7 p93



The Basic Rule applies when r(x) is a single term. The Basic Rule applies when r(x) is a single term. The Modification Rule helps in the indicated case, and to The Modification Rule helps in the indicated case, and to recognize such a case, we have to solve the homogeneous recognize such a case, we have to solve the homogeneous ODE first. ODE first. The Sum Rule follows by noting that the sum of two solutions The Sum Rule follows by noting that the sum of two solutions of (1) with r = of (1) with r =rr11and r = and r =rr22(and the same left side!) is a (and the same left side!) is a solution of (1) with r = solution of (1) with r =rr11+ r + r2. (Verify!) 2. (Verify!) The method is self‐correcting. A false choice for y or one The method is self‐correcting. A false choice for yppor one with too few terms will lead to a contradiction. A choice with with too few terms will lead to a contradiction. A choice with too many terms will give a correct result, with superfluous too many terms will give a correct result, with superfluous coefficients coming out zero. coefficients coming out zero.

Section 2.7 p94



Table 2.1 Method of Undetermined Coefficients Term in r(x) keγx

Choice for yp(x) Ceγx

kxn (n = 0, 1,… ) Knxn + Kn−1xn−1 + … + K1x + K0 k cos ωx k sin ωx keαx cos ωx keαx

Section 2.7 p95

sin ωx

}Kcos ωx + Msin ωx }eαx(Kcos ωx + Msin ωx) Advanced Engineering Mathematics, 10/e by Erwin Kreyszig Copyright 2011 by John Wiley & Sons. All rights reserved.


EXAMPLE 2 Application of the Modification Rule (b)

Solve the initial value problem Solve the initial value problem −1.5x, y(0) = 1, y’(0) = 0. (6) y” + 3y’ + 2.25y = − 10e (6) y” + 3y’ + 2.25y = −10e−1.5x, y(0) = 1, y’(0) = 0. Solution. Step 1. General solution of the homogeneous ODE. Solution. Step 1. General solution of the homogeneous ODE. The characteristic equation of the homogeneous ODE is The characteristic equation of the homogeneous ODE is 2 = 0. Hence the homogeneous ODE λλ22+ 3λ + 2.25 = (λ + 1.5) + 3λ + 2.25 = (λ + 1.5)2 = 0. Hence the homogeneous ODE has the general solution has the general solution −1.5x. yyh ==(c(c1 + c x)e −1.5x. 2 h 1 + c2x)e

Section 2.7 p96



EXAMPLE 2

(continued) Application of the Modification Rule (b)

Solution. Solution. (continued) (continued) Step 2. Solution y p of the nonhomogeneous ODE. Step 2. Solution y p of the nonhomogeneous ODE. −1.5x The function e −1.5xon the right would normally require the The function e on the right would normally require the −1.5x choice Ce choice Ce−1.5x. But we see from y . But we see from yhhthat this function is a that this function is a solution of the homogeneous ODE, which corresponds to a solution of the homogeneous ODE, which corresponds to a double root of the characteristic equation. Hence, according to double root of the characteristic equation. Hence, according to the Modification Rule we have to multiply our choice the Modification Rule we have to multiply our choice 2. That is, we choose function by x function by x2. That is, we choose 2 e−1.5x. yp = Cx yp = Cx2 e−1.5x . −1.5x 2 2) Then y ’ = C(2x − 1.5x ) e , y ” = C(2 − 3x − 3x + 2.25x 2 −1.5x 2) p p Then y ’ = C(2x − 1.5x ) e , y ” = C(2 − 3x − 3x + 2.25x p p ee−1.5x −1.5x. .

Section 2.7 p97



EXAMPLE 2


Solution. Solution. (continued) (continued) Step 2. Step 2. (continued) (continued) We substitute these expressions into the given ODE and omit We substitute these expressions into the given ODE and omit ‐1.5x. This yields the factor e the factor e‐1.5x. This yields 2) + 3C(2x − 1.5x2) + 2.25Cx2 = 10. − 6x + 2.25x C(2 2) + 2.25Cx2 = 10. − 1.5x C(2 − 6x + 2.25x2) + 3C(2x 2, x, x0 gives Comparing the coefficients of x Comparing the coefficients of x2, x, x0 gives 0 = 0, 0 = 0, 2C = −−5. 0 = 0, 0 = 0, 2C =−−10, hence C = 10, hence C = 5. 2 −1.5x This gives the solution This gives the solution yypp==−5x −5x2ee−1.5x. . Hence the given ODE has the general solution Hence the given ODE has the general solution 2e−1.5x. yy==yyh ++yyp ==(c(c1 ++cc2x)x)ee−1.5x 5x −1.5x− − 5x2e−1.5x. h

Section 2.7 p98

p

1

2



EXAMPLE 2


Solution. Solution. (continued) (continued) Step 3. Solution of the initial value problem. Step 3. Solution of the initial value problem. Setting x = Setting x =0 in y and using the first initial condition, 0 in y and using the first initial condition, we obtain y(0) = c we obtain y(0) = c11= 1. Differentiation of y gives = 1. Differentiation of y gives −1.5x − 10x e−1.5x + 7.5x2e−1.5x. − 1.5c − 1.5c x)e y’y’= (c 2 1 2 = (c2 − 1.5c1 − 1.5c2x)e−1.5x − 10x e−1.5x + 7.5x2e−1.5x. From this and the second initial condition we have From this and the second initial condition we have y’y’(0) = c (0) = c22−−1.5c 1.5c11= 0. = 0.Hence c Hence c22= 1.5c = 1.5c11= 1.5. = 1.5. This gives the answer (Fig. 51) This gives the answer (Fig. 51) −1.5x − 5x2 e−1.5x = (1 + 1.5x − 5x2) e−1.5x. y = (1 + 1.5x) e y = (1 + 1.5x) e−1.5x − 5x2 e−1.5x = (1 + 1.5x − 5x2) e−1.5x. The curve begins with a horizontal tangent, crosses the x‐axis The curve begins with a horizontal tangent, crosses the x‐axis 2 = 0) and approaches the − 5x at x = 0.6217 (where 1 + 1.5x at x = 0.6217 (where 1 + 1.5x − 5x2 = 0) and approaches the axis from below as x increases. axis from below as x increases. Section 2.7 p99



EXAMPLE 2


Solution. Solution. (continued) (continued) Step 3. Step 3. (continued) (continued)

Section 2.7 p100


2.8 2.8 Modeling: Modeling: Forced Forced

Oscillations. Oscillations. Resonance Resonance

Section 2.8 p101


2.8 Modeling: Forced Oscillations. Resonance

In Sec. 2.4 we considered vertical motions of a mass–spring In Sec. 2.4 we considered vertical motions of a mass–spring system (vibration of a mass m on an elastic spring, as in Figs. system (vibration of a mass m on an elastic spring, as in Figs. 33 and 53) and modeled it by the 33 and 53) and modeled it by the homogeneous homogeneous linear ODE linear ODE (1) my” (1) my”+ cy’ + cy’+ ky = + ky =0.0. Here y(t) as a function of time t is the Here y(t) as a function of time t is the displacement of the body of mass m displacement of the body of mass m from rest. from rest.

Section 2.8 p102



The mass–spring system of Sec. 2.4 exhibited only free motion. The mass–spring system of Sec. 2.4 exhibited only free motion. This means no external forces (outside forces) but only This means no external forces (outside forces) but only internal forces controlled the motion. internal forces controlled the motion. The internal forces are forces within the system. They are the The internal forces are forces within the system. They are the force of inertia my”, the damping force cy’ force of inertia my”, the damping force cy’(if c > (if c >0), and the 0), and the spring force ky, a restoring force. spring force ky, a restoring force.

Section 2.8 p103



We now extend our model by including an additional force, We now extend our model by including an additional force, that is, the external force r(t), on the right. Then we have that is, the external force r(t), on the right. Then we have my” (2*) (2*) my”+ cy’ + cy’+ ky = r(t). + ky = r(t). Mechanically this means that at each instant t the resultant of Mechanically this means that at each instant t the resultant of the internal forces is in equilibrium with r(t). the internal forces is in equilibrium with r(t). The resulting motion is called a The resulting motion is called a forced motion forced motion with with forcing forcing function function r(t), which is also known as r(t), which is also known as input input or or driving force driving force, , and the solution y(t) to be obtained is called the output or the or the and the solution y(t) to be obtained is called the output response response of the system to the driving force. of the system to the driving force.

Section 2.8 p104



Of special interest are periodic external forces, and we shall Of special interest are periodic external forces, and we shall consider a driving force of the form consider a driving force of the form r(t) = F r(t) = F00cos ωt cos ωt Then we have the nonhomogeneous ODE Then we have the nonhomogeneous ODE

(F (F00> 0, ω > 0, ω> 0). > 0).

(2) my” (2) my”+ cy’ + cy’+ ky = F + ky = F00cos ωt. cos ωt. Its solution will reveal facts that are fundamental in Its solution will reveal facts that are fundamental in engineering mathematics and allow us to model resonance. engineering mathematics and allow us to model resonance.

Section 2.8 p105



Solving the Nonhomogeneous ODE (2) From Sec. 2.7 we know that a general solution of (2) is the From Sec. 2.7 we know that a general solution of (2) is the sum of a general solution sum of a general solution yyhhof the homogeneous ODE (1) of the homogeneous ODE (1) plus any solution y of (2). To find ypp, we use the method , we use the method plus any solution yppof (2). To find y of undetermined coefficients (Sec. 2.7), starting from of undetermined coefficients (Sec. 2.7), starting from yyp(t) = a cos ωt + b sin ωt. (t) = a cos ωt + b sin ωt. p

(3) (3)

k − mω 2 ωc a = F0 , b F . = 0 2 2 2 2 2 2 ( k − mω ) + ω c ( k − mω ) + ω c

Section 2.8 p106



Solving the Nonhomogeneous ODE (2) (continued) If we set k / m = ω0 ( > 0) as in Sec. 2.4, then k = mω0 2 and we obtain

m(ω0 2 − ω 2 ) ωc , b = F0 2 2 . a = F0 2 2 (5) 2 2 2 2 2 2 (5) m (ω0 − ω ) + ω c m (ω0 − ω ) + ω c We thus obtain the general solution of the We thus obtain the general solution of the nonhomogeneous ODE (2) in the form nonhomogeneous ODE (2) in the form (6) y(t) = y (6) y(t) = yhh(t) + y (t) + ypp(t). (t). Here y Here yhhis a general solution of the homogeneous ODE (1) is a general solution of the homogeneous ODE (1) and y and yp is given by (3) with coefficients (5). is given by (3) with coefficients (5). p

Section 2.8 p107



Case 1. Undamped Forced Oscillations. Resonance If the damping of the physical system is so small that its If the damping of the physical system is so small that its effect can be neglected over the time interval considered, effect can be neglected over the time interval considered, 2 − ω2)] and 0. Then (5) reduces to a = F we can set c = /[m(ω 2 − ω2)] and 0 0 we can set c = 0. Then (5) reduces to a = F /[m(ω 0 0 2 = k/m) b = 0 Hence (3) becomes (use ω b = 0 Hence (3) becomes (use ω0 2 = k/m) 0

F0 F0 y p (t ) = cos ωt = cos ωt. 2 2 2 m(ω0 − ω ) k[1 − (ω / ω0 ) ] 2 ≠ ω 2; physically, the Here we must assume that ω Here we must assume that ω2 ≠ ω002; physically, the frequency ω/(2π) [cycles/sec] of the driving force is frequency ω/(2π) [cycles/sec] of the driving force is different from the natural frequency ω different from the natural frequency ω00/(2π) of the system, /(2π) of the system, which is the frequency of the free undamped motion which is the frequency of the free undamped motion [see (4) in Sec. 2.4]. [see (4) in Sec. 2.4].

(7) (7)

Section 2.8 p108



Case 1. Undamped Forced Oscillations. Resonance (continued 1) From (7) and from (4*) in Sec. 2.4 we have the general From (7) and from (4*) in Sec. 2.4 we have the general solution of the “undamped system” solution of the “undamped system” (8) (8)

F0 y(t ) = C cos (ω0t − δ ) + cos ωt. 2 2 m(ω0 − ω )

We see that this output is a We see that this output is a superposition of two harmonic superposition of two harmonic oscillations oscillations of the frequencies just mentioned. of the frequencies just mentioned.

Section 2.8 p109



Case 1. Undamped Forced Oscillations. Resonance (continued 2) Resonance. Resonance. We discuss (7). We see that the maximum We discuss (7). We see that the maximum amplitude of y is (put cos ωt =1) 1) amplitude of yppis (put cos ωt = F0 1 (9) a0 = ρ where ρ = . 2 (9) k 1 − (ω / ω0 ) and ω →→ωω0 , ,then ρρ and a aa0 depends on ω 0. If ω 0 tend to depends on ω and ω . If ω then and a 0 0 0 0 tend to infinity. This excitation of large oscillations by matching infinity. This excitation of large oscillations by matching input and natural frequencies (ω ) is called resonance resonance. . input and natural frequencies (ω= ω = ω00) is called ρρis called the is called the resonance factor resonance factor (Fig. 54), and from (9) we (Fig. 54), and from (9) we see that see that ρρ/k = /k =aa00/F /F00is the ratio of the amplitudes of the is the ratio of the amplitudes of the particular solution y particular solution yppand of the input F and of the input F00cos ωt. We shall cos ωt. We shall see later in this section that resonance is of basic see later in this section that resonance is of basic importance in the study of vibrating systems. importance in the study of vibrating systems. Section 2.8 p110



Case 1. Undamped Forced Oscillations. Resonance (continued 3) Resonance. Resonance. (continued) (continued)

Section 2.8 p111



Case 1. Undamped Forced Oscillations. Resonance (continued 4) In the case of resonance the nonhomogeneous ODE (2) In the case of resonance the nonhomogeneous ODE (2) becomes becomes F0 2 (10) y′′ + ω0 y = cos ω0t. (10) m Then (7) is no longer valid, and from the Modification Rule Then (7) is no longer valid, and from the Modification Rule in Sec. 2.7, we conclude that a particular solution of (10) is in Sec. 2.7, we conclude that a particular solution of (10) is of the form of the form yyp(t) = t(a cos ω (t) = t(a cos ω0t + t +b sin ω b sin ω0t). t). p

Section 2.8 p112

0

0



Case 1. Undamped Forced Oscillations. Resonance (continued 5) By substituting this into (10) we find a = By substituting this into (10) we find a =0 and 0 and b = /(2mω00). Hence (Fig. 55) ). Hence (Fig. 55) b =FF00/(2mω F0 (11) y p (t ) = t sin ω0t. (11) 2 mω0

Section 2.8 p113



Case 1. Undamped Forced Oscillations. Resonance (continued 6)

We see that, because of the factor t, the amplitude of the We see that, because of the factor t, the amplitude of the vibration becomes larger and larger. Practically speaking, vibration becomes larger and larger. Practically speaking, systems with very little damping may undergo large systems with very little damping may undergo large vibrations that can destroy the system. vibrations that can destroy the system. Section 2.8 p114



Case 1. Undamped Forced Oscillations. Resonance (continued 7) Beats. Beats. Another interesting and highly important type of Another interesting and highly important type of oscillation is obtained if ω oscillation is obtained if ωis close to ω is close to ω00. Take, for example, . Take, for example, the particular solution [see (8)] the particular solution [see (8)]

(12) (12)

F0 y( t ) = (cos ωt − cos ω0t ) (ω ≠ ω0 ). 2 2 m(ω0 − ω )

Using (12) in App. 3.1, we may write this as Using (12) in App. 3.1, we may write this as 2 F0 ω0 + ω ω0 − ω y(t ) = t )sin( t ). sin( 2 2 2 2 m(ω0 − ω )

Section 2.8 p115



Case 1. Undamped Forced Oscillations. Resonance (continued 8) Beats. Beats. (continued) (continued) Since is ω Since is ωclose to ω close to ω00, the difference ω , the difference ω00−−ωωis small. Hence is small. Hence the period of the last sine function is large, and we obtain the period of the last sine function is large, and we obtain an oscillation of the type shown in Fig. 56, the dashed an oscillation of the type shown in Fig. 56, the dashed curve resulting from the first sine factor. This is what curve resulting from the first sine factor. This is what musicians are listening to when they musicians are listening to when they tune tune their instruments. their instruments.

Section 2.8 p116



Case 1. Undamped Forced Oscillations. Resonance (continued 9) Beats. Beats. (continued) (continued)

Section 2.8 p117



Case 2. Damped Forced Oscillations If the damping of the mass–spring system is not negligibly If the damping of the mass–spring system is not negligibly small, we have c > small, we have c >0 and a damping term cy’ 0 and a damping term cy’in (1) and (2). in (1) and (2). Then the general solution y Then the general solution yhhof the homogeneous ODE (1) of the homogeneous ODE (1) approaches zero as t goes to infinity, as we know from approaches zero as t goes to infinity, as we know from Sec. 2.4. Sec. 2.4. Practically, it is zero after a sufficiently long time. Practically, it is zero after a sufficiently long time. Hence the “ Hence the “transient solution transient solution””(6) of (2), given by (6) of (2), given by y = y =yyhh++yypp, approaches the “ , approaches the “steady‐state solution steady‐state solution””yypp. . This proves the following theorem. This proves the following theorem.

Section 2.8 p118



THEOREM 1 Steady-State Steady-StateSolution Solution After a sufficiently long time the output of a damped vibrating After a sufficiently long time the output of a damped vibrating system under a purely sinusoidal driving force [see (2)] will system under a purely sinusoidal driving force [see (2)] will practically be a harmonic oscillation whose frequency is that of practically be a harmonic oscillation whose frequency is that ofthe the input. input.

Section 2.8 p119



Amplitude Amplitudeof ofthe theSteady-State Steady-StateSolution. Solution. Practical PracticalResonance Resonance

Whereas in the undamped case the amplitude of y Whereas in the undamped case the amplitude of ypp approaches infinity as ω approaches infinity as ωapproaches ω approaches ω00, this will not happen , this will not happen in the damped case. In this case the amplitude will always be in the damped case. In this case the amplitude will always be finite. But it may have a maximum for some ω finite. But it may have a maximum for some ωdepending on depending on the damping constant c. the damping constant c. This may be called practical resonance. . This may be called practical resonance It is of great importance because if c It is of great importance because if cis not too large, then is not too large, then some input may excite oscillations large enough to damage or some input may excite oscillations large enough to damage or even destroy the system. even destroy the system.

Section 2.8 p120



To study the amplitude of y To study the amplitude of yppas a function of ω, we write (3) in as a function of ω, we write (3) in the form the form −−η). (13) yyp(t) = C* cos (ωt (t) = C* cos (ωt η). (13) p C* is called the C* is called the amplitude amplitude of y of yppand η and ηthe the phase angle phase angle or or phase lag phase lag because it measures the lag of the output behind the because it measures the lag of the output behind the input. According to (5), these quantities are input. According to (5), these quantities are C * (ω ) = a + b = 2

(14) (14)

F0

2

m2 (ω0 2 − ω 2 )2 + ω 2 c 2

b ωc tanη (ω ) = = . 2 2 a m(ω0 − ω )

Section 2.8 p121



(16) (16)

C * (ωmax ) =

2 mF0

c 4 m2ω0 2 − c 2 ) is always finite when c > 0. We see that C*(ω ) is always finite when c > 0. We see that C*(ωmax max Furthermore, since the expression Furthermore, since the expression 2ω 2 − c4 = c2(4mk − c2) cc224m 4m2ω002 − c4 = c2(4mk − c2) 2 in the denominator of (16) decreases monotone to zero as c in the denominator of (16) decreases monotone to zero as c2 (0). 0). Now in an actual circuit, R From this it follows that From this it follows that Ih Ih approaches zero, theoretically approaches zero, theoretically t t → →∞ ∞, as but practically after a relatively short time. , as but practically after a relatively short time. Hence the transient current Hence the transient current I = I =IIhh++IpIptends to the steady‐ tends to the steady‐ state current , and after some time the output will state current IpIp, and after some time the output will practically be a harmonic oscillation, which is given by (5) practically be a harmonic oscillation, which is given by (5) and whose frequency is that of the input (of the and whose frequency is that of the input (of the electromotive force). electromotive force). Section 2.9 p134


2.9 Modeling: Electric Circuits

Analogy of Electrical and Mechanical Quantities Entirely different physical or other systems may have the Entirely different physical or other systems may have the same mathematical model. same mathematical model. For instance, we have seen this For instance, we have seen this from the various applications of the ODE ky in Chap. 1. in Chap. 1. from the various applications of the ODE yy’ ’= = ky Another impressive demonstration of this Another impressive demonstration of this unifying power unifying power of mathematics of mathematics is given by the ODE (1) for an electric is given by the ODE (1) for an electric RLC RLC‐‐ circuit and the ODE (2) in the last section for a mass–spring circuit and the ODE (2) in the last section for a mass–spring system. Both equations system. Both equations 1 LI ′′ + RI ′ + I = E′(t ) = E0ω cos ωt C and my and my””+ + cy cy’ ’+ + ky = ky =FF0 cos cos ωt ωt 0

are of the same form. are of the same form. Section 2.9 p135



Analogy of Electrical and Mechanical Quantities (continued 1) Table 2.2 shows the analogy between the various quantities Table 2.2 shows the analogy between the various quantities involved. The inductance involved. The inductance L L corresponds to the mass corresponds to the mass m m and, and, indeed, an inductor opposes a change in current, having an indeed, an inductor opposes a change in current, having an “inertia effect” R “inertia effect”similar to that of a mass. The resistance similar to that of a mass. The resistance R corresponds to the damping constant corresponds to the damping constant cc, and a resistor , and a resistor causes loss of energy, just as a damping dashpot does. And causes loss of energy, just as a damping dashpot does. And so on. so on. This analogy is This analogy is strictly quantitative strictly quantitative in the sense that in the sense that to a given mechanical system we can construct an electric to a given mechanical system we can construct an electric circuit whose current will give the exact values of the circuit whose current will give the exact values of the displacement in the mechanical system when suitable scale displacement in the mechanical system when suitable scale factors are introduced. factors are introduced. Section 2.9 p136



Analogy of Electrical and Mechanical Quantities (continued 2) The The practical importance practical importance of this analogy is almost of this analogy is almost obvious. The analogy may be used for constructing an obvious. The analogy may be used for constructing an “electrical model” “electrical model”of a given mechanical model, resulting of a given mechanical model, resulting in substantial savings of time and money because electric in substantial savings of time and money because electric circuits are easy to assemble, and electric quantities can be circuits are easy to assemble, and electric quantities can be measured much more quickly and accurately than measured much more quickly and accurately than mechanical ones. mechanical ones.

Section 2.9 p137



Analogy of Electrical and Mechanical Quantities (continued 3)

Section 2.9 p138


2.10 2.10 Solution Solution by by Variation Variation of of Parameters Parameters

Section 2.10 p139


2.10 Solution by Variation of Parameters

We continue our discussion of nonhomogeneous linear ODEs, We continue our discussion of nonhomogeneous linear ODEs, that is that is (1) yy””+ + pp(x (1) (x)y )y’ ’+ + qq(x (x))y = y =rr((xx).). To obtain To obtain yyp p when when r(x) r(x)is not too complicated, we can often use is not too complicated, we can often use the the method of undetermined coefficients. method of undetermined coefficients. However, since this method is restricted to functions However, since this method is restricted to functions rr(x (x)) whose derivatives are of a form similar to whose derivatives are of a form similar to rr((xx))itself (powers, itself (powers, exponential functions, etc.), it is desirable to have a method exponential functions, etc.), it is desirable to have a method valid for more general ODEs (1), which we shall now develop. valid for more general ODEs (1), which we shall now develop. It is called the method of variation of parameters and is and is It is called the method of variation of parameters credited to Lagrange (Sec. 2.1). Here credited to Lagrange (Sec. 2.1). Here pp, , q, r q, r in (1) may be in (1) may be variable (given functions of variable (given functions of xx), but we assume that they are ), but we assume that they are continuous on some open interval continuous on some open interval I.I. Section 2.10 p140



Lagrange’s method gives a particular solution Lagrange’s method gives a particular solution yyppof (1) on of (1) on I I in the form in the form (2) (2)

y2 r y1r y p ( x) = − y1 ∫ dx + y2 ∫ dx W W

where where yy11, , yy22form a basis of solutions of the corresponding form a basis of solutions of the corresponding homogeneous ODE homogeneous ODE (3) yy””+ + pp(x (3) (x)y )y’ ’+ + qq(x (x)y = )y = 00 on I, and W W is the Wronskian of is the Wronskian of yy1, , yy2. . on I, and 1

(4) (4)

Section 2.10 p141

2

W = W =yy11yy2’2’−−yy22yy1’1’

(see Sec. 2.6). (see Sec. 2.6).



EXAMPLE 1 Method of Variation of Parameters Method of Variation of Parameters Solve the nonhomogeneous ODE Solve the nonhomogeneous ODE 1 ′′ y + y = sec x = . cos x Solution. Solution. A basis of solutions of the homogeneous ODE on A basis of solutions of the homogeneous ODE on any interval is any interval is yy11= cos = cos xx, , yy22= sin = sin x. x. This gives the This gives the Wronskian Wronskian W W( ( yy1, 1, yy22) = cos ) = cos x x cos cos x x −−sin sin x x (− (−sin sin xx) = 1. ) = 1. From (2), choosing zero constants of integration, we get the From (2), choosing zero constants of integration, we get the particular solution of the given ODE particular solution of the given ODE y p = − cos x ∫ sin x sec x dx + sin x ∫ cos x sec x dx = cos x ln cos x + x sin x Section 2.10 p142

(Fig. 70) Engineering Mathematics, 10/e by Erwin Kreyszig (Fig. 70)Advanced Copyright 2011 by John Wiley & Sons. All rights reserved.


EXAMPLE 1 (continued) Solution. Solution. (continued 1) (continued 1) Figure 70 shows Figure 70 shows yyppand its first term, which is small, so that and its first term, which is small, so that x x sin sin x x essentially determines the shape of the curve essentially determines the shape of the curve of of yypp. From . From yyp p and the general solution and the general solution yyhh==cc11yy11+ + cc22yy22of the of the homogeneous ODE we obtain the a nswer homogeneous ODE we obtain the answer y = + ln|cos xx|) cos |) cos x + x +(c(c22+ + xx) sin ) sin xx. . y =yyhh++yypp==(c(c11+ ln|cos Had we included integration constants Had we included integration constants −−cc11, , cc22in (2), in (2), then (2) would have given the additional then (2) would have given the additional cc1cos x + cc2sin x = cc1yy1 + + cc2yy2, that is, a general solution of the cos x + sin x = 1 2 1 1 2 2, that is, a general solution of the given ODE directly from (2). This will always be the case. given ODE directly from (2). This will always be the case.

Section 2.10 p143



EXAMPLE 1 (continued) Solution. Solution. (continued 2) (continued 2)

Section 2.10 p144


SUMMARY SUMMARYOF OFCHAPTER CHAPTER 2 2

Second-Order Second-Order ODEs ODEs

Section 2.Summary p145


SUMMARY OF CHAPTER

2

Second-Order ODEs

Second‐order linear ODEs are particularly important in Second‐order linear ODEs are particularly important in applications, for instance, in mechanics (Secs. 2.4, 2.8) and applications, for instance, in mechanics (Secs. 2.4, 2.8) and electrical engineering (Sec. 2.9). A second‐order ODE is electrical engineering (Sec. 2.9). A second‐order ODE is called linear if it can be written if it can be written called linear (1) yy””+ + pp(x (Sec. 2.1). (1) (x)y )y’ ’+ + qq(x (x))y = y =rr((xx)) (Sec. 2.1). (If the first term is, say, (If the first term is, say, f(fx (x))yy”, divide by ”, divide by f(fx (x) to get the ) to get the ““standard form standard form””(1) with (1) with yy””as the first term.) Equation (1) as the first term.) Equation (1) is called homogeneous if if rr((xx) is zero for all ) is zero for all x x considered, considered, is called homogeneous usually in some open interval; this is written usually in some open interval; this is written r r ≣ ≣0. Then 0. Then (2) yy””+ + pp(x (2) (x)y )y’ ’+ + qq(x (x))y = y =00 Equation (1) is called Equation (1) is called nonhomogeneous nonhomogeneous if if rr(x (x) ) ≣ ≣0 0 (meaning (meaning rr(x (x) is not zero for some ) is not zero for some x x considered). considered). Section 2.Summary p146


SUMMARY OF CHAPTER (continued 1)

2

Second-Order ODEs

For the homogeneous ODE (2) we have the important For the homogeneous ODE (2) we have the important superposition principle superposition principle (Sec. 2.1) that a linear combination (Sec. 2.1) that a linear combination y = y =ky ky11+ + lyly22of two solutions of two solutions yy11, , yy22is again a solution. is again a solution. Two Two linearly independent linearly independent solutions solutions yy11, , yy22of (2) on an open of (2) on an open interval interval I I form a form a basis basis (or (or fundamental system fundamental system) of solutions ) of solutions on cc1, , cc2 a a 2 with arbitrary constants on I,I,and and y = y =cc11yy11+ + cc2y y with arbitrary constants 2 2 1 2 general solution of (2) on I . From it we obtain a particular general solution of (2) on I. From it we obtain a particular solution solution if we specify numeric values (numbers) for if we specify numeric values (numbers) for cc11and and cc2, usually by prescribing two initial conditions , usually by prescribing two initial conditions 2 (3) (3) yy(x (x00) = ) = KK00, , yy’(’(xx00) = ) = KK11 (x (x00, , KK00, , KK11given numbers; Sec. 2.1). given numbers; Sec. 2.1). (2) and (3) together form an (2) and (3) together form an initial value problem initial value problem. . Similarly for (1) and (3). Similarly for (1) and (3). Section 2.Summary p147



2

Second-Order ODEs

For a nonhomogeneous ODE (1) a For a nonhomogeneous ODE (1) a general solution general solution is of is of the form the form (4) y = (Sec. 2.7). (4) y =yyhh++yypp (Sec. 2.7). Here Here yyhhis a general solution of (2) and is a general solution of (2) and yyppis a particular is a particular solution of (1). Such a can be determined by a general solution of (1). Such a yyp p can be determined by a general method ( method (variation of parameters variation of parameters, Sec. 2.10) or in many , Sec. 2.10) or in many practical cases by the practical cases by the method of undetermined coefficients method of undetermined coefficients. . The latter applies when (1) has constant coefficients The latter applies when (1) has constant coefficients p p and and qq, , and and rr(x (x) is a power of ) is a power of xx, sine, cosine, etc. (Sec. 2.7). Then we , sine, cosine, etc. (Sec. 2.7). Then we write (1) as write (1) as yy””+ + ay (Sec. 2.7). (5) ay’ ’+ + by = by =rr((xx)) (Sec. 2.7). (5) The corresponding homogeneous ODE yy’ ’+ + ay ’ ’+ + by = 0 has The corresponding homogeneous ODE ay by = 0 has λx solutions where λλis a root of solutions y = y =eeλx where is a root of 2 (6) λλ2+ + aλ (6) aλ+ + b = b =0.0. Section 2.Summary p148


SUMMARY OF CHAPTER

2

Second-Order ODEs

(continued 3)

Hence there are three cases (Sec. 2.2): Hence there are three cases (Sec. 2.2):

Here * is used since ωωis needed in driving forces. is needed in driving forces. Here ωω* is used since Important applications of (5) in mechanical and electrical Important applications of (5) in mechanical and electrical engineering in connection with engineering in connection with vibrations vibrations and and resonance resonance are discussed in Secs. 2.4, 2.7, and 2.8. are discussed in Secs. 2.4, 2.7, and 2.8.




Second-Order ODEs

Another large class of ODEs Another large class of ODEssolvable “algebraically” solvable “algebraically” consists of the Euler–Cauchy equations consists of the Euler–Cauchy equations 2 (7) xx2yy””+ + axy (Sec. 2.5). (7) axy’ ’+ + by = by =00 (Sec. 2.5). m These have solutions of the form These have solutions of the form y = y =xxm, where , where m m is a is a solution of the auxiliary equation solution of the auxiliary equation 2 + (a − 1)m + b = 0. (8) m (8) m2 + (a − 1)m + b = 0. Existence and uniqueness Existence and uniqueness of solutions of (1) and (2) is of solutions of (1) and (2) is discussed in Secs. 2.6 and 2.7, and reduction of order discussed in Secs. 2.6 and 2.7, and reduction of order in Sec. 2.1. in Sec. 2.1.


2
