Some practice multiple choice questions

Math 285 Section D2 Practice multiple choice questions (with answers) Problem 1. Let y1 , y2 , y3 : R → R be functions which are linearly dependent on (−∞, ∞). For each of the following statements indicate if it is true or false: (1) There exist i 6= j, where i, j ∈ {1, 2, 3}, and a number c ∈ R such that yj (x) = cyi (x) for all x ∈ R. (2) For every x ∈ R we have W (y1 , y2 , y3 ) = 0. (3) We have W (y1 , y2 , y3 )(0) = 0. (4) If c1 , c2 , c3 ∈ R are such that c1 y1 (x) + c2 y2 (x) + c3 y3 (x) = 0, then c1 = c2 = c3 = 0. (5) If c1 , c2 , c3 ∈ R are such that for all x ∈ R c1 y1 (x) + c2 y2 (x) + c3 y3 (x) = 0, then for some i ∈ {1, 2, 3} ci 6= 0. (6) There exist c1 , c2 , c3 ∈ R are such that c21 + c22 + c23 6= 0 and that for all x ∈ R c1 y1 (x) + c2 y2 (x) + c3 y3 (x) = 0. (7) The functions y1 , y2 are linearly dependent on (−∞, ∞). (8) The functions y1 , y2 are linearly independent on (−∞, ∞). (9) For any functions y4 : R → R the functions y1 , y2 , y3 , y4 are linearly dependent on (−∞, ∞). Answers: (1) False. For example, y1 = x, y2 = x2 and y3 = x + x2 are linearly dependent on R since y1 + y2 − y3 ≡ 0. However, none of the functions x, x2 , x + x2 is a scalar multiple of one of the others. (2) False. For example, for y1 = y2 = y3 = |x| these functions are linearly dependent on R. However, W (y1 , y2 , y3 )(0) does not exist since |x| is not differentiable at x = 0. (3) False, for the same reason as in (2). (4) False, since the functions y1 , y2 , y3 are assumed to be linearly dependent on (−∞, ∞). (5) False. For example, if we take c1 = c2 = c3 = 0, then c1 y1 + c2 y2 + c3 y3 ≡ 0, regardless of which functions y1 , y2 , y3 we use. (6) True, by definition of linear dependence. (7) False. For example, y1 = x, y2 = x2 and y3 = x + x2 are linearly dependent on R, but the functions y1 = x, y2 = x2 are linearly independent on R. (8) False. For example, if we take y1 = y2 = y3 = x, then y1 , y2 , y3 are linearly dependent on R and y1 , y2 are linearly dependent on R. (9) True. Since y1 , y2 , y3 are assumed to be linearly dependent on R, there exist c1 , c2 , c3 ∈ R such that at least one of c1 , c2 , c3 is 6= 0 but c1 y1 + c2 y2 + c3 y3 ≡ 0 on R. For any function y4 we can take c4 = 0 and then 1

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c1 y1 + c2 y2 + c3 y3 ≡ 0 on R, so that y1 , y2 , y3 , y4 are linearly dependent on R Problem 2. Let f : R → R be the 20-periodic function such that ( −1, if − 10 < t ≤ 3 f (t) = 50 t , if 3 < t ≤ 10 and let

 ∞  a0 X πnt πnt f (t) ∼ + an cos + bn sin 2 10 10 n=1

be the General Fourier Series of f . For each of the following statements indicate if it is true or false: (1) For every t ∈ (−20, 55) the series  ∞  πnt πnt a0 X + + bn sin an cos 2 10 10 n=1

converges. (2) For every integer n ≥ 1 we have Z 97 πnt 1 f (t) sin dt bn = 10 77 10 (3) For every integer n ≥ 1 we have Z 10 2 πnt bn = dt f (t) sin 10 0 10 (4) We have  ∞  a0 X 3πn 3πn −1 = + an cos + bn sin . 2 10 10 n=1

(5) We have  ∞  350 − 1 3πn a0 X 3πn = + an cos + bn sin . 2 2 10 10 n=1

(6) We have  ∞  a0 X 42πn 42πn −1 = + an cos + bn sin . 2 10 10 n=1

(7) For every t ∈ (−2, 3) we have   ∞ X πn πnt πnt 0= −an sin + bn cos . 10 10 10 n=1

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Answers: (1) True. Since the function f is piecewise-smooth, by the Convergence Theorem the General Fourier Series of f converges for every t ∈ R, and, in particular, for every t ∈ (−20, 55). (2) True. If g(t) is a piecewise-continuous p-periodic function (where R a+p R b+p p > 0) then for any a, b ∈ R we have a g(t) dt = b g(t) dt. In our case the function f (t) sin πnt 10 is 20-periodic and hence Z 10 Z 97 πnt πnt 1 1 f (t) sin f (t) sin bn = dt = dt 10 −10 10 10 77 10 (3) False, since the function f (t) in this example is not odd. (4) False. The function f (t) is discontinuous at t = 3, with f (3+) = 350 and f (3−) = −1. Therefore, by the Convergence Theorem,  ∞  a0 X 3πn 3πn 350 − 1 = + an cos + bn sin . 2 2 10 10 n=1

(5) True; see explanation for (4) above. (6) True. We have 42 = 2+2·20. Therefore, by the Convergence Theorem,  ∞  a0 X 42πn 42πn + an cos + bn sin = f (42) = f (2) = −1. 2 10 10 n=1

(7) True, by the Termwise Differentiation Theorem, since f 0 (t) = 0 for all t ∈ (−10, 3), and, in particular, for all t ∈ (−2, 3). Problem 3. Consider the initial value problem dy 1 (∗) = sin(y 2 exy + 2xy − 5), y(1) = 55. dx 4−x For each of the following statements indicate if it is true or false: (1) The problem is guaranteed to have a unique solution on the interval (0, 4). (2) The problem is guaranteed to have a solution on the interval (0, 4). (3) There exists  > 0 such that the problem has a solution on the interval (1 − , 1 + ). (4) There exists  > 0 such that the problem has at most one solution on the interval (1 − , 1 + ). (5) For some  > 0 the problem has a solution on the interval (1 − 2, 1 + 3). Answers (1) False. Theorem 1 on p. 24 never guarantees the existence of a solution on a specific interval, such as the interval (0, 4).

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(2) False, for the same reason as in (1). (3) True, by Theorem 1 on p. 24. (4) True, by Theorem 1 on p. 24. (5) True, by Theorem 1 on p. 24. Problem 4. Consider the problem  ytt = 25yxx for t > 0, 0 < x < π,    y(0, t) = y(π, t) = 0, for t > 0, (∗∗) 2  y(x, 0) = x − πx for 0 < x < π,    yt (x, 0) = 0, for 0 < x < π. For each of the following statements indicate if it is true or false: (1) The function y(x, t) = 21 [(x + 5t)2 + (x + 5t) + (x − 5t)2 + (x − 5t)], where t ≥ 0, 0 ≤ x ≤ π, is a solution of (∗∗). (2) If y(x, t) is the solution of (∗∗) then y(1, 1) =

23 . 2

(3) Problem (∗∗) has a solution of the form ∞ X

y(x, t) =

cn cos(5nt) sin(nx)

n=1

for some choice of constants cn ∈ R, n = 1, 2, 3, . . . (4) Problem (∗∗) has the solution ∞ X

y(x, t) =

cn cos(5nt) sin(nx)

n=1

where π

Z

(x2 − πx) sin(nx) dx

cn = 0

for n = 1, 2, 3, . . . . (5) Problem (∗∗) has the solution ∞ X

y(x, t) =

cn cos(5nt) sin(nx)

n=1

where 1 cn = π for n = 1, 2, 3, . . . .

Z

π

−π

(x2 − πx) sin(nx) dx

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(6) Problem (∗∗) has the solution y(x, t) =

∞ X

cn cos(5nt) sin(nx)

n=1

where

Z 1 π fO (x) sin(nx) dx π −π for n = 1, 2, 3, . . . , and where fO (x) is the 2π-periodic odd extension of the function f (x) = x2 − πx, 0 < x < π. (7) Problem (∗∗) has the solution cn =

y(x, t) =

∞ X

cn cos(5nt) sin(nx)

n=1

where

Z 1 10+2π fO (x) sin(nx) dx π 10 for n = 1, 2, 3, . . . , and where fO (x) is the 2π-periodic odd extension of the function f (x) = x2 − πx, 0 < x < π. cn =

Answers: (1) False. Note that for f (x) = x2 −πx, 0 ≤ x ≤ π, we have f (0) = f (π) = 0. Therefore the odd 2π-periodic extension fO (x) of f is continuous on R. The D’Alambert’s form of the solution of (∗∗) is y = 12 [fO (x+5t)+fO (x−5t)]. However, fO (x) is defined by different formulas from x2 − πx outside of the interval [0, π] and it is not true that for arbitrary x ∈ R and t ≥ 0 1 1 2 2 2 [fO (x + 5t) + fO (x − 5t)] = 2 [(x + 5t) − π(x + 5t) + (x − 5t) − (x − 5t)]. (2) False. As explained in part (1) above, by D’Alambert’s formula, we have y = 1 1 [f 2 O (x + 5t) + fO (x − 5t)]. In particular, y(1, 1) = 2 [fO (6) + fO (−4)]. Since fO is 2π-periodic and −π < 6 − 2π < 0, we have fO (6) = fO (6−2π) = −f (2π−6) = −(2π−6)2 +π(2π−6) = −2π 2 +18π−36. Also, 0 < −4 + 2π < π and hence fO (−4) = fO (−4+2π) = f (−4+2π) = (2π−4)2 −π(2π−4) = 2π 2 −12π+16. Therefore 23 1 y(1, 1) = [fO (6) + fO (−4)] = −2π 2 + 3π − 10 6= . 2 2 (3) True. (4) False. The factor of π2 in front of the integral sign in the formula for cn is missing. (5) False. See item (6) below. Note that fO (x) = −x2 + πx for −π < x < 0.

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(6) True. Since fO is odd, we have Z π Z Z 1 2 π 2 π 2 fO (x) sin(nx) dx = fO (x) sin(nx) dx = (x −πx) sin(nx) dx. π −π π 0 π 0 (7) True. Since fO is 2π-periodic, we have 1 π

Z

π

1 fO (x) sin(nx) dx = π −π

Z

10+2π

fO (x) sin(nx) dx. 10

Problem 5. Consider the equation (†)

y 000 + 3y 00 + 3y 0 + y = x2 e−x + 5xe−x cos(2x)

on the interval (−∞, ∞). For each of the following statements indicate if it is true or false: (1) Equation (†) has a particular solution of the form y = (A + Bx + Cx2 )e−x + (D + Ex)e−x cos(2x) for some constants A, B, C, D, E. (2) Equation (†) has a particular solution of the form y = x2 (A + Bx + Cx2 )e−x + (D + Ex)e−x cos(2x) for some constants A, B, C, D, E. (3) Equation (†) has a particular solution of the form y = x3 (A + Bx + Cx2 )e−x + (D + Ex)e−x cos(2x) for some constants A, B, C, D, E. (4) Equation (†) has a particular solution of the form y = x3 [(A + Bx + Cx2 )e−x + (D + Ex)e−x cos(2x)] for some constants A, B, C, D, E. (5) Equation (†) has a particular solution of the form y = x3 (A + Bx + Cx2 )e−x + (D + Ex)e−x cos(2x) + (F + Gx)e−x sin(2x) for some constants A, B, C, D, E, F, G. (6) For every A, B, C, D, E, F, G ∈ R the function y = x3 (A + Bx + Cx2 )e−x + (D + Ex)e−x cos(2x) + (F + Gx)e−x sin(2x) is a solution of equation (†). (7) The general solution of of equation (†) is y = (c1 +c2 x+c3 x2 )e−x +x3 (A+Bx+Cx2 )e−x +(D+Ex)e−x cos(2x)+(F +Gx)e−x sin(2x) where c1 , c2 , c3 , A, B, C, D, E, F, G ∈ R are arbitrary constants.

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(8) There exist A0 , B0 , C0 , D0 , E0 , F0 , G0 ∈ R such that the general solution of of equation (†) is y = (c1 +c2 x+c3 x2 )e−x +x3 (A0 +B0 x+C0 x2 )e−x +(D0 +E0 x)e−x cos(2x)+(F0 +G0 x)e−x sin(2x) where c1 , c2 , c3 ∈ R are arbitrary constants. Answers: (1) False. Duplication with terms from yc is not yet eliminated, and the terms involving e−x sin(2x) are missing. (2) False, for similar reasons to (1). (3) False. Duplication with terms from yc is eliminated, but the terms involving e−x sin(2x) are still missing. (4) False. See (5) below for the correct form of yp . (5) True. (6) False. To make the statement in (5) true we must replace the opening phrase “For every” by “For some”. (7) False, for similar reasons to (6). The constants A, B, C, D, E, F, G are not allowed to be arbitrary. (8) True.