MathQuest: Differential Equations

Transcription

1 MathQuest: Differential Equations Laplace Tranforms 1. True or False The Laplace transform method is the only way to solve some types of differential equations. (a) True, and I am very confident (b) True, but I am not very confident (c) False, but I am not very confident (d) False, and I am very confident. Which of the following differential equations would be impossible to solve using the Laplace transform? (a) dc dr = 1c+sin(r)+8cos (r +) (b) d f dx 100 df dx = 18 ln(x) (c) g (b) = 1 g (d) p (q) = 4 q +96p 1p. Suppose we know that zero is an equilibrium value for a certain homogeneous linear differential equation with constant coefficients. Further, suppose that if we begin at equilibrium and we add the nonhomogenous driving term f(t) to the equation, then the solution will be the function y(t). True or False: If instead we add the nonhomogeous driving term f(t), then the solution will be the function y(t). (a) True, and I am very confident (b) True, but I am not very confident (c) False, but I am not very confident (d) False, and I am very confident 4. Suppose we know that zero is an equilibrium value for a certain homogeneous linear differential equation with constant coefficients. Further, suppose that if we begin at equilibrium and we add a nonhomogenous driving term f(t) to the equation that acts onlyforaninstant, att = t 1, thenthesolutionwillbethefunctiony(t). True or False: Even if we know nothing about the differential equation itself, using our knowledge of f(t) and y(t) we can infer how the system must behave for any other nonhomogeneous driving function g(t) and for any other initial condition. 1

2 (a) True, and I am very confident (b) True, but I am not very confident (c) False, but I am not very confident (d) False, and I am very confident. The Laplace transform of a function y(t) is defined to be a function Y(s) so that L[y(t) = Y(s) = y(t)e st dt. If we have the function y(t) = e t, then what is its 0 Laplace transform L[y(t) = Y(s)? (a) L[y(t) = 1 s (b) L[y(t) = 1 s (c) L[y(t) = 1 s (d) L[y(t) = 1 s if s >. if s <. if s >. if s <. 6. What is the Laplace transform of the function y(t) = 1? (a) L[y(t) = 1 if s > 0. s (b) L[y(t) = 1 if s < 0. s (c) L[y(t) = 1 if s > 0. s (d) L[y(t) = 1 if s < 0. s (e) This function has no Laplace transform. 7. What is L[e t? (a) L[e t = 1 s (b) L[e t = 1 s (c) L[e t = 1 s+ (d) L[e t = 1 s+ if s <. if s >. if s >. if s <. 8. Suppose we have the Laplace transform Y(s) = 1 s y(t). What is L 1[ 1 s? (a) L 1[ 1 = e t s = e t (b) L 1[ 1 s (c) This cannot be done and we want the original function

3 9. Suppose we know that the Laplace transform of a particular function y(t) is the function Y(s) so that L[y(t) = Y(s). Now, suppose we multiply this function by and take the Laplace transform. What can we say about L[y(t)? (a) L[y(t) = Y(s). (b) L[y(t) = Y(s). (c) L[y(t) = 1 Y(s). (d) L[y(t) = 1 e Y(s). (e) None of the above 10. Suppose we know that the Laplace transform of a particular function f(t) is the function W(s) and that the Laplace transform of another function g(t) is the function X(s) so that L[f(t) = W(s) and L[g(t) = X(s). Now, suppose we multiply these two functions together and take the Laplace transform. What can we say about L[f(t)g(t)? (a) L[f(t)g(t) = W(s)X(s). (b) L[f(t)g(t) = W(s) X(s). (c) L[f(t)g(t) = W(s)+X(s). (d) L[f(t)g(t) = W(s) X(s). (e) None of the above 11. Suppose we know that the Laplace transform of a particular function f(t) is the function W(s) and that the Laplace transform of another function g(t) is the function X(s) so that L[f(t) = W(s) and L[g(t) = X(s). Now, suppose we add these two functions together and take the Laplace transform. What can we say about L[f(t)+g(t)? (a) L[f(t)+g(t) = W(s)X(s). (b) L[f(t)+g(t) = W(s) X(s). (c) L[f(t)+g(t) = W(s)+X(s). (d) L[f(t)+g(t) = W(s) X(s). (e) None of the above 1. SupposewehaveY(s) = 1 s + s+6 andwewantthefunctiony(t). WhatisL 1[ 1 s + (a) L 1[ 1 s + s+6 (b) L 1[ 1 s + s+6 (c) L 1[ 1 s + s+6 = e t +e 6t/ = e t +e 6t = e t e 6t s+6?

4 (d) L 1[ 1 + s s+6 = e t e 6t (e) This cannot be done 1. Suppose we know that L[y(t) = Y(s) and we want to take the Laplace transform of the derivative of y(t), L [ dy dt. To get started, recall the method of integration by parts, which tells us that udv = uv v du. (a) L [ dy dt = y(t)e st +sy(s). (b) L [ dy dt = y(t)e st sy(s). (c) L [ dy dt = y(0)+sy(s). (d) L [ dy dt = y(0) sy(s). (e) L [ dy dt = y(0)+sy(s). 14. We know that L[y(t) = Y(s) = y(t)e st dt and that L [ dy 0 dt = y(0)+sy(s). What dowegetwhenwetakethelaplacetransformofthedifferential equation dy = dt y+e t with y(0) =? (a) +sy(s) = Y(s)+ (b) +sy(s) = 1 Y(s)+ s 1 (c) +sy(s) = Y(s)+ (d) +sy(s) = Y(s) (e) sy(s) = 1 Y(s)+ 1 s 1. We know that L[y(t) = Y(s) and that L [ dy dt = y(0) + sy(s). Take the Laplace transform of the differential equation dy = dt y+e t with y(0) = 4 and solve for Y(s). (a) Y(s) = ( s )(s ) + 4 s (b) Y(s) = + 4 ( s )(s ) s+ (c) Y(s) = (s+)(s ) + 4 s (d) Y(s) = (s+)(s+) + 4 s+ 16. Suppose that after taking the Laplace transform of a differential equation we solve for the function Y(s) =. This is equivalent to which of the following? (a) Y(s) = /8 s+ + /8 s 6 (s+)(s 6) 4

5 (b) Y(s) = 8/ s+ + 8/ s 6 (c) Y(s) = /8 + /8 s+ s 6 (d) Y(s) = 8/ + 8/ s+ s Find L 1 [ (a) L 1 [ (b) L 1 [ (c) L 1 [ (d) L 1 [ (e) L 1 [ (s+)(s 6) (s+)(s 6) (s+)(s 6) (s+)(s 6) (s+)(s 6) (s+)(s 6). = e t e 6t = 8 e t + 8 e6t = 8 et 8 e 6t = e /8 e t +e /8 e 6t = 8 et + 8 e 6t 18. Solve the differential equation dy dt = 4y + e t with y(0) = using the method of Laplace transforms: First take the Laplace transform of the entire equation, then solve for Y(s), use the method of partial fractions to simplify the result, and take the inverse transform to get the function y(t). (a) y(t) = 9 e4t + 1 e t (b) y(t) = 7 e4t + e t (c) y(t) = 9 e 4t + 1 et (d) y(t) = 7 e 4t + et 19. Consider the Heaviside function, which is defined as { 0 if t < a u a (t) = 1 if t a Suppose a =. What is the Laplace transform of this L[u (t)? Hint: It may be helpful to break this integral into two pieces, from 0 to and from to. (a) L[u (t) = 1 s es + 1 s (b) L[u (t) = 1 s e s 1 s (c) L[u (t) = 1 s e s (d) L[u (t) = 1 s e s (e) L[u (t) = 1 s es

6 0. Find the Laplace transform of the function u (t)e (t ). (a) L[u (t)e (t ) = e (b) L[u (t)e (t ) = e s (c) L[u (t)e (t ) = e e s (d) L[u (t)e (t ) = 1 1. Find the Laplace transform of the function u 4 (t)e (t 4). (a) L[u 4 (t)e (t 4) = e s s 4 (b) L[u 4 (t)e (t 4) = es s+4 (c) L[u 4 (t)e (t 4) = e 4s s+ (d) L[u 4 (t)e (t 4) = e 4s s (e) L[u 4 (t)e (t 4) = e4s s. Find L 1 [ (a) L 1 [ (b) L 1 [ (c) L 1 [ (d) L 1 [ (e) L 1 [ (f) L 1 [ e s. s+ e s s+ e s s+ e s s+ e s s+ e s s+ e s s+ = u (t)e (t ) = u (t)e (t ) = u (t)e (t ) = u (t)e (t ) = u (t)e (t ) = u (t)e (t ). Use Laplace transforms to solve the differential equation dy dt = y + 4u (t) with y(0) = 1. (a) y(t) = e t +u (t)e (t ) u (t)e (t ) (b) y(t) = e t +4u (t)e (t ) 4u (t)e (t ) (c) y(t) = e t +u (t) u (t)e (t ) (d) y(t) = e t +4u (t) 4u (t)e (t ) 6

7 4. What is the Laplace transform of the second derivative, L (a) L (b) L [ d y dt = y (0)+sL[y [ d y dt = y (0) sy(0)+s L[y [ d y (c) L = y (0) sy(0)+sl[y dt [ d (d) L y = y (0)+sy(0) s L[y dt [ d y dt?. We know that the differential equation y = y with the initial condition y(0) = 0 and y (0) = 1 is solved by the function y = sint. By taking the Laplace transform of this equation find an expression for L[sin t. (a) L[sint = 1 s +1 (b) L[sint = 1 s 1 (c) L[sint = s s +1 (d) L[sint = s s 1 6. Find an expression for L[sinωt, by first using your knowledge of the second order differential equation and initial conditions y(0) and y (0) that are solved by this function, and then taking the Laplace transform of this equation. (a) L[sinωt = ω s +1 (b) L[sinωt = 1 s ω +ω (c) L[sinωt = 1 s +ω (d) L[sinωt = ω s +ω (e) L[sinωt = ω s +ω 7. Find an expression for L[cosωt, by first using your knowledge of the second order differential equation and initial conditions y(0) and y (0) that are solved by this function, and then taking the Laplace transform of this equation. (a) L[cosωt = s s +ω (b) L[cosωt = sω s +ω (c) L[cosωt = sω s ω (d) L[cosωt = ω s +ω (e) L[cosωt = ω s ω 7

8 8. Find L 1[ s 4+s. (a) L 1[ s 4+s = cost. (b) L 1[ s 4+s = cos4t. (c) L 1[ s 4+s = 4cost. (d) L 1[ s 4+s = sint. (e) L 1[ s 4+s = sin4t. (f) L 1[ s 4+s = 4sint. 9. Find L 1[ s +6. (a) L 1[ s +6 = sin t. (b) L 1[ s +6 = sin 6t. (c) L 1[ s +6 = sin t. (d) L 1[ s +6 = sin 6t. (e) L 1[ s +6 = sin t. 0. We know that the function f(t) has a Laplace transform F(s), so that L[f(t) = 0 f(t)e st dt = F(s). What can we say about the Laplace transform of the product L[e at f(t)? (a) L[e at f(t) = e at F(s). (b) L[e at f(t) = e as F(s). (c) L[e at f(t) = F(s a). (d) L[e at f(t) = F(s+a). (e) We cannot make a general statement about this Laplace transform without knowing f(t). 1. Find L[e 4t sint. (a) L[e 4t sint = (s 4) +. (b) L[e 4t sint = 1 (s 4) +. (c) L[e 4t sint = 1 (s+4) +. (d) L[e 4t sint = (s 4) +. 8

9 (e) L[e 4t sint = (s+4) +.. Find L 1 [ (a) L 1 [ (b) L 1 [ (c) L 1 [ (d) L 1 [ 8 (s+) (s+) (s+) (s+) (s+) +16. = e t sin4t = e 4t sint = 8e t sin4t = e t sin4t. s +6 is equivalent to which of the following? (a) s +6 = (s+) +11 (b) s +6 = (s+) +6 (c) s +6 = (s+) 10 (d) s +6 = (s+6) 1 4. Find L 1[ s+6 s +6, by first completing the square in the denominator. (a) L 1[ s+6 s +6 = e t sin 6t (b) L 1[ s+6 s +6 = e t cos 6t (c) L 1[ s+6 s +6 = e t cos 6t (d) L 1[ s+6 s +6 = e t sin 6t (e) L 1[ s+6 s +6 = e 6t cos 6t (f) L 1[ s+6 s +6 = e 6t cos 6t. Find L 1[ s+8 s +6, by first completing the square in the denominator. (a) L 1[ s+8 s +6 = e t cos 6t (b) L 1[ s+8 s +6 = e t cos 6t+ (c) L 1[ s+8 s +6 = e t cos 6t+e t sin 6t (d) L 1[ s+8 s +6 = e t cos 6t+ 6 e t sin 6t 9

10 6. Suppose that after taking the Laplace transform of a differential equation we obtain the function Y(s) =. This is equivalent to which of the following? (a) Y(s) = s/19 + s/19 s +16 s + (b) Y(s) = s/19 + s/19 s +16 s + s (s +16)(s +) (c) Y(s) = s/1 + s/1 s +16 s + (d) Y(s) = s/1 + s/1 s +16 s + 7. Find L 1 [ (a) L 1 [ (b) L 1 [ (c) L 1 [ (d) L 1 [ s (s +16)(s +) s (s +16)(s +) s (s +16)(s +) s (s +16)(s +) s (s +16)(s +). = 1 cos4t+ 1 cos t. = 1 cos4t 1 cos t. = sin4t+ 1 sin t. = sin4t 1 sin t. 8. Use Laplace transforms to solve the differential equation y = y+4sint if we know that y(0) = y (0) = 0. (a) y = 6 7 sin t+ 4 7 sint (b) y = 6 7 sin t 4 7 sint (c) y = 1 7 sin t 1 7 sint (d) y = 1 7 cos t+ 1 7 cost 9. The Dirac delta function δ a (t) is defined so that δ a (t) = 0 for all t a, and if we integrate this function over any integral containing a, the result is 1. What would be 10 (δ (t)+δ 6 (t) δ 8 (t)+δ 11 (t))dt? (a) 1 (b) 0 (c) 4 (d) (e) 0 (f) This integral cannot be determined. 10

11 40. Find 10 (δ 6 (t) δ 7 (t) δ 8 (t) δ 9 (t))dt? (a) 1 (b) 0 (c) 1 (d) 4 (e) (f) This integral cannot be determined. 41. Find L[δ 7 (t). (a) 1 (b) 7 (c) e 7s (d) e 7t (e) This integral cannot be determined. 4. Find L[δ 0 (t). (a) 0 (b) 1 (c) e s (d) e t (e) This integral cannot be determined. 4. We have a system modeled as an undamped harmonic oscillator, that begins at equilibrium and at rest, so y(0) = y (0) = 0, and that receives an impulse force at t = 4, so that it is modeled with the equation y = 9y +δ 4 (t). Find y(t). (a) y(t) = u 4 (t)sint (b) y(t) = 1 u 4(t)sint (c) y(t) = u 4 (t)sin(t 4) (d) y(t) = 1 u 4(t)sin(t 4) 44. We have a system modeled as an undamped harmonic oscillator, that begins at equilibrium and at rest, so y(0) = y (0) = 0, and that receives an impulse forcing at t =, so that it is modeled with the equation y = 4y +δ (t). Find y(t). 11

12 (a) y(t) = u (t)sint (b) y(t) = 1u (t)sin(t ) (c) y(t) = 1u (t)sin(t ) (d) y(t) = 1u (t)sin(t ) (e) y(t) = 1u (t)sin(t ) 4. For many differential equations, if we know how the equation responds to a nonhomogeneous forcing function that is a Dirac delta function, we can use this response function to predict how the differential equation will respond to any other forcing function and any initial conditions. For which of the following differential equations would this be impossible? (a) p (q) = 4 q +96p 1p (b) d m 100 dm nm = dn dn ln(n) (c) d f dx 100 df dx = 18 ln(x) (d) dc dr = 1c+sin(r)+8cos (r +) 46. The convolution of two functions f g is defined to be t f(t u)g(u)du. What is the 0 convolution of the functions f(t) = and g(t) = e t? (a) f g = e t 1 (b) f g = e t (c) f g = 4e t 4 (d) f g = 1 et 1 (e) This convolution cannot be computed. 47. What is the convolution of the functions f(t) = u (t) and g(t) = e t? (a) f g = 1 e (t ) 1 (b) f g = 1 e t 1 (c) f g = 1 e (t ) + 1 (d) f g = 1 e t + 1 (e) This convolution cannot be computed. 48. Let ζ(t) be the solution to the initial-value problem d y + p dy + qy = δ dt dt 0(t), with y(0) = y (0) = 0. Find an expression for L[ζ. 1

13 (a) L[ζ = 0 (b) L[ζ = s +ps+q (c) L[ζ = 1 s +ps+q (d) L[ζ = e s s +ps+q 49. Let ζ(t) = e t be the solution to an initial-value problem d y +qy = δ dt 0(t), with y(0) = y (0) = 0 1 for some specific values of p and q, so that L[ζ =. What s +ps+q will be the solution of d y +p dy +qy = 0 if y(0) = 0 and dt dt y (0) =? (a) y = e t (b) y = 10e t (c) y = e t (d) y = e 1t (e) It cannot be determined from the information given. dt +p dy 0. Let ζ(t) be the solution to the initial-value problem d y + p dy + qy = δ dt dt 0(t), with y(0) = y (0) = 0. Now suppose we want to solve this problem for some other forcing function f(t), so that we have d y +p dy +qy = f(t). Which of the following is a correct dt dt expression for L[y? (a) L[y = 0. (b) L[y = L[fL[ζ. (c) L[y = L[f L[ζ. (d) L[y = L[fζ. (e) None of the above 1. Let ζ(t) = e t be the solution to an initial-value problem d y +p dy +qy = δ dt dt 0(t), with y(0) = y (0) = 0. Now suppose we want to solve this problem for the forcing function f(t) = e t, so that we have d y +p dy +qy = f(t). Find y(t). dt dt (a) y(t) = e t e t (b) y(t) = e t e t (c) y(t) = e t e t (d) y(t) = et e t 1