Question

Problem #16; Use the Laplace transform to solve the following initial value problem y2y35 t-4), y(0) = 0. y'(0) 0 The solution is of the form Ug(t] h(t). (a) Enter the function g(f) into the answer box below. (b) Enter the function h(t) into the answer box below Enter your answer as a symbolic function of t. as in these Problem #16(a): examples Enter your answer as a symbolic function of t. as in these examples Problem 16(b): Submit Problem # 16 for Grading Just Save Attempt #4 Attempt #1 Problem #16 Attempt #3 Attempt #5 Attempt#2 Your Answer: 16(a) 16(a) 16(b) 16(a) 16(b) 16(a) 16(b) 16(a) 16(b) 16(b) Your Mark: 16(a) 16(a) 16(b) 16(a) 16(b) 16(a) 16(b) 16(a) 16(b) 16(b)

Answer 1

y'' - 2y' - 35y = delta(t - 4), y(0) = 0, y'(0) = 0 Taking laplace transform on both side alpha(y'' - 2y' - 35y) = alpha(delta(t - 4)) implies s^2 y(s) - sy(0) - y'(0) - 23 times (s) + 2y(0) - 35y(8) = e^- 4s implies y(s) (s^2 - 2s - 3s) = e^- 4s (because y(0) = 0, y'(0) = 0 implies y(s) = e^- 4s/(s^2 - 2s - 35) = e^- 4s/(s^2 - 7s + 5s - 35) = e^- 4s/(s - 7)(s + 5) y(t) = alpha^- 1(e^- 4s/(s - 7)(s + 4) = alpha^- 1(1/11 e^- 4s/(s - 7)) - alpha^- 1(1/11 e^- 4s/s + 4) = 1/11(alpha^- 1(e^- 4s/s - 7) - alpha^- 1(e^- 4s/(s + 4)) = 1/11 u_x (t) e^7(t - 4) - 1/11 u_4(t) e^-4(t - 4) y(t) = 1/11 u_4 (t) (e^7t - 28 - e^- 4t + 16) (a) g(t) = e^7t - 28 - e^- 4t + 16 (b) h(t) = 1/11 u_4(t)