Question

Problem #12: The inverse Laplace transform f( = £"{F(s)} of the function ei(-3s +8) F(s) s249 is of the form g(i) U[h{f)]. (a) Enter the function g(t) into the answer box below (b) Enter the function A(t) into the answer box below. Enter your answer as a symbolic function of t. as in these examples Problem #12(a): Enter your answer as a symbolic function of t, as in these examples Problem 12(b) Submit Problem #12 for Grading Just Save Problem #12 Attempt#1 Attempt #2 Attempt #3 Attempt#5 Attempt #4 Your Answer: 12(a) 12(b) 12(a) 12(b) 12(a) 12(b) 12(a) 12(b) 12(a) 12(b) Your Mark: 12(a) 12(b) 12(a) 12(b) 12(a) 12(b) 12(a) 12(b) 12(a) 12(b)

Answer 1

F(S) = e^-2S (-3S + 8)/S^2 + 49 F(S) = e^-2S [-3 S/S^2 + 7^2 + 8 1/S^2 + 7^2] F(S) = [-3 e^-2S (S/S^2 + 7^2) + 8 e^-2S (1/S^2 + 7^2)] {therefore L^-1 (S/S^2 + 7^2) = Cos 7t, L^-1 (1/S^2 + 7^2) = Sin 7t] L^-1(F(S)) = -3 L^-1[e^-2S (S/S^2 + 7^2) + 8L^-1(e^-2S 1/S^2 + 7^2)] f(t) = 1/7 (-21 Cos (7t - 14) + 8 Sin (7t - 14) U(t - 2) f(t) = 1/7 [-21 Cos(7t - 14) + 8 Sin (7t - 14) U_2(t) [Therefore L^-1(e^-as F(S)) = U_a(t) f(t - a)] comparing with g(t) U[h(t)] we get g(t) = 1/7[8 Sin (7t - 14) - 21 Cos (7t - 14)] and h(t) = t - 2