Question

LaTech Zill8e Ch7Sec5: Problem 2 Previous Problem Problem List Next Problem 1 point) Use the Laplace transform to solve the following initial value problem: Notation for the step function is ul-C-le (1).

Answer 1

"Given ODE y""- 3y' - 10y = S(t - 2) y(0) = 0, y'(0) = 0 -(0) L(y') = phi^2 y(s) - s y(0) - y'(0) L(y') = s y(s) - y(0) Taking Laplace transformation of Both sides L(y""- 3y' - 12y) = l(s(t - 2)) (s^2 y(s) - 3s y(s) - theta y(s)) - sx_0 - 0 -0 = l(s(t - 2)) implies (s^2 - 3s - 1 theta) y(s) = L(s(t - 2) = L({0 for t < 2 s(t - 2) for t greaterthanorequalto 2}) y(s) = 1/s^2 - 3s - theta) L(s(t - 2) = 1/s^2 - 3s - 1 theta e^-2s delta(s) y(s) = 1/(s - 6)(s - 2) (e^-2s/s) = 1/s(s - 6) (s - 1) e^-2s -(*) partial fraction 1/s(s - 6) (s - 2) = A/s + B/s - 3 + c/s - 6 1 = A(s - 2)(s - 6) + B s(s - 6) + c s(s - 2) for A = 0: A = 1/18 for B = s = 3, B = -1/3 for c = s = 6 c = 1/18 So y(s) = (1/18 s - 1/9(s "