Question

Solve the Ordinary Diferential Equation using Laplace Transformation y’’’ – 3y’’ + 3y’ – y = t^2 e^t when y(0) = 1, y’(0) = 0 and y’’(0) = 2.

(a) \(2e^t  \frac{t^5}{720}+e^t+2e^t  \frac{t}{6}+4e^t  \frac{t^2}{24} \)

(b) \(e^t  \frac{t^5}{720}+2e^{-t}+2e^t  \frac{t}{6}+4e^t  \frac{t^2}{24} \)

(c) \(e^{-t}  \frac{t^5}{720}+e^{-t}+2e^{-t}  \frac{t}{6}+4e^{-t}  \frac{t^2}{24} \)

(d) \(2e^{-t}  \frac{t^5}{720}+e^{-t}+2e^{-t}  \frac{t}{6}+4e^{-t}  \frac{t^2}{24} \)

I have been asked this question in an international level competition.

My enquiry is from Solution of DE With Constant Coefficients using the Laplace Transform topic in division Laplace Transform of Engineering Mathematics

Answer 1

Right answer is (a) \(2e^t  \frac{t^5}{720}+e^t+2e^t  \frac{t}{6}+4e^t  \frac{t^2}{24} \)

The explanation: L[y’’’ – 3y’’ + 3y’ – y = t^2 e^t]

s^3 Y(s) – s^2 y(0) – sy'(0) – y”(0) – 3s^2 Y(s) + 3sy(0) + 3y'(0) + 3sY(s) – 3y(0) – Y(s) = \(\frac{2}{(s-1)^3} \)

\(Y(s) = \frac{2}{(s-1)^6} +\frac{(s^2+3s+5)}{(s-1)^3} \)

\(y(t) = 2e^t  \frac{t^5}{720}+e^t+2e^t  \frac{t}{6}+4e^t  \frac{t^2}{24}\).