Question

Inverse laplace transform of \(\frac{1}{(s-1)^2 (s+5)}\) is?

(a) ^1⁄6 e^ – t – ^1⁄36 e^t + ^1⁄36 e^-5t

(b) ^1⁄6 e^tt – ^1⁄36 e^t + ^1⁄36 e^-5t

(c) ^1⁄6 e^-tt^2 – ^1⁄36 e^-t + ^1⁄36 e^5t

(d) ^1⁄6 e^-t t-^1⁄36 e^-t + ^1⁄36 e^5t

This question was addressed to me during an online interview.

Question is taken from Laplace Transform by Properties topic in portion Laplace Transform of Engineering Mathematics

Answer 1

The correct answer is (a) ^1⁄6 e^ – t – ^1⁄36 e^t + ^1⁄36 e^-5t

Easiest explanation: Given, \(F(s)=\frac{1}{(s-1)^2 (s+5)}=\frac{1}{(s-1)} \left [\frac{1}{(s-1)(s+5)}\right ]\)

=\(\frac{1}{(s-1)} \left [\frac{1}{6(s-1)}-\frac{1}{6(s+5)}\right ]\)

=\(\frac{1}{6} \left [\frac{1}{(s-1)^2}-\frac{1}{(s-1)(s+5)}\right ]\)

=\(\frac{1}{6} \left [

\frac{1}{(s-1)^2} – \frac{1}{6} \left [\frac{1}{(s-1)} – \frac{1}{(s+5)}\right ]\right ]\)

=\(\frac{1}{6(s-1)^2}-\frac{1}{36(s-1)}+\frac{1}{36(s+5)}\)

Inverse Laplace transform is \(f(t)=\frac{1}{6} e^t t-\frac{1}{36} e^t+\frac{1}{36} e^{-5t}\)