Question

Find the inverse laplace transform of \(\frac{1}{(s^2+1)(s – 1)(s + 5)}\).

(a) ^1⁄12 e^t – ^1⁄13 Cos(-t) – ^1⁄12 Sin(-t) – ^1⁄156 e^-5t

(b) ^1⁄12 e^-t – ^1⁄13 Cos(t) – ^1⁄12 Sin(t) – ^1⁄156 e^5t

(c) ^1⁄12 e^t – ^1⁄13 Cos(t) – ^1⁄12 Sin(t) – ^1⁄156 e^-5t

(d) ^1⁄12 e^t + ^1⁄13 Cos(t) + ^1⁄12 Sin(t) + ^1⁄156 e^-5t

I had been asked this question by my college professor while I was bunking the class.

This is a very interesting question from Laplace Transform by Properties in section Laplace Transform of Engineering Mathematics

Answer 1

Correct choice is (c) ^1⁄12 e^t – ^1⁄13 Cos(t) – ^1⁄12 Sin(t) – ^1⁄156 e^-5t

The best explanation: Given , F(s)=\(\frac{1}{(s^2+1)(s-1)(s+5)}\)

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

=\(\frac{1}{6} \left [\frac{1}{2*(s – 1)}-\frac{1}{2} \frac{s+1}{(s^2+ 1)}\right ]-\frac{1}{6}\left [\frac{1}{26*(s + 5)}-\frac{1}{26} \frac{s-5}{(s^2+1)}\right ]\)

=\(\frac{1}{12(s – 1)}-\frac{1}{26} \frac{2s+3}{(s^2+ 1)}-\frac{1}{156(s + 5)}\)

=\(\frac{1}{12(s – 1)}-\frac{1}{13} \frac{s}{(s^2+ 1)}-\frac{1}{12} \frac{1}{(s^2+ 1)}-\frac{1}{156(s + 5)}\)

=\(\frac{1}{12} e^t-\frac{1}{13} Cos(t)-\frac{1}{12} Sin(t)-\frac{1}{156}e^{-5t}\)