Question

Find the \(L^{-1} \left (\frac{s+1}{(s-1)(s+2)^2}\right )\).

(a) \(\frac{2}{7} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}×t\)

(b) \(\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}×t\)

(c) \(\frac{2}{9} e^t-\frac{2}{9} e^{-3t}+\frac{1}{3} e^{-2t}×t\)

(d) \(\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}\)

This question was addressed to me in exam.

The question is from General Properties of Inverse Laplace Transform in chapter Laplace Transform of Engineering Mathematics

Answer 1

The correct choice is (b) \(\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}×t\)

The explanation: In the given question,

\(L^{-1} \left (\frac{s+1}{(s-1)(s+2)^2}\right )\)

Using properties of partial fractions-

s+1=A(s+2)^2+B(s-1)(s+2)+C(s-1)

At s=1, A=\(\frac{2}{9}\)

At s=2, C=\(\frac{1}{3}\)

At s=0, B=\(\frac{-2}{9}\)

Re substituting all these values in the original fraction,

=\(L^{-1} \left (\frac{2}{9(s-1)} + \frac{-2}{9(s+2)} + \frac{1}{3(s+2)^2}\right)\)

=\(\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}×t\).