Question

Find Laplace transform of :
$\frac{1}{(s+2)(s^2+4s+9)}$

Answer 1

$L^{-1}\frac{1}{(s+2)(s^2+4s+9)}$

Let $\overline{f}\left(s\right)=\frac{1}{s+2}$

$\therefore$ $f\left(s\right)=e^{-2t}$

Let $\overline{g}\left(s\right)=\frac{1}{s^2+4s+9}=\frac{1}{{(s+2)}^2+5}=\frac{1}{\sqrt{5}}\frac{\sqrt{5}}{{(s+2)}^2+5}$

$\therefore$ $g\left(s\right)=e^{-2t}.\frac{1}{\sqrt{5}}sin\sqrt{5}t$

By convolution,

${\int .}^{-1}\frac{1}{(s+2)(s^2+4s+9)}={\int }_0^t{e}^{-24}.\frac{1}{\sqrt{5}}\sin \sqrt{54}.e^{-2(t-u)}du$

$=\frac{1}{\sqrt{5}}{\int }_0^t\sin \sqrt{5}u.e^{-2u}.e^{-2t}.e^{2u}du$

$=\frac{1}{\sqrt{5}}.e^{-2t}{\int }_0^t\sin \sqrt{5}udu=\frac{1}{\sqrt{5}}{e}^{-2t}\left[\frac{-(\cos \sqrt{5}t-\cos 0)}{\sqrt{5}}\right]$

$=\frac{e^{-2t}}{5}(1-\cos \sqrt{5}t)$