Leonard Dyer

## Answered question

2022-11-22

Find Inverse Laplace Transform of $\frac{1}{\left({s}^{2}+1\right)\left({s}^{2}-2s+7\right)}.$

### Answer & Explanation

ysik92ASw

Beginner2022-11-23Added 13 answers

Hint. By a partial fraction decomposition, one may avoid convolution, getting
$\frac{20}{\left({s}^{2}+1\right)\left({s}^{2}-2s+7\right)}=\frac{s+3}{\left({s}^{2}+1\right)}-\frac{s+1}{\left({s}^{2}-2s+7\right)}$
then, by using the Inverse Laplace Transform, we have
${\mathcal{L}}^{-1}\left(\frac{s+3}{\left({s}^{2}+1\right)}\right)\left(t\right)=3\mathrm{sin}\left(t\right)+\mathrm{cos}\left(t\right),$
${\mathcal{L}}^{-1}\left(\frac{s+1}{\left(s-1{\right)}^{2}+6}\right)\left(t\right)=\frac{{e}^{t}}{\sqrt{6}}\cdot \left(2\mathrm{sin}\left(\sqrt{6}t\right)+\sqrt{6}\mathrm{cos}\left(\sqrt{6}t\right)\right).$

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