klupko5HR

2022-11-25

Find the inverse Laplace transform of $\frac{{s}^{2}-4s-4}{{s}^{4}+8{s}^{2}+16}$

Kozuh5F7

Expert

Note that
${\mathcal{L}}^{-1}\left[\frac{a}{{s}^{2}+{a}^{2}}\right]=\mathrm{sin}at$
and
${\mathcal{L}}^{-1}\left[\frac{{s}^{2}-{a}^{2}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\right]=t\mathrm{cos}at,$
then
$\frac{a}{{s}^{2}+{a}^{2}}-a\cdot \frac{{s}^{2}-{a}^{2}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}=\frac{2{a}^{3}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}.$
Hence
${\mathcal{L}}^{-1}\left[\frac{2{a}^{3}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\right]=\mathrm{sin}at-at\mathrm{cos}at$
and
${\mathcal{L}}^{-1}\left[\frac{8}{\left({s}^{2}+4{\right)}^{2}}\right]=\frac{1}{2}\left(\mathrm{sin}2t-2t\mathrm{cos}2t\right).$

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