illusiia

2021-09-28

Solue using partial fractions
$F\left(S\right)=\frac{-4s}{{\left({s}^{2}+4\right)}^{2}}$

Step 1
$F\left(s\right)=\frac{-4s}{{\left({s}^{2}+4\right)}^{2}}$
Partial form:
$\frac{-4s}{{\left({s}^{2}+4\right)}^{2}}=\frac{As+B}{{\left({s}^{2}+4\right)}^{2}}$
$-4s=As+B$

Laplace Properties:
${L}^{-1}\left[f\left(s\right)\right]=F\left(t\right)$
${L}^{-1}\left[f\left(s\right)\right]=\frac{d}{dt}F\left(t\right)+F\left(g\right)$
${L}^{-1}\left[\frac{-4s}{{\left({s}^{2}+4\right)}^{2}}\right]=-{L}^{-1}\left[s\left\{\frac{4}{{\left({s}^{2}+4\right)}^{2}}\right\}\right]$
$=-{L}^{-1}\left[\mathsf{s}\right]=-\left[\frac{d}{dt}\left\{F\left(t\right)\right\}+F\left(g\right)\right]$
$=-\frac{d}{dt}{L}^{-1}\left[\frac{4}{{\left({s}^{2}+4\right)}^{2}}\right]+{L}^{-1}\left[\frac{4}{{\left({s}^{2}+4\right)}^{2}}\right]\left\{0\right\}$
Step 2
$={L}^{-1}\left[\frac{4}{{\left({s}^{2}+4\right)}^{2}}\right]=\frac{1}{4}\left[\mathrm{sin}2t-2t\mathrm{cos}2t\right]\left\{0\right\}$
$\frac{d}{dt}\left[{L}^{-1}\left(\frac{4}{{\left({s}^{2}+4\right)}^{2}}\right)\right]=\frac{1}{4}\frac{d}{dt}\left[\mathrm{sin}2t-2t\mathrm{cos}2t\right]$