Use the convolution theorem to find the inverse Laplace transform of the given function.

$\frac{\mathbf{s}}{{\left(s^2+1\right)}^{\mathbf{2}}}$

Let $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}$and $\mathit{g}\mathbf{\left(}\mathit{t}\mathbf{\right)}$be piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$and of exponential order $\alpha$and set role="math" localid="1664879363851" $\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{}\mathbf{=}\mathcal{L}\mathbf{\left\{}\mathit{f}\mathbf{\right\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{G}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathcal{L}\mathbf{\left\{}\mathit{g}\mathbf{\right\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$, then,

role="math" localid="1664879377536" $\mathcal{L}\mathbf{\{}\mathit{f}\mathbf{*}\mathit{g}\mathbf{\}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{=}\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathit{G}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{,}$

or

role="math" localid="1664879391067" ${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\mathbf{\left\{}\mathit{F}\mathbf{\right(}\mathit{s}\mathbf{\left)}\mathit{G}\mathbf{\right(}\mathit{s}\mathbf{\left)}\mathbf{\right\}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathit{f}\mathbf{*}\mathit{g}\mathbf{)}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

Consider the given function,$\frac{s}{{\left(s^2+1\right)}^2}$

Let,

$y\left(s\right)=\frac{s}{{\left(s^2+1\right)}^2}$

Take inverse Laplace transform,

${\mathcal{L}}^{-1}\left[y\right(s\left)\right]={\mathcal{L}}^{-1}\left[\frac{s}{{\left(s^2+1\right)}^2}\right]$

Hence, the convolution formula is,${\mathcal{L}}^{-1}\left[f\right(s)·g(s\left)\right]=f\star g={\int }_0^{t}f(t-v)g\left(v\right)dv$, where

$f\left(s\right)=\frac{s}{s^2+1}andf\left(t\right)=\cos t$

$$\begin{gathered} g\left(s\right)=\frac{1}{s^2+1} \\ g\left(t\right)=\sin t \end{gathered}$$

Thus, the equationcan be written as,

$$\begin{gathered} y\left(t\right)={\int }_{}^{}c\mathrm{os}(t-v)·\sin vdv \\ ={\int }_0^t\frac{{\displaystyle 1}}{{\displaystyle 2}}[\sin t+\sin (2v-t\left)\right]dv \end{gathered}$$

Use the formula,$\sin (A+B)+\sin (A-B)=2\sin A·\cos B$

$$\begin{gathered} y\left(t\right)={\int }_0^t\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin tdv+{\int }_0^t\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin (2v-t)dv \\ =\frac{1}{2}\sin t[v{]}_0^t-\frac{1}{4}[\cos (2v-t){]}_0^t \\ =\frac{1}{2}t\sin t-\frac{1}{4}[\cos t-\cos (-t\left)\right] \\ =\frac{1}{2}t\sin t \because \left[\cos (-x)=\cos x\right] \end{gathered}$$

Hence, the inverse Laplace transform for the given function is.

$y\left(t\right)=\frac{1}{2}t\sin t$