inverse laplace transform of $\frac{s^3}{(s^2+4)^2}$

Question

Using partial fractions gives $\frac{s}{s^2+4}$ - $\frac{4s}{(s^2+4)^2}$

Inverse laplace transform of the first member ($\frac{s}{s^2+4}$) is cos(2t). Can't figure out how to transform $\frac{4s}{(s^2+4)^2}$ . Help needed.

Answer 1

Hint:

Notice that \begin{align} \frac{s^3}{(s^2+4)^2}&=\frac{s(s^2+4)-4s}{(s^2+4)^2}=\frac{s}{s^2+4}-\frac{4s}{(s^2+4)^2} \end{align}

Also notice $\mathscr{L}\left\{\sin (2t)\right\}=\frac{2}{s^2+4}\,\,$ and $\,\,\frac{\mathrm d}{\mathrm ds}\left(\frac{2}{s^2+4}\right)=-\frac{4s}{s^2+4}$. Then use the fact that $$\mathscr{L}\left\{tf(t)\right\}=-\frac{\mathrm dF(s)}{\mathrm d s}$$ where $F(s)=\mathscr{L}\left\{f(t)\right\}$.