How do I solve Inverse Laplace transform of $F(s)=\frac{4(e^{-2s}-1)}{(s)(s^2+4)}$?

Question

$$F(s)=\frac{4(e^{-2s}-1)}{(s)(s^2+4)}$$

My work:

Breaking them into partial fractions

$$\frac{1}{s(s^2+4)}=\frac{A}{s}+\frac{B}{s^2+4}$$

$$F(s)=\frac{e^{-2s}-1}{s}-\frac{e^{-2s}-1}{s^2+4}$$

$$\mu_{2}(t)=0, 0<t<2 , \mu_{2}(t)=1,t>2 $$

$$f(t)=\begin{cases}\frac{1}{2}\sin 2t-1 &, 0<t<2\\ \frac{1}{2}\sin2t-\frac{1}{2}\sin(2t-4) & t>2\end{cases}$$

I need some help. It does not seem right to me.

Answer 1

I think that after the partial fraction decomposition, it should be $$F(s)=\frac{e^{-2s}-1}{s}-\frac{(e^{-2s}-1)s}{s^2+4}.$$ Hence $$f(t)=\begin{cases}\cos( 2t)-1 &\mbox{for $0<t<2$,}\\ \cos(2t)-\cos(2(t-2)) &\mbox{for $t\geq 2$.}\end{cases}$$