Question

In Problems$37-48$find either $F\left(s\right)$or$f\left(t\right)$, as indicated.$L^{-1}\left\{\frac{e^{-2s}}{s^2(s-1)}\right\}$

Short answer

The inverse is$L^{-1}\left\{\frac{e^{-2e}}{s^2(s-1)}\right\}=-\cos \left(2t\right)U\left(t-\frac{\pi }{2}\right)$

Step-by-step solution

Formula to be used 

${\text{L}}^{-1}\left[e^{-as}F\left(s\right)\right]=f(t-a)U(t-a)$

We find F(s)

$$\begin{gathered} F\left(s\right)=\frac{s}{s^2+4} \\ a=\frac{\pi }{2} \\ f\left(t\right)=\cos \left(2t\right) \end{gathered}$$

Substituting the equation in step 2 with the formula from step 1

$$\begin{gathered} L^{-1}\left[e^{-as}F\left(s\right)\right]=f(t-a)U(t-a) \\ {L}^{-1}\left[e^{-as}F\left(s\right)\right]=\cos \left(2\left(t-\frac{\pi }{2}\right)\right)U\left(t-\frac{\pi }{2}\right) \\ {L}^{-1}\left[e^{-as}F\left(s\right)\right]=\cos \left(2t-\pi \right)U\left(t-\frac{\pi }{2}\right) \\ {L}^{-1}\left[e^{-as}F\left(s\right)\right]=-\cos \left(2t\right)U\left(t-\frac{\pi }{2}\right) \end{gathered}$$

Hence, the answer is $L^{-1}\left\{\frac{e^{-2e}}{s^2(s-1)}\right\}=-\cos \left(2t\right)U\left(t-\frac{\pi }{2}\right)$