Question

$L^{-1}\left\{\frac{e^{-2s}}{s^3}\right\}$

Short answer

The Laplace transform is$\frac{1}{2!}{\left(t-2\right)}^{2}U(t-2)$

Step-by-step solution

Second Translation Theorem

If$F\left(s\right)=L\left\{f\right(t\left)\right\}$and$a>0,$then$L\left\{f\right(t-a\left)U\right(t-a\left)\right\}=e^{-as}F\left(s\right)$

Apply Laplace transform

Here, $a=2$we only need to manipulate, $\frac{1}{s^3}$to make it look like something we know.

Thus,

$$\begin{gathered} L^{-1}\left\{\frac{e^{-2s}}{s^3}\right\}=L^{-1}\left\{\frac{1}{2!}\frac{2!}{s^3}{e}^{-2s}\right\} \\ Y\left(t\right)=\frac{1}{2!}{\left(t-2\right)}^{2}U(t-2) \end{gathered}$$

Hence, the solution is$Y\left(t\right)=\frac{1}{2!}{\left(t-2\right)}^{2}U(t-2)$