Inverse Laplace Step Function

Question

I've been asked to find the inverse Laplace Transform of: $$\mathcal L^{-1}\left\lbrace e^{-2s}\over {s^3}\right\rbrace$$
I lost my notes, so I'm going off of examples I have found online. I got stuck on the last step:
$$\mathcal L^{-1}\left\lbrace e^{-2s}\cdot {1\over {s^3}}\right\rbrace$$ $$=u(t-2)\cdot\frac 12t^2$$
That's unfortunately as far as I have gotten... If someone could help me figure out the last part that would be nice!

Answer 1

Here's one way to approach the problem: $$ \begin{align} \mathcal L^{-1}\left\lbrace e^{-2s}\cdot {1\over {s^3}}\right\rbrace &= \mathcal L^{-1}\left\{e^{-2s}\right\} * \mathcal L^{-1}\left\{1/s^3\right\}\\&= \delta(t-2)*\frac 12 t^2u(t) \\&= \frac 12 (t-2)^2u(t-2) \end{align} $$ Where $*$ represents convolution.

Answer 2

You can solve with the Laplace's time translation property. $$\mathscr{L}\{f(t-a)\}=e^{-sa}F(s)$$ And the transformation: $$\mathscr{L}\{\frac{1}{s^3}\}=\frac{t^2}{2}$$

So, from: $$F(s)=\frac{t^2}{2}$$ You immediately obtain: $$\frac{(t-2)^2}{2}$$