What is the inverse Laplace transform of $\frac{e^{-2x}}{s^{2}}$ $\frac{e^{-2x}}{s^{2}}$
$\frac{dy}{dt}=u_{2}(t)$ , $y(0)=3$
$\mathscr{L}\left(\frac{dy}{dt}\right)=\mathscr{L}\left(u_{2}(t)\right)$
$\vdots$
$y=\mathscr{L}^{-1}\left(\frac{e^{-2x}}{s^{2}}\right)+\mathscr{L}^{-1}\left(\frac{3}{s}\right)$
the right term of the RHS of the equation is 3. But is there a laplace formula that i can follow when i have to take the inverse laplace transform of the heaviside function over a term to the nth degree? thanks. 

It ought to be $(t-2)u_2(t)$ .
From #27 on this list:

