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. 

I am going to use a Table of Laplace Transforms, but if you've learned 
complex variables, you can use the formal definition for the inverse Laplace.
The original problem is
$$ \tag 1  \dfrac{dy}{dt}=u_2(t),~~ y(0)=3$$
We want to solve this using Laplace transforms.
$\mathscr{L}\left(\dfrac{dy}{dt}\right)= s y(s) - y(0) = s y(s) - 3$, and $\mathscr{L}\left(u_{2}(t)\right) = \dfrac{e^{-2 s}}{s}$
Substituting and simplifying $(1)$, we get
$$y(s) = \dfrac{e^{-2s}}{s^2} + \dfrac{3}{s}$$ 
Using Table item $1$, we have
$$3\mathscr{L}^{-1}\left(\dfrac{1}{s}\right) = 3$$
Using Table items $3$ and $27$, we have
$$\mathscr{L}^{-1}\left(\dfrac{e^{-2s}}{s^2}\right)= (t-2)u_2(t)$$
I would also recommend reviewing this nice set of notes.
