Compute Heaviside Laplace transform, then use this to solve initial value problem I've been stuck on this problem for a while, and can't really seem to find where I should go with it, or where I went wrong if I made a mistake.
Let $L(x)$ denote the Laplace transform of x.
Q: Compute $L(g)(s)$ and then use the method of Laplace transform to find the solution of the initial value problem $$y''-y'=g(t)$$ with $$y(0) = y'(0) = 0$$ where $$g(t) = \begin{cases} 1, & \text{if } & 0<=t<2 \\ 0, & \text{if} & t>=2   \end{cases}$$
So I have found that$$g(t) = 1 - u_2(t)$$ and that $$L(g)(s) = \frac{1-e^{-2s}}{s}$$
Now, I after doing the Laplace transform of the IVP, I got $$s^2Y(s)-sy(0)-y'(0)-sY(s)+y(0) = {1-e^{-2s}\over s}$$ After plugging in initial conditions and solving for $Y(s)$, I get $$Y(s) = {1-e^{-2s} \over s^2(s-1)}$$This is where I am stuck. I know I need to take the inverse Laplace transform of this, but I can't figure out how to. My usual strategy for a question like this would be partial fraction decomposition, but the $e^{-2s}$in the numerator makes that tricky.
Any sort of help is appreciated! Thanks!

I figured it out finally. I didn't think that you could do a fractional decomposition when the numerator was 1, which I was wrong because if you split out $$Y(s)={1-e^{-2s}\over s^2(s-1)}$$ you get $$Y(s)={1\over s^2(s-1)} - {e^{-2s}\over s^2(s-1)}$$After this you can do the fractional decomposition of the first term (the second term is $e^{-2s}$ times the first) and get $${1\over s^2(s-1)} = {A\over s} + {B\over s^2} + {C\over s-1}$$ So $A=-1, B=-1, C=1$ which means $${1\over s^2(s-1)} = {-1\over s} - {1\over s^2} + {1\over s-1}$$ After this, we can find the inverse Laplace transform of each term, which is $-1, -t,$ and $e^t$ respectively. 
Finally we can use the fractional decomposition multiplied by $e^{-2s}$ and use rules 25&27 on a transform table (http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf).
Our final answer is $$y(t)=e^t-t-1+u_2(t)(-e^{t-2}+t-1)$$
