I'm working on a differential equations homework problem which I have to solve with laplace transforms: $$y''-y'-2y=4x^2, y(0)=-1, y'(0)=1$$ And I'm having trouble specifically because of the forcing part, $4x^2$, which causes me to have two variables in the transformed equation $Y(s)$.
$$Y(s) = \frac{-s+(8/r^3)}{s^2-s-2}$$
Is this correct, and then how do I get $Y(s)$ into a form that I can take the inverse laplace of?
It should be $8/s^3$. You either use $r$ or $s$ as a variable. If you are using $s$ as the variable for Laplace transform, you have $\mathcal L[f(x)]=\int_0^\infty f(x)e^{-st}dt$, so that $\mathcal L[4x^2]=8/s^3$.