$\displaystyle \frac{d^2y}{dx^2}+\frac{dy}{dx}-2y=-3e^x$

given y=2 when x=0, and y=0 when x=2

I found that $\displaystyle y= Ae^x+Be^{-2x}-3xe^x$

then $\displaystyle 2=A+B$

and $\displaystyle 0=Ae^2+Be^{-4}-6e^2$

is this correct so far?

my answer is different to the answer given.

thanks