I think I'm doing this right, but I'm not getting the right answer it seems.

Use variation of parameters to find a particular solution of the non-homogeneous Cauchy-Euler problem:

$\displaystyle x^2y''-2xy'+2y=x^4e^x$

I found the roots to be 1,2 so my general solution is $\displaystyle y=c_{1}x+c_{2}x^2+y_{p}$

Then I used variation of parameters. I found $\displaystyle W$, the Wronskian of $\displaystyle y_{1}$ and $\displaystyle y_{2}$, to be $\displaystyle x^2$ and $\displaystyle W_{1}$ and $\displaystyle W_{2}$ to be $\displaystyle -x^{6}e^x$ and $\displaystyle x^{5}e^x$ respectively.

Dividing $\displaystyle W_{1}$ and $\displaystyle W_{2}$ by $\displaystyle W$ I get $\displaystyle u_{1}'$ as $\displaystyle -x^{4}e^x$ and $\displaystyle u_{2}'$ as $\displaystyle x^{3}e^x$

Integrating I get $\displaystyle u_{1}=-e^{x}(x^4-4x^3+12x^2-24x+24)$ and $\displaystyle u_{2}=e^{x}(x^3-3x^2+6x-6)$

So my $\displaystyle y_{p}$ should be $\displaystyle -xe^{x}(x^4-4x^3+12x^2-24x+24)+x^{2}e^{x}(x^3-3x^2+6x-6)$

Whenever I put this in, it says it's not correct. If there's something I'm missing or that I did incorrectly, any help would be appreciated.