1. ## non-homogeneous Cauchy-Euler problem

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:

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

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

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

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

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

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

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.

2. For the method to work, the coefficient of the second derivative has to be unity. You have to start the calculation of the P.I. by dividing throughout by $x^2.$

3. Thanks! It worked now.