What do I do wrong solving this differential equation? $$x^2 y''-2xy'+2y=2+x$$
With
$$t=\ln(x), x=et$$
I have 
$$y_H= c_1e^t +c_2e^{2t}$$ . Going on: $y_p$ should be rather easy to calculate, yet I fail to do it correctly. I'm calculating the determinant of the following matrices
 $$W=\begin{bmatrix} e^t & e^{2t} \\
e^t & 2e^{2t}\end{bmatrix} $$ 
$$ W_1=\begin{bmatrix} 0 & e2t \\
2+et & 2e2t\end{bmatrix}$$
$$ W_2=\begin{bmatrix} e^t & 0\\
e^t & 2+e^t\end{bmatrix} $$
$$y_p=u_1e^t +u_2e^{2t}$$ , where $$u_i=\frac{|W_i|}{|W|}$$
I integrate them, substitute it and get the wrong answer (should be -$x\ln(x)+1$, but I get $ -x\ln(x)+x+3$).
Can you please help me figure it out, how to do it correctly?
Thanks! 
 A: I suspect you have made some sort of sign error during the computation of the determinant or integration, since I see it is very easy to make a mistake there. However, it is difficult to see exactly where you have done so, since you have omitted both of these steps in your post. Hence, I provide my own solution below using your method (Variation of Parameters using Cramer's rule), so that you can pinpoint where your mistake was.

Note that:
$$W=\begin{vmatrix} e^t & e^{2t} \\ e^t & 2e^{2t} \end{vmatrix}=2e^{3t}-e^{3t}=e^{3t}$$
And:
$$\quad W_1=\begin{vmatrix} 0 & e^{2t} \\ 2+e^t & 2e^{2t} \end{vmatrix}=-2e^{2t}-e^{3t}, \qquad W_2=\begin{vmatrix} e^t & 0 \\ e^t & 2+e^t \end{vmatrix}=2e^t+e^{2t}$$
Thus, by Cramer's rule:
$$u_1'=\frac{W_1}{W}=\frac{-2e^{2t}-e^{3t}}{e^{3t}}=-2e^{-t}-1 \implies u_1=2e^{-t}-t$$
$$u_2'=\frac{W_2}{W}=\frac{2e^t+e^{2t}}{e^{3t}}=2e^{-2t}+e^{-t} \implies u_2=-e^{-2t}-e^{-t}$$
Substituting this into $y_p$ gives:
$$y_p=(2e^{-t}-t)e^t-(e^{-2t}+e^{-t})e^{2t}=2-te^t-1-\color{red}{e^t}\implies y_p=1-te^t$$
Notice that for $y_p$, we may omit the solution $e^t$, since it is already part of the homogeneous solution. Hence, we obtain the correct particular solution after substituting back for $t=\ln(x)$.
