<s>[INST] The general solution of $2x^2y''-xy'+y=x$ 
Find the general solution of $2x^2y''-xy'+y=x$.

My attempt:
Consider its homogeneous version: $2x^2y''-xy'+y=0$. Letting $y=x^r$ and substituting this and its first- and second- order derivatives in the equation, I have the characteristic equation $2r^2-3r+1=0$. This gives $r=1,\frac{1}{2}$. It follows that the complementary solution is

$y_c(x)=C_1x+C_2x^{\frac{1}{2}}$

Now back to the original ODE. Dividing the whole equation by $2x^2$,

$y''-\frac{1}{2x}y'+\frac{1}{2x^2}y=\frac{1}{2x}=g(x)$

Then consider the Wronskian

$W= \begin{vmatrix} {y_1} & {y_2 }\\\ {y_1'} & {y_2'} \end{vmatrix}= \begin{vmatrix} x & x^{\frac{1}{2}}\\\ 1 & (2x^{\frac{1}{2}})^{-1} \end{vmatrix}=-\frac{x^{\frac{1}{2}}}{2}$
$W_1= \begin{vmatrix} 0 & y_2\\\ g(x) & y_2' \end{vmatrix}=\begin{vmatrix} 0 & x^\frac{1}{2}\\\ \frac{1}{2x} & (2x^\frac{1}{2})^{-1}\end{vmatrix}=(-2x^{\frac{1}{2}})^{-1}$
$u_1'=\frac{W_1}{W}=\frac{1}{x} \implies u_1=ln|x|+C$
$W_2=\begin{vmatrix} y_1 & 0\\\ y_1' & g(x)\end{vmatrix}=\begin{vmatrix} x & 0\\\ 1 & \frac{1}{2x}\end{vmatrix}=\frac{1}{2}$
$u_2'=\frac{W_2}{W}=-x^{-\frac{1}{2}} \implies u_2=-2x^{\frac{1}{2}}+C$

So, we have the particular solution

$y_p(x)=u_1y_1+u_2y_2=x\ln|x|-2x$

Hence, our solution is

$y(x)=y_c(x)+y_p(x)=C_1x+C_2x^\frac{1}{2}+x(\ln|x|-2)$

I think I've solved the problem, but I'm currently reviewing this material for computational PDE course, so I'm not sure. Is something wrong with this?
 [/INST] Your solution looks correct here is another approach:
$$2x^2y''-xy'+y=x$$
$$2y''-\dfrac {y'}{x}+\dfrac y{x^2}=\dfrac 1x$$
$$2y''-\left (\dfrac {y}{x}\right)'=\dfrac 1x$$
Integrate.
$$2y'-\dfrac {y}{x}=\ln x+ C$$
$$\left (\dfrac {2y}{\sqrt x}\right )'=\dfrac 1 {\sqrt x}(\ln x+ C)$$
$$\dfrac {2y}{\sqrt x} =2C \sqrt x +C_2+ \int \dfrac {\ln x} {\sqrt x}dx$$
$$y( x) =C_0  x +C_1\sqrt x+ x {\ln x} $$
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