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?

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.
$$2y'-\dfrac {y}{x}=\ln x+ C$$
$$\left (\dfrac {2y}{\sqrt x}\right )'=\dfrac 1 {\sqrt x}(\ln x+ C)$$
$$\dfrac {2y}{\sqrt x} =2C \sqrt x +C_2+ \int \dfrac {\ln x} {\sqrt x}dx$$
$$y( x) =C_0  x +C_1\sqrt x+ x {\ln x} $$
