Here's this ODE: $$x(x-1)y'' -(2x-1)y' + 2y = 2x^3 -3x^2$$

and $$y\_1 =x^2$$

I know that I have to consider the homogenous part of the ODE first, which is $$x(x-1)y'' -(2x-1)y' + 2y = 0$$

If one solution is already known, then the second one can be calculated as:

$$y\_2(x) = y\_1(x)\int \frac{e^{-\int P(x)dx}}{y\_1^2(x)}dx$$, after which the solution of the homogenous part is $$C\_1y\_1(x) + C\_2y\_2(x)$$

After I divided my equation by $x(x-1)$ I solved the first part and got $$C\_1x^2 + C\_2(-x+\frac{1}{2})$$

Now I need to find the rest of the solution (By the way, the full solution is: $$C\_1x^2 + C\_2(-x+\frac{1}{2}) +x^3 -\frac{x^2}{2}+x-\frac{1}{2}$$

I tried the variation of constants by forming the following system of equations:

$$C\_1'(x)x^2 + C\_2'(x)(-x+\frac{1}{2})=0$$
$$C\_1'(x)2x - C\_2'(x)=\frac{x(2x-3)}{x-1}$$

I used Cramer's method of solving the system and got that

$C\_1' = \frac{-(2x-3)(-2x+1)}{2(x-1)^2}$ and
$C\_2' = \frac{x^2(2x-3)}{(x-1)^2}$

Now if I integrate both I get:
$C\_1 = 2x + \frac{1}{2(x-1)} -2$
and

$C\_2 = \frac{x^3 -3x +3}{x-1}$

However, when I add these two $C\_1$ and $C\_2$ I don't get anything that resembles the $x^3 -\frac{x^2}{2}+x-\frac{1}{2}$ part of the solution.

Can anyone please help me with this? Have I correctly calculated $C\_1$ and $C\_2$? What do I do with them?