$xy''-(2x^2+1)y'=x^5e^{x^2}$ $xy''-(2x^2+1)y'=x^5e^{x^2}$

Hi there,
needs help with the second order ODE.
I've tried (for the homogenous part) to substitute $\frac{dy}{dx} =u$ to get $ u'-(2x+\frac{1}{x})u=0$
solving for the first non trivial solution
$y_1=C_1 e^{x^2}$
however with $y_2$ and the particular solution i stuck.
asks for help.

$$xy''-(2x^2+1)y'=x^5e^{x^2}$$
$$(xy''-y')-2x^2y'=x^5e^{x^2}$$
$$\dfrac {(xy''-y')}{x^2}-2y'=x^3e^{x^2}$$
$$\left (\dfrac {y'}{x}\right)'-2y'=x^3e^{x^2}$$
Better to substitute $u=x^2$:
$$\left (2 {y'}\right)'-2y'=\dfrac 12ue^{u}$$
$$\left ( {y'e^{-u}}\right)'=\dfrac u4$$
$$y'e^{-u}=\dfrac {u^2}8+c_1$$
$$y'=\dfrac {u^2}8e^u+c_1e^u$$
$$y(u)=\dfrac 18\int  {u^2}e^u \, du+c_1e^u+c_2$$

Edit1
$$\left (\dfrac {y'}{x}\right)'-2y'=x^3e^{x^2}$$
Note that:
$$y'=\dfrac {dy}{dx}=\dfrac {dy}{dx^2}\dfrac {dx^2}{dx}=2x\dfrac {dy}{du}$$
$$\implies \left (\dfrac {y'}{x}\right)=2\dfrac {dy}{du}$$
And:
$$\dfrac {d}{dx}\left (\dfrac {y'}{x}\right)=2\dfrac {d}{dx}\dfrac {dy}{du}$$
$$\dfrac {d}{dx}\left (\dfrac {y'}{x}\right)=2\dfrac {d}{du}\dfrac {dy}{du}\dfrac {du}{dx}$$
$$\dfrac {d}{dx}\left (\dfrac {y'}{x}\right)=2\dfrac {d^2y}{du^2}\dfrac {du}{dx}$$
And $\dfrac {du}{dx}=2x$ since $u=x^2$.
Now we have: $$
4x\dfrac {d^2y}{du^2}-4x\dfrac {dy}{du}=x^3e^{x^2}$$
Divide the equation by $x$ you have:
$$
4\dfrac {d^2y}{du^2}-4\dfrac {dy}{du}=x^2e^{x^2}$$
Substitute $u=x^2$
$$
4\dfrac {d^2y}{du^2}-4\dfrac {dy}{du}=ue^{u}$$
$$
\dfrac {d^2y}{du^2}e^{-u}-\dfrac {dy}{du}e^{-u}=\dfrac u 4$$
So that the differential equation is simply:
$$
\boxed {(y'e^{-u})'=\dfrac u 4}$$
And it's easy to solve it.
