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

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.

$$y'(x)=u(x)$$
$$x u'(x)-\left(2 x^2+1\right) u(x)=0$$
$$u(x)=c_1 e^{x^2} x$$
$$y(x)=\int c_1 e^{x^2} x\,dx=\frac{c_1 e^{x^2}}{2}+c_2$$
To solve the non homogeneous equation $$xy''-(2x^2+1)y'=x^5e^{x^2}$$
we guess the particular solution as $y_1(x)=(a x^4+b x^3+c x^2+d x+e)e^{x^2}$
$$y_1'(x)=2 e^{x^2} x \left(a x^4+b x^3+c x^2+d x+e\right)+e^{x^2} \left(4 a x^3+3 b x^2+2 c x+d\right)=\\=e^{x^2} \left(2 a x^5+4 a x^3+2 b x^4+3 b x^2+2 c x^3+2 c x+2 d x^2+d+2 e x\right)$$
$$y_1''(x)=2 e^{x^2} \left(2 a x^6+9 a x^4+6 a x^2+2 b x^5+7 b x^3+3 b x+2 c x^4+5 c x^2+c+2 d x^3+3 d x+2 e x^2+e\right)$$
So plugging this in the equation we get
$$8 a x^5+6 b x^4+ x^3 (8 a+4 c)+x^2 (3 b+2 d)-d\equiv x^5e^{x^2}$$
$a=\frac18;\;c=-\frac{1}{4}$ so the particular solution is
$$y_1=\frac{1}{8} e^{x^2} x^2 \left(x^2-2\right)$$
The general solution of the DE is
$$y=\frac{c_1 e^{x^2}}{2}+c_2+\frac{1}{8} e^{x^2} x^2 \left(x^2-2\right)$$
$$y=\frac{1}{8} e^{x^2} \left(x^4-2 x^2+4 c_1\right)+c_2$$
