How to solve the ordinary differential equation $ x^2 y'' - 2 x y' + 2y = x^4 \mathrm{e}^x $ Please tell me how to solve this differential equation.
$$  x^2 y'' - 2 x y' + 2y = x^4 \mathrm{e}^x  $$
I tried to solve it but finally I stuck tell me how to go further more if anyone gives me a hint(I want to know how to find paticular integral?).I will appreciate about it.i posted my way below please tell me any hint. Here's my work:


Operators and first and second integrals are two methods to find a solution. This solution uses the series solution. 
Let 
$$y(x) = \sum_{n=0}^{\infty} a_{n} \, x^n$$
which easily leads to
\begin{align}
x^2 \, y'' - 2 x y' + 2 y &= x^4 \, e^{x} \\
\sum_{n=0}^{\infty} a_{n} \, (n(n-1) - 2n + 2) \, x^n &= \sum_{n=0}^{\infty} \frac{x^{n+4}}{n!} \\
\sum_{n=0}^{\infty} (n-1)(n-2) \, a_{n} \, x^n &= \sum_{n=4}^{\infty} \frac{x^n}{(n-4)!} \\
a_{1} \, x + a_{2} \, x^2 + \sum_{n=4}^{\infty} (n-1)(n-2) \, a_{n} \, x^n &= \sum_{n=4}^{\infty} \frac{x^n}{(n-4)!}
\end{align}
This leads to $a_{1}$ and $a_{2}$ along with $$a_{n} = \frac{n(n-3)}{n!} \hspace{5mm} \text{for} \hspace{5mm} n \geq 4.$$
Now,
\begin{align}
y(x) &= a_{1} \, x + a_{2} \, x^2 + \sum_{n=0}^{\infty} \frac{n(n-3) \, x^n}{n!} \\
&= a_{1} \, x + a_{2} \, x^2 + x \, \frac{d}{dx} \, \sum_{n=0}^{\infty} (n-3) \, \frac{x^n}{n!} \\
&= a_{1} \, x + a_{2} \, x^2 + x \, \frac{d}{dx} \left[ x \, \frac{d}{dx} e^x - 3 \, e^x \right] \\
&= a_{1} \, x + a_{2} \, x^2 + x \, \frac{d}{dx} \left( (x-3) \, e^x \right) \\
&= a_{1} \, x + a_{2} \, x^2 + x(x-2) \, e^x.
\end{align}
