Particular solution of $x^2y''-4xy'+3y=e^x$ I tried to assume a function which gives me  the particular solution but I could't get it.
 A: Suppose that $\displaystyle y(x)=\sum_{n=0}^{\infty}a_n x^n$ for some $a_n$, then:
$$y'=\sum_{n=1}^{\infty}na_nx^{n-1}$$
$$y''=\sum_{n=2}^{\infty}n(n-1)x^{n-2}$$
$$-4y'x=\sum_{n=1}^{\infty}-4na_nx^{n}$$
$$y''x^2=\sum_{n=2}^{\infty}n(n-1)x^{n}$$
So:
$$y''x^2-4xy'+3y=\sum_{n=2}^{\infty}n(n-1)x^{n}+\sum_{n=1}^{\infty}-4na_nx^{n}+\sum_{n=0}^{\infty}a_nx^n=e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
Next:
$$\sum_{n=2}^{\infty}n(n-1)x^{n}+\sum_{n=1}^{\infty}-4na_nx^{n}+\sum_{n=0}^{\infty}a_nx^n=\sum_{n=2}^{\infty}n(n-1)x^{n}+-4a_1x+\sum_{n=2}^{\infty}-4na_nx^{n}+\sum_{n=2}^{\infty}a_nx^n+a_0+a_1x$$
So we must have:
$$a_0=1$$
$$a_1=\frac{-1}{3}$$
For $n>1$:
$$a_n(n(n-1)-4n+3)=\frac{1}{n!}$$
A: A CAS effectively found a quite nasty expression in which appears the incomplete gamma function but the expression can simplify to $$c_1 x^{\frac{1}{2} \left(5-\sqrt{13}\right)}+c_2 x^{\frac{1}{2} \left(5+\sqrt{13}\right)}+\frac{E_{\frac{1}{2}
   \left(7-\sqrt{13}\right)}(-x)-E_{\frac{1}{2}
   \left(7+\sqrt{13}\right)}(-x)}{\sqrt{13}}$$ where appear  exponential integral functions.
Hoping that this could help.
