Let $y=\sum\limits_{n=0}^\infty a_nx^n$ ,
Then $y'=\sum\limits_{n=0}^\infty na_nx^{n-1}=\sum\limits_{n=1}^\infty na_nx^{n-1}$
$y''=\sum\limits_{n=1}^\infty n(n-1)a_nx^{n-2}=\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}$
$\therefore\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=1}^\infty na_nx^{n-1}-x\sum\limits_{n=0}^\infty a_nx^n=0$
$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=1}^\infty na_nx^{n-1}-\sum\limits_{n=0}^\infty a_nx^{n+1}=0$
$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=2}^\infty(n-1)a_{n-1}x^{n-2}-\sum\limits_{n=3}^\infty a_{n-3}x^{n-2}=0$
$a_1+2a_2+\sum\limits_{n=3}^\infty(n(n-1)a_n+(n-1)a_{n-1}-a_{n-3})x^{n-2}=0$
$\therefore\begin{cases}a_1+2a_2=0\\n(n-1)a_n+(n-1)a_{n-1}-a_{n-3}=0\end{cases}$
$\therefore y''+y'-xy=0$ has a three-term recursion formula.
However, when we really to solve it, we will not handle directly the above recursion formula as it is too complicated to solve it.
So we will try this approach:
Let $y=e^{ax}u$ ,
Then $y'=e^{ax}u'+ae^{ax}u$
$y''=e^{ax}u''+ae^{ax}u'+ae^{ax}u'+a^2e^{ax}u=e^{ax}u''+2ae^{ax}u'+a^2e^{ax}u$
$\therefore e^{ax}u''+2ae^{ax}u'+a^2e^{ax}u+e^{ax}u'+ae^{ax}u-xe^{ax}u=0$
$e^{ax}u''+(2a+1)e^{ax}u'+(a^2+a-x)e^{ax}u=0$
$u''+(2a+1)u'+(a^2+a-x)u=0$
Choose $a=-\dfrac{1}{2}$ , the ODE becomes $u''-\left(x+\dfrac{1}{4}\right)u=0$
Let $u=\sum\limits_{n=0}^\infty b_n\left(x+\dfrac{1}{4}\right)^n$ ,
Then $u'=\sum\limits_{n=0}^\infty nb_n\left(x+\dfrac{1}{4}\right)^{n-1}=\sum\limits_{n=1}^\infty nb_n\left(x+\dfrac{1}{4}\right)^{n-1}$
$u''=\sum\limits_{n=1}^\infty n(n-1)b_n\left(x+\dfrac{1}{4}\right)^{n-2}=\sum\limits_{n=2}^\infty n(n-1)b_n\left(x+\dfrac{1}{4}\right)^{n-2}$
$\therefore\sum\limits_{n=2}^\infty n(n-1)b_n\left(x+\dfrac{1}{4}\right)^{n-2}-\left(x+\dfrac{1}{4}\right)\sum\limits_{n=0}^\infty b_n\left(x+\dfrac{1}{4}\right)^n=0$
$\sum\limits_{n=2}^\infty n(n-1)b_n\left(x+\dfrac{1}{4}\right)^{n-2}-\sum\limits_{n=0}^\infty b_n\left(x+\dfrac{1}{4}\right)^{n+1}=0$
$\sum\limits_{n=2}^\infty n(n-1)b_n\left(x+\dfrac{1}{4}\right)^{n-2}-\sum\limits_{n=3}^\infty b_{n-3}\left(x+\dfrac{1}{4}\right)^{n-2}=0$
$2b_2+\sum\limits_{n=3}^\infty(n(n-1)b_n-b_{n-3})\left(x+\dfrac{1}{4}\right)^{n-2}=0$
$\therefore\begin{cases}2b_2=0\\n(n-1)b_n-b_{n-3}=0\end{cases}$
$\begin{cases}b_2=0\\b_n=\dfrac{b_{n-3}}{n(n-1)}\end{cases}$
$\therefore\begin{cases}b_0=b_0\\b_{3n}=\dfrac{b_0}{(2\times3)(5\times6)(8\times9)......((3n-1)3n)}\forall n\in\mathbb{N}\\b_1=b_1\\b_{3n+1}=\dfrac{b_1}{(3\times4)(6\times7)(9\times10)......(3n(3n+1))}\forall n\in\mathbb{N}\\b_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$
$\begin{cases}b_0=b_0\\b_{3n}=\dfrac{(4\times7\times10\times......(3n+1))b_0}{2\times3\times4\times5\times6\times7\times8\times9\times10\times......(3n-1)3n(3n+1)}\forall n\in\mathbb{N}\\b_1=b_1\\b_{3n+1}=\dfrac{(2\times5\times8\times......(3n-1))b_1}{2\times3\times4\times5\times6\times7\times8\times9\times10\times......(3n-1)3n(3n+1)}\forall n\in\mathbb{N}\\b_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$
$\begin{cases}b_0=b_0\\b_{3n}=\dfrac{\biggl(\prod\limits_{k=1}^n(3k+1)\biggr)b_0}{(3n+1)!}\forall n\in\mathbb{N}\\b_1=b_1\\b_{3n+1}=\dfrac{\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)b_1}{(3n+1)!}\forall n\in\mathbb{N}\\b_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$
$\begin{cases}b_{3n}=\dfrac{\biggl(\prod\limits_{k=0}^n(3k+1)\biggr)b_0}{(3n+1)!}\forall n\in\mathbb{Z}^*\\b_1=b_1\\b_{3n+1}=\dfrac{\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)b_1}{(3n+1)!}\forall n\in\mathbb{N}\\b_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$
$\therefore y=C_1e^{-\frac{x}{2}}\sum\limits_{n=0}^\infty\dfrac{\biggl(\prod\limits_{k=0}^n(3k+1)\biggr)\left(x+\dfrac{1}{4}\right)^{3n}}{(3n+1)!}+C_2e^{-\frac{x}{2}}\biggl(x+\dfrac{1}{4}+\sum\limits_{n=1}^\infty\dfrac{\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)\left(x+\dfrac{1}{4}\right)^{3n+1}}{(3n+1)!}\biggr)$
$y'=C_1e^{-\frac{x}{2}}\biggl(\sum\limits_{n=1}^\infty\dfrac{3n\biggl(\prod\limits_{k=0}^n(3k+1)\biggr)\left(x+\dfrac{1}{4}\right)^{3n-1}}{(3n+1)!}-\sum\limits_{n=0}^\infty\dfrac{\biggl(\prod\limits_{k=0}^n(3k+1)\biggr)\left(x+\dfrac{1}{4}\right)^{3n}}{2(3n+1)!}\biggr)+C_2e^{-\frac{x}{2}}\biggl(\dfrac{7}{8}-\dfrac{x}{2}+\sum\limits_{n=1}^\infty\dfrac{\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)\left(x+\dfrac{1}{4}\right)^{3n}}{(3n)!}-\sum\limits_{n=1}^\infty\dfrac{\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)\left(x+\dfrac{1}{4}\right)^{3n+1}}{2(3n+1)!}\biggr)$
For $y_1$ , $C_1$ and $C_2$ are the solution of $\begin{cases}C_1\sum\limits_{n=0}^\infty\dfrac{\prod\limits_{k=0}^n(3k+1)}{4^{3n}(3n+1)!}+C_2\biggl(\dfrac{1}{4}+\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{4^{3n+1}(3n+1)!}\biggr)=1\\C_1\biggl(\sum\limits_{n=1}^\infty\dfrac{3n\prod\limits_{k=0}^n(3k+1)}{4^{3n-1}(3n+1)!}-\sum\limits_{n=0}^\infty\dfrac{\prod\limits_{k=0}^n(3k+1)}{2^{6n+1}(3n+1)!}\biggr)+C_2\biggl(\dfrac{7}{8}+\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{4^{3n}(3n)!}-\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{2^{6n+3}(3n+1)!}\biggr)=0\end{cases}$
For $y_2$ , $C_1$ and $C_2$ are the solution of $\begin{cases}C_1\sum\limits_{n=0}^\infty\dfrac{\prod\limits_{k=0}^n(3k+1)}{4^{3n}(3n+1)!}+C_2\biggl(\dfrac{1}{4}+\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{4^{3n+1}(3n+1)!}\biggr)=0\\C_1\biggl(\sum\limits_{n=1}^\infty\dfrac{3n\prod\limits_{k=0}^n(3k+1)}{4^{3n-1}(3n+1)!}-\sum\limits_{n=0}^\infty\dfrac{\prod\limits_{k=0}^n(3k+1)}{2^{6n+1}(3n+1)!}\biggr)+C_2\biggl(\dfrac{7}{8}+\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{4^{3n}(3n)!}-\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{2^{6n+3}(3n+1)!}\biggr)=1\end{cases}$