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 $\begin{cases}C_1\sum\limits_{n=0}^\infty\dfrac{\prod\limits_{k=0}^n(3k+1)}{4^{3n}(3n+1)!}+C_2\biggl(\dfrac{1}{4}+\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{4^{3n+1}(3n+1)!}\biggr)=0\\C_1\biggl(\sum\limits_{n=1}^\infty\dfrac{3n\prod\limits_{k=0}^n(3k+1)}{4^{3n-1}(3n+1)!}-\sum\limits_{n=0}^\infty\dfrac{\prod\limits_{k=0}^n(3k+1)}{2^{6n+1}(3n+1)!}\biggr)+C_2\biggl(\dfrac{7}{8}+\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{4^{3n}(3n)!}-\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{2^{6n+3}(3n+1)!}\biggr)=1\end{cases}$
Best Answer
We consider \begin{align*} A(x)=\sum_{k=0}^\infty a_kx^k\qquad\qquad A^{n}(x)=\sum_{k=0}^\infty c_k^{(n)}x^k\qquad n\geq 1\tag{1} \end{align*}
Merging $c_m^{(n)}$ with the sum we show the following is valid for $n\geq 1$: \begin{align*} \color{blue}{c_0^{(n)} }&\color{blue}{= a_0^n}\\ \color{blue}{\sum_{k=0}^m(kn-m+k)a_kc_{m-k}^{(n)}}&\color{blue}{=0\qquad\qquad m\geq 1}\tag{2} \end{align*} It is convenient to use the coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ in a series. This way we can write for instance \begin{align*} [x^m]A(x)=[x^m]\sum_{k=0}^\infty a_kx^k=a_m\tag{3} \end{align*}
Comment:
In (4.1) we use the coefficient of operator (3) to work with $c_{m-k}^{(n)}$.
In (4.2) we use the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.
In (4.3) we split the sum and we also set the upper limit of the sums to $\infty$ without changing anything, since the coefficient of operator skips terms with powers of $x$ greater $m$.
In (4.4) we write the sum using the differential operator $\frac{d}{dx}$.
In (4.5) we recall the rule $\frac{d}{dx}A^{n+1}(x)=(n+1)A^{n}(x)\frac{d}{dx}A(x)$.
In (4.6) we write the series again using the notation from (1).
In (4.7) and the line that follows we select the coefficient of $[x^m]$ resp. $[x^{m-1}]$.