Verify that the equation $y''+y'-xy=0$ has a three-term, recursion
formula and find its series solutions $y_1$ and $y_2$ such that$a)$ $y_1(0)=1$, $y_1'(0)=0$;
$b)$ $y_2(0)=0$, $y_2'(0)=1$.
Using the following theorem, of which it guarantees that both series converge at ever $x\in \mathbb{R}$, let $\displaystyle\sum_{j=0}^{ \infty } a_j$ and $\displaystyle\sum_{j=0}^{ \infty } b_j$ be two absolutely converfent series which converge to limits $\alpha$ and $\beta$, respectively. Define the seies as $\displaystyle\sum_{j=0}^{ \infty } c_m$ with summands $c_m=\displaystyle\sum_{n=0}^{ m } a_j\cdot b_{m-j}$. Then the series $\displaystyle\sum_{m=0}^{ \infty } c_m$ converges to $\alpha \cdot \beta$.
Best Answer
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 $\begin{cases}C_1\sum\limits_{n=0}^\infty\dfrac{\prod\limits_{k=0}^n(3k+1)}{4^{3n}(3n+1)!}+C_2\biggl(\dfrac{1}{4}+\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{4^{3n+1}(3n+1)!}\biggr)=0\\C_1\biggl(\sum\limits_{n=1}^\infty\dfrac{3n\prod\limits_{k=0}^n(3k+1)}{4^{3n-1}(3n+1)!}-\sum\limits_{n=0}^\infty\dfrac{\prod\limits_{k=0}^n(3k+1)}{2^{6n+1}(3n+1)!}\biggr)+C_2\biggl(\dfrac{7}{8}+\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{4^{3n}(3n)!}-\sum\limits_{n=1}^\infty\dfrac{\prod\limits_{k=1}^n(3k-1)}{2^{6n+3}(3n+1)!}\biggr)=1\end{cases}$