# Solving a Second Order Linear Equation with Power Series

$$y''+y'+xy=0$$ I can't seem to get $y_1$ or $y_2$ to have any sort of pattern. I understand the technique to it but have no idea what the general solution is.

The equation I have for the coefficients is $C_{k+2}=(-(k+1)\cdot C_{k+1}-C_{k-1})/((k+2)\cdot(k+1))$ I'm pretty sure that is right, but not certain.

Also, $C_{2}=-C_{1}/2$

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Hi, Welcome to MSE =) I am not sure I corrected your equation for your coefficients correctly, perhaps you should make sure I didn't make any typo. It wasn't quite clear what was written there before! – Patrick Da Silva Nov 30 '11 at 5:00
Suggestion: pick some simple values for $C_0$ and $C_1$, calculate the next few $C_i$ and, if you don't see a pattern, post the numbers here and maybe someone else will see it. Alternatively, multiply all the values you got by the LCM of their denominators, so you have an integer sequence, which you can look up in the Online Encyclopedia of Integer Sequences. – Gerry Myerson Dec 1 '11 at 6:29

$$y=\sum a_kx^k$$

$$xy=\sum a_kx^{k+1}=\sum a_{k-1}x^k$$

$$y'=\sum ka_kx^{k-1}=\sum(k+1)a_{k+1}x^k$$

$$y''=\sum k(k+1)a_{k+1}x^{k-1}=\sum(k+1)(k+2)a_{k+2}x^k$$

$$0=\sum(a_{k-1}+(k+1)a_{k+1}+(k+1)(k+2)a_{k+2})x^k$$

So I get $$a_{k-1}+(k+1)a_{k+1}+(k+1)(k+2)a_{k+2}=0$$

EDIT: Which is what OP got, so we have independent confirmation, always important in the experimental sciences.

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Always note that for solving ODEs by power series or frobenius method approach, only the ODEs of the form $(x+k)^2(a_1(x+k)^n+a_2)y''+(x+k)(b_1(x+k)^n+b_2)y'+(c_1(x+k)^n+c_2)y=0$ can find all of the coefficients most nicely, since they only involves two terms recurrence relations with variable coefficients.

$y''+y'+xy=0$ is obviously not belongs to an ODE of the form $(x+k)^2(a_1(x+k)^n+a_2)y''+(x+k)(b_1(x+k)^n+b_2)y'+(c_1(x+k)^n+c_2)y=0$ , so it should be better try to convert the ODE to an ODE of the form $(x+k)^2(a_1(x+k)^n+a_2)y''+(x+k)(b_1(x+k)^n+b_2)y'+(c_1(x+k)^n+c_2)y=0$ first by some variable transformations.

Let $y=e^{ax}u$ ,

Then $y'=e^{ax}u'+ae^{ax}u$

$y''=e^{ax}u''+ae^{ax}u'+ae^{ax}u'+a^2e^{ax}u=e^{ax}u''+2ae^{ax}u'+a^2e^{ax}u$

$\therefore e^{ax}u''+2ae^{ax}u'+a^2e^{ax}u+e^{ax}u'+ae^{ax}u+xe^{ax}u=0$

$e^{ax}u''+(2a+1)e^{ax}u'+(x+a^2+a)e^{ax}u=0$

$u''+(2a+1)u'+(x+a^2+a)u=0$

Choose $a=-\dfrac{1}{2}$ , the ODE becomes $u''+\left(x-\dfrac{1}{4}\right)u=0$ , which belongs to an ODE of the form $(x+k)^2(a_1(x+k)^n+a_2)y''+(x+k)(b_1(x+k)^n+b_2)y'+(c_1(x+k)^n+c_2)y=0$

Let $u=\sum\limits_{n=0}^\infty a_n\left(x-\dfrac{1}{4}\right)^n$ ,

Then $u'=\sum\limits_{n=0}^\infty na_n\left(x-\dfrac{1}{4}\right)^{n-1}=\sum\limits_{n=1}^\infty na_n\left(x-\dfrac{1}{4}\right)^{n-1}$

$u''=\sum\limits_{n=1}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}=\sum\limits_{n=2}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}$

$\therefore\sum\limits_{n=2}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}+\left(x-\dfrac{1}{4}\right)\sum\limits_{n=0}^\infty a_n\left(x-\dfrac{1}{4}\right)^n=0$

$\sum\limits_{n=2}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}+\sum\limits_{n=0}^\infty a_n\left(x-\dfrac{1}{4}\right)^{n+1}=0$

$\sum\limits_{n=2}^\infty n(n-1)a_n\left(x-\dfrac{1}{4}\right)^{n-2}+\sum\limits_{n=3}^\infty a_{n-3}\left(x-\dfrac{1}{4}\right)^{n-2}=0$

$2a_2+\sum\limits_{n=3}^\infty(n(n-1)a_n+a_{n-3})\left(x-\dfrac{1}{4}\right)^{n-2}=0$

$\therefore\begin{cases}2a_2=0\\n(n-1)a_n+a_{n-3}=0\end{cases}$

$\begin{cases}a_2=0\\a_n=-\dfrac{a_{n-3}}{n(n-1)}\end{cases}$

$\therefore\begin{cases}a_0=a_0\\a_{3n}=\dfrac{(-1)^na_0}{(2\times3)(5\times6)(8\times9)......((3n-1)3n)}\forall n\in\mathbb{N}\\a_1=a_1\\a_{3n+1}=\dfrac{(-1)^na_1}{(3\times4)(6\times7)(9\times10)......(3n(3n+1))}\forall n\in\mathbb{N}\\a_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$

$\begin{cases}a_0=a_0\\a_{3n}=\dfrac{(-1)^n(4\times7\times10\times......(3n+1))a_0}{2\times3\times4\times5\times6\times7\times8\times9\times10\times......(3n-1)3n(3n+1)}\forall n\in\mathbb{N}\\a_1=a_1\\a_{3n+1}=\dfrac{(-1)^n(2\times5\times8\times......(3n-1))a_1}{2\times3\times4\times5\times6\times7\times8\times9\times10\times......(3n-1)3n(3n+1)}\forall n\in\mathbb{N}\\a_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$

$\begin{cases}a_0=a_0\\a_{3n}=\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k+1)\biggr)a_0}{(3n+1)!}\forall n\in\mathbb{N}\\a_1=a_1\\a_{3n+1}=\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)a_1}{(3n+1)!}\forall n\in\mathbb{N}\\a_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$

$\begin{cases}a_{3n}=\dfrac{(-1)^n\biggl(\prod\limits_{k=0}^n(3k+1)\biggr)a_0}{(3n+1)!}\forall n\in\mathbb{Z}^*\\a_1=a_1\\a_{3n+1}=\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)a_1}{(3n+1)!}\forall n\in\mathbb{N}\\a_{3n+2}=0~\forall n\in\mathbb{Z}^*\end{cases}$

$\therefore y=C_1e^{-\frac{1}{2}\left(x-\frac{1}{4}\right)}\sum\limits_{n=0}^\infty\dfrac{\biggl((-1)^n\prod\limits_{k=0}^n(3k+1)\biggr)\left(x-\dfrac{1}{4}\right)^{3n}}{(3n+1)!}+C_2e^{-\frac{1}{2}\left(x-\frac{1}{4}\right)}\biggl(x-\dfrac{1}{4}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)\left(x-\dfrac{1}{4}\right)^{3n+1}}{(3n+1)!}\biggr)=C_1\biggl(\sum\limits_{n=0}^\infty\dfrac{(-1)^n}{2^nn!}\left(x-\dfrac{1}{4}\right)^n\biggr)\sum\limits_{n=0}^\infty\dfrac{\biggl((-1)^n\prod\limits_{k=0}^n(3k+1)\biggr)\left(x-\dfrac{1}{4}\right)^{3n}}{(3n+1)!}+C_2\biggl(\sum\limits_{n=0}^\infty\dfrac{(-1)^n}{2^nn!}\left(x-\dfrac{1}{4}\right)^n\biggr)\biggl(x-\dfrac{1}{4}+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\biggl(\prod\limits_{k=1}^n(3k-1)\biggr)\left(x-\dfrac{1}{4}\right)^{3n+1}}{(3n+1)!}\biggr)$

Expand it you will get the power series solution centered at $\dfrac{1}{4}$

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