Question

Find a power series solution to the differential equation at the point x0. (2+x2)y′′-xy′+4y=0

Answer 1

$(x^2+2)y''-xy'+4y=0$

I'll assume you are looking for a series centered at $x=0$, which is an ordinary point for the ODE. Substituting

$y=\displaystyle\sum_{n\ge0}a_nx^n$

into the ODE, we can rewrite it as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^n+2\sum_{n\ge2}n(n-1)a_nx^{n-2}$
$\,\,\,\,\,\,\,\,-\displaystyle\sum_{n\ge1}na_nx^n+4\sum_{n\ge0}a_nx^n=0$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^n+2\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$
$\displaystyle\,\,\,\,\,\,\,\,-\sum_{n\ge1}na_nx^n+4\sum_{n\ge0}a_nx^n=0$

$\displaystyle4(a_0+a_2)+3(a_1+4a_3)x+\sum_{n\ge2}\bigg[2(n+2)(n+1)a_{n+2}+(n^2-2n+4)a_n\bigg]x^n=0$

so the coefficients of the power series are defined by the recurrence

$\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=-\dfrac{(n-3)^2+3}{2n(n-1)}a_{n-2}&\text{for }n\ge2\end{cases}$

For even $n$, i.e. $n=2k$ $(k\ge0)$, we have

$a_0=a_0$
$a_2=-a_0$
$a_4=-\dfrac4{4\times3\times2}a_0=\dfrac4{4!}a_0$
$a_6=-\dfrac{12}{2\times6\times5}a_4=-\dfrac{12\times4}{2\times6!}a_0$
$a_8=-\dfrac{28}{2\times8\times7}a_6=\dfrac{28\times12\times4}{2^2\times8!}a_0$

and so on. The pattern in the denominator is pretty clear, but we can also find a compact form for the numerator. When $k=5$, we can write it as

$4\times12\times28\times52=4^4(1\times3\times7\times13)$
$=4^4\bigg(1\times(1+2)\times(1+2+4)\times(1+2+4+6)\bigg)$
$=4^4\bigg(1\times(1+2(1))\times(1+2(1+2))\times(1+2(1+2+3)\bigg)$
$=4^4\displaystyle\prod_{i=1}^3\left(1+2\sum_{j=1}^{i-1}j\right)$
$=4^4\displaystyle\prod_{i=1}^3(i^2-i+1)$

So in general we have

$a_{2k}=\dfrac{(-1)^k2^k\displaystyle\prod_{i=1}^{k-1}(i^2-i+1)}{(2k)!}$

We can treat the odd-indexed terms similarly. For $n=2k-1$ $(k\ge1)$ we have

$a_1=a_1$
$a_3=-\dfrac14a_1=-\dfrac3{2\times3\times2}a_1=-\dfrac3{2\times3!}a_1$
$a_5=-\dfrac7{2\times5\times4}a_3=\dfrac{7\times3}{2^2\times5!}a_1$
$a_7=-\dfrac{19}{2\times7\times6}a_5=-\dfrac{19\times7\times3}{2^3\times7!}a_1$

and so on. Again, the pattern in the denominator is simple. For $k=5$, we would get a numerator of

$3\times7\times19\times39=3\times(3+4)\times(3+4+12)\times(3+4+12+20)$
$=3\times(3+4(1))\times(3+4(1+3))\times(3+4(1+3+5))$
$=\displaystyle\prod_{i=1}^4\left(3+4\sum_{j=1}^{i-1}(2j-1)\right)$
$=\displaystyle\prod_{i=1}^4(4i^2-8i+7)$

and in general we'd have

$a_{2k-1}=\dfrac{(-1)^{k+1}\displaystyle\prod_{i=1}^{k-1}(4i^2-8i+7)}{2^{k-1}(2k-1)!}$

Thus the power series solution to this ODE is

$y(x)=\displaystyle\sum_{k\ge0}a_{2k}x^{2k}+\sum_{k\ge1}a_{2k-1}x^{2k-1}$

Attached below is a plot of a numerical solution (blue) compared to the first 9 terms $(0\le n\le8)$ and first 21 terms $(0\le n\le21)$ of the series solution over the interval $|x|\le3$, assuming initial values of $y(0)=y'(0)=1$ $(a_0=a_1=1)$.