$$x^3y'' + xy′ + 2y = 0 $$
Find a number $r \in \mathbb{R}$ and coefficients $a_n$ such that $y(x) = x^r \sum_{n=0}^{\infty}a_n x^n$ is a non-constant solution of the equation above. I am having trouble with this problem. I don't even know how to start. Any hints or help would be greatly appreciated.
On
$0$ is an irregular singular point of this differential equation, so you don't get two linearly independent series solutions. But there is one.
Substitute $y(x) = \sum_{n} a_j x^{r+n}$ into the DE and you'll get a relation betwen $a_n$ and $a_{n+1}$. Find $r$ so that for $n=-1$ this doesn't require an $a_{-1}$ term.
$$\begin{align} x^3y''+xy'+2y&=0\\ x^3\left(\sum_{n=0}^{\infty}a_n x^{n+r}\right)''+x\left(\sum_{n=0}^{\infty}a_n x^{n+r}\right)'+2\left(\sum_{n=0}^{\infty}a_n x^{n+r}\right)&=0\\ \end{align}$$
Carry out the differentiation. Reindex any sums as needed so that you can combine them into a single sum grouping by $x^{n+r}$. Deduce what can be deduced.