Consider the ODE $$ xy''=y $$ in the interval $(0,\infty)$. By plugging $$ y=\sum_{n=0}^{\infty}a_nx^n $$ in the ODE, it is not hard to see that $a_0=0$ and $a_{n}=\frac{a_{n-1}}{(n-1)n}$ for all $2\leq n$. Hence, we obtain the solutions $$ y=a_1\sum_{n=1}^{\infty}\frac{n}{(n!)^2}x^n $$ where $a_1$ is arbitrary.
Since the ODE is of second order, it must have another independent solution. How can I find this solution? Thanks.