Solution of differential equation with power series $x^{2}y''+xy'+y=0$
I tried to solve this with the power series method but I can't make the $a_{n+1}$ appear in my new equation..
Here's my solution:
$t^2\sum_{n=2}^{\infty}n(n-1)a_{n} t^{n-2}+t\sum_{n=1}^{\infty}na_{n}t^{n-1}+\sum_{n=0}^{\infty}a_{n}t^n=0$
$\sum_{n=2}^{\infty}n(n-1)a_{n} t^{n}+\sum_{n=1}^{\infty}na_{n}t^{n}+\sum_{n=0}^{\infty}a_{n}t^n=0$
If I try to change the index from 2 and 1 to zero the $t^{n}$ gets messed up
If I add sepperately the sum for $n=0,n=1$ to shift the index of the sums, the $a_{n+1}$ doesn't appear
$\sum_{n=2}^{\infty}n(n-1)a_{n} t^{n}+a_{1}t+\sum_{n=2}^{\infty}na_{n}t^{n}+a_{0}+a_{1}t+\sum_{n=2}^{\infty}a_{n}t^n=0$
What do I do?

You need a shifted power series
$$
y(x)=\sum a_n x^{n+r}
$$
Then you should get the condition that $r^2+1=0$ and $a_n=0$ for $n>0$ so that a solution basis is $x^{\pm i}$ or $\cos(\ln x), \sin(\ln x)$.
You could also directly identify the equation as a Euler differential equation and start with the appropriate trial solution $y=x^m$.
