power series solution for $xy''+y'+xy=0$ I need to solve 
$$xy''+y'+xy=0$$
at $x_0 =1$
so I supposed:
$$y = \sum_{n=0}^{\infty}a_n(x-1)^n\implies$$
$$y'= \sum_{n=1}^{\infty}n\cdot a_n\cdot (x-1)^{n-1}\implies$$
$$y'' = \sum_{n=2}^{\infty}n\cdot (n-1)\cdot (x-1)^{n-2}$$
Then I did:
$$(x-1+1)\sum_{n=2}^{\infty}n\cdot (n-1)\cdot (x-1)^{n-2}+\sum_{n=1}^{\infty}n\cdot a_n\cdot (x-1)^{n-1}+(x-1+1) \sum_{n=0}^{\infty}a_n(x-1)^n = $$
$$\sum_{n=2}^{\infty}n\cdot (n-1)\cdot (n-2)(x-1)^{n-1}+\sum_{n=2}^{\infty}n\cdot (n-1)\cdot (x-1)^{n-2}+\sum_{n=1}^{\infty}n\cdot a_n\cdot (x-1)^{n-1}+\sum_{n=0}^{\infty}a_n(x-1)^{n+1}+\sum_{n=0}^{\infty}a_n(x-1)^n$$
as you can see, I get this $(x-1)^{n+1}$ term. When it happens, I normally shift the index to get $(x-1)^n$, but in this case I'm already in the index $0$. What should I do? 

Rewrite as $$(x-1)y''+(x-1)y+y+y'+y''=0 $$
The coefficient of $(x-1)^n$ 


*

*in $y$ is $a_n$

*in $(x-1)y$ is $a_{n-1} $ (with $a_{-1}=0$ understood)

*in $y'$ is $(n+1)a_{n+1}$

*in $y''$ is $(n+2)(n+1)a_{n+2}$

*in $(x-1)y''$ is $(n+1)na_{n+1}$


Thus you get the equation
$$(n+1)na_{n+1} + a_{n-1}+a_n+(n+1)a_{n+1}+(n+2)(n+1)a_{n+2}=0$$
or the recursion
$$a_{n+2}=-\frac{(n+1)^2a_{n+1} + a_{n-1}+a_n}{(n+2)(n+1)} $$
