Solve $(x^2 + 1)y'' - 6xy' + 10y =0$ using series method 
Use series methods to solve: $(x^2 + 1)y'' - 6xy' + 10y =0$
a) Give the recursion formula
b) Give the first two non-zero terms of the solution corresponding to $a_0 = 1$ and $a_1 = 0$
c) Give the first three non-zero terms of the solution corresponding to $a_0 = 0$ and $a_1 = 1$

Here is what I have so far, then I get stuck on the series when I am trying to get all of my $n = 0$ within each series.
$y = \sum^{\infty}_{n=0} na_nX^n$
$y' = \sum^{\infty}_{n=1} na_nX^{n-1}$
$y'' = \sum^{\infty}_{n=2} n(n-1)a_nX^{n-2}$
so we have: $(x^2+1)[\sum^{\infty}_{n=2} n(n-1)a_nX^{n-2}] - (6x) \sum^{\infty}_{n=1} na_nX^{n-1} + 10\sum^{\infty}_{n=0} na_nX^n $
After doing some series manipulation I got:
$= \sum^{\infty}_{n=2} n(n-1)a_nX^n + \sum^{\infty}_{n=0} (n+2)(n+1)a_{n+2}X^n - \sum^{\infty}_{n=0} 6na_nX^n + \sum^{\infty}_{n=0} 10na_nX^n$
Now I'm stuck on what to do with the first series that has $n=2$ because I don't want to mess up the powers.
 A: I do not know how this is teached but I think that you make the problem difficult changing the starting value of the index in the summation. Let me try $$y=\sum_{n=0}^{\infty}a_n x^n$$ $$y'=\sum_{n=0}^{\infty}n a_n x^{n-1}$$ $$y''=\sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}$$ Rewrite the equation as $$(x^2 + 1)y'' - 6xy' + 10y =x^2y''+y''-6xy'+10y=0$$ and include the summations. So, $$\sum_{n=0}^{\infty}n(n-1) a_n x^{n}+\sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}-6\sum_{n=0}^{\infty}n a_n x^{n}+10\sum_{n=0}^{\infty}a_n x^n=0$$ Now consider a given power $m$; this gives $$m(m-1)a_m+(m+2)(m+1)a_{m+2}-6ma_m+10a_m=0$$ So, after grouping the terms and simplifying you have $$(m+1)(m+2)a_{m+2}+(m-2)(m-5)a_m=0$$ or $$a_{m+2}=-\frac{(m-2)(m-5)}{(m+1)(m+2)}a_m$$ from which some interesting features appear after a particular value of $m$ (I let you finding it).
A: UPDATE!
The correct solutions for part b) and part c) are:
b)
$a_0 = 1$
$a_1 = 0$
$a_2 = -2/3$
$a_3 = 0$
$a_4 = 0$
$\therefore$ $y_1 = 1 + (-\frac{2}{3}x^2) + ... + ...$ 
c)
$a_0 = 0$
$a_1 = 1$
$a_2 = 0$
$a_3 = -2/3$
$a_4 = 0$
$a_5 = 1/15$
$\therefore$ $y_2 = x + (-\frac{2}{3}x^2) + \frac{1}{15}x^5 + ...$
