Solution to ODE using Power Series I'm currently trying to wrap my head around how to solve an ODE with series. 
The problem I am working on is this:
Find the indicated coefficients of the power series solution about x=0 of the differential equation:
$(x^2 + 1)y'' - xy' + y = 0$
$y(0) = 3, y'(0) = -8$
The answer blanks are as follows:
$y = 3-8x + BLANKx^2 + BLANKx^4 + BLANKx^6 + BLANKx^8 + 0(x^9)$
From my notes, I can discern that because $x=0$, 
$y = \sum_{n=0}^{\infty}a_n x^{n}$
$y' = \sum_{n=1}^{\infty}na_n x^{n-1}$
$y'' = \sum_{n=2}^{\infty}n(n-1)a_n x^{n-2}$
From here, I'm unsure how to proceed. If someone could walk me step by step through these types of problems (I have several to complete), I would be very grateful.
From what I can tell, I need to manipulate the equation with those sums in place to get "formulas" for $a_2$, $a_3$ etc...but I don't know how I would go about that. 
Thank you in advance for your help!
 A: Here are the steps.. I'll do some of the working, but I'll mainly leave it up to you to solve.


*

*Substitute your power series for $y$, $y'$, $y''$ into your equation $(x^{2} + 1)y'' -xy' + y = 0$ i.e.


$$\begin{align}
(x^{2} + 1) \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} - x \sum_{n = 1}^{\infty} n a_n x^{n - 1} + \sum_{n = 0}^{\infty} a_n x^{n} &= 0 \\
\implies \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n} + \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} - \sum_{n = 1}^{\infty} n a_n x^{n} + \sum_{n = 0}^{\infty} a_n x^{n} &= 0 \\
\end{align}$$


*Shift your power series for those series that don't have $x^{n}$ so that they do have $x^{n}$ and then replace the old series with the new one i.e.


$$\implies \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} = \sum_{n = 0}^{\infty} (n + 2)(n + 1) a_{n + 2} x^{n}$$


*Notice that you have series that start at $0, 1, 2$ so evaluate those at $0, 1$ so that all the series then start at $2$ i.e.


$$\sum_{n = 1}^{\infty} n a_n x^{n} = a_1 x^{1} + \sum_{n = 2}^{\infty} n a_n x^{n}$$


*Collect summation terms under a single summation starting at $n = 2$, with individual terms (like the $a_1 x$ above) not written in the summation i.e.


$$ a_1 x + ... + \sum_{n = 2}^{\infty} \bigg[ a_n - n a_n + ... \bigg] x^{n} = 0$$ 
Once you have done that, comment below and then we can do the next steps (Or if you need any help at all, just comment below).
EDIT
In step $1$, just expand the $(x^{2} + 1)y''$ term i.e.
$$\begin{align}
(x^{2} + 1)y'' &= x^{2}y'' + y'' \\
&= x^{2} \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} + \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} \\
\end{align}$$
Now, the $x^{2}$ term can be taken inside the summation (or you can take the $x^{-2}$ term out of the summation, it's the same thing), because we are not summing over that term. Hence
$$\begin{align}
x^{2} \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} + \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} &= \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n} + \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} \\
\end{align}$$
Similarly, for $xy'$
$$\begin{align}
xy' &= x \sum_{n = 1}^{\infty} n a_n x^{n - 1} \\
&= \sum_{n = 1}^{\infty} n a_n x^{n} \\
\end{align}$$
