Best methods for solving ODE with series I am looking for some tips and suggestions in regard to the following problem (post below). I am not sure if I am on the right track, so if anyone could let me know that would be greatly appreaciated.
It is just a rather simple problem,
that is, solve
$$(x^2+4)y''-xy'+y=0$$ using series centred around $x=0$ with IC $y(0)=8$ and $y'(0)=4$
What I have done so far;
Suppose we have a solution $$y(x)=\sum_{n=0}^{\infty}a_nx^n$$
then $$ y'(x)= \sum_{n=1}^{\infty}(n)a_nx^{n-1}$$
and $$ y''(x)= \sum_{n=2}^{\infty}(n-1)(n)a_nx^{n-2}$$
by plugging these into the equation I get to the form;
$$(x^2+4) \sum_{n=2}^{\infty}(n-1)(n)a_nx^{n-2}-x\sum_{n=1}^{\infty}(n)a_nx^{n-1}+\sum_{n=0}^{\infty}a_nx^n=0$$
Now , If I am correct I believe I need to get this into the form such that all series start at the same index, and also that all series have the same exponent on the x.
So I tried expanding and simplifying to get
$$\sum_{n=2}^{\infty}(n-1)(n)a_nx^{n}+4\sum_{n=2}^{\infty}(n-1)(n)a_nx^{n-2}-(a_1+\sum_{n=2}^{\infty}(n)a_nx^{n})+(a_0+a_1+\sum_{n=2}^{\infty}a_nx^{n})$$
But I am not sure if it is correct, and also how I should handle the term with the 4 out front. Because I think I managed to get the same index, and almost the same exponent but I don't know how else the standard approach should go?  Any comments, suggestions, answers, etc?
thanks
 A: hints
change index on the two summations so you index over $[0,\infty)$ and then since the equality to 0 must hold, you must have it for each power of $x$. This will yield a recurrence relationship for $a_n$ only in terms of other $a_n$-like terms.
A: I think that changing the index from $0$ to $1$ and then to $2$ makes life more complex than required. 
Let us set $$y=\sum_{n=0}^{\infty}a_nx^n$$ $$y'=\sum_{n=0}^{\infty}na_nx^{n-1}$$ $$y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}$$ You notice that all the summations start at $n=0$ even if they would really start at $n=1$ for $y'$ and at $n=2$ for $y''$ but it does not matter since $0\times a_0 x^{-1}=0$ (this is for  $y'$) and $0\times(0-1)\times a_0 x^{-2}+1\times(1-1)\times a_1x^{-1}=0$ (this is for  $y''$).
So, now rewrite $$(x^2+4)y''-xy'+y=x^2 y''+4y''-xy'+y$$ and replace. The rhs of the last expression then write $$\sum_{n=0}^{\infty}n(n-1)a_nx^{n}+4\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}-\sum_{n=0}^{\infty}na_nx^{n}+\sum_{n=0}^{\infty}a_nx^{n}=0$$ Now, select a power $m$; for this power, the equation write $$m(m-1)a_mx^m+4(m+2)(m+1)a_{m+2}x^m-ma_mx^m+a_mx^m=0$$ and simplify to get $$4(m+2)(m+1)a_{m+2}+(m-1)^2a_m=0$$ that is to say $$a_{m+2}=-\frac{(m-1)^2}{4(m+2)(m+1)}a_m$$ The initial conditions give $a_0=8$ and $a_1=4$.
