Finding series solution about zero $y''+x^2y'+4y=1-x^2$
To find a power series, one substitutes in $y= \sum_0^\infty a_nx^n$. So after substitution, I've gotten 
$\sum_0^\infty (n+1)(n+2)a_{n+2}x^n + \sum_1^\infty (n-1)a_{n-1}x^n + 4\sum_0^\infty a_nx^n$
The recurrence relation would then be in terms of $a_{n-2}$ since it's the lowest term. But how do I solve this if there are two other unknown terms? And I've never worked with a power series set to equal anything other than zero. I am just lost.
 A: For the equation
$$ y'' + x^2 \, y' + 4 \, y = 1 - x^2$$
use the solution
$$y = \sum_{n=0}^{\infty} a_{n} \, x^n$$
to obtain:
\begin{align}
\sum_{n} (n)(n-1) \, a_{n} \, x^{n-2}  + \sum_{n} n \, a_{n} \, x^{n+1} + 4 \, \sum_{n} a_{n} &= 1 - x^2 \\
\sum_{n=0} [(n+1)(n+2) \, a_{n+2} + 4 \, a_{n}] \, x^{n} + \sum_{n=1} n \, a_{n} \, x^{n+1} &= 1 - x^2 \\
\sum_{n=0} [(n+1)(n+2) \, a_{n+2} + 4 \, a_{n}] \, x^{n} + \sum_{n=0} (n+1) \, a_{n+1} \, x^{n+2} &= 1 - x^2 \\
\end{align} 
Now equating coefficients leads to:
\begin{align}
2 \, a_{2} + 4 \, a_{0} &= 1 \\
6 \, a_{3} + 4 \, a_{1} &= 0 \\
12 \, a_{4} + a_{1} + 4 \, a_{2} &= -1 \\
(n+4)(n+5) \, a_{n+5} + (n+2) \, a_{n+2} + 4 \, a_{n+3} &= 0.
\end{align}
From these equations it is determined that:
\begin{align}
a_{2} &= \frac{1}{2} \, (1- 4 \, a_{0}) \\
a_{3} &= - \frac{4}{6} \, a_{1} \\
a_{4} &= \frac{1}{12} \, ( -3 + 8 \, a_{0} - a_{1} ) \\
a_{5} &= \frac{1}{60} \, (-3 + 12 \, a_{0} + 8 \, a_{1} ),
\end{align}
where $a_{0}$ and $a_{1}$ are determined by the initial conditions. 
