Prove that a power series satifies that $x^2y''-4xy'+(x^2+6)y=0$ Consider the power series
$$
\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}z^{2n+3}
$$
I have shown that f is differentiable and that
$$
f'(x) = \sum_{n=0}^\infty \frac{(-1)^n(2n+3)}{(2n+1)!}x^{2n+2}
$$
and
$$
f''(x) = \sum_{n=0}^\infty \frac{(-1)^n(2n+3)(2n+2)}{(2n+1)!}x^{2n+1}
$$
and are now asked to prove that $f$ is a solution to the differential equation
$$
x^2y''-4xy'+(x^2+6)y=0
$$
on the interval $]-\infty, \infty[$. I just really have a hard time starting these kind of questions. I have probably tried for 45 minuts and still can't make it work. This is some of what I have gotten. Let $S = x^2y''$. Then 
\begin{align*}
S & = x^2 \sum_{n=0}^\infty \frac{(-1)^n(2n+3)(2n+2)}{(2n+1)!}x^{2n+1} \\
& = \sum_{n=0}^\infty \frac{(-1)^n(2n+3)(2n+2)}{(2n+1)!}x^{2n+3} \\
& = 6f(x) + \sum_{n=0}^\infty \frac{(-1)^n(4n^2+10n)}{(2n+1)!} x^{2n+3}
\end{align*}
and then I got stucked and tried some other things but still couldn't get any further in regards what I want to be shown. Do you mind helping me? 
 A: It is plain that the power series is convergent for all $z$ because the factorial denominator grows much faster than any power of $z$.  Convergent power series may be differentiated term by term.
It is then a question - as you have started - of working it through. 
Let $S = z^2 f''(z) - 4zf'(z) + (z^2 + 6) f(z)$, so that
$$\begin{align}
S&=\sum_{n \geqslant 0} \frac{(-1)^n}{(2n+1)!} \Big\{ (2n+3)(2n+2) z^{2n+3} -4 (2n+3) z^{2n+3} + z^{2n+5} + 6z^{2n+3} \Big\} \\
&= \sum_{n\geqslant 0}\frac{(-1)^n}{(2n+1)!}\Big\{
(2n+1)(2n)z^{2n+3} + z^{2n+5}
\Big\}
\end{align}
$$
We can now shift the indexation of the $z^{2n+3}$ terms.  First, peel off the $z^3$ term arising when $n=0$, and the  re-index the remaining terms so that powers of $z$ are aligned,  to get,
$$\begin{align}
S&= 0 \cdot z^{2n+3} + \sum_{n \geqslant 1} \frac{(-1)^n}{(2n+1)!}(2n+1)(2n) z^{2n+3} + \sum_{n\geqslant 0} \frac{(-1)^n}{(2n+1)!} z^{2n+5} \\
&= \sum_{n\geqslant 0}\Big\{ \frac{(-1)^{n+1}(2n+3)(2n+2)}{(2n+3)!} z^{2n+5} + \frac{(-1)^n}{(2n+1)!} z^{2n+5} \Big\}  \\
&= \sum_{n\geqslant 0}\Big\{ \frac{(-1)^{n+1}}{(2n+1)!} z^{2n+5} + \frac{(-1)^n}{(2n+1)!} z^{2n+5} \Big\} 
\end{align}
$$
Plainly the terms cancel and we find $S = 0$, which is what we required,
In fact, in case you have not noticed, this is the series for $z^2 \sin z$.
