Show power series is solution to differential equation I am currently studying for my analysis 2 course, and I've run into a couple of questions from old exams regarding power series as solutions for differential equations which I have a trouble completing.
We have a specific power series i.e.
$$f(x)=\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}x^{4n+3}$$
And we have to show this solves the differential equation,
$$x^2y''-3xy'+(3-4x^4)y=0$$
I have done some work to the left hand site, and gotten it to
$$\frac{-2x^{11}}{3}-7x^7+3x^3+\sum_{n=2}^{\infty}((\frac{16n^2+8n-4x^4}{(2n+1)!})x^{4n+3})$$
However I dont know how to continue from here in showing that the equation is equal to $0$. Of course we have the trivial solution for $x=0$, however how would I proceed for $x\neq0$?
 A: I will get you started. The goal is to get an expression
$$
x^2y''-3xy'+(3-4x^2)y=\sum_{n=0}^\infty a_n x^n
$$
where $a_n$ is a function of $n$ alone. You sort of got that in your last equation, but you have $x^4$ in your expression for the coefficient of $x^{4n+3}$. If all goes well, you should have $a_n=0$ for all $n$. It helps to work piece by piece:
\begin{align}
x^2y''
&=\sum_n \frac{(4n+3)(4n+2)}{(2n+1)!} x^{4n+3}
\\
-3xy' &=-3\sum_n \frac{(4n+3)}{(2n+1)!} x^{4n+3}
\\
(3-4x^4)y
&=3\sum_n\frac1{(2n+1)!}x^{4n+3}-4\sum_{n=0}^\infty\frac1{(2n+1)!} x^{4n+7} \\
&=3\sum_n\frac1{(2n+1)!}x^{4n+3}-4\sum_{n=1}^\infty\frac1{(2n-1)!} x^{\color{blue}{4n+3}} 
\end{align}
Notice how I re-indexed the last series so that everything was a coefficient times $x^{4n+3}$. This lets you combine everything together, except for the $n=0$ of the first three series, which will have to be pulled out and dealt with separately. The result is
\begin{align}
&x^2y''-3xy'+(3-4x^2)y=\\
&\big(\frac{3\cdot 2}{1!}-3\cdot \frac{3}{1!}+3\frac{1}{1!}\big)x^{4\cdot 0+3}+
\\&\sum_{n=1}^\infty \underbrace{\Big(\frac{(4n+3)(4n+2)}{(2n+1)!}-3\cdot \frac{(4n+3)}{(2n+1)!}+3\cdot \frac1{(2n+1)!}-4\frac1{(2n-1)!}\Big)}_{a_n}x^{4n+3}
\end{align}
Now you have a big complicated expression for $a_n$ you need to show is zero.
A: If$$f(x)=\sum_{n=0}^\infty\frac1{(2n+1)!}x^{4n+3},$$then$$f'(x)=\sum_{n=0}^\infty\frac{4n+3}{(2n+1)!}x^{4n+2}\quad\text{and}\quad f''(x)=\sum_{n=0}^\infty\frac{(4n+3)(4n+2)}{(2n+1)!}x^{4n+1}.$$It follows that$$x^2f''(x)=\sum_{n=0}^\infty\frac{(4n+3)(4n+2)}{(2n+1)!}x^{4n+3}\quad\text{and}\quad-3xf'(x)=\sum_{n=0}^\infty\frac{-12n-9}{(2n+1)!}x^{4n+3}.$$Therefore,$$x^2f''(x)-3xf'(x)=\sum_{n=0}^\infty\frac{16n^2+8n-3}{(2n+1)!}x^{4n+3}.\tag1$$On the other hand,$$3f(x)=\sum_{n=0}^\infty\frac3{(2n+1)!}x^{4n+3}$$and\begin{align}-4x^4f(x)&=-\sum_{n=0}^\infty\frac4{(2n+1)!}x^{4n+7}\\&=-\sum_{n=1}^\infty\frac4{(2n-1)!}x^{4n+3}.\end{align}So,\begin{align}(3-4x^2)f(x)&=3x^3+\sum_{n=1}^\infty\left(\frac3{(2n+1)!}-\frac4{(2n-1)!}\right)x^{4n+3}\\&=3x^3+\sum_{n=1}^\infty\frac{-16n^2-8n+3}{(2n+1)!}x^{4n+3}.\end{align}Since the RHS of $(1)$ is $-3x^3$ when $n=0$, we're done.
