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 $x\neq0$?
Best Answer
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.