Where did I go wrong on this approach?

Question

I have the following DE $y'' + xy=0$, then I have to use power series method to break apart my problem: leading me to then do the following \begin{align}\sum_{n=2}^\infty n(n-1)c_nx^{n-2}+x\sum_{n=0}^\infty c_nx^n&=0\\ 2c_2+\sum_{n=3}^\infty n(n-1)c_nx^{n-2}+\sum_{n=0}^\infty c_nx^{n+1}&=0 \\ 2c_2+\sum_{k=0}^\infty(k+3)(k+2)c_{k+3}x^{k+1}+\sum_{k=0}^\infty c_kx^{k+1}&=0 \end{align} To top it off I got the following zeroes for the equation: \begin{equation}c_2=0\end{equation}\begin{align} c_{k+3}=\frac{-c_k}{(k+3)(k+2)}\end{align} My question is my approach right and will I got a nice looking factorial in a summation for the end answer or will it be just a infinite sum? I was doing the work and I keep getting two products like for $c_6=\frac{c_0}{6*5*3*2*1}$ I noted that right there I had a 4 missing so I thought if it would be a factorial over a factorial any thoughts I would be grateful.

Answer 1

Following from your work on can then follow the power series solution method, and by that I mean: \begin{equation}c_0+c_1x+c_2x^2+c_3x^3+c_4x^4\dots\end{equation} Then you just have to group your solution sets into two solutions, which would be the at most expected from a second order differential equation: \begin{align}y(x)=c_0(1-\frac{x^3}{3\cdot2}+\frac{x^6}{6\cdot5\cdot3\cdot2}\dots)+c_1(x-\frac{x^4}{4\cdot3}+\frac{x^7}{7\cdot6\cdot4\cdot3 }\dots)\end{align} This would be the indicated solution without using any Products, and factorials.