Consider the differential equation

The solution to this differential equation has a power-series expansion

Using the method of undetermined coefficients obtain a recursion formula relating the terms to the terms . Give an explicit formula for for each and find the sum of the series.

First, we differentiate twice,

From the initial conditions and we have

Now, we plug the expressions for , and back into the given differential equation,

Then, we use the fact from above that and to get

Since this sum is equal to 0, we know that every coefficient of every power of must be equal to 0. First, we solve for and ,

Then, we establish the recursive relationship between and ,

for all . Then since we have for all (since for every odd integer the formula for has is multiplied by , but each of these will be 0). For the even terms we have for ,

This means all of the remaining even terms will be 0 as well. So we have

Hence,

Big mistake here… the remaining terms do not ALL turn to zero after a4. Every other term will be zero from that point on, given that the recursive relation is between m and m+2… this leaves m+1, m+3, etc… to be nonzero.

I don´t know if I understood your point, but the terms become zero as a_1=0 so a_3,a_5,… are zero and a_4=0 so a_6,a_8,… become zero too.