# Homework Help: Differential Equations - Power Series problem with e^t

1. Jul 15, 2013

### skyturnred

1. The problem statement, all variables and given/known data

The problem is to solve:

y''+ty'+e$^{t}$y=0, y(0)=0 and y'(0)=-1

2. Relevant equations

3. The attempt at a solution

My main issue is the following: I normally find the recursion relation, and then factor out the t$^{whatever}$ and I know that the coefficient to this must equal 0. However, the addition of the e$^{t}$ means I can't factor out ALL of the t$^{whatever}$ so I can't exactly equate the coefficient to 0.

So specifically with the question above, I get the following:

ty'=$\sum$$^{inf}_{k=1}$ka$_{k}$t$^{k}$

e$^{t}$=$\sum$$^{inf}_{k=0}$t$^{k}$/k!

e$^{t}$y=$\sum$$^{inf}_{k=0}$t$^{2k}$a$_{k}$/k!

y''=$\sum$$^{inf}_{k=2}$k(k-1)a$_{k}$t$^{k-2}$

Then I plug those into the DE, write out the first few terms so that they all start at k=2, and put them all under the same sum to get as follows:

a$_{0}$+a$_{1}$t+a$_{1}$t$^{2}$+2a$_{2}$+6a$_{3}$t+$\sum$$^{inf}_{k=2}$[(k+2)(k+1)a$_{k+2}$+ka$_{k}$+a$_{k}$t$^{k}$]t$^{k}$=0

From this you find that a$_{2}$=-a$_{0}$/2

and

a$_{1}$=0

and

a$_{3}$=0

Like I said, normally at this point I find the recursion relation and factor out a t$^{k}$ so that I know that the entire coefficient is 0. But in this case, after factoring out the t$^{k}$ one of the terms in the coefficient still have a t$^{k}$, so I don't know if I can't just equate this whole thing to 0.

Can anyone help?

Thank-you

2. Jul 15, 2013

### LCKurtz

You don't need to put the itex tags around every item. Just put them around the whole equation. Right click on the above expression and see how I changed it for you.

You don't multiply two series like that. You need the Cauchy Product. See:
http://planetmath.org/cauchyproduct