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