# Solving DE using power series (w/initial conditions)

1. Oct 24, 2009

### Fuzedmind

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

Solve the initial value problem y'' = y' + y where y(0) = 0 and y(1) = 1

derive the power series solution y(x) = $$\ \ \sum_{n=1}^{\infty}{(F_{n}x^n)/n!} \ \$$ where {Fn} is the sequence 0,1,1,2,3,5,8,13.... of Fibonacci numbers defined by F0 = 0 and F1 = 1

2. Relevant equations

3. The attempt at a solution

I plugged in the series and got this equation:

$$\ \ \sum_{n=2}^{\infty}{(n)(n-1)c_{n}x^(n-2)} \ \ + \ \ \sum_{n=1}^{\infty}{c_{n}x^(n-1)} \ \ + \ \ \sum_{n=0}^{\infty}{c_{n}x^n} \ \ = x$$

from which i got the recurrence equation:

Fn = Fn-1/n + Fn-2/n(n-1)

Problem is I don't know how to get F0 and F1 from the initial conditions. My teacher never taught us how to solve initial condition power series ODE's nor are there any examples in the book. Can anyone explain it to me?

2. Oct 24, 2009

### LCKurtz

You are assuming y = c0 + c1x + c2x2+...

y(0) is just c0. And y'(0) would be c1 , which would be the usual conditions for an IVP. Are you sure you have the second boundary condition written correctly? My guess it is supposed to be y'(0) = 1.

3. Oct 24, 2009

### Fuzedmind

no y(0) = 0 and y(1) = 1, I am looking at it right now

4. Oct 24, 2009

### LCKurtz

I don't have time to work through the solution right now, but what you have then is a boundary value problem, not an initial value problem as claimed. I'm suspecting a misprint. You might just try using y'(0)=1 and working it through to see if you get the correct answer.

5. Oct 24, 2009

### LCKurtz

I had a little time later in the day. Your text indeed has a misprint. If you take your equation with the initial conditions y(0) = 0 and y'(0) = 1, which give you the values of c0 and c1 and work out the series solution, you get exactly what the problem claims you will get.

6. Oct 25, 2009

### Fuzedmind

Ok thanks a bunch man. I did that and it worked out for me too.