# Homework Help: Series solution up to a term, power series

1. Jul 2, 2011

### dp182

1. The problem statement, all variables and given/known data
consider the initial value problem (1-x)y,,+xy,-2y=0 find the series solution up to the term with x6

2. Relevant equations
(1-x)y,,+xy,-2y=0

3. The attempt at a solution
assuming the answer has the form $\Sigma$anxn
that gives y,,=$\Sigma$nanxn-1 and y,,=$\Sigma$n(n-1)anxn-2 then plugging these back in and getting rid of the xn-2 and xn-1 you get the equation $\Sigma$(n+2)(n+1)an+2xn-$\Sigma$n(n+1)an+1xn+$\Sigma$nanxn-2$\Sigma$anxn so what I'm wondering is how do you replace the an+1 with an an so you can solve for an+2

2. Jul 2, 2011

### tiny-tim

hi dp182!
just subtract 1 from everything

(eg ∑an+1xn = ∑anxn-1)

3. Jul 2, 2011

### vela

Staff Emeritus
You don't. The recurrence relation involves an, an+1, and an+2. There's nothing wrong with that.

4. Jul 2, 2011

### id the sloth

It makes sense that the recurrence relation for a_n+2 involves a_n+1 and a_n. It would require you to define a_0 and a_1 and Second order differential equations require two boundary/initial conditions for a solution.

5. Jul 2, 2011

### dp182

So If I solve for an+2 I get an+2=n(n+1)an+1+an(2-n)/(n+2)(n+1) so to get my terms up to x6 I will just input values of n, so my series will be in terms of a0and a1?

6. Jul 2, 2011

### vela

Staff Emeritus
Yup, but you made an algebra error solving for an+2 (or omitted very needed parentheses).