# Power series solution to degree 2 ODE

1. Mar 11, 2015

### Panphobia

1. The problem statement, all variables and given/known data
(x+1)y'' - (x-1)y' - y = 0
centred around x=1
y(1) = 2, y'(1) = 3

3. The attempt at a solution
I know I am supposed to get two power series, one with a0 and one with a1 but when I am trying to figure out a pattern, I keep getting both a0 and a1 in all of my terms.

So I end up with a2 = (1/4)*a0
and an+2 = (an*n - n*(n+1)*an+1)/(2*(n+1)*(n+2))
for n = 1 I got a3 = (2a1-(1/2)*a0)/12

for n = 2 I got a4 = ((3/4)*a0-6*((2a1-(1/2)*a0)/12)/24

I did this for n=3 and n=4 and n=5 and I still can't come up with a pattern for two different series, one with a0 and one with a1

2. Mar 11, 2015

### fourier jr

My recurrence relation is a little different. Did you replace (x+1) with [(x-1) + 2] in the first term? Then my recurrence relation turned out to be $a_{n+2} = \frac{a_{n} - na_{n+1}}{2(n+2)}$ & I think I noticed a $2^{n}(n+2)!$ in the denominators but I'm having trouble with the numerators. & $a_{0}=a_{2}=0$ since everything on the left-hand side has to add to zero. I could show you how I got that if you really want but these series solution problems are a lot of shovelling. (enough shovelling that I could have easily made a mistake )

3. Mar 11, 2015

### Panphobia

Yea I got that and now I got a3 = a1/6 - a0/24 and a4 = a0/24-a1/24 and that doesn't make sense since the a0 is basically the same in both. ugh