# Finding a recursion relation

1. May 7, 2010

### Jamin2112

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

I just have a problem with series solutions when I get to the point of needing to find the recursion relation........and a few other problems.

2. Relevant equations

Assume y can be written ∑anxn

3. The attempt at a solution

So, on y'' + xy' + 2y = 0, I got to the point of

an+2=-an/(n+1). I wrote out a few n's but still can't figure out exactly what the relation is. A little help, maybe?

On another problem, (1+x2)y'' - 4xy' + 6y = 0, I got the point of

2a2 + 6a3x - 4a1x + 6a0 + 6a1x + [n=2 to ∞]∑( (n+1)(n+2)an+2 + (n-1)nan - 4nan + 6an)xn = 0.

Now what? Detailed help would be much appreciated.

2. May 7, 2010

### tiny-tim

Hi Jamin2112!
Hint: 7*5*3*1 = 7*6*5*4*3*2*1/(23*3*2*1)
General method is to have onen=1(…)xn or ∑n=2(…)xn (whichever seems to work), from which you get a general recursion relation for an for n ≥ 1 (or n ≥ 2 or whichever), together with some stray terms that don't fit inside the ∑, and which give you the initial terms.

3. May 7, 2010

### Jamin2112

So the stray terms are the initial conditions. They get set to zero, along with everything inside the sigma, right?

4. May 7, 2010

Yup!