# Recurrance relaltion

1. Aug 27, 2010

### vipertongn

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

Find the recurrence relation for the following diﬀerential equation. You
do not need to solve the rest of the way.

y" − 2xy = 0

3. The attempt at a solution

I'm not exactly sure how to do this problem but from the example of the book I tried this:

y"-2xy= $$\sum$$ cnn(n-1)xn-2- 2x$$\sum$$ cnxn+1 and replaced k=n-2 and k=n+1 simultaneously and kinda ended up with
(k+1)(k+2)ck+2+2ck-1= 0 k=1,2,3...

Is that correct?

2. Aug 28, 2010

### lanedance

not that you have to, but i usually keep the starting points of the sum & maybe use different dummy indicies to be clear, until you get the final relation

so assume
$$y= \sum_{n=0}c_nx^n$$

differentiating twice gives
$$y''= \sum_{m=2}c_m m(m-1)x^{m-2}$$

combining in the DE
$$y'' - 2xy = 0$$

$$= \sum_{m=2} c_m m(m-1) x^{m-2} + 2x( \sum_{n=0} c_n x^n)$$

$$= \sum_{m=2} c_m m(m-1) x^{m-2} + \sum_{n=0} 2 c_n x^{n+1}$$

so now to compare terms make the substitution m -2 = n+1 which gives
m = n+3, n=m-3, when m = 2, n = -1

$$= \sum_{n=-1} c_{n+3} (n+3)(n+2)x^{n+1} + \sum_{n=0}2c_nx^{n+1}$$

removing the n=-1 case we can then combine them
$$c_{2} (2)(1) + \sum_{n=0}(c_{n+3} (n+3)(n+2)+ 2c_n)x^{n+1} = 0$$

hopefully i didn't make any mistakes