Recurrence Relation: Solving y''-2xy=0

[vipertongn]

Homework Statement

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

y" − 2xy = 0

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 c_nn(n-1)x^n-2- 2x\sum c_nxⁿ⁺¹ and replaced k=n-2 and k=n+1 simultaneously and kinda ended up with
(k+1)(k+2)c_k+2+2c_k-1= 0 k=1,2,3...

Is that correct?

 

[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