1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Recurrance relaltion

  1. Aug 27, 2010 #1
    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= [tex]\sum[/tex] cnn(n-1)xn-2- 2x[tex]\sum[/tex] 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. jcsd
  3. Aug 28, 2010 #2


    User Avatar
    Homework Helper

    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
    [tex]y= \sum_{n=0}c_nx^n[/tex]

    differentiating twice gives
    [tex]y''= \sum_{m=2}c_m m(m-1)x^{m-2}[/tex]

    combining in the DE
    [tex]y'' - 2xy = 0[/tex]

    [tex]= \sum_{m=2} c_m m(m-1) x^{m-2} + 2x( \sum_{n=0} c_n x^n) [/tex]

    [tex]= \sum_{m=2} c_m m(m-1) x^{m-2} + \sum_{n=0} 2 c_n x^{n+1} [/tex]

    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

    [tex]= \sum_{n=-1} c_{n+3} (n+3)(n+2)x^{n+1} + \sum_{n=0}2c_nx^{n+1}[/tex]

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

    hopefully i didn't make any mistakes
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Recurrance relaltion Date
Showing a limit of the recurrence relation Feb 22, 2018
Legendre recurence relation Nov 5, 2017
Recurrence Stability Oct 2, 2017