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Homework Help: 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


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    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
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