Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Series solution for y +x*y=0

  1. Jul 21, 2012 #1
    Series solution for y"+x*y=0

    Working on recurance realtion.
    Get to (sum(n=2))n*(N-1)*a(n)*X^(n-2)+(sum(n=0))a(n)*x^(n)
    Try several things but not sure if their correct.
     
  2. jcsd
  3. Jul 21, 2012 #2
    Re: Series solution for y"+x*y=0

    Try y = Ʃ a(n)*x^n
    where n goes from 0 to infinity

    Now y'' = Ʃ n(n-1)*a(n)*x^(n-2)
    where n goes from 2 to infinity

    and x*y = Ʃ a(n)*x^(n+1)
    where n goes from 0 to infinity

    We want these to have the same power of x i.e. n-2, so let's write them as follows:

    y'' = 2*a(2) + Ʃ n(n-1)*a(n)*x^(n-2)
    where n goes from 3 to infinity
    (we just pulled the first term out of the summation)

    x*y = Ʃ a(n-3)*x^(n-2)
    where n goes from 3 to infinity
    (index shift)

    Thus we have y'' + x*y = 2*a(2) + Ʃ [ n(n-1)*a(n) + a(n-3) ]*x^(n-2)
    where n goes from 3 to infinity.

    We want this to equal zero, so we have that:
    y = Ʃ a(n)*x^n, where
    a(0) and a(1) are arbitrary,
    a(2) = 0,
    and a(n) = -a(n-3) / [n(n-1)]


    Hope that helps :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook