1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

Homework Help: Series Solution of ODE

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data
    (1 - x)y'' + xy' + xy = 0

    Find the first 3 nonzero coefficients of the power series expansion about x = 0 if y(0) = -1 and y'(0) = 0


    2. Relevant equations



    3. The attempt at a solution

    [itex]y = \sum^{∞}_{n = 0}c_{n}x^{n}[/itex]

    From above, I can say that [itex]y(0) = 1 = c_{0}[/itex] and [itex]y'(0) = 0 = c_{1}[/itex]

    [itex]L_{y} = (1 - x)\sum^{∞}_{n = 2}c_{n}n(n-1)x^{n-2} + \sum^{∞}_{n = 1}c_{n}nx^{n} + \sum^{∞}_{n = 0}c_{n}x^{n + 1} = 0[/itex]

    Some indices manipulation so all the power series are for [itex]x^{m-2}[/itex]...

    [itex]L_{y} = \sum^{∞}_{n = 3}[c_{m}m(m-1) - c_{m - 1}(m - 1)(m - 2) + c_{m - 2}(m - 2) + c_{m - 3}]x^{m - 2} + 2c_{2} = 0[/itex]

    [itex]c_{2} = arbitrary number[/itex] but I can't assume that it's 0, right?

    [itex]c_{m}m(m-1) - c_{m - 1}(m - 1)(m - 2) + c_{m - 2}(m - 2) + c_{m - 3} = 0 for m ≥ 3[/itex]

    m = 3:
    [itex] 6c_{3} - 2c_{2} = -1[/itex]

    m = 4:
    [itex] 12c_{4} - 6c_{3} + 2c_{2} = 0[/itex]

    [itex]c_{4} = -1/12[/itex]

    recurrence equation:
    [itex]c_{m} = \frac{c_{m-1}(m - 1)(m - 2) - c_{m-2}(m - 2) - c_{m-3}}{m(m - 1)}[/itex]

    I can't seem to find the 3rd non-zero constant...
     
    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 21, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You don't have to assume ##c_2## is zero. You have proved it is by equating the constant terms on both sides of that equation.
     
  4. Jan 21, 2012 #3
    Ah, right. Got it.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...