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