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Series solution of an ODE

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve for [tex] y' = x^2y [/tex]

    3. The attempt at a solution

    There's something that's been really bothering me about this question and similar ones.

    We assume that the solution to the ODE will take the form

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

    After finding y', plugging in the expressions into the ODE, and distributing x2,

    [tex] \sum_{n=1}na_nx^{n-1} - \sum_{n=0}a_nx^{n+2} = 0 [/tex]

    I want to eventually combine the summations, so first I make the exponents match.

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

    I need to make the indices match without manipulating exponents, so I just peel off the first two terms (n = 0 and n = 1) from the first summation. After doing so and combining the summations:

    [tex] \sum_{n=2}((n+1)(a_{n+1} - a_{n-2}))x^n + a_1 + 2a_3x = 0 [/tex]

    Now here's where I start becoming uncertain...

    The summation begins at n = 2, so the smallest power of x that can come out is x2. Since the RHS is 0,

    [tex] a_1 = 0 [/tex]
    [tex] 2a_3x = 0 [/tex]

    This means that the coefficients a_1 and a_3 must be equal to 0. Also, since the coefficients for all powers of x on the RHS are zero,

    [tex] (n+1)a_{n+1} - a_{n-2} = 0 [/tex]

    Rearranging gives me the recursion formula for n ≥ 2:

    [tex] a_{n+1} = \frac{c_{n-2}}{n+1} [/tex]

    My end goal in here is to write the sum in a way that doesn't involve any past terms so that there's no recursion. I start plugging in numbers and hope that I see some pattern...

    n = 2, [itex] a_3 = \frac{a_0}{3} [/itex]

    n = 3, [itex] a_4 = \frac{a_1}{5} = 0 [/itex]

    n = 4, [itex] a_5 = \frac{a_2}{6} [/itex]

    n = 5, [itex] a_6 = \frac{a_3}{7} = 0 [/itex]

    n = 6, [itex] a_7 = \frac{a_4}{8} = \frac{a_1}{5*8} [/itex]

    At this point it's clear that the odd n will give a zero and that I only have to worry about even n.

    What bothers me is the term I get when n = 4. I'm not given any information as far as I know about a2 and since the ODE is first-order, there's only one arbitrary constant which I assume is a0 (if I don't assume it's arbitrary then I have no idea what to say about a0 either). So I'm stumped. How do I write a non-recursive formula for the sum with the mystery number a2? What am I missing in here?

    Apologies for the lengthy post

    EDIT: Disregard the thing I said about odd and even n. n = 7 doesn't equal 0.
     
    Last edited: Mar 9, 2013
  2. jcsd
  3. Mar 10, 2013 #2
    I guess there's a typo there. You should have a1 = a2 = 0.
     
  4. Mar 10, 2013 #3

    vela

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    Because of the error clamtrox pointed out, you had a3=0, and the n=2 equation then tells you a0=0. So a2 would be the arbitrary constant.

    Since you're really supposed to have ##a_1=a_2=0##, the non-vanishing coefficients are ##a_{3k}## (k=0, 1, 2, …), and a0 plays the role of the arbitrary constant.
     
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