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Power series solution for 1st-order ODE

  1. Jun 13, 2004 #1
    I am trying to find the power series solution of

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

    but dont know how to arrive at the answer of [tex]y = a_0exp(x^3/3)[/tex]. [I know that it's an easily solved separable equation, I'm just trying to figure out how to find the power series solution]

    My solution so far:

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

    then
    [tex] y' = \sum_{n=1}^\infty na_nx^{n-1} [/tex]

    giving:

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

    changing the index for the LHS to give [tex]x^n [/tex]

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

    changing the index for the RHS to give [tex]x^n[/tex]:

    [tex] \sum_{n=2}^\infty a_{n - 2}x^{n} [/tex]

    Then taking the first two terms out of the LHS sum, so that both sums start from the same point:

    [tex] a_1 + 2a_2x + \sum_{n=2}^\infty (n + 1)a_{n + 1}x^n = \sum_{n=2}^\infty a_{n - 2}x^{n} [/tex]

    I don't know what to do after this (I'm not entirely sure if what I've done so far is right, either).

    If the y term in the initial equation didn't have the [tex]x^2[/tex] in front of it, it would be easy to equate the coefficients of [tex]x^n[/tex] to get the recursion formula. But having the terms [tex]a_1[/tex] and [tex]2a_2x[/tex] in front of the sum on the LHS throws me - can anyone explain clearly to me the correct steps required to solve the problem?
     
    Last edited: Jun 13, 2004
  2. jcsd
  3. Jun 13, 2004 #2
    We know that [tex]y'(0) = a_1[/tex] and [tex]y''(0) = 2 a_2[/tex]

    We also know that [tex]y'(x) = x^2 \ y(x)[/tex] so that [tex]y'(0) = 0 = a_1[/tex]

    Differentiating both sides we get [tex]y''(x) = 2x \ y(x) + x^2 \ y'(x)[/tex] and it is seen that [tex]y''(0) = 0 = 2 a_2[/tex] so now we know that [tex]a_1=a_2=0[/tex].

    Hope this helps!
     
  4. Jun 13, 2004 #3
    Thanks, that was a great help.
     
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