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Power series solution for ODE

  1. Nov 28, 2013 #1
    1. The problem statement, all variables and given/known data

    I am trying to find the recursion relation for the coefficients of the series around x=0 for the ODE: [itex]y'''+x^2y'+xy=0[/itex]

    3. The attempt at a solution
    Therefore letting:

    [itex]y=\sum_{m=0}^\infty y_mx^m[/itex]

    [itex]\therefore y'=\sum_{m=1}^\infty my_mx^{m-1}[/itex]

    [itex]\therefore y''=\sum_{m=2}^\infty m(m-1)y_mx^{m-2}[/itex]

    [itex]\therefore y'''=\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}[/itex]

    Subbing this back in gives:

    [itex]\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}+x^2\sum_{m=1}^\infty my_mx^{m-1}+x\sum_{m=0}^\infty y_mx^m=0[/itex]

    Fixing [itex]y'''[/itex]:

    [itex]\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}=\sum_{m=2}^\infty (m+1)(m+2)(m+3)y_{m+3}x^{m}+6y_3+24xy_4[/itex]

    Fixing [itex]y'[/itex]:

    [itex]x^2\sum_{m=1}^\infty my_mx^{m-1}=\sum_{m=2}^\infty (m-1)y_{m-1}x^m[/itex]

    Fixing [itex]y[/itex]:

    [itex]x\sum_{m=0}^\infty y_mx^m=\sum_{m=2}^\infty y_{m-1}x^m+xy_0[/itex]

    Therefore combining these terms gives:


    Therefore I have two equations which sum to 0, with one of them being this:


    [itex]\therefore y_3=0[/itex] and [itex]y_4=-\frac{1}{24}y_0[/itex]

    Using the other equation:


    [itex]\therefore (m+1)(m+2)(m+3)y_{m+3}+y_{m-1}((m-1)+1)=0[/itex]

    [itex]\therefore y_{m+3}=-\frac{y_{m-1}m}{ (m+1)(m+2)(m+3)}[/itex]

    This gives me:

    The problem is I am not sure how to relate these together to solve the recursion! Any help is much appreciated.
    Last edited: Nov 28, 2013
  2. jcsd
  3. Nov 28, 2013 #2


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    Science Advisor

    It is rare that one can solve recursive equations for a "closed form" solution. Do you have reason to believe you can here?
  4. Nov 28, 2013 #3
    Hm I am thinking then that there is a limited number of free parameters? So to answer your question, I have no good reason to believe a closed form solution exists. So how would I comment on the number of free parameters? Does this mean anything?
  5. Nov 28, 2013 #4


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    Homework Helper

    You have
    [tex]y_{m+3} = - \frac{y_{m-1} m}{(m+ 1)(m + 2)(m + 3)}[/tex]
    [tex]y_{m+4} = - \frac{y_m (m+1)}{(m+2)(m+ 3)(m+ 4)}[/tex]

    Thus you have four independent sequences [itex]y_{4n + k}[/itex] for [itex]k = 0,1,2,3[/itex]. Setting [itex]m = 4n + k[/itex] then gives
    y_{4(n+1)+k} = - \frac{y_{4n + k} (4n + k +1)}{(4n + k +2)(4n + k + 3)(4n + k + 4)}
    Now, instead of having [itex]m[/itex] increasing by 4 at each step, we have [itex]n[/itex] increasing by 1, and to make things clearer we can set [itex]y_{4n + k} = a_n(k)[/itex], so that
    a_{n+1}(k) = - \frac{a_n(k) (4n + k +1)}{(4n + k +2)(4n + k + 3)(4n + k + 4)}

    One of the convenient aspects of homogenous linear recurrences is that we can deal with each factor multiplying [itex]a_{n}[/itex] separately. The -1 is the easiest: we get a factor of [itex](-1)^n[/itex]. The factors which are linear in [itex]n[/itex] are more difficult, and
    b_{n+1} = (An + B)b_n
    doesn't have a solution other than the not closed-form
    b_n = b_0\prod_{r = 0}^{n-1} (Ar + B)
    and similarly
    b_{n+1} = \frac{b_n}{An + B}
    doesn't have a solution other than
    b_n = b_0\prod_{r = 0}^{n-1} \frac{1}{Ar + B}
    (and we obviously require that there not exist any integer [itex]r \geq 0[/itex] for which [itex]Ar + B = 0[/itex]).

    If [itex]A = 1[/itex] (which in your problem it isn't) these products can be expressed in terms of factorials or gamma functions. If [itex]A[/itex] were an integer and there were [itex]A[/itex] factors [itex](Ar + B)(Ar + B + 1) \cdots (Ar + B + A)[/itex] (or [itex](Ar + B)^{-1}(Ar + B + 1)^{-1} \cdots (Ar + B + A)^{-1}[/itex]) then again there would be a solution in terms of factorials or gamma functions, but your problem doesn't have sufficient factors.

    However, you can say that
    y(x) = \sum_{m = 0}^{\infty} y_m x^m = \sum_{n=0}^{\infty} x^{4n} (a_n(0) + a_n(1) x + a_n(2) x^2)
    since the initial condition [itex]y_3 = 0[/itex] requires that [itex]a_n(3) = 0[/itex] for all [itex]n[/itex].
  6. Nov 28, 2013 #5
    Thank you so much, was a good read :smile:
    Last edited: Nov 28, 2013
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