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Homework Help: Cubic ODE: Series solution

  1. Jun 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the solution to the ODE via the power series:
    [tex] y = \Sigma_{i=0} a_j x^{2j + m} [/tex]

    2. Relevant equations

    [tex] y' - y^3 = 0 [/tex]


    3. The attempt at a solution
    I get
    [tex] \Sigma_{i=0} a_j (2j+m) x^{2j+m-1} - \Sigma_{i=0} (a_j)^3 x^{3(2j + m)} [/tex] = 0
    I don't know how to deal with the cubic part. No matter where I start my series I can't get the recursion relation without x
     
  2. jcsd
  3. Jun 2, 2010 #2

    Mark44

    Staff: Mentor

    The summation on the right above is incorrect. y^3 is the cube of the summation, not the summation of the cubes of its terms.

    It will probably be helpful to work with the a few terms of the expanded series, rather than work with the closed forms of the series. The equation above should be written as follows.

    [tex] \Sigma_{i=0} a_j (2j+m) x^{2j+m-1} - \left(\Sigma_{i=0} (a_j) x^{2j + m}\right)^3 [/tex]
     
  4. Jun 2, 2010 #3
    Thanks for the correction.
    So it looks like I would have to expand the cubic part to just a couple terms?
    Where do you go from there?
     
  5. Jun 2, 2010 #4

    Mark44

    Staff: Mentor

    Expand both parts and combine like terms. You'll probably need to expand to more than just two terms, though. I haven't worked this problem, but that's the tack I would take.

    The equation is separable, so you can solve it using that technique as a check of your work using series.
     
  6. Jun 3, 2010 #5
    I don't see how expanding and combining like terms will set me up for a recursion relation.
     
  7. Jun 3, 2010 #6

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Once you do that, you know that the series is supposed to be zero, so each coefficient in the new series has to be zero as well.
     
  8. Jun 3, 2010 #7
    I'm not sure this is what you want, but it's the only way I know how to express that DE in terms of a power series. ll do the square via Cauchy product:

    [tex]\begin{aligned}
    \sum_{n=0}^{\infty} a_n x^n \sum_{n=0}^{\infty} a_n x^n &=\sum_{n=0}^{\infty}\sum_{k=0}^n a_k a_{n-k} x^n \\
    &=\sum_{n=0}^{\infty} b_n x^n,\quad b_n=\sum_{k=0}^{n} a_k a_{n-k}
    \end{aligned}
    [/tex]

    Now, how would express:

    [tex]\sum_{n=0}^{\infty} a_n x^n \sum_{n=0}^{\infty} b_n x^n [/tex]
     
  9. Jun 14, 2010 #8
    So I am still having problems
    for m = 0 I get:
    [tex]
    a_{i} i x^{i-1} - (a_{i} x^i)^3 =0
    [/tex]
    which becomes
    [tex]
    a_{i+1} (i+1) x^{i} - (a_{i} x^i)^3 =0
    [/tex]
    so for i = 0
    [tex]
    a_1= a_0
    [/tex]
    for i =1
    [tex]
    a_2(2)x - (a_{1}^3 x^3) = 0
    [/tex]
    which is
    [tex]
    a_2(2) - (a_{1}^3 x^2) = 0
    [/tex]
    Is this right? having an x in the coefficent for a2?
     
  10. Jun 14, 2010 #9

    Mark44

    Staff: Mentor

    You should not be comparing terms with different powers of x.
     
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