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Power Series to solve Second order Differential Equations

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data

    When solving a D.E. with power series, I've encountered something along the lines of:

    [itex](2 - r)^{2}g'' = -2[/itex]

    2. Relevant equations

    Power Series

    3. The attempt at a solution

    I know I am just supposed to assume such a series exists, and work from there. But I'm really getting fudged up when it comes to factoring the r's back into the series, which makes the r's in the summation 2 powers higher than just 'n'.

    I've reached the end where I have a recursive definition of an+1 in terms of an, but there are ridiculous fractions on the coefficients that I don't think I can generalise for values of n. They don't seem to have a pattern, and that's..a problem.

    Is there an easier way to solve this, leaving the r-polynomial as it is? I expanded it and applied g'' to each term, then subbed in the second derivative of the series that represents g(x) [the standard assumed series, anrn]
     
  2. jcsd
  3. Nov 4, 2012 #2

    haruspex

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    This is just g = g(r), right?
    Do you have to use power series? Looks more straightforward to get it into the form g'' = ... and integrate twice.
    If you wish to persevere with power series, please post your working.
     
  4. Nov 4, 2012 #3
    It's an assignment, I'm much more than capable of solving this with more traditional means, but I can't submit that work.

    I don't have my notebook at the moment but from memory, when using power series, you assume

    [itex]y = \sum_{n=0}^{\infty} a_{n}x^{n}[/itex]

    Which means

    [itex]y' = \sum_{n=1}^{\infty} na_{n}x^{n-1}[/itex]
    [itex]y'' = \sum_{n=2}^{\infty} n(n-1)a_{n}x^{n-2}[/itex]

    And particularly for the second derivative, re-write the sum so the power of x is the same as for y

    [itex]y'' = \sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}[/itex]

    Now, after putting the sum forms of y and its derivatives back into the equation, I think I fiddled around with it by expanding whatever quantity was multiplied with y'', and replaced all occurrences of y and its derivatives with the appropriate sums.

    I guess you could re-write the r's as x's since I wrote x's in the sums instead of r's, so you'd need to factor that variable into the sums when appropriate, which messes with the powers of x again, even though I just fixed them before.

    This also leads to problematic situations where different sums are -beginning- at different indices of n. I know I'm supposed to pop out the first few terms til the sum starts at the same index as the others, but I'm unsure about how to do this procedure.

    I know the end goal is to get all sums starting at the same index, and all x's in the sums to have the same power, so I can factor out the x's and have one large sum of coefficients.

    My problem is, I don't exactly know how to shift indices correctly (because of the a's within the sums) and I don't really know how to take out the first term or two to get the starting index to match with the other sums. I also don't know what I'm supposed to actually do with those coefficients once I get them out of their mother sums.
     
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