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Math Help - find the solution...

  1. #1
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    find the solution...

    ok im not sure how to start this problem....

    (1-x^{2})y''-2xy'+90y=0

    the term in front of the y'' is messing me up...any thoughts? anybody? thanks in advance
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    ok im not sure how to start this problem....

    (1-x^{2})y''-2xy'+90y=0

    the term in front of the y'' is messing me up...any thoughts? anybody? thanks in advance
    This is the Legendre differential equation. Use a power series solution.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    This is the Legendre differential equation. Use a power series solution.

    -Dan
    ok i found the recursion formula, can you check it, to make sure its right?


    a_{n+2}=a_{n}\frac{(n-9)(n+10)}{(n+2)(n+1)}

    i have the work for it, if you want to see it, just let me know...thanks in advance.
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  4. #4
    Forum Admin topsquark's Avatar
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    I'm not getting your recursion relation. Let me briefly plot out my own work. I am putting l(l + 1) = 90.

    \displaystyle y = \sum_{n = 0}^{\infty}a_n x^n

    \displaystyle y' = \sum_{n = 0}^{\infty}n a_n x^{n - 1}

    \displaystyle y'' = \sum_{n = 0}^{\infty}n(n - 1) a_n x^{n - 2}

    So
    \displaystyle \sum_{n = 0}^{\infty}n(n - 1) a_n x^{n - 2} - \sum_{n = 0}^{\infty}n(n - 1) a_n x^n - \sum_{n = 0}^{\infty}2n a_n x^n + \sum_{n = 0}^{\infty}l(l + 1) a_n x^n = 0

    Change the dummy variable in the first term:
    \displaystyle \sum_{m = -2}^{\infty}(m + 2)(m + 1) a_{m + 2} x^m - \sum_{n = 0}^{\infty}n(n - 1) a_n x^n - \sum_{n = 0}^{\infty}2n a_n x^n + \sum_{n = 0}^{\infty}l(l + 1) a_n x^n = 0

    (Note that the first two terms of the first sum are 0. So relabeling...)
    \displaystyle \sum_{n = 0}^{\infty}(n + 2)(n + 1) a_{n + 2} x^n - \sum_{n = 0}^{\infty}n(n - 1) a_n x^n - \sum_{n = 0}^{\infty}2n a_n x^n + \sum_{n = 0}^{\infty} l (l + 1) a_n x^n = 0


    Or (Thank God for cut and paste!)
    \displaystyle \sum_{n = 0}^{\infty} \left [ (n + 2)(n + 1) a_{n + 2} + \left \{ -n(n - 1) - 2n + l(l + 1) \right \} a_n \right ] x^n = 0

    Giving
    \displaystyle a_{n + 2} = \frac{n(n + 1) - l (l + 1)}{(n + 2)(n + 1)}a_n

    -Dan
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  5. #5
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    ok i was kinda close..im gonna go back over my work and try to see where i went wrong....
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  6. #6
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    find the solution revisited....

    ok so in this problem i think i am getting it, i just need help simplifying the problem down..

    here is the problem..



    this is what i got so far....




    i need to get this problem down to where i can solve for

    [IMG]a_{n+2}[/IMG]

    any help would be greatly appreciated...
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  7. #7
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    ok so in this problem i think i am getting it, i just need help simplifying the problem down..

    here is the problem..



    this is what i got so far....




    i need to get this problem down to where i can solve for

    [IMG]a_{n+2}[/IMG]

    any help would be greatly appreciated...
    Let's suppose to search a solution in the form...

    (1)

    ... and to know the 'initial conditions' y(0)=a_{0} and y'(0)=a_{1}... the others a_{n} are found from the DE as follows...





    ... and so one...

    Please revise my computation [and also try to procede...] because an 'old wolf' like me doesn't excell in computation ...

    Kind regards

    \chi \sigma
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    ok so in this problem i think i am getting it, i just need help simplifying the problem down..

    here is the problem..



    this is what i got so far....




    i need to get this problem down to where i can solve for

    [IMG]a_{n+2}[/IMG]

    any help would be greatly appreciated...
    Is there something wrong with your original thread?

    Please do not double post. If you have more questions about a problem then post them in the original thread.

    -Dan

    [MODERATOR EDIT]: I have merged the threads.
    Last edited by Ackbeet; April 18th 2011 at 05:14 AM. Reason: Merged threads.
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