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  1. #1
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    series solution...

    Find a series solution y(x) of the IVP

    y'' -xy' + 5y = 0, y(0) = 0, y'(0) = 1

    Show that for the given initial conditions the series terminates and hence the solution reduces to a fifth order polynomial.


    I seriously do not have a clue where to start on this problem! Can anyone please give me any help at all?!

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by hunkydory19 View Post
    Find a series solution y(x) of the IVP

    y'' -xy' + 5y = 0, y(0) = 0, y'(0) = 1

    Show that for the given initial conditions the series terminates and hence the solution reduces to a fifth order polynomial.


    I seriously do not have a clue where to start on this problem! Can anyone please give me any help at all?!

    Thanks in advance!
    Let y=\sum_{n=0}^{\infty}a_n x^n be a solution.

    Then, \sum_{n=2}^{\infty}n(n-1)a_n x^{n-2} - x\sum_{n=1}^{\infty} na_nx^{n-1} + 5\sum_{n=0}^{\infty} a_n x^n = 0

    Change index on the first, \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n - \sum_{n=1}^{\infty} na_n x^n + \sum_{n=0}^{\infty} 5a_n x^n=0

    Evaluate 1st and 3rd sums at n=0 and combine, 2a_2 + 5a_0 + \sum_{n=1}^{\infty} [(n+2)(n+1)a_{n+2} - (n-5)a_n]x^n = 0.

    This means,
    (i) 2a_2 + 5a_0 = 0
    (ii) (n+2)(n+1)a_{n+2} - (n-5)a_n = 0 for n\geq 1.

    Note that (i) is implied from (ii) if we let n=0.
    Thus, we get the relation, (n+2)(n+1)a_{n+2} - (n-5)a_n = 0 \implies a_{n+2} = \frac{(n-5)a_n}{(n+2)(n+1)} for n\geq 0.
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  3. #3
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    Thank you SO much, genius!!
    Last edited by ThePerfectHacker; January 21st 2008 at 01:10 PM.
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