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Math Help - xy''+(1-x)y'+y=0

  1. #1
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    Jun 2005
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    Zagreb, Croatia
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    xy''+(1-x)y'+y=0

    I'm supposed to solve differential equation:

    xy''(x)+(1-x)y'(x)+y(x)=0

    by Frobenius method.
    The first solution is y1=c(1-x), where c is a constant. I am not sure if the second solution is correct. I got:
    y2=c(1-x)ln(x)+c(3x-\frac{x^2}{4}-\frac{x^3}{36}-\frac{2x^4}{576}-...)
    or if I put it in the sum:
    y2=c(1-x)ln(x)+c(3x-\sum_{n=0}^{\infty}\frac{n!}{((n+2)!)^2}x^{(n+2)})

    So, please if someone could check and tell me if this is OK. If not, I would appreciate if you could write the procedure how you got the second solution different from mine.
    Last edited by Ene Dene; September 6th 2005 at 06:18 AM.
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  2. #2
    Junior Member
    Joined
    Jun 2005
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    Zagreb, Croatia
    Posts
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    I know the solution now... so no need to answer now, this one is correct.
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