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Math Help - Power series method problem

  1. #1
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    Power series method problem

    Hi,

    Can you please check this for me?

    Thanks,
    James

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  2. #2
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    Quote Originally Posted by james121515 View Post
    Hi,

    Can you please check this for me?

    Thanks,
    James

    Dear James,

    You have done most of it correctly, except in writing the final answer. Since you have taken the trial solution to be y(x)=\sum_{n=0}^{\infty}c_{n}x^{n}

    The general solution would be,

    y(x)=\sum_{n=0}^{\infty}c_{n}x^{n}=\sum^{\infty}_{  k=0}c_{2k}x^{2k}+\sum_{k=0}^{\infty}c_{2k+1}x^{2k+  1}

    You have found out that, c_{2k}=\frac{(-1)^k c_{0}}{(2k)!} and c_{2k+1}=\frac{(-1)^{k}(c_{1}-1)}{(2k+1)!} for k\geq{2}

    (Note: You have written the c_{2k+1} term incorrectly)

    Therefore, y(x)=c_{0}+c_{2}x^{2}+\sum_{k=2}^{\infty}c_{2k}x^{  2k}+c_{1}x+c_{3}x^{3}+\sum_{k=2}^{\infty}c_{2k+1}x  ^{2k+1}

    y(x)=c_{0}+\left(\frac{-c_{0}}{2}\right)x^{2}+\sum_{k=2}^{\infty}\frac{(-1)^k c_{0}}{(2k)!}x^{2k}+c_{1}x+\left(\frac{-(c_{1}-1)}{6}\right)x^{3}+\sum_{k=2}^{\infty}\frac{(-1)^{k}(c_{1}-1)}{(2k+1)!}x^{2k+1}

    Hope this will help you.
    Last edited by Sudharaka; May 12th 2011 at 06:07 PM.
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