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Math Help - Second order diff. equation. Series solution

  1. #1
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    Second order diff. equation. Series solution

    Hi there. I have to solve: y''+xy'-2y=e^x
    I thought I should use a series solution, but I'm not sure if what I'm doing is ok. So, I've considered

    y(x)=\sum_0^{\infty}a_n x^n

    y'(x)=\sum_1^{\infty}n a_n x^{n-1}

    y''(x)=\sum_2^{\infty}n(n-1) a_n x^{n-1}

    And e^x=\sum_0^{\infty}\frac{x^n}{n!}

    Then, using that m=n-2 for y'' and then replacing in the diff. eq:
    \sum_0^{\infty}(n+2)(n+1)a_{n+2} x^n+x\sum_1^{\infty}n a_n x^{n-1}-2\sum_0^{\infty}a_n x^n=\sum_0^{\infty}\frac{x^n}{n!}

    So:
    2a_2-2a_0-1+\sum_1^{\infty}\left [(n+2)(n+1)a_{n+2}+n a_n -2a_n -\frac{1}{n!} \right ] x^n=0

    Then: 2a_2-2a_0-1=0\rightarrow a_2=1/2+a_0

    And: (n+2)(n+1)a_{n+2}+n a_n -2a_n -\frac{1}{n!}=0\rightarrow a_{n+2}=a_n\frac{(2-n)}{(n+2)(n+2)}

    Now I have to find the recurrence for a_n
    Is this ok?
    Last edited by Ulysses; March 12th 2012 at 05:40 PM.
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  2. #2
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    Re: Second order diff. equation. Series solution

    I think I should solve the homogeneus equation first.
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  3. #3
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    Re: Second order diff. equation. Series solution

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  4. #4
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    Re: Second order diff. equation. Series solution

    Thanks, but I have to solve it using series.
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