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Thread: Exact LDE factorizing

  1. #1
    Junior Member
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    Exact LDE factorizing

    Hi all,
    I'm currently working on a question that I can't fully work out. Can someone look over my working and see where I may have possibly gone wrong and maybe point me in the right direction?
    The question is,
    Given a LDE $\displaystyle F=b_{0}y''+b_{1}y'+b_{2}y=0$ and the relationship $\displaystyle b_{0}''-b_{1}'+b_{2}=0$, show that the factorization of the LDE is $\displaystyle F=D[b_{0}D+(b_{1}-b_{0}')]y$.

    From the relationship, I can use $\displaystyle b_{2}=b_{1}'-b_{0}''$ and I can factorize that into $\displaystyle b_{2}=D[b_{1}-b_{0}']$.
    Factorizing: F = $\displaystyle [b_{0}D^2+ b_{1}D+b_{0}]y$

    Therefore, $\displaystyle F= D[b_{0}D+ b_{1}+(b_{1}-b_{0})]y$.
    I have that $\displaystyle b_{1}$ that I'm not sure how to get rid of.
    Thank you for your time.
    Last edited by Silverflow; Mar 28th 2010 at 01:13 AM. Reason: Mispelling
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  2. #2
    Super Member
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    Expand $\displaystyle F$:

    $\displaystyle F=D[b_0 D+(b_1-b_0^{'})]y$

    $\displaystyle F=D[b_0y'+(b_1-b_0^{'})y]$

    $\displaystyle F=b_0 y''+y'b_0^{'}+(b_1-b_0^{'})y'+y(b_1^{'}-b_0^{''})$

    now let $\displaystyle b_2=b_1^{'}-b_0^{''}$
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  3. #3
    Junior Member
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    Thank you, I never thought about working backwards.
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