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Thread: Just need help simplifying

  1. #1
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    Just need help simplifying

    This is for a reduction of order problem. I won't post the whole problem because I only need help simplifying this..

    $\displaystyle (x-1)(e^{x}v''+e^{x}2v'+ve^{x})-x(e^{x}v'+ve^{x})+ve^{x}=0$

    So what the heck does this simplify out to because I don't think I'm doing it right
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  2. #2
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    Hi Steph, expand it all out and group together in terms of $\displaystyle v, v'$ and $\displaystyle v''$
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  3. #3
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    That's what I'm doing but I don't think I am doing it correctly
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  4. #4
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    I can get it down to a point where I get $\displaystyle xv''e^{x}+2xv'e^{x}-v''e^{x}-2v'e^{x}-xv'e^{x}=0$

    After this I don't know how to simplify it anymore..help please?
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  5. #5
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    Quote Originally Posted by steph3824 View Post
    I can get it down to a point where I get $\displaystyle xv''e^{x}+2xv'e^{x}-v''e^{x}-2v'e^{x}-xv'e^{x}=0$
    This is not correct as your $\displaystyle v$ term is no longer present.

    Have another go at expanding there should be 9 terms. You can then group by $\displaystyle v, v'$ and $\displaystyle v"$

    This is how you do it with your previous answer

    $\displaystyle xv''e^{x}+2xv'e^{x}-v''e^{x}-2v'e^{x}-xv'e^{x}=0$

    $\displaystyle xv''e^{x}-v''e^{x}+2xv'e^{x}-2v'e^{x}-xv'e^{x}=0$

    $\displaystyle (xe^{x}-e^{x})v"+(2xe^{x}-2e^{x}-xe^{x})v'=0$

    $\displaystyle (xe^{x}-e^{x})v"+(xe^{x}-2e^{x})v'=0$
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