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Math Help - Another differential equation

  1. #1
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    Another differential equation

    Hi,

    I'm solving this exercise, but i can not continue it, please, could you show me what to do with log.
    thank you in advance.


    y(0) = 0
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by JrShohin View Post
    Hi,

    I'm solving this exercise, but i can not continue it, please, could you show me what to do with log.
    thank you in advance.


    y(0) = 0
    I would leave it as \frac{1}{4}\log\left|4y+1\right|=\frac{x^2}{2}+C

    Now, multiply both sides by 4 and then get rid of the log:

    \log|4y+1|=2x^2+C\implies e^{\log|4y+1|}=e^{2x^2+C}\implies 4y+1=Ce^{2x^2}

    Now solve for y. Then apply the initial condition.

    Can you take it from here?

    --Chris
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  3. #3
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    Grazie, i solved it.
    But i have a question, (minus) which near 1/4 doesn't effect to the solution? You just left it... sorry for a crazy question.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by JrShohin View Post
    Grazie, i solved it.
    But i have a question, (minus) which near 1/4 doesn't effect to the solution? You just left it... sorry for a crazy question.
    Note that -\frac{1}{4}\log\left|\frac{1}{4y+1}\right|=\frac{1  }{4}\log|4y+1|

    So I just used \frac{1}{4}\log|4y+1|, rather than -\frac{1}{4}\log\left|\frac{1}{4y+1}\right|

    --Chris
    Last edited by Chris L T521; October 28th 2008 at 08:46 AM.
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