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Math Help - First-order nonlinear DE

  1. #1
    Junior Member Greg98's Avatar
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    First-order nonlinear DE

    Hello,
    I tried searching similar DE-problems, but I couldn't find one. The problem is following differential equation:
    e^{x}y'+xe^{-y}=0

    So, it seems to be first-order nonlinear DE. But, I don't know how to solve it. I couldn't separe it, though I managed to separate simpler equations, but I think that's the right way to go.

    Just got it to form,
    e^{-y}(e^{y+x}y'+x)=0,

    which obviously doesn't help anything...

    Any help is appreciated. Thank you!
    Last edited by Greg98; November 25th 2010 at 07:44 AM. Reason: Brrr... I made horrible typo (xe^-x). It's now correct (xe^-y).
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  2. #2
    MHF Contributor harish21's Avatar
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    y'=-\dfrac{xe^{-x}}{e^x}=-\dfrac{x}{e^{2x}}
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  3. #3
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    Quote Originally Posted by Greg98 View Post
    Hello,
    I tried searching similar DE-problems, but I couldn't find one. The problem is following differential equation:
    e^{x}y'+xe^{-x}=0

    So, it seems to be first-order nonlinear DE. But, I don't know how to solve it. I couldn't separe it, though I managed to separate simpler equations, but I think that's the right way to go.

    Just got it to form,
    e^{-y}(e^{y+x}y'+x)=0,

    which obviously doesn't help anything...

    Any help is appreciated. Thank you!
    Dear Greg98,

    e^{x}y'+xe^{-x}=0

    \frac{dy}{dx}=-xe^{-2x}

    y=-\int{xe^{-2x}}dx

    This integration could be done by integration by part method.....

    Hope this helps.
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  4. #4
    Junior Member Greg98's Avatar
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    Thanks for the help! Because of my omnipotent typing skills, I still need some advice (see original post). Sorry... I tried some logarithms to eliminate e^{-y}, but that didn't help.
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  5. #5
    MHF Contributor harish21's Avatar
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    e^{x}y'+xe^{-y}=0

    e^{x}y'+\dfrac{x}{e^{y}}=0

    divide both sides by e^x

    y'+\dfrac{x}{e^{x}\;e^{y}}=0

    multiply both sides by e^y

    y'\;e^y +\dfrac{x}{e^x}=0

    e^y \frac{dy}{dx}=-\dfrac{x}{e^x}

    e^y\;dy=-xe^{-x}\;dx

    finish...
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