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Math Help - ODE

  1. #1
    Member Miss's Avatar
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    ODE

    hello
    I have big problem with this one:

    \displaystyle{ \frac{dy}{dx}=1+e^\left( e^x + y \right) }
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  2. #2
    MHF Contributor
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    Mathematica says the solution is

    y = x - \ln{\left(-C - e^{e^x}\right)}.
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  3. #3
    Super Member General's Avatar
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    \frac{dy}{dx}=1+e^{e^x} \cdot e^y

    multiply both sides by e^{-y}

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

    Substitute u=e^{-y}, to get:

    \frac{du}{dx}=-u-e^{e^x}

    \frac{du}{dx}+u=-e^{e^x}

    Find your integrating factor and continue . . .
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  4. #4
    Member Miss's Avatar
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    Thanks..
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