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Thread: ODE

  1. #1
    Member Miss's Avatar
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    ODE

    hello
    I have big problem with this one:

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

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

    multiply both sides by $\displaystyle e^{-y}$

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

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

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

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

    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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