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Math Help - Solve the following ODE .. #9

  1. #1
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    Solve the following ODE .. #9

    Hello

    I've this equation using inspection :

    \displaystyle \dfrac{dy}{dx} = 1 + e^{y+e^x}


    ok
    I couldn't solve it using any method !!
    how in earth can i solve this one using inspection !!!!
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  2. #2
    A Plied Mathematician
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    Well, I don't know about inspection, but the substitution u=y-x renders the equation separable:

    u'=y'-1, and hence the DE becomes

    u'=e^{u+x+e^{x}}=e^{u}e^{x+e^{x}}.
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  3. #3
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    Thanks !!
    it is really hard
    thanks you man
    But am really intrested for the "inspection solution" for it since it is in the inspection section's exercises
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  4. #4
    Super Member General's Avatar
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    Hello,

    Here is what do you want:

    We have \dfrac{dy}{dx}=1+e^y \cdot e^{e^x}

    \implies dy=dx+e^y \cdot e^{e^x} \, dx

    \implies e^x \, dy = e^x dx + e^x e^y e^{e^x} \, dx

    \implies \dfrac{e^x}{e^y} \, dy = \dfrac{e^x}{e^y} \, dx + e^x e^{e^x} \, dx

    \implies e^{x-y} \, dy - e^{x-y} dx = e^x e^{e^x} \, dx

    \implies e^{x-y} \, ( dy - dx ) = e^x e^{e^x} \, dx

    \implies e^{x-y} \, d(y-x) = e^x e^{e^x} \, dx

    \implies e^{-(y-x)} \, d(y-x) = e^x e^{e^x} \, dx ... (1)

    Your equation is ready for integrating now.
    Again, equation (1) is equivalent to the last equation in Ackbeet's reply.

    Another way ,if you are intersted , :

    We have \dfrac{dy}{dx}=1+e^y \cdot e^{e^x} ... (2)

    Multiplying (2) by e^{-y} and substituting t=e^{-y} will give you a linear equation.
    Last edited by General; February 12th 2011 at 08:47 AM. Reason: Replacing (1) by (2) in the last line.
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  5. #5
    A Plied Mathematician
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    Quote Originally Posted by Liverpool View Post
    Thanks !!
    it is really hard
    thanks you man
    But am really intrested for the "inspection solution" for it since it is in the inspection section's exercises
    You're quite welcome. I'm not smart enough to "see" a "by inspection solution" for this DE. In looking at the results I get using my method, the solution is much too complicated for me to have "seen it".
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  6. #6
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    Thanks all.
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