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Math Help - Differential equation

  1. #1
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    Differential equation

    Hey guys i'm banging my head against the wall with this question.

    dy/dx = (e^(x+y) ) / y-1

    i've tried multiplying top & bottom by e^-y and seperating, but I cant seem to match the answer in the back of the book which is e^x + ye^-y = C

    The fact that the equation is equal to C gave me the idea that it might be exact, but when I tried the partial derivative test I couldn't get a single variable by itself (I might have screwed up there somewhere also)

    Thanks in advance
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  2. #2
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    Hello, ubique!

    You started off great . . .


    \frac{dy}{dx} \:=\:\frac{e^{x+y}}{y-1}

    i've tried multiplying top & bottom by e^{-y} and separating . Good!
    but I can't seem to match the answer in the book: . e^x + ye^{-y} \:=\: C

    We have: . \frac{dy}{dx} \:=\:\frac{e^x\cdot e^y}{y-1}

    Separate variables: . \frac{y-1}{e^y}\,dy \;=\;e^x\,dx .[1]


    On the left, we can integrate by parts:

    . . \begin{array}{ccccccc}u &=& y-1 & & dv &=& e^{-y}\,dy \\ du &=& dy & & v &=& \text{-}e^{-y} \end{array}

    And we have: . -e^{-y}(y-1) + \int e^{-y}\,dy \;=\;-e^{-y}(y-1) - e^{-y}+C_1\;=\;-ye^{-y} +C_1


    Then [1] becomes: . -ye^{-y} +C_1\;=\;e^x+ C_2 \quad\Rightarrow\quad\boxed{ e^x + ye^{-y} \:=\:C}

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  3. #3
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    Thanks a lot Soroban
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