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Thread: Solving dy/dx = e^(x+y)

  1. #1
    M.R
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    Solving dy/dx = e^(x+y)

    Hi,

    I am trying to solve the following differential equation:

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

    Now:

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

    $\displaystyle \frac {1}{e^y} dy = e^x dx$

    $\displaystyle \int \frac {1}{e^y} dy = \int e^x dx$

    $\displaystyle -e^{-y} = e^x + C$

    $\displaystyle ln(e^{-y^{-1})} = ln(e^x + C)$

    $\displaystyle \frac{-1}{y} = x + ln(C)$

    $\displaystyle y = \frac {1}{-x - ln(C)}$

    But the answer in the book is shown as:

    $\displaystyle e^{x+y}+Ce^y+1=0$

    Where am I going wrong?
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  2. #2
    Master Of Puppets
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    Re: Differential equation

    Maybe you are correct, at this step $\displaystyle -e^{-y} = e^x + C$ , multiply both sides through by $\displaystyle e^{y} $

    What do you get?
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  3. #3
    Super Member ILikeSerena's Avatar
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    Re: Differential equation

    Quote Originally Posted by M.R View Post
    $\displaystyle ln(e^{-y^{-1})} = ln(e^x + C)$

    $\displaystyle \frac{-1}{y} = x + ln(C)$
    This is not right.
    ln(ab) = ln(a) + ln(b)
    ln(a + b) ≠ ln(a) + ln(b)


    Quote Originally Posted by M.R View Post
    $\displaystyle -e^{-y} = e^x + C$
    You already had this.
    What do you get if you multiply left and right with $\displaystyle e^y$?
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  4. #4
    M.R
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    Re: Differential equation

    Thank you
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