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Math Help - 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:

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

    Now:

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

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

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

    -e^{-y} = e^x + C

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

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

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

    But the answer in the book is shown as:

    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 -e^{-y} = e^x + C , multiply both sides through by 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
    ln(e^{-y^{-1})} = ln(e^x + C)

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

    Thank you
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