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    Given that y=1 and x=0, express y in terms of x if dx/dy=e^x/e^y
    The answer I get is y=-ln(e^-x+e^1,-1) however the answer in my book says y=x+1 -ln(e^-x+e^1.-1).
    My question is where does the x+1 come from?
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    I get the same answer as you.
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    (Original post by DFranklin)
    I get the same answer as you.
    Surely it must be a mistake in the book?
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    (Original post by gooner1010)
    Surely it must be a mistake in the book?
    I reckon. (I'm also not 100% sure I do get the same answer as you - what exactly y=-ln(e^-x+e^1,-1) is supposed to mean (specifically e^1,-1) is unclear).

    FWIW, I'll post my working:

    \dfrac{dx}{dy} =  e^x / e^y

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

    \implies e^{-x} + C = e^{-y}. When x = 0, y = 1, so 1 + C = 1/e, so C = 1/e -1

    take logs: \ln(e^{-x} + \frac{1}{e} - 1) = -y, or y = -\ln(e^{-x} + \frac{1}{e} -1)
 
 
 
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Updated: February 12, 2017
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