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Math Help - Help solving y'=e^(x-y)

  1. #1
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    Help solving y'=e^(x-y)

    Prompt: y'=e^(x-y); y(0)=1

    I understand what to do for the particular solution but am having trouble finding the general equation.
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  2. #2
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    Re: Help solving y'=e^(x-y)

    Quote Originally Posted by Lamont View Post
    Prompt: y'=e^(x-y); y(0)=1

    I understand what to do for the particular solution but am having trouble finding the general equation.
    The equation is seperable!

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

    Now just plug in your point to get

    e=e^{0}+C \implies C=e-1

    This gives

    e^y=e^x+e-1 \implies y=\ln(e^x+e-1)
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  3. #3
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    Re: Help solving y'=e^(x-y)

    Thanks!
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