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Math Help - Seperation of Variables

  1. #1
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    Seperation of Variables

    Hello, I was wondering if you guys could quickly check over my work to make sure I did this question correctly. We're solving differential equations using the seperation of variables:
    y(0)=e
    dy/dt = e^(t-y)
    ln(dy/dt)=t-y
    y+ln(dy)=t+ln(dt)
    e^(y+ln(dy))=e^(t+ln(dt))
    e^y(dy)=e^t(dt)
    (Take integrals of both sides)
    e^y=(e^t)+C
    ln(e^y)=ln((e^t)+C)
    y=ln(e^t+C)
    y(0)=e
    e=ln(e^0+C)
    e=ln(1+C)
    e^e=1+C
    C=e^e-1
    y=ln(e^t+e^e-1)

    I'm not sure if this is right, I wasn't sure how to isolate dy/dt so I put both sides to the power of e, I was just wondering if that was the correct thing to do. All help is appreciated!
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  2. #2
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    I didnt check your answer but I would say,

    \displaystyle \frac{dy}{dt} = e^{t-y}

    \displaystyle \frac{dy}{dt} = \frac{e^t}{e^y}

    \displaystyle e^y~dy = e^t~dt

    \displaystyle \int e^y~dy = \int e^t~dt
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  3. #3
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    Quote Originally Posted by Dudealadude View Post
    Hello, I was wondering if you guys could quickly check over my work to make sure I did this question correctly. We're solving differential equations using the seperation of variables:

    e^y(dy)=e^t(dt)
    (Take integrals of both sides)
    e^y=(e^t)+C
    ln(e^y)=ln((e^t)+C)
    y=ln(e^t+C)
    y(0)=e
    e=ln(e^0+C)
    e=ln(1+C)
    e^e=1+C
    C=e^e-1
    y=ln(e^t+e^e-1)

    I'm not sure if this is right, I wasn't sure how to isolate dy/dt so I put both sides to the power of e, I was just wondering if that was the correct thing to do. All help is appreciated!
    Correct.
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