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Math Help - ODE

  1. #1
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    ODE



    Does anyone know how to find the general solution for the above ODE?

    I've tried an integrating factor and I can't seem to get anywhere with it. Any pointers would be much appreciated!
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  2. #2
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    Quote Originally Posted by cdlegendary View Post


    Does anyone know how to find the general solution for the above ODE?

    I've tried an integrating factor and I can't seem to get anywhere with it. Any pointers would be much appreciated!
    The equation is first order linear and seperable

    (1+e^{t})\frac{dy}{dt}=-e^{t}y \iff \frac{dy}{y}=-\frac{e^{t}}{1+e^{t}}dt

    Now just integrate...
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  3. #3
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    ln(y) = \int -e^t/(1+e^t) dt

    ln(y) = -ln(1+e^t) + c

    y = -e^{1+e^t} * e^c

    y = -Ce^{1+e^t}

    Alright, so I've integrated the equation and solved for y. It doesn't seem to be correct though. Is there any error in my calculations?
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  4. #4
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    Quote Originally Posted by cdlegendary View Post
    ln(y) = \int -e^t/(1+e^t) dt

    ln(y) = -ln(1+e^t) + c

    y = -e^{1+e^t} * e^c

    y = -Ce^{1+e^t}

    Alright, so I've integrated the equation and solved for y. It doesn't seem to be correct though. Is there any error in my calculations?
    You have made an algebra error

    This is correct but to simplify calculations add \ln(c)=c
    for the arbitary constant
    ln(y) = -ln(1+e^t) + \ln(c) \iff \ln(y)=\ln\left( \frac{1}{1+e^{t}}\right)+\ln(c)=\ln\left( \frac{c}{1+e^{t}} \right)

    y=\frac{c}{1+e^{t}}
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