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Math Help - Solve dU/dT = 1(1-U) using seperation of variables.

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    Solve dU/dT = 1(1-U) using seperation of variables.

    Hi all, just a problem I'm having trouble with - thanks

    1) consider the differential equation dU/dT = 1(1-U) for t greater than or equal to 0 to infinity.

    Using the method of separation of variables show that the general solution of the differential equation is

    U(t) = e^t / (A + e^t)
    where A is an arbitrary constant.

    My attempt: I trouble the 1(1-U) over with the dU and using partial fractions solved for log |U| + log |1-U| C(1) = t + C(2)

    This is as far as i got, please help!
    Last edited by mr fantastic; November 17th 2011 at 06:00 PM. Reason: Re-titled.
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  2. #2
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    Re: Differential Equations

    Quote Originally Posted by andrew2322 View Post
    Hi all, just a problem I'm having trouble with - thanks

    1) consider the differential equation dU/dT = 1(1-U) for t greater than or equal to 0 to infinity. ?

    Using the method of separation of variables show that the general solution of the differential equation is

    U(t) = e^t / (A + e^t)
    where A is an arbitrary constant.

    My attempt: I trouble the 1(1-U) over with the dU and using partial fractions solved for log |U| + log |1-U| C(1) = t + C(2)

    This is as far as i got, please help!
    I assume you mean ...

    \frac{du}{dt} = u(1-u)

    \frac{du}{u(1-u)} = dt

    \int \left(\frac{1}{u} + \frac{1}{1-u}\right) \, du = \int dt

    \ln|u| - \ln|1-u| = t + C

    \ln\left|\frac{u}{1-u}\right| = t+C

    \frac{u}{1-u} = e^{t+C}

    take it from here?
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  3. #3
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    Re: Differential Equations

    that's exactly where I got stuck, I know it's simple algebra but ive been up all night working for my exam and my brain is just a fuzz right now

    could you please point it out
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    Re: Differential Equations

    \frac{u}{1-u} = e^{t+C} = e^t \cdot e^C = Be^t , where B = e^C

    u = (1-u)Be^t

    u = Be^t - uBe^t

    u + uBe^t = Be^t

    u(1 + Be^t) = Be^t

    u = \frac{Be^t}{1 + Be^t}

    let the constant A = \frac{1}{B}

    u = \frac{Be^t}{1 + Be^t} \cdot \frac{A}{A}

    u = \frac{e^t}{A + e^t}
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