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

**andrew2322** · November 17th 2011, 04:40 PM

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!

**skeeter** · November 17th 2011, 04:53 PM

Originally Posted by andrew2322

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?

**andrew2322** · November 17th 2011, 05:07 PM

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

**skeeter** · November 17th 2011, 05:33 PM

$\frac{u}{1-u} = e^{t+C} = e^t \cdot e^C = Be^t$ , 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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Solve dU/dT = 1(1-U) using seperation of variables.

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