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

• Nov 17th 2011, 04:40 PM
andrew2322
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)

• Nov 17th 2011, 04:53 PM
skeeter
Re: Differential Equations
Quote:

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)

I assume you mean ...

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

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

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

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

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

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

take it from here?
• Nov 17th 2011, 05:07 PM
andrew2322
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
• Nov 17th 2011, 05:33 PM
skeeter
Re: Differential Equations
$\displaystyle \frac{u}{1-u} = e^{t+C} = e^t \cdot e^C = Be^t$ , where $\displaystyle B = e^C$

$\displaystyle u = (1-u)Be^t$

$\displaystyle u = Be^t - uBe^t$

$\displaystyle u + uBe^t = Be^t$

$\displaystyle u(1 + Be^t) = Be^t$

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

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

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

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