# Social Diffusion (Integration)

• Sep 18th 2012, 08:32 AM
Preston019
Social Diffusion (Integration)
In a sufficiently large population the number x who have the information is treated as a differentiable function of time t The rate of diffusion, dx/dt, is assumed to be proportional to the number of people who have the information times the number of people who don't. This leads to the differential equation dx/dt = Kx(N-x), where N is the number of people in the population.

Suppose t is measured in days, K=1/25, and 4 people start a rumor a time t=0 in a population N=100 people.

a) Find x as a function of t by integrating both sides of the equation (1/(x(N-x))) dx = K dt
b) When will half the population have heard the rumor?
c) When will the rumor be spreading the fastest?

I'm stuck and don't know where to go.
• Sep 18th 2012, 11:45 AM
johnsomeone
Re: Social Diffusion (Integration)
Quote:

Originally Posted by Preston019
I'm stuck and don't know where to go.

Do you need help translating the word problem? Or is it the integral you don't know how to do? Or is it all just a blur?

Here's the set up, after translating the information in the word problem:

$\displaystyle x(t)$ = number of people hearing the rumor at time t in days.

$\displaystyle x(0) = 4$.

$\displaystyle dx/dt = (1/25)(x)(100-x) = 4x - x^2/25$.

a) FIND $\displaystyle x(t)$. Hint: $\displaystyle \int \frac{1}{x(100-x)} dx = \int (1/25) dt$.

b) FIND $\displaystyle t_0$ such that $\displaystyle x(t_0) = 50$.

c) FIND $\displaystyle t_1$ such that $\displaystyle \frac{dx}{dt}(t_1) = \max \{ \frac{dx}{dt}(t) | t \ge 0 \}$.