# Rumour problem (logistic differential equations)

• Oct 14th 2008, 11:10 PM
sqleung
Rumour problem (logistic differential equations) - somewhat urgent
Hello. I'm currently stuck on a problem that I can't seem to work out. It's quite complex and difficult (for me anyway) and it would be great if you could help me out. Below is the background required:

Consider a population $\displaystyle N$ where $\displaystyle R$ members know the rumour and meet one of $\displaystyle (N - R)$ members who doesn't know the rumour.

The rate of change in the population who know the rumour is hence, given by:

$\displaystyle \frac{dR}{dt} = k(N - R)R$ where $\displaystyle k$ is some constant.

---------------------------

Here's the question I'm stuck with:

A town has N residents. At 8 am, a rumour begins with 2 people (spread by a logistic growth). At 9 am, 22 know the rumour. At 12 pm, half the population know the rumour. Find the population and k.

First, I transformed it into a differential equation and proceeded to try and solve it:

$\displaystyle \frac{dR}{dt} = k(N - R)R$

$\displaystyle \int \! \frac{1}{R(N - R)} \, dR$ = $\displaystyle \int \! k \, dt$

$\displaystyle \int \! \frac{1}{N}\left (\frac{1}{R} + \frac{1}{N - R}\right ) \, dR$ = $\displaystyle \int \! k \, dt$

$\displaystyle \frac{1}{N}\int \! \left (\frac{1}{R} + \frac{1}{N - R}\right ) \, dR$ = $\displaystyle \int \! k \, dt$

$\displaystyle \frac{1}{N}(ln |R| - ln |N - R|)$ = $\displaystyle kt + c$

$\displaystyle ln \left |\frac{R}{N - R}\right |$ = $\displaystyle Nkt + c$

Now I'm stuck...It sounds as easy as plugging in the values where:

$\displaystyle t = 0$
$\displaystyle R = 2$

$\displaystyle t = 1$
$\displaystyle R = 22$

$\displaystyle t = 4$
$\displaystyle R = 0.5N$

And solve it simultaneously. However, I can't seem to get it (did I do something wrong?)

If you could assist me from this point on, it would definitely be appreciated.

The answers given by the textbook is:

k = 0.00008169
Popluation = 29360

Thanks.
• Oct 15th 2008, 04:29 AM
shawsend
I think it should be $\displaystyle \ln\left(\frac{R}{N-R}\right)=nkt+nc$

but even so, I end up with the equations:

$\displaystyle c=1/N\ln\left(\frac{2}{N-2}\right)$
$\displaystyle k+c=1/N\ln\left(\frac{22}{N-22}\right)$
$\displaystyle c=-4k$

or:

$\displaystyle -4k=1/N\ln\left(\frac{2}{N-2}\right)$
$\displaystyle -3k=1/N\ln\left(\frac{22}{N-22}\right)$

which looks to me to be best solved numerically. Maybe another way though.