# Rumour problem (logistic differential equations)

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• 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 $N$ where $R$ members know the rumour and meet one of $(N - R)$ members who doesn't know the rumour.

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

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

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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:

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

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

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

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

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

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

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

$t = 0$
$R = 2$

$t = 1$
$R = 22$

$t = 4$
$R = 0.5N$

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 $\ln\left(\frac{R}{N-R}\right)=nkt+nc$

but even so, I end up with the equations:

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

or:

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

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