# Differential equation word problem

• Nov 3rd 2010, 06:10 AM
SyNtHeSiS
Differential equation word problem
There are 500 students enrolled for a certain course. A rumour starts to spread
among these students that the next class test has been cancelled. Suppose that the rate at which this rumour spreads is proportional to the product of the fraction of the students in the course who have heard the rumour and the fraction of students in the course who have not yet heard it. At 8:00 am, 40 students have heard the rumour, and by 12:00 noon half the students have heard it. By what time would 90% of the students have heard the rumour?

I checked the answer and I didnt understand why the derivative used was:

$\displaystyle \frac{dp}{dt} = kp(1 - p)$

p = % of students who have heard the rumour
t = time at which these % of students hear the rumour

Which part on the right-hand side of this equation refers to:

*product of the fraction of the students in the course who have heard the rumour?
*fraction of students in the course who have not yet heard it?
• Nov 3rd 2010, 08:04 AM
Ackbeet
Technically, it could be either! However, most likely, p refers to those students who have heard the rumor. In this case, k will be positive. In the other case, k would turn out negative. In either case, you'll have 0 < p < 1. Does that answer your question?
• Nov 3rd 2010, 03:07 PM
SyNtHeSiS
Im still a bit confused. It makes a bit more sense to me if the equation was:

$\displaystyle \frac{dp}{dt} = kp$

since as the rate increases, so does the % of students that have heard the rumour.
• Nov 3rd 2010, 03:32 PM
Ackbeet
But then the rate would only depend on the percentage of people who had heard it.
• Nov 6th 2010, 12:06 AM
SyNtHeSiS
Why is it that the rate would also depend on the % of students who havent heard the rumour? I mean the aim is to find the % of people who have heard the rumour, so why would you need to include it (i.e. multiply it to % who have heard the rumour).
• Nov 6th 2010, 01:14 AM
Ackbeet
Well, think of it this way: you have a finite group of people who are subject to hearing the rumor. It seems reasonable to suppose that as p grows and gets closer to the total student population, the rate of growth would slow down (you'll have "loner" students who don't interact so much with the other students). Finally, once every student knows the rumor, you'd better have a zero growth rate! If you leave out the 1-p term, then the rumor will keep growing exponentially even after every student knows the rumor! That's hardly realistic.

Make sense?
• Nov 6th 2010, 09:23 AM
SyNtHeSiS
Thanks for the good explanation, it makes sense now.

I got a function:

$\displaystyle \frac{p}{p -1} = Ae^{kt}$ and subbed in the initial condition p(0) = 8% to get:

$\displaystyle \frac{0.8}{0.8 - 1} = A$
$\displaystyle A = -4$
but my memo said A = $\displaystyle \frac{2}{23}$. I cant see anything wrong with what I did though.
• Nov 6th 2010, 12:05 PM
skeeter
Quote:

Originally Posted by SyNtHeSiS
I got a function:

$\displaystyle \frac{p}{p -1} = Ae^{kt}$

how did you get that function from the DE

$\displaystyle \displaystyle \frac{dp}{dt} = kp(1-p)$ ?
• Nov 9th 2010, 03:17 AM
SyNtHeSiS
Quote:

Originally Posted by skeeter
how did you get that function from the DE

$\displaystyle \displaystyle \frac{dp}{dt} = kp(1-p)$ ?

Here is my working:

$\displaystyle \int \frac{1}{p(1 - p)}dp = \int k dt$

$\displaystyle \int \frac{1}{p} + \frac{1}{1 - p} dp = kt + C$ (by partial fractions)

$\displaystyle \int \frac{1}{p} - \frac{1}{p - 1} dp = kt + C$

$\displaystyle ln|p| - ln|p - 1| = kt + C$

$\displaystyle ln|\frac{p}{p - 1}| = kt + C$

$\displaystyle \frac{p}{p - 1} = e^{kt}e^c$

$\displaystyle \frac{p}{p - 1} = Ae^{kt}$

using p(0) = 8%:

$\displaystyle \frac{0.8}{0.8 - 1} = A$

$\displaystyle A = -4$
• Nov 9th 2010, 04:45 AM
Ackbeet
p(0) = 0.08, not 0.8. That will get you the right magnitude. We've still got a sign error somewhere, though. I think the problem is here:

$\displaystyle \ln\left|\dfrac{p}{p-1}\right|=kt+C.$

$\displaystyle \dfrac{p}{p-1}=Ae^{kt}.$

I think it needs to be this:

$\displaystyle \ln\left|\dfrac{p}{p-1}\right|=\ln\left(\dfrac{p}{1-p}\right)=kt+C.$

Hence,

$\displaystyle \dfrac{p}{1-p}=Ae^{kt}.$

Do you follow?
• Nov 9th 2010, 04:58 AM
SyNtHeSiS
Yeah I do.

The only place where I made a change of sign was in the following steps:

$\displaystyle \int \frac{1}{p} + \frac{1}{1 - p} dp = kt + C$ (by partial fractions)

$\displaystyle \int \frac{1}{p} - \frac{1}{p - 1} dp = kt + C$, since later when I integrate $\displaystyle \frac{1}{1 - p}$, it must be in the form of $\displaystyle -\frac{1}{p - 1}$ for the log
• Nov 9th 2010, 05:16 AM
Ackbeet
Your integration was totally fine. I'm telling you that since the logarithm's argument is an absolute magnitude, and since 0 < p < 1 always, you need to take the form of the absolute magnitude that I just outlined.