# Exponential Growth Differential Equation

A population of buffalo grows exponentially (the rate of growth is determined by the population itself) but has a carrying capacity. Its population (in tens of thousands) at a time t ( in years ) is governed by the differential equation $\frac{dP}{dt} = kP(1-P)$ where $k$ the growth rate is yet to be determined.

What is the carrying capacity?

Solve the differential equation for $P(t)$ if we have that $P(0) = \frac23$

Suppose that $P(1) = \frac{8}{10}$ Find k.

How many buffalo will be alive when $t = 2\text{ years}$

I dont know how to solve for the carrying capacity first of all because what I think needs to be done is solve the equation for 0 but then I get P = 0 or 1 for an answer and I don't know if that really makes sense.... unless its 10,000 buffalo

And solving the differential is confusing because I don't know how to separate this equation, so some help on that would be appreciated.

• hold on still editting, i put it up so people could read it Jul 29, 2014 at 20:46
• This equation is called logistic equation, if you plot the function using the derivatives it's really easy to get the result you want: en.wikipedia.org/wiki/Logistic_function Jul 29, 2014 at 20:47
• Hit : $k dt = \frac{dP}{P(1-P)}$ Jul 29, 2014 at 21:02

In light of JJacquelin's hint, you may play with the equation as follows:

$$kdt=\frac{dP}{P(1-P)}=\left(\frac{1}P+\frac{1}{1-P}\right)dP$$ And a simple integration from both sides, gives us:

$$kt+C=\ln|P|-\ln|1-P|=\ln|\frac{P}{1-P}|$$ So we have: $$P(t)=\frac{\exp(kt+C)}{1+\exp(kt+C)},~~\text{or}~~P(t)=\frac{\exp(kt+C)}{-1+\exp(kt+C)}$$

• thank you, i have completed this question :) Jul 29, 2014 at 22:14
• This shows that $P\to 1$, but it is useful to know (1) how to read that off quickly from the differential equation without solving it, and (2) why that carrying capacity is the reason why the differential equation was written as it is. Jul 29, 2014 at 22:16
• @Panthy: I see that you did it. If I were you, I would think of the way the other post by M. Hardy instead. Indeed, I just solve the equation as you wanted but he made a complete reference without solving the equation completely. Think about his post again. :-) Jul 29, 2014 at 22:26

If $P=0\text{ or }1$ then the growth rate is $0$, so the population does not change. If $P$ is between $0$ and $1$ then the growth rate is positive, so the population is getting bigger. So it approaches $P=1$ and then stops growing.

So $P=1$ is the carrying capacity i.e. 10,000 buffalo.

• Yes the two point are very reasonable here. I, also, think that the book wants reader to reflect the problem without solving it. Jul 29, 2014 at 22:20