Exponential Growth Differential Equation

Question

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.

Answer 1

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

Answer 2

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.