Problem Statement
The population $P(t)$ of a colony of animals is modeled by the logistic differential equation
$$\frac{dP}{dt} = 0.05 P \left(1 - \frac{P}{2000}\right)$$
where $t$ is measured in months, with an initial population $P(0) = 400$.
(a) Solve the differential equation by separation of variables to express $P(t)$ explicitly in terms of $t$. [5 marks]
(b) Find the time $t$ when the population reaches half of its carrying capacity ($P = 1000$). [2 marks]
(c) State the carrying capacity of the environment. [1 mark]
Verified Solution & Marking Scheme
Part (a): Separate variables and partial fractions
$$\frac{2000}{P(2000 - P)} \, dP = 0.05 \, dt$$
Resolve into partial fractions: $\frac{2000}{P(2000 - P)} = \frac{1}{P} + \frac{1}{2000 - P}$.
$$\int \left( \frac{1}{P} + \frac{1}{2000 - P} \right) dP = \int 0.05 \, dt$$
$$\ln|P| - \ln|2000 - P| = 0.05 t + C \implies \ln\left|\frac{P}{2000 - P}\right| = 0.05 t + C$$
Determine constant C and isolate P(t)
At $t = 0, P = 400$: $\ln\left(\frac{400}{1600}\right) = C \implies C = \ln(1/4) = -\ln 4$.
Exponentiating: $\frac{P}{2000 - P} = \frac{1}{4}e^{0.05 t} \implies 4P = (2000 - P)e^{0.05 t} = 2000e^{0.05 t} - P e^{0.05 t}$.
$$P(4 + e^{0.05 t}) = 2000e^{0.05 t} \implies P(t) = \frac{2000}{1 + 4e^{-0.05 t}}$$
Part (b): Calculate time to reach half carrying capacity
Set $P(t) = 1000$:
$$1000 = \frac{2000}{1 + 4e^{-0.05 t}} \implies 1 + 4e^{-0.05 t} = 2 \implies 4e^{-0.05 t} = 1$$
$$e^{-0.05 t} = \frac{1}{4} \implies -0.05 t = -\ln 4 \implies t = \frac{\ln 4}{0.05} = 20\ln 4 \approx 27.7\text{ months}$$
Part (c): Carrying capacity
As $t \to \infty$, $e^{-0.05 t} \to 0$, so $\lim_{t \to \infty} P(t) = 2000$ animals.