Differential equations: DHS 2023 Promo Q11

(a)

For increasing population, $\dfrac{\,\mathrm{d} P}{\,\mathrm{d} t} > 0$:

$$\begin{aligned} 2P - 0.02P^2 &> 0\\ P(1 - 0.01P) &> 0\\ 0 < P &< 100. \end{aligned}$$

(b)

$$\begin{aligned} \frac{\,\mathrm{d} P}{\,\mathrm{d} t} &= 0.02\!\left(100P - P^2\right)\\ \int \frac{1}{P(100 - P)}\,\mathrm{d} P &= \int 0.02\,\mathrm{d} t\\ \frac{1}{100}\int\left(\frac{1}{P} + \frac{1}{100 - P}\right)\,\mathrm{d} P &= \int 0.02\,\mathrm{d} t\\ \frac{1}{100}\Big(\ln|P| - \ln|100 - P|\Big) &= 0.02t + C\\ \ln\left|\frac{P}{100 - P}\right| &= 2t + 100C\\ \frac{P}{100 - P} &= A\mathrm{e}^{2t}, \qquad A = \pm\mathrm{e}^{100C}. \end{aligned}$$ Solving for $P$: $$\begin{aligned} P &= A\mathrm{e}^{2t}(100 - P)\\ P + A\mathrm{e}^{2t}P &= 100 A\mathrm{e}^{2t}\\ P &= \frac{100 A\mathrm{e}^{2t}}{1 + A\mathrm{e}^{2t}}. \end{aligned}$$ When $t = 0$, $P = P_0$: $$\begin{aligned} P_0 &= \frac{100A}{1 + A}\\ A &= \frac{P_0}{100 - P_0}. \end{aligned}$$ Therefore $$\begin{aligned} P &= \frac{100\left(\frac{P_0}{100 - P_0}\right)\mathrm{e}^{2t}}{1 + \left(\frac{P_0}{100 - P_0}\right)\mathrm{e}^{2t}} = \frac{100 P_0 \mathrm{e}^{2t}}{100 - P_0 + P_0 \mathrm{e}^{2t}}. \quad\text{((shown))} \end{aligned}$$

(c)

As $t \to \infty$, $\mathrm{e}^{2t} \to \infty$: $$\begin{aligned} P &= \frac{100 P_0 \mathrm{e}^{2t}}{100 - P_0 + P_0 \mathrm{e}^{2t}} = \frac{100 P_0}{\dfrac{100 - P_0}{\mathrm{e}^{2t}} + P_0} \to 100. \end{aligned}$$ Hence the population of rabbits approaches $100\,000$ eventually.

(d)