一位流行病学家正在研究蚊子传播的疾病——登革热在 A 镇的传播情况。$P$ 定义为研究开始 $t$ 年后感染人数（以千人为单位）。该流行病学家预测 $P$ 的增长率与已感染人数和未感染人数的乘积成正比。已知 A 镇有 1 万人，其中最初有 1 千人被感染。

1. Identify the quantities involved and translate the rate of increase and proportionality into a mathematical equation.

The number of infected people is $P$ (in thousands).
The total population is $10$ (in thousands), so the number of uninfected people is $10 - P$.
The rate of increase of $P$ is $\frac{\mathrm{d}P}{\mathrm{d}t}$.
Since it is proportional to the product of infected and uninfected people:
$\frac{\mathrm{d}P}{\mathrm{d}t} = kP(10 - P)$
where $k$ is a constant of proportionality (with $k > 0$).

1. State the differential equation formulated in part (i) and separate the variables.

From the problem description in (i), the rate of increase of $P$ is proportional to the product of infected ($P$) and uninfected ($10 - P$) people.
$\frac{\mathrm{d}P}{\mathrm{d}t} = kP(10 - P)$
Separating the variables:
$\int \frac{1}{P(10-P)} \, \mathrm{d}P = \int k \, \mathrm{d}t$

2. Integrate both sides using partial fractions.

Using partial fractions for the integrand:
$\frac{1}{P(10-P)} = \frac{1}{10}\left(\frac{1}{P} + \frac{1}{10-P}\right)$
Integrating both sides:
$\frac{1}{10} \int \left(\frac{1}{P} + \frac{1}{10-P}\right) \, \mathrm{d}P = \int k \, \mathrm{d}t$
$\frac{1}{10} (\ln P - \ln(10-P)) = kt + C$
$\frac{1}{10} \ln\left(\frac{P}{10-P}\right) = kt + C$

3. Apply the initial condition to find the constant of integration.

Initially, 1000 people are infected. Since $P$ is in thousands, $P = 1$ when $t = 0$:
$\frac{1}{10} \ln\left(\frac{1}{10-1}\right) = k(0) + C$
$C = \frac{1}{10} \ln\left(\frac{1}{9}\right)$
Substituting $C$ back into the equation:
$\frac{1}{10} \ln\left(\frac{P}{10-P}\right) = kt + \frac{1}{10} \ln\left(\frac{1}{9}\right)$
Multiplying by 10 and combining the logarithms:
$\ln\left(\frac{9P}{10-P}\right) = 10kt$

4. Apply the given condition at $t = 2$ to find the proportionality constant $k$.

When $t = 2$, half the town is infected, so $P = 5$:
$\ln\left(\frac{9(5)}{10-5}\right) = 10k(2)$
$\ln 9 = 20k \implies 10k = \frac{1}{2}\ln 9 = \ln 3$

5. Rearrange the equation to express $P$ explicitly in terms of $t$.

Substitute $10k = \ln 3$ into the equation:
$\ln\left(\frac{9P}{10-P}\right) = t \ln 3$
$\frac{9P}{10-P} = e^{t \ln 3} = 3^t$
$9P = 10(3^t) - P(3^t)$
$P(9 + 3^t) = 10(3^t)$
$P = \frac{10(3^t)}{9 + 3^t}$

1. Identify the relevant equation, intercept, and asymptote from the context.

From part (ii), the solution to the differential equation is $P = \frac{10(3^t)}{9+3^t}$.
When $t = 0$, $P = \frac{10(1)}{9+1} = 1$, giving the $P$-intercept $(0, 1)$ as defined by the initial population.
As $t \to \infty$, $3^t \to \infty$. Rewriting $P$ by dividing the numerator and denominator by $3^t$:
$P = \frac{10}{9(3^{-t}) + 1}$
As $t \to \infty$, $3^{-t} \to 0$, so $P \to \frac{10}{0 + 1} = 10$. This indicates a horizontal asymptote at $P = 10$.

2. Describe the shape of the graph to be sketched.

Because $P$ models an increasing number of infected people, the curve is strictly increasing for $t \ge 0$.
It starts at $(0, 1)$, concaves upwards initially, then curves downwards (inflection) as the rate of infection slows, flattening out to approach the asymptote $P = 10$.

1. Integrate the given differential equation to find the general solution for $P$.

$P = \int \frac{2\cos t}{(2-\sin t)^2} \,\mathrm{d}t$
Recognising the form $\int \mathrm{f}'(t)[\mathrm{f}(t)]^n \,\mathrm{d}t$ where $\mathrm{f}(t) = 2-\sin t$ and $\mathrm{f}'(t) = -\cos t$:
$P = \int -2(2-\sin t)^{-2} (-\cos t) \,\mathrm{d}t$
$P = \frac{-2(2-\sin t)^{-1}}{-1} + C$
$P = \frac{2}{2-\sin t} + C$

2. Use the initial conditions to find the constant of integration, $C$.

From the problem context, 1000 people were initially infected, and $P$ is the number of infected people in thousands. Therefore, when $t = 0$, $P = 1$.
$1 = \frac{2}{2-\sin(0)} + C$
$1 = \frac{2}{2} + C$
$1 = 1 + C \implies C = 0$

3. State the specific expression for $P$.

$P = \frac{2}{2-\sin t}$

1. Solve the differential equation from part (iv) to find the explicit expression for $P(t)$.

$P = \int \frac{2\cos t}{(2-\sin t)^2} \, \mathrm{d}t = \frac{2}{2-\sin t} + C$
Using the initial condition $P=1$ when $t=0$:
$1 = \frac{2}{2-\sin 0} + C \implies C = 0$
$\therefore P = \frac{2}{2-\sin t}$

2. Use the bounded range of the sine function to determine the greatest and least values of $P$.

For $t \ge 0$, the value of $\sin t$ fluctuates between $-1$ and $1$.
$-1 \le \sin t \le 1$
To find the greatest value of $P$, we must minimise the denominator:
$\text{Min denominator} = 2 - 1 = 1 \implies \text{Greatest } P = \frac{2}{1} = 2$
To find the least value of $P$, we must maximise the denominator:
$\text{Max denominator} = 2 - (-1) = 3 \implies \text{Least } P = \frac{2}{3}$

1. Link the periodic nature of the alternative model to a real-world environmental factor.

The alternative model yields an expression with trigonometric functions, indicating that the number of infected people fluctuates periodically. A possible contributing factor is the seasonal variation in weather (e.g., changing seasons or rainfall), which directly affects the breeding and population of mosquitoes, thus causing the spread of the disease to rise and fall periodically over time.