S-I 模型用于研究不同情景下传染病的传播。在本题中，假设在整个考虑期间，我们研究的是一个规模恒定为 $N$ 的封闭社区中的均匀人群。人群被分为两组：一组是患有该疾病的已感染个体，另一组是健康且易感的个体，后者在与已感染个体进行某种形式的接触时会被感染。分别用 $x$ 和 $y$ 表示在时间 $t$ 时的感染个体数量和健康且易感的个体数量，我们可以使用以下微分方程对这种情况进行建模
$\frac{\text{d}x}{\text{d}t} = kxy, \quad\quad\quad\quad \text{(I)}$
其中 $k$ 是一个正常数。

1. Express $y$ in terms of $N$ and $x$, and rewrite the differential equation.

Since the population size is constant at $N$, the total number of individuals is the sum of infected and healthy individuals:
$x + y = N$
$y = N - x$
Substituting this into equation (I):
$\frac{\text{d}x}{\text{d}t} = kx(N - x)$

2. Separate the variables to prepare for integration.

$\frac{1}{x(N - x)} \frac{\text{d}x}{\text{d}t} = k$
$\int \frac{1}{x(N - x)} \,\text{d}x = \int k \,\text{d}t$

3. Use partial fractions to split the integrand.

Let $\frac{1}{x(N - x)} = \frac{P}{x} + \frac{Q}{N - x}$
$1 = P(N - x) + Qx$
Comparing coefficients or substituting values for $x$ gives $P = \frac{1}{N}$ and $Q = \frac{1}{N}$.
$\frac{1}{N} \int \left( \frac{1}{x} + \frac{1}{N - x} \right) \,\text{d}x = \int k \,\text{d}t$

4. Integrate both sides and combine logarithmic terms.

$\frac{1}{N} \left( \ln x - \ln(N - x) \right) = kt + C$
Note: Modulus signs are not strictly necessary as $0 < x < N$ (the number of infected is positive and cannot exceed the total population).
$\frac{1}{N} \ln \left( \frac{x}{N - x} \right) = kt + C$
$\ln \left( \frac{x}{N - x} \right) = Nkt + NC$

5. Convert to exponential form to isolate the fraction.

$\frac{x}{N - x} = \text{e}^{Nkt + NC}$
$\frac{x}{N - x} = D\text{e}^{Nkt}$
where $D = \text{e}^{NC}$ is an arbitrary constant.

6. Substitute the initial condition to find the constant $D$.

Given that $x = \alpha$ when $t = 0$:
$\frac{\alpha}{N - \alpha} = D\text{e}^{0}$
$D = \frac{\alpha}{N - \alpha}$
Substituting $D$ back into the equation:
$\frac{x}{N - x} = \left( \frac{\alpha}{N - \alpha} \right) \text{e}^{Nkt}$

7. Rearrange the equation to make $x$ the subject in the required format.

Taking the reciprocal on both sides:
$\frac{N - x}{x} = \left( \frac{N - \alpha}{\alpha} \right) \text{e}^{-Nkt}$
$\frac{N}{x} - 1 = \left( \frac{N}{\alpha} - 1 \right) \text{e}^{-Nkt}$
$\frac{N}{x} = 1 + \left( \frac{N}{\alpha} - 1 \right) \text{e}^{-Nkt}$
$x = \frac{N}{1 + \left( \frac{N}{\alpha} - 1 \right)\text{e}^{-Nkt}}$

8. State the values of constants $A$ and $B$.

Comparing with $x = \frac{A}{1 + (B - 1)\text{e}^{-Nkt}}$:
$A = N$
$B - 1 = \frac{N}{\alpha} - 1 \implies B = \frac{N}{\alpha}$

1. Determine the values of constants A, B, and the exponent coefficient from the given data.

From part (i), $A = N$ and $B = \frac{N}{\alpha}$.
Given $\alpha = 1$ and $N = 206$:
$A = 206$
$B = \frac{206}{1} = 206$
The coefficient for $t$ is $Nk = 206 \times (8.1835 \times 10^{-4}) = 0.1685801 \approx 0.169$ (to 3 s.f.).
The equation of the solution curve is:
$x = \frac{206}{1 + 205\text{e}^{-0.169t}}$

2. Identify key features to sketch the relevant part of the solution curve.

The context considers time $t \ge 0$.
At $t = 0$, $x = 1$. The curve starts at $(0, 1)$.
As $t \to \infty$, $\text{e}^{-0.169t} \to 0$, so $x \to 206$. There is a horizontal asymptote at $x = 206$.
Since $x$ models cumulative infections, it is strictly increasing.
The sketch will have a horizontal axis labeled $t$ and a vertical axis labeled $x$. It will show a logistic 'S' shape in the first quadrant, starting at $(0, 1)$ and flattening out as it approaches the dotted line $x = 206$.

3. State the long-term behavior of $x$ for large $t$.

As $t \to \infty$, the term $\text{e}^{-0.169t}$ tends to $0$.
$x = \frac{206}{1 + 0} = 206$
For large values of $t$, $x$ approaches $206$. (The entire population eventually becomes infected.)