在一个大城镇中，在首次发现某种病毒 $t$ 天后，感染该病毒的人数为 $x$。假设感染率与 $x$ 成正比。起初有 5 人感染该病毒，在首次发现该病毒 30 天后，有 5120 人感染。

1. Formulate a differential equation representing the proportional rate of infection.

The rate of infection is given by $\frac{\mathrm{d}x}{\mathrm{d}t}$. Since it is proportional to $x$:

$\frac{\mathrm{d}x}{\mathrm{d}t} = kx$

where $k$ is a constant.

2. Solve the separable differential equation to find $x$ in terms of $t$.

$\int \frac{1}{x} \,\mathrm{d}x = \int k \,\mathrm{d}t$

$\ln |x| = kt + C$

Since the number of infected people $x \ge 5 > 0$, we can drop the absolute value:

$\ln x = kt + C$

$x = e^{kt+C} = Ae^{kt}$

where $A = e^C$.

3. Apply the initial condition to find the constant $A$.

Initially, there are $5$ people infected, so when $t = 0$, $x = 5$.

$5 = Ae^{k(0)} \implies A = 5$

So, the equation becomes:

$x = 5e^{kt}$

4. Apply the second condition to evaluate the exponential term.

After 30 days, 5120 people are infected, so when $t = 30$, $x = 5120$.

$5120 = 5e^{30k}$

$e^{30k} = \frac{5120}{5}$

$e^{30k} = 1024$

5. Manipulate logarithms and indices to express $x$ in the required form.

Recognising that $1024 = 2^{10}$:

$e^{30k} = 2^{10}$

Taking the natural logarithm on both sides:

$30k = \ln(2^{10}) = 10 \ln 2$

$k = \frac{10}{30} \ln 2 = \frac{1}{3} \ln 2$

Substitute $k$ back into the expression for $x$:

$x = 5e^{\left(\frac{1}{3} \ln 2\right) t}$

$x = 5\left(e^{\ln 2}\right)^{\frac{t}{3}}$

$x = 5(2)^{\frac{t}{3}} \text{ (Shown)}$

1. Formulate the differential equation and separate the variables.

$\frac{\mathrm{d}y}{\mathrm{d}p} = k(6400y - y^2)$

$\frac{\mathrm{d}y}{\mathrm{d}p} = ky(6400 - y)$

$\int \frac{1}{y(6400 - y)} \, \mathrm{d}y = \int k \, \mathrm{d}p$

2. Use partial fractions to integrate the rational expression.

Let $\frac{1}{y(6400 - y)} = \frac{A}{y} + \frac{B}{6400 - y}$

$1 = A(6400 - y) + By$

Substituting $y = 0 \implies A = \frac{1}{6400}$

Substituting $y = 6400 \implies B = \frac{1}{6400}$

$\frac{1}{6400} \int \left( \frac{1}{y} + \frac{1}{6400 - y} \right) \, \mathrm{d}y = \int k \, \mathrm{d}p$

$\frac{1}{6400} (\ln y - \ln(6400 - y)) = k p + c$

$\ln \left( \frac{y}{6400 - y} \right) = 6400kp + 6400c$

Let $M = 6400k$ and $C = 6400c$.

$\ln \left( \frac{y}{6400 - y} \right) = Mp + C$

3. Determine the constant of integration using the initial condition.

At $p = 0$, the number of infected people $y$ is the number of people infected 30 days after the virus was originally discovered. From the stem, $y = 5120$.

$\ln \left( \frac{5120}{6400 - 5120} \right) = M(0) + C$

$\ln \left( \frac{5120}{1280} \right) = C \implies C = \ln 4$

$\ln \left( \frac{y}{6400 - y} \right) = Mp + \ln 4$

4. Use the condition at $p=30$ to find the proportionality constant and convert the relationship to exponential form.

Given at $p = 30$, $y = 3200$:

$\ln \left( \frac{3200}{6400 - 3200} \right) = 30M + \ln 4$

$\ln(1) = 30M + \ln 4 \implies 0 = 30M + \ln 4 \implies M = -\frac{\ln 4}{30}$

Substitute $M$ back into the equation:

$\ln \left( \frac{y}{6400 - y} \right) = -\left(\frac{\ln 4}{30}\right)p + \ln 4$

$\frac{y}{6400 - y} = e^{\ln 4 - \frac{p}{30}\ln 4} = e^{\ln 4} \cdot (e^{\ln 4})^{-\frac{p}{30}}$

$\frac{y}{6400 - y} = 4 \cdot 4^{-\frac{p}{30}} = 2^2 \cdot (2^2)^{-\frac{p}{30}} = 2^{2 - \frac{p}{15}}$

5. Rearrange to make $y$ the subject and deduce the value of $H$.

$y = (6400 - y) 2^{2 - \frac{p}{15}}$

$y = 6400(2^{2 - \frac{p}{15}}) - y(2^{2 - \frac{p}{15}})$

$y(1 + 2^{2 - \frac{p}{15}}) = 6400(2^{2 - \frac{p}{15}})$

$y = \frac{6400(2^{2 - \frac{p}{15}})}{1 + 2^{2 - \frac{p}{15}}}$

Divide numerator and denominator by $2^{2 - \frac{p}{15}}$:

$y = \frac{6400}{\frac{1}{2^{2 - \frac{p}{15}}} + 1} = \frac{6400}{1 + 2^{\frac{p}{15} - 2}}$

Comparing with $y = \frac{6400}{1 + 2^{\left(\frac{p}{15} + H\right)}}$, we get $H = -2$.

1. Evaluate the long-term number of infected people by taking the limit as $p \to \infty$.

From part (ii), the solution with the determined constant $H = -2$ is $y = \frac{6400}{1 + 2^{\left(\frac{p}{15} - 2\right)}}$.
In the long term, $p \to \infty$.
As $p \to \infty$, the exponent $\left(\frac{p}{15} - 2\right) \to \infty$, which means $2^{\left(\frac{p}{15} - 2\right)} \to \infty$.
Therefore, the denominator becomes infinitely large:
$1 + 2^{\left(\frac{p}{15} - 2\right)} \to \infty$
Hence, $y \to 0$.
The number of people infected in the long term is 0.

2. Comment on the effectiveness of the cure and vaccine based on the mathematical result.

Since the number of infected people eventually reaches 0, the virus will be completely eradicated.
This implies that the cure and vaccine administered are highly effective.