在一个大城镇中，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，而在首次发现病毒 30 天后，有 5120 人感染该病毒。 在发现病毒 30 天后，人们发现了该病毒的治疗方法和疫苗并对人群进行了接种。治愈的人没有再次感染的风险。接种治疗方法和疫苗 $p$ 天后感染该病毒的人数用 $y$ 表示。据信，从那时起新的感染率与 $6400y - y^2$ 成正比。已知在接种治疗方法或疫苗 30 天后，仍有 3200 人感染该病毒。

1. Formulate the differential equation from the given rate of infection.

$\frac{dx}{dt} = kx$

2. Separate variables and integrate to find the general solution.

$\int \frac{1}{x} \, dx = \int k \, dt$
$\ln x = kt + C$
$x = e^{kt+C} = Ae^{kt} \quad (\text{where } A = e^C)$

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

When $t = 0$, $x = 5$:
$5 = Ae^{0}$
$A = 5 \implies x = 5e^{kt}$

4. Apply the condition at $t = 30$ to find the proportionality constant $k$.

When $t = 30$, $x = 5120$:
$5120 = 5e^{30k}$
$e^{30k} = \frac{5120}{5} = 1024$
$30k = \ln(1024)$
$30k = \ln(2^{10}) = 10\ln 2$
$k = \frac{1}{3}\ln 2$

5. Substitute $k$ back into the equation and manipulate it to the required form.

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

1. Formulate the differential equation relating $y$ and $p$.

Let the rate of infection be $\frac{\mathrm{d}y}{\mathrm{d}p}$.
Given that the rate of infection is proportional to $6400y - y^2$, we can write:
$\frac{\mathrm{d}y}{\mathrm{d}p} = k(6400y - y^2)$
where $k$ is a constant of proportionality.

2. Separate the variables and use partial fractions to prepare for integration.

Separate the variables:
$\int \frac{1}{y(6400 - y)} \mathrm{d}y = \int k \mathrm{d}p$
Use partial fractions for the integrand on the left. Let $\frac{1}{y(6400 - y)} = \frac{A}{y} + \frac{B}{6400 - y}$.
$1 = A(6400 - y) + By$
When $y = 0$, $6400A = 1 \implies A = \frac{1}{6400}$.
When $y = 6400$, $6400B = 1 \implies B = \frac{1}{6400}$.
Thus, the equation becomes:
$\frac{1}{6400} \int \left( \frac{1}{y} + \frac{1}{6400 - y} \right) \mathrm{d}y = \int k \mathrm{d}p$

3. Integrate both sides of the differential equation.

Multiply by 6400 and integrate:
$\ln |y| - \ln |6400 - y| = 6400kp + C$
Since the number of infected people $y$ is initially 5120 and drops to 3200, $0 < y < 6400$. We can drop the absolute value signs and combine the logarithms:
$\ln \left( \frac{y}{6400 - y} \right) = cp + C$
where $c = 6400k$ is a new constant.

4. Use the given boundary conditions to evaluate the constants $C$ and $c$.

From the context, 30 days after the virus was discovered, the cure was administered. The number of people infected at this time was 5120. Thus, at $p = 0$, $y = 5120$.
$\ln \left( \frac{5120}{6400 - 5120} \right) = c(0) + C$
$\ln \left( \frac{5120}{1280} \right) = C \implies C = \ln 4$
It is given that at $p = 30$, $y = 3200$.
$\ln \left( \frac{3200}{6400 - 3200} \right) = 30c + \ln 4$
$\ln(1) = 30c + \ln 4 \implies 30c = -\ln 4 = -2\ln 2$
$c = -\frac{2\ln 2}{30} = -\frac{\ln 2}{15}$

5. Substitute the constants and simplify the exponential terms into base 2.

Substitute $C$ and $c$ back into the equation:
$\ln \left( \frac{y}{6400 - y} \right) = -\frac{p}{15}\ln 2 + \ln 4$
Express in exponential form:
$\frac{y}{6400 - y} = \mathrm{e}^{-\frac{p}{15}\ln 2 + \ln 4}$
$\frac{y}{6400 - y} = \left( \mathrm{e}^{\ln 2} \right)^{-\frac{p}{15}} \cdot \mathrm{e}^{\ln 4}$
$\frac{y}{6400 - y} = 2^{-\frac{p}{15}} \cdot 4 = 2^2 \cdot 2^{-\frac{p}{15}} = 2^{2 - \frac{p}{15}}$

6. Rearrange the equation to express $y$ in the required format and state $H$.

Take the reciprocal of both sides:
$\frac{6400 - y}{y} = \frac{1}{2^{2 - \frac{p}{15}}}$
$\frac{6400}{y} - 1 = 2^{\frac{p}{15} - 2}$
$\frac{6400}{y} = 1 + 2^{\frac{p}{15} - 2}$
$y = \frac{6400}{1 + 2^{\left(\frac{p}{15} - 2\right)}}$
Comparing this with $y = \dfrac{6400}{1+2^{\left(\frac{p}{15}+H\right)}}$, we have successfully shown the form with $H = -2$.

1. Evaluate the number of infected people as time $p$ approaches infinity.

From part (ii), $y = \dfrac{6400}{1+2^{\left(\frac{p}{15}+H\right)}}$ where $H = -2$.

As $p \to \infty$, the exponent $\left(\frac{p}{15} - 2\right) \to \infty$.
This implies that $2^{\left(\frac{p}{15} - 2\right)} \to \infty$, meaning the denominator grows infinitely large.

$\lim_{p \to \infty} y = 0$

Hence, the number of people infected in the long term is $0$.

2. Comment on the effectiveness of the cure and vaccine based on the long-term prediction.

Since the model predicts that the number of infected people will eventually drop to $0$, the cure and vaccine administered are highly effective in successfully eliminating the virus from the population.