In a large town, the number of people infected by a particular virus $t$ days after the virus was first discovered is $x$. It is assumed that the rate of infection is proportional to $x$. Initially there are 5 people who are infected by the virus, and there are 5120 people who are infected by the virus 30 days after the virus was first discovered. A cure and vaccine for the virus were discovered and administered to the population 30 days after the virus was discovered. Individuals who were cured are not at risk of reinfection. The number of people infected by the virus $p$ days after the cure and vaccine were administered is represented by $y$. It is believed that the new rate of infection from then on is proportional to $6400y - y^2$. It is given that 30 days after the cure or vaccine was administered, 3200 people remain infected with the virus.

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.