This problem describes an epidemic model where 100 students are susceptible to an infection. The goal is to demonstrate that the number of infected students at time $t$ follows the given expression:

$I(t) = \frac{100 e^{100kt}}{99 + e^{100kt}}$

Given Information:

• There are 100 students.
• At $t = 0$, one student is infected.
• The rate of change in the number of infected students $I$ is proportional to the number of infected and non-infected students. This is a typical logistic growth model.

Steps to Demonstrate the Equation:

1. Formulation of the Differential Equation:

Let $I(t)$ represent the number of infected students at time $t$. The rate of change of $I(t)$ is proportional to both the number of infected students and the number of uninfected students:

$\frac{dI}{dt} = k I(t) \left( N - I(t) \right)$

Where:

• $N = 100$ (total number of students),
• $k$ is a constant (infection rate),
• $I(t)$ is the number of infected students at time $t$.

So, the equation becomes:

$\frac{dI}{dt} = k I(t) (100 - I(t))$

This is a standard logistic differential equation.

2. Solving the Logistic Equation:

To solve this equation, we separate the variables:

$\frac{1}{I(t)(100 - I(t))} \, dI = k \, dt$

Now we integrate both sides. This step involves partial fraction decomposition on the left-hand side.

3. Partial Fraction Decomposition:

We decompose $\frac{1}{I(t)(100 - I(t))}$:

$\frac{1}{I(t)(100 - I(t))} = \frac{A}{I(t)} + \frac{B}{100 - I(t)}$

Multiplying both sides by $I(t)(100 - I(t))$, we get:

$1 = A(100 - I(t)) + B I(t)$

This simplifies to:

$1 = 100A + (B - A)I(t)$

From here, we can solve for $A$ and $B$:

• $100A = 1 \implies A = \frac{1}{100}$,
• $B - A = 0 \implies B = A = \frac{1}{100}$.

So, the partial fractions are:

$\frac{1}{I(t)(100 - I(t))} = \frac{1}{100 I(t)} + \frac{1}{100(100 - I(t))}$

4. Integrating Both Sides:

We now integrate both sides:

$\int \left( \frac{1}{100 I(t)} + \frac{1}{100(100 - I(t))} \right) \, dI = \int k \, dt$

This gives:

$\frac{1}{100} \ln |I(t)| - \frac{1}{100} \ln |100 - I(t)| = kt + C$

Multiplying by 100:

$\ln \left| \frac{I(t)}{100 - I(t)} \right| = 100kt + C'$

Exponentiating both sides:

$\frac{I(t)}{100 - I(t)} = e^{100kt + C'} = A e^{100kt}$

Where $A = e^{C'}$ is a constant.

5. Solving for $I(t)$:

Now, solve for $I(t)$:

$I(t) = \frac{A e^{100kt} (100 - I(t))}{1}$

At $t = 0$, $I(0) = 1$, so we use this condition to determine $A$.

[ I(0) = \frac{Ae^{100k(0)} \cdot 100}{99 + e^{