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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^{

Un albergue universitario aloja a 100 estudiantes, cada uno de los cuales es susceptible a contraer cierta infección viral. Un modelo matemático simple supone que durante el curso de una epidemia la tasa de cambio con respecto al tiempo del número de estudiantes contagiados I es proporcional al número de estudiantes contagiados y también al número de alumnos no contagiados. Si en el tiempo t = 0 un solo estudiante está contagiado, demuestre que el número de estudiantes contagiados en el tiempo t es: I = (100e^{100kt})/(99 + e^{100kt})

This problem involves 100 students in a university dormitory who are susceptible to a viral infection. Using a logistic growth model, learn how to demonstrate that the number of infected students follows the equation I = (100e^{100kt}) / (99 + e^{100kt}) given the initial condition where only one student is infected at t = 0.

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