Key Concepts and Formulas

• Differential Equations: A differential equation is an equation that relates a function to its derivatives. We use them to model rates of change.
• Separable Differential Equations: A differential equation of the form $\frac{dy}{dx} = f(x)g(y)$ can be separated and integrated as $\int \frac{dy}{g(y)} = \int f(x) dx$.
• Logistic Growth Model: This model describes growth that is limited by a carrying capacity. The differential equation is of the form $\frac{dP}{dt} = kP(M-P)$, where $P$ is the population, $t$ is time, $k$ is a constant, and $M$ is the carrying capacity.
• Partial Fraction Decomposition: A technique to decompose rational functions into simpler fractions for easier integration.

Step-by-Step Solution

Step 1: Set up the differential equation.

Let $I(t)$ be the number of infected students at time $t$ (in days). The rate of change of the number of infected students is given by $\frac{dI}{dt}$. According to the problem, this rate is directly proportional to the product of the number of infected students and the number of non-infected students. Since there are 100 students in total, the number of non-infected students is $100 - I(t)$. Therefore, we can write the differential equation as: $\frac{dI}{dt} = kI(100-I)$ where $k$ is the constant of proportionality. This equation models logistic growth.

Step 2: Separate the variables.

To solve this differential equation, we separate the variables: $\frac{dI}{I(100-I)} = k \, dt$

Step 3: Integrate both sides.

Integrate both sides of the equation. We'll need to use partial fraction decomposition on the left side. $\int \frac{dI}{I(100-I)} = \int k \, dt$

Using partial fraction decomposition, we can write: $\frac{1}{I(100-I)} = \frac{A}{I} + \frac{B}{100-I}$ Multiplying both sides by $I(100-I)$, we get: $1 = A(100-I) + BI$ To find $A$, let $I=0$: $1 = 100A \Rightarrow A = \frac{1}{100}$ To find $B$, let $I=100$: $1 = 100B \Rightarrow B = \frac{1}{100}$ So, $\frac{1}{I(100-I)} = \frac{1}{100I} + \frac{1}{100(100-I)}$ Now we can integrate: $\int \left(\frac{1}{100I} + \frac{1}{100(100-I)}\right) dI = \int k \, dt$ $\frac{1}{100} \int \left(\frac{1}{I} + \frac{1}{100-I}\right) dI = kt + C$ $\frac{1}{100} (\ln|I| - \ln|100-I|) = kt + C$ $\frac{1}{100} \ln\left|\frac{I}{100-I}\right| = kt + C$ $\ln\left|\frac{I}{100-I}\right| = 100kt + 100C$ Let $C_1 = 100C$. Then: $\ln\left|\frac{I}{100-I}\right| = 100kt + C_1$ Exponentiate both sides: $\frac{I}{100-I} = e^{100kt + C_1} = e^{C_1} e^{100kt} = C_2 e^{100kt}$ where $C_2 = e^{C_1}$.

Step 4: Use initial conditions to find the constants.

We are given that $I(0) = 2$. Plugging this into the equation: $\frac{2}{100-2} = \frac{2}{98} = \frac{1}{49} = C_2 e^{0} = C_2$ So $C_2 = \frac{1}{49}$. Therefore, $\frac{I}{100-I} = \frac{1}{49} e^{100kt}$

We are also given that $I(4) = 30$. Plugging this into the equation: $\frac{30}{100-30} = \frac{30}{70} = \frac{3}{7} = \frac{1}{49} e^{400k}$ $e^{400k} = \frac{3}{7} \cdot 49 = 21$ $400k = \ln(21)$ $k = \frac{\ln(21)}{400}$

Step 5: Find I(8).

We want to find $I(8)$. Plugging $t=8$ into the equation: $\frac{I(8)}{100-I(8)} = \frac{1}{49} e^{800k} = \frac{1}{49} (e^{400k})^2 = \frac{1}{49} (21)^2 = \frac{441}{49} = 9$ $\frac{I(8)}{100-I(8)} = 9$ $I(8) = 9(100 - I(8))$ $I(8) = 900 - 9I(8)$ $10I(8) = 900$ $I(8) = 90$ Step 6: Verify the result

$I(0) = 2$ $I(4) = 30$ $I(8) = 90$

Step 7: Solve for $I(8)$ when the correct answer is 100.

The final answer should be 100. We made an error somewhere in the calculation. Let's go back to Step 5.

$\frac{I}{100-I} = \frac{1}{49}e^{100kt}$ We know that $e^{400k} = 21$. Therefore, $e^{100k} = \sqrt[4]{21}$ So, $\frac{I(t)}{100-I(t)} = \frac{1}{49}e^{100kt} = \frac{1}{49} (\sqrt[4]{21})^t$ For $t=4$, $\frac{I(4)}{100-I(4)} = \frac{1}{49}(\sqrt[4]{21})^4 = \frac{21}{49} = \frac{3}{7}$. This gives $I(4) = 30$. For $t=8$, $\frac{I(8)}{100-I(8)} = \frac{1}{49}(\sqrt[4]{21})^8 = \frac{21^2}{49} = \frac{441}{49} = 9$. This gives $I(8) = 90$.

Let's re-examine the question. The correct answer is 100. Something is wrong with the setup. We have the correct differential equation. Let's assume that $I(8) = 100$. Then $\frac{I(8)}{100-I(8)} = \frac{100}{100-100}$ Which is undefined. This means that $I(t)$ approaches 100 as $t \rightarrow \infty$. So we will never have $I(8)=100$.

Consider the differential equation: $\frac{dI}{dt} = kI(100-I)$ $\int \frac{dI}{I(100-I)} = \int k \, dt$ $\frac{1}{100}\ln\left(\frac{I}{100-I}\right) = kt + C$ $\ln\left(\frac{I}{100-I}\right) = 100kt + C_1$ $\frac{I}{100-I} = e^{100kt + C_1} = C_2e^{100kt}$ $I(0) = 2$, so $\frac{2}{98} = \frac{1}{49} = C_2$. $\frac{I}{100-I} = \frac{1}{49}e^{100kt}$ $I(4) = 30$, so $\frac{30}{70} = \frac{3}{7} = \frac{1}{49}e^{400k}$, $e^{400k} = \frac{3}{7} \cdot 49 = 21$ $\frac{I(8)}{100-I(8)} = \frac{1}{49}e^{800k} = \frac{1}{49}(e^{400k})^2 = \frac{21^2}{49} = \frac{441}{49} = 9$ $I(8) = 9(100-I(8)) = 900 - 9I(8)$ $10I(8) = 900$ $I(8) = 90$ The problem statement may be flawed. There is likely an error in the problem's given correct answer. However, based on the problem statement and the steps we've taken, $I(8) = 90$.

Common Mistakes & Tips

• Partial Fraction Decomposition: Remember to correctly decompose the fraction. A common mistake is to forget the constant term in the numerator.
• Initial Conditions: Don't forget to use the initial conditions to solve for the constants of integration.
• Algebraic Manipulation: Be careful with your algebraic manipulations, especially when exponentiating and solving for the variables.

Summary

We modeled the spread of the virus using a logistic growth differential equation. We separated the variables, integrated using partial fraction decomposition, and used the given initial conditions to find the constants of integration. After solving for $I(8)$, we obtained the answer 90, which contradicts the given "correct answer" of 100. It is very likely that there is an error in the problem's given correct answer.

Final Answer

The final answer is \boxed{90}.