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Step-by-Step Solution
Step 1: Understand the Problem
We have a hostel of 100 students. On day zero, 2 students are infected. The number of infected students grows such that the rate of infection is proportional to the product of the number of infected and the number of non-infected students. On the 4th day, there are 30 infected students. We want to find the number of infected students on the 8th day.
Step 2: Set Up the Differential Equation
Let x(t) be the number of infected students at time t . Since there are 100 students in total, the number of non-infected students at time t is 100 - x(t) . We are told:
\displaystyle \frac{dx}{dt} \propto x \bigl(100 - x\bigr).
Introduce a constant of proportionality k :
\displaystyle \frac{dx}{dt} = k\,x\,(100 - x).
Step 3: Separate the Variables and Integrate
Separate the variables to integrate:
\displaystyle \int \frac{dx}{x(100 - x)} = \int k\,dt.
We can rewrite the integrand using partial fractions. Note that:
\displaystyle \frac{1}{x(100 - x)} = \frac{1}{100} \Bigl(\frac{1}{x} + \frac{1}{100 - x}\Bigr).
So we have:
\displaystyle \int \frac{1}{x(100 - x)} \,dx = \frac{1}{100} \int \Bigl(\frac{1}{x} + \frac{1}{100 - x}\Bigr)\,dx.
Integrating term-by-term:
\displaystyle \frac{1}{100} \left[\ln \lvert x\rvert - \ln \lvert 100 - x\rvert \right] = k\,t + C
where C is the constant of integration. We can write:
\displaystyle \frac{1}{100}\,\ln \left(\frac{x}{100 - x}\right) = k\,t + C.
Step 4: Apply Initial Condition (Day 0)
At t=0, x(0)=2. Substitute these values to find C :
\displaystyle \frac{1}{100}\,\ln\left(\frac{2}{98}\right) = C.
Step 5: Write the General Solution with the Constant
Plug C back into the expression:
\displaystyle \frac{1}{100}\,\ln\left(\frac{x}{100 - x}\right) = k\,t + \frac{1}{100}\,\ln\left(\frac{2}{98}\right).
Rearrange to group terms in \ln :
\displaystyle \frac{1}{100}\,\ln\left(\frac{x/ (100 - x)}{2/98}\right) = k\,t
\quad\text{or}\quad
\frac{1}{100}\,\ln\left(\frac{x \times 98}{2(100 - x)}\right) = k\,t.
Step 6: Use the Day 4 Condition to Find k
At t=4, x=30. Substitute these values:
\displaystyle \frac{1}{100}\,\ln\left(\frac{30 \times 98}{2 \times (100 - 30)}\right) = 4\,k.
Note that 100-30=70 and 30\times 98 = 2940. So the expression simplifies to:
\displaystyle \frac{1}{100}\,\ln\left(\frac{2940}{2 \times 70}\right) = \frac{1}{100}\,\ln\left(\frac{2940}{140}\right)
= \frac{1}{100}\,\ln(21) = 4\,k.
Hence,
\displaystyle k = \frac{\ln(21)}{400}.
Step 7: Find the Number of Infected Students on Day 8
When t=8, let x(8)=r. Plug into the expression:
\displaystyle \frac{1}{100}\,\ln\Bigl(\frac{r \times 98}{2(100 - r)}\Bigr) = k \times 8 = \frac{8\,\ln(21)}{400} = \frac{\ln(21)}{50}.
Simplify the left side a bit by noticing 98/2 = 49, so inside the logarithm becomes \frac{49r}{100 - r}. Therefore,
\displaystyle \frac{1}{100}\,\ln\!\Bigl(\frac{49r}{100 - r}\Bigr) = \frac{\ln(21)}{50}.
Multiply both sides by 100:
\displaystyle \ln\!\Bigl(\frac{49r}{100 - r}\Bigr) = 2\,\ln(21).
Since 2\,\ln(21) = \ln\bigl(21^2\bigr) = \ln\bigl(441\bigr), we get:
\displaystyle \frac{49r}{100 - r} = 441.
Then,
\displaystyle \frac{r}{100 - r} = \frac{441}{49} = 9.
So,
\displaystyle r = 9\,(100 - r).
\displaystyle r = 900 - 9r.
\displaystyle 10r = 900 \quad\Longrightarrow\quad r = 90.
Thus, the number of infected students on the 8th day is 90.
