Question

Suppose a student carrying a flu virus returns to an isolated college campus of 1000 students. If it is assumed that the rate at which the virus spreads is proportional not only to the number x of infected students but also to the number of students not infected, and assume that no one leaves the campus throughout the duration of the disease, determine the number of infected students after 6 days if it is further observed that after 4 days x(4) = 50.

Answer 1

Solution

determine the number of infected students after 6 days

x is the number of infected students among 100 student then (1000-x) is the number of student that are not infected since, the rate of the virus spreads is proportional not only to the number x but also the number of students not infected.

$\therefore\hspace{5mm}\frac{dx}{dt}=kx(100-x)$ (k is constant)

$\implies \frac{1}{k}\frac{dx}{x(100-x)}=dt$

$\implies \frac{1}{1000k}\Big(\frac{1}{x}+\frac{1}{1000-x}\Big)dx=dt$

$\implies \frac{1}{1000k}\int\Big(\frac{1}{x}+\frac{1}{1000-x}\Big)dx=\int dt$

$\implies \frac{1}{1000k}ln\Big|\frac{x}{1000-x}\Big|=t+C_1,\quad C_1\in \mathbb{R}$

$\implies ln\Big|\frac{x}{1000-x}\Big|=1000kt+1000kC_1$

$\implies \Big|\frac{x}{1000-x}\Big|=e^{1000kt+1000kC_1}$

$\implies \frac{x}{1000-x}=Ce^{1000kt}\quad$we know that x(0)=1

$\implies \frac{1}{1000-1}=C\implies C=\frac{1}{999}$

Then $999x=1000e^{kt}-xe^{1000kt}$

$\iff x(999+e^{1000kt})=1000e^{1000kt}\implies x=\frac{1000e^{1000kt}}{999+e^{1000kt}}$

Since, x(4)=50$\implies 50=\frac{1000e^{4000k}}{999+e^{4000k}}$

$\iff 50=\frac{1000}{999e^{-4000k}+1}$

$\iff 49950e^{-4000k}+50=1000\implies e^{-4000k}=\frac{950}{49950}$

$\implies -4000k=ln\Big(\frac{950}{49950}\Big)$

$\implies k=\frac{-3.96}{-4000}=0.00099 \implies x(t)=\frac{1000e^{0.99t}}{999+e^{0.99t}}$

Then $x(6)=\frac{1000e^{0.99\times 6}}{999+e^{0.99\times 6}}=275.63\approx 276$

Therefore. 276  students will be infected after 6 day.