An influenza epidemic is spreading throughout a population of $50,000$ people at a rate that is proportional to the product of the number of infected people and the number of noninfected people. Suppose that $100$ people were infected initially and $1000$ were infected after $10$ days.

Solution (a) If $y=y( t)$ is the number of people infected at time $t$, then $$\begin{equation*} \dfrac{dy}{dt}=ky( 50,000-y) \tag{7} \end{equation*}$$

where $k$ is the constant of proportionality. As given in equation (5), the solution is $$\begin{eqnarray*} y(t)&=&\frac{100(50,000)}{100+(50,000-100) e^{-50,000kt}}\qquad \color{#0066A7}{R=100,\quad M=50,000}\\ &=&\frac{100(50,000)}{100+49,900e^{-50,000kt}} =\frac{50,000}{1+499e^{-50,000kt}} \end{eqnarray*}$$

We find $k$ from the boundary condition $y(10)=1000$. Then $$\begin{eqnarray*} 1000& =&\frac{50,000}{1+499e^{-500,000k}} \\ 499e^{-500,000k}& =&49 \\ -500,000k& =&\ln \left( \frac{49}{499}\right) \\ k& \approx &0.00000464=4.64 \times 10^{-6} \end{eqnarray*}$$

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So, $$\begin{equation*} y(t)=\frac{50,000}{1+499e^{-0.232t}} \end{equation*}$$

Half the population is infected when $y(t)=25,000$. $$\begin{eqnarray*} 25,000& =&\dfrac{50,000}{1+499e^{-0.232t}} \\ 1+499e^{-0.232t}& =&2 \\ e^{-0.232t}& =&\frac{1}{499} \\ t& =&\frac{\ln \dfrac{1}{499}}{-0.232}\approx 27 \hbox{days} \end{eqnarray*}$$

Half the population is infected in approximately $27$ days.

(b) The time $t$ we seek is the inflection point of $y=y( t)$. We need to find $t$ so that $y^{\prime \prime} ( t) =0$. $$\begin{eqnarray*} y^{\prime} ( t) &=&ky( 50,000-y) =k( 50,000y-y^{2})\qquad \color{#0066A7}{{(6)}} \\ y^{\prime \prime} &=&k( 50,000y^{\prime} -2yy^{\prime} ) =0 \\ 2y &=&50,000 \\ y &=&25,000 \end{eqnarray*}$$

From (a), $y=25,000$ when $t=27$. That is, on Day $27$, the rate of infection stops increasing and begins to decrease.