Solving a Logistic Equation

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.

1. Find the number of infected people at any time \(t\). How long will it take for half the population to be infected?
2. Find the time \(t\) at which the rate of spread of infection stops increasing and begins to decrease.

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.