Posted on by

202212071606 Solution to 1976-AL-AMATH-I-8

During an epidemic in a country, the rate of spread of the disease is proportional to the number of people who are healthy; the rate of cure is proportional to the number currently sick. Let P be the total population:

(a) Show that

\displaystyle{\frac{\mathrm{d}W}{\mathrm{d}t}=kP-(k+k_1)W}

where W is the number of sick people; k and k_1 are constants.
(b) Let the total population be growing at a steady rate, so that (independent of the epidemic)

P=a+bt

where a and b are positive. Show that the number of sick people is given by

W=\alpha +\beta t+ce^{\mu t}.

Calculate \alpha, \beta, and \mu in terms of the given constants a, b, k, and k_1.
What further information would you require to determine c?
(c) After a very long time, what is the proportion of people who will be in a state of sickness?

Roughwork.

\begin{aligned} \frac{\mathrm{d}W}{\mathrm{d}t}+(k+k_1)W & = k(a+bt) \\ \dots\,\textrm{ by integrating factor } I=e^{\int (k+k_1)\,\mathrm{d}t} & =e^{(k+k_1)t}\,\dots \\ e^{(k+k_1)t}\bigg(\frac{\mathrm{d}W}{\mathrm{d}t}\bigg) + \Big( e^{(k+k_1)t} (k+k_1)\Big) W & = e^{(k+k_1)t} k(a+bt) \\ e^{(k+k_1)t}W & = \int e^{(k+k_1)t}k(a+bt)\,\mathrm{d}t \\ \end{aligned}

\begin{aligned} \textrm{RHS} & = \int e^{(k+k_1)t}k(a+bt)\,\mathrm{d}t \\ & = ka\int e^{(k+k_1)t}\,\mathrm{d}t + kb\int te^{(k+k_1)t}\,\mathrm{d}t \\ & = \frac{ka}{k+k_1}\Big( e^{(k+k_1)t}\Big) + \frac{kb}{(k+k_1)^2}\int t'e^{t'}\,\mathrm{d}t' \\ & = \frac{ka}{k+k_1}\Big( e^{(k+k_1)t}\Big) + \frac{kb}{(k+k_1)^2}\Big( e^{(k+k_1)t}((k+k_1)t-1)+C\Big) \\ W &= \textrm{RHS}\Big/ e^{(k+k_1)t} \\ & = \bigg(\frac{ka}{k+k_1}-\frac{kb}{(k+k_1)^2}\bigg) + \bigg(\frac{kb}{k+k_1}\bigg) t + \bigg(\frac{kbC}{(k+k_1)^2}\bigg) e^{(-(k+k_1))t} \\ & = \alpha +\beta t+ce^{\mu t} \\ \end{aligned}

\begin{aligned} \frac{W}{P} & = \frac{\alpha +\beta t+ce^{\mu t}}{a+bt} \\ \lim_{t\to \infty}\frac{W}{P} & = \lim_{t\to\infty}\frac{\beta t}{a+bt} \\ & = \frac{\beta}{b} \\ & = \frac{k}{k+k_1} \\ \end{aligned}

This problem is not to be attempted.