Question

The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli in 1760 on smallpox. In more recent years many mathematical models have been proposed and studied for many different diseases. The following problem deals with a few of the simpler models and the conclusions that can be drawn from them. Similar models have also been used to describe the spread of rumors and of consumer products. Some diseases (such as typhoid fever) are spread largely by carriers, individuals who can transmit the disease but who exhibit no overt symptoms. Let x and y denote the proportions of susceptibles and carriers, respectively, in the population. Suppose that carriers are identified and removed from the population at a rate β, so dy/dt = −βy.
(i) Suppose also that the disease spreads at a rate proportional to the product of x and y; thus dx/dt = −αxy.
(ii)
(a) Determine y at any time t by solving Eq. (i) subject to the initial condition y(0) = y0.
y(t) =
(b) Use the result of part (a) to find x at any time t by solving Eq. (ii) subject to the initial condition x(0) = x0.
x(t) =
(c) Find the proportion of the population that escapes the epidemic by finding the limiting value of x as t → [infinity].

Answer 1

Answer:

a

   $y(t) = y_o e^{\beta t}$

b

      $x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

c

      $\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

Step-by-step explanation:

From the question we are told that

    $\frac{dy}{y} = -\beta dt$

Now integrating both sides

     $ln y = \beta t + c$

Now taking the exponent of both sides

       $y(t) = e^{\beta t + c}$

=>     $y(t) = e^{\beta t} e^c$

Let  $e^c = C$

So

      $y(t) = C e^{\beta t}$

Now  from the question we are told that

      $y(0) = y_o$

Hence

        $y(0) = y_o = Ce^{\beta * 0}$

=>     $y_o = C$

So

        $y(t) = y_o e^{\beta t}$

From the question we are told that

      $\frac{dx}{dt} = -\alpha xy$

substituting for y

      $\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })$

=>   $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

         $lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

        $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

=>     $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let  $e^c = A$

=>  $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now  from the question we are told that

      $x(0) = x_o$

So  

      $x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

=>    $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side  by    $(K * x_o)$

=>    $K = x_o e^{\frac {\alpha y_o }{\beta } }$

So

    $x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=>   $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=>    $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as  t tends to infinity ,  $e^{- \beta t}$ tends to zero  

so

    $\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$