Question

In an isolated environment, a disease spreads at a rate proportional to the product of the infected and non-infected populations. Let I(t) denote the number of infected individuals. Suppose that the total population is 2000, the proportionality constant is 0.0001, and that 1% of the population is infected at time t-0, write down the intial value problem and the solution I(t).
dI/dt =
1(0) =
I(t) =
symbolic formatting help

Answer 1

Answer:

dI/dt = 0.0001(2000 - I)I

I(0) = 20

$I(t)=\frac{2000}{1+99e^{-0.2t}}$

Step-by-step explanation:

It is given in the question that the rate of spread of the disease is proportional to the product of the non infected and the infected population.

Also given I(t) is the number of the infected individual at a time t.

$\frac{dI}{dt}\propto \textup{ the product of the infected and the non infected populations}$

Given total population is 2000. So the non infected population = 2000 - I.

$\frac{dI}{dt}\propto (2000-I)I\\\frac{dI}{dt}=k (2000-I)I, \ \textup{ k is proportionality constant.}\\\textup{Since}\ k = 0.0001\\ \therefore \frac{dI}{dt}=0.0001 (2000-I)I$

Now, I(0) is the number of infected persons at time t = 0.

So, I(0) = 1% of 2000

            = 20

Now, we have dI/dt = 0.0001(2000 - I)I  and  I(0) = 20

$\frac{dI}{dt}=0.0001(2000-I)I\\\frac{dI}{(2000-I)I}=0.0001 dt\\\left ( \frac{1}{2000I}-\frac{1}{2000(I-2000)} \right )dI=0.0001dt\\\frac{dI}{2000I}-\frac{dI}{2000(I-2000)}=0.0001dt\\\textup{Integrating we get},\\\frac{lnI}{2000}-\frac{ln(I-2000)}{2000}=0.0001t+k \ \ \ (k \text{ is constant})\\ln\left ( \frac{I}{I-222} \right )=0.2t+2000k$

$\frac{I}{I-2000}=Ae^{0.2t}\\\frac{I-2000}{I}=Be^{-0.2t}\\\frac{2000}{I}=1-Be^{-0.2t}\\I(t)=\frac{2000}{1-Be^{-0.2t}}\textup{Now we have}, I(0)=20\\\frac{2000}{1-B}=20\\\frac{100}{1-B}=1\\B=-99\\ \therefore I(t)=\frac{2000}{1+99e^{-0.2t}}$