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 0.0002, and that 1% of the population is infected at time t=0. Write down the intial value problem and the solution I(t).

Answer 1

Answer:

Expression: N = C·L·l(t)· T + 20

The initial value problem and solution are expressed as a first order differential equation.

Step-by-step explanation:

First, gather the information:

total population, N = 2 000

Proportionality constant, C = 0.0002

l(t) number of infected individuals = l(t)

healthy individuals = L

The equation is given as follows:

N = C·L·l(t)

However, there is a change with time, so the expression will be:

$\frac{dN}{dt}$ = C·L·l(t)

multiplying both sides  by dt gives:

dN  =   C·L·l(t)

Integrating both sides gives:

$\int\limits^a_b {dN} \, dt$ = $\int\limits^a_b {CLl(t)} \, dt$

N = C·L·l(t)· T + K

initial conditions:

T= 0, N₀ = (0.01 ₓ 2 000)  = 20

to find K, plug in the values:

N₀ = K

20 = K

At any time T, the expression will be:

N = C·L·l(t)· T + 20 Ans