Question

Suppose a student carrying a flu virus returns to an isolated college campus of 6000 students. Determine a differential equation governing the number of students x(t) who have contracted the flu if the rate at which the disease spreads is proportional to the number of interactions between students with the flu and students who have not yet contracted it. (Use k > 0 for the constant of proportionality and x for x(t).)

Answer 1

Answer:

$\dfrac{dx}{dt}= kx[6000-x], x(0)=0, k>0$

Step-by-step explanation:

Total Number of Students =6000

Number of students who have contracted the flu =x(t)

Number of students who have not contracted the flu =6000- x(t)

Now, the rate at which the disease spreads is proportional to the number of interactions between students with the flu and students who have not yet contracted it.

$\dfrac{dx(t)}{dt}\propto x(t)[6000-x(t)] \\ Introducing the constant of proportionality, k, we have:\\\dfrac{dx}{dt}= kx[6000-x]$

Initially, the campus is uninfected, therefore: x(0)=0

Therefore, a differential equation governing the number of students x(t) who have contracted the flu is:

$\dfrac{dx}{dt}= kx[6000-x], x(0)=0, k>0$