Question

Suppose a student carrying a flu virus returns to an isolated college campus of 2000 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:

x(t) = 2000 - e^(-k*t)

Step-by-step explanation:

Interpretation:

No . infected student = x

Total student = 2000

rate of infected students = dx / dt

not-infected student = 200 - x

The general rate at which student are infected can be expressed as below:

dx / dt = k * ( 2000 - x )

To develop an expression of x(t) we integrate the above expression by separating variables:

dx / (2000 - x ) = k * dt

Now integrate:

$\int\limits{\frac{1}{(2000-x)} } \, dx = k * \int\limits{dt}\\\\- ln (2000-x) = k*t + C$

@ t = 0 , infected students x = 0

Hence,

C = - ln (2000)

$- ln (2000-x) = k*t + - ln (2000)\\\\ln (\frac{2000-x}{2000}) = -k*t\\\\ 2000 - x = e^(-kt)\\\\x(t) = 2000 - e ^ (-k*t)$