Question

suppose a student carrying a flu virus returns to an isolated college campus of 1000 students. It is assumed
that the rate at which the virus spreads is proportional not only to the number x of infected students but also to
the number of students not infected. It is observed that after 4 days, 30 students are infected.
(a) Write the initial value problem that is associated with this scenario.
(b) Solve the IVP and determine the number of infected students after 9 days.
(c) At what time is the rate of infection the greatest?

Answer 1

Answer:

Step-by-step explanation:

Let's denote:

   x(t)x(t) as the number of infected students at time tt.

   N=1000N=1000 as the total number of students.

   kk as the constant of proportionality for the rate of spread of the virus.

The differential equation representing the rate of change of infected students with respect to time can be formulated as follows:

dxdt=kx(t)(N−x(t))dtdx​=kx(t)(N−x(t))

This equation reflects the assumption that the rate of spread is proportional not only to the number of infected students (x(t)x(t)) but also to the number of students who are not infected (N−x(t)N−x(t)).

Now, let's write the initial condition based on the given information:

At t=0t=0, there is one infected student, so x(0)=1x(0)=1.

So, the initial value problem (IVP) is:

dxdt=kx(t)(N−x(t)),x(0)=1dtdx​=kx(t)(N−x(t)),x(0)=1

To solve this differential equation, we can separate variables and integrate. But first, let's find the value of kk using the given information.

Given that after 4 days, 30 students are infected, we can substitute t=4t=4 and x(t)=30x(t)=30 into the equation:

30=1000k(1000−30)30=1000k(1000−30)

30=1000k(970)30=1000k(970)

k=301000×970k=1000×97030​

Now, let's solve this initial value problem:

dxx(N−x)=k dt1x(N−x)dx​=1kdt​

Integrate both sides:

∫dxx(N−x)=∫k dt∫x(N−x)dx​=∫kdt

This can be solved using partial fraction decomposition, but I'll spare you the details and directly provide the solution:

x(t) = \frac{{N}}{{1 + \left(\frac{{N}}{{x(0)}} - 1\right)e^{-kNt}}

Now, we can plug in the values to find the number of infected students after 9 days (t=9t=9):

x(9) = \frac{{1000}}{{1 + \left(\frac{{1000}}{{1}} - 1\right)e^{-\frac{{30}}{{1000 \times 970}} \times 9}}

x(9) = \frac{{1000}}{{1 + 999e^{-\frac{{30}}{{1000 \times 970}} \times 9}}

x(9)≈10001+999e−0.02914×9x(9)≈1+999e−0.02914×91000​

x(9)≈10001+999e−0.26226x(9)≈1+999e−0.262261000​

x(9)≈10001+999×0.769245x(9)≈1+999×0.7692451000​

x(9)≈10001+769.476555x(9)≈1+769.4765551000​

x(9)≈1000770.476555x(9)≈770.4765551000​

x(9)≈1.2976≈1.3x(9)≈1.2976≈1.3

So, after 9 days, approximately 1.3 students are infected.

To find out when the rate of infection is greatest, we need to find the maximum of the function x(t)x(t). Since x(t)x(t) is in the form of exponential decay, the rate of infection is greatest at t=0t=0 and then decreases over time. So, the rate of infection is greatest at the beginning, when the first student gets infected.