Logistic Problem

Question

A rumor spreads through a school. Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

The school has 1000 students in total. At 8 a.m., 106 students have heard the rumor, and by noon, half the school has heard it. Using the logistic model explained above, determine how much time passes before 90% of the students will have heard the rumor.

90% of the students have heard the rumor after about () hours.

Answer 1

dy/dt = ky/(1 - y); dy/(y - 1) = kdt; ∫[1/(y - 1) - 1/y]dy = ∫kdt; lnI1 - 1/yI = kt + lnC; I1 - 1/yI = Ce^kt;

Let t = 0 at 8 a.m. y = 0.106. So, I1 - 1/0.106I = C; C = 8.434; I1 - 1/yI = 8.434e^kt.

At noon t = 4 and y = 0.5; 1 = 8.434e^4k; k = ln(1/8.434)/4 = - 0.533.

Now I1 - 1/yI = 8.434e^-0.533t.

Let y = 0.9, then I1 - 1/0.9I = 8.434e^-0.533t; t = ln((1/9)/8.434)/(-0.533) = 8.123 hours

After about 8 hours (at about 4 hours p.m.) y = 90 % = 0.9