One model for the spread of a rumor is that the rate of spread is proportional t

Question

One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction y of the population who have heard the rumor and the fraction who have not heard the rumor. (a) Write a differential equation that is satisfied by y. (Use k for the constant of proportionality.) dy/dt = (b) Solve the differential equation. Assume y(0) = c. y = (c) A small town has 1000 inhabitants. At 8 AM, 120 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor? (Do not round k in your calculation. Round the final answer to one decimal place.) hours after the beginning

Explanation / Answer

Solution:

(a)

differential equation dy/dt = ky(1-y)

(b)
This is a separable equation

dy/(y*(1-y)) = k dt

Expand the left hand side in terms of partial fractions

(1/y - 1/(y-1)) dy = k dt

Integrate:

ln(y) - ln(y-1) - ln(c) + ln(c-1) = k*t

where y(0) = c.

Combine the log terms:

ln(y*(c-1)/(c*(y-1))) = kt

y/(y-1) = (c/(c-1))*e^kt

y/(y-1) = ce^kt / (c-1)

(c-1)/ce^kt = (y-1)/y => (c-1)/ce^kt = 1 - 1/y

1/y = 1 - {(c-1)/ce^kt} => 1/y = {ce^kt - (c-1)} / ce^kt

y = ce^kt / {ce^kt - (c-1)}
y = ce^kt / {c(e^kt - 1) +1}

This is the solution to the DE.

Now use some of the information to solve for the constant of proportionality, k. It is actually easier to do this if we go back to the form of the solution:

ln(y*(c-1)/(c*(y-1))) = kt

Let 8am be t = 0, so c = 120/1000 = 3/25 At noon, t = 4 hr, and y = 1/2, so:

ln((1/2)*(3/25 - 1)/((3/25)*(-1/2)) = k*4

ln(22/3) = k*4

k = (ln(22/3))/4 = 0.498/hr

Now you can solve for the time at which y = 0.9 (90%):

ln((9/10)*(3/25 - 1)/((3/25)*(-1/10)) = 0.498*t

ln(66) = 0.498*t

t = 8.4 hr

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.