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

(b) Solve the differential equation. Assume y(0) = c.
y =

(c) A small town has 1300 inhabitants. At 8 AM, 80 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

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

This is a separable equation:

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

Expand the left hand side in terms of partial fractions (see source if you don't know how to do this):

(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))) = k*t

y/(y-1) = (c/(c-1))*exp(k*t)

y(t) = c*exp(k*t))/[c*(exp(k*t) - 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))) = k*t

Let 8am be t = 0, so c = 100/1300 = 1/13 At noon, t = 4 hr, and y = 1/2, so:

ln((1/2)*(1/13 - 1)/((1/13)*(-1/2)) = k*4 hr

ln(12) = k*4hr

k = (ln(12))/4hr = 0.621/hr

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

ln((9/10)*(1/13 - 1)/((1/13)*(-1/10))*hr = 0.621*t

t = 7.537 hr

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