Question

A rumor spreads through a small town. 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.

a. Write the differential equation satisfied by y in terms of proportionality k.
b. Find k (in units of day−1, assuming that 10% of the population knows the rumor at time t=0 and 40% knows it at time t=2 days.
c. Using the assumptions in part (b), determine when 75% of the population will know the rumor.
d. Plot the direction field for the differential equation and draw the curve that fits the solution y(0)=0.1 and y(0)=0.5.

Answer 1

The answer is shown below

Explanation:

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.

a)

`(dy)/(dt)\ \alpha\ y(1-y)`

`(dy)/(dt)=ky(1-y)`

where k is the constant of proportionality, dy/dt = rate at which the rumor spreads

b)

`(dy)/(dt)=ky(1-y)\\(dy)/(y(1-y))=kdt\\\int\limits {(dy)/(y(1-y))} \, =\int\limit {kdt}\\\int\limits {(dy)/(y)} +\int\limits {(dy)/(1-y)} =\int\limit {kdt}\\\\ln(y)-ln(1-y)=kt+c\\ln((y)/(1-y)) =kt+c\\taking \ exponential \ of\ both \ sides\\(y)/(1-y) =e^(kt+c)\\(y)/(1-y) =e^(kt)e^c\\let\ A=e^c\\(y)/(1-y) =Ae^(kt)\\y=(1-y)Ae^(kt)\\y=(Ae^(kt))/(1+Ae^(kt)) \\at \ t=0,y=10\%\\0.1=(Ae^(k*0))/(1+Ae^(k*0)) \\0.1=(A)/(1+A) \\A=(1)/(9) \\`

`y=((1)/(9) e^(kt))/(1+(1)/(9) e^(kt))\\y=(1)/(1+9e^(-kt))`

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

`y=(1)/(1+9e^(-0.8959t))\\0.75=(1)/(1+9e^(-0.8959t))\\t=3.68\ days`