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PtichkaEL [24]
3 years ago
9

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 t

hat 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.
Mathematics
1 answer:
sladkih [1.3K]3 years ago
5 0

Answer:

The answer is shown below

Step-by-step 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)

\frac{dy}{dt}\ \alpha\  y(1-y)

\frac{dy}{dt}=ky(1-y)

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

b)

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

y=\frac{\frac{1}{9} e^{kt}}{1+\frac{1}{9} e^{kt}}\\y=\frac{1}{1+9e^{-kt}}

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

y=\frac{1}{1+9e^{-0.8959t}}\\0.75=\frac{1}{1+9e^{-0.8959t}}\\t=3.68\ days

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