Answer
(a) $\frac{dy}{dt} = ky(1-y)$
(b) $y(t) = \frac{y_0}{y_0+ (1-y_0)e^{-kt}}$
(c) 90% of the population will have heard the rumor by 3:35 pm
Work Step by Step
(a) $y$ is the fraction of people who have heard the rumor
Then (1-y) is the fraction of people who have not heard the rumor.
We can write a differential equation according to the information in the question:
$\frac{dy}{dt} = ky(1-y)$
(b) A logistic differential equation has this form:
$\frac{dP}{dt} = kP(1-\frac{P}{M})$
Consider the differential equation in part a:
$\frac{dy}{dt} = ky(1-y) = ky(1 - \frac{y}{1})$
The solution has this form:
$y(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-y_0}{y_0}$
We can find $A$:
$A = \frac{M-y_0}{y_0} = \frac{1-y_0}{y_0}$
We can write the solution:
$y(t) = \frac{1}{1+ \frac{1-y_0}{y_0}e^{-kt}} = \frac{y_0}{y_0+ (1-y_0)e^{-kt}}$
(c) Let $t$ be in units of minutes. We can find the value of $k$:
$P(240) = \frac{y_0}{y_0+(1-y_0)e^{-240k}} = 0.5$
$P(240) = \frac{0.08}{0.08+(1-0.08)e^{-240k}} = 0.5$
$0.08+(0.92)e^{-240k} = \frac{0.08}{0.5}$
$0.08+(0.92)e^{-240k} = 0.16$
$0.92e^{-240k} = 0.08$
$e^{-240k} = \frac{0.08}{0.92}$
$-240k = ln(\frac{0.08}{0.92})$
$k = \frac{1}{-240}ln(\frac{0.08}{0.92})$
$k = 0.0102$
We can find the time $t$ when $y = 0.9$:
$\frac{y_0}{y_0+(1-y_0)e^{-0.0102t}} = 0.9$
$\frac{0.08}{0.08+(1-0.08)e^{-0.0102t}} = 0.9$
$0.08+(0.92)e^{-0.0102t} = \frac{0.08}{0.9}$
$0.92e^{-0.0102t} = 0.00889$
$e^{-0.0102t} = \frac{0.00889}{0.92}$
$-0.0102t = ln(\frac{0.00889}{0.92})$
$t = \frac{1}{-0.0102}ln(\frac{0.00889}{0.92})$
$t = 455~minutes$
$t = 7~h~35~minutes$
90% of the population will have heard the rumor by 3:35 pm
