0

Hi I'm having a lot of trouble modelling the equation described by the following problem, mostly because I don't really understand the first paragraph.

A rumor spreads through a school. Let () be the fraction of the population that has heard the rumor at time and assume that the rate at which the rumor spreads is proportional to the product of the fraction of the population that has heard the rumor and the fraction 1− that has not yet heard the rumor.

The school has 1000 students in total. At 8 a.m., 107 students have heard the rumor, and by noon, half the school has heard it. Using the logistic model explained above, determine how much time passes before 90% of the students will have heard the rumor.

Thank you!

Alex.G
  • 139

1 Answers1

1

Let's set up a differential equation for $y(t)$.

Let $y(t)$ be the fraction of the population that has heard the rumor at time $t$

So keep in mind that $y$ is a fraction of students, not the number of students.

the rate at which the rumor spreads

that's $y'(t)$

is proportional to

that means "is equal to some constant times..."

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.

that's $y \cdot (1-y)$, or really $y(t)(1-y(t))$ since $y$ here describes the fraction that has heard the rumor at time $t$.

Putting that all together, $y'(t)=ry(t)(1-y(t))$ for some constant $r$. You can solve this differential equation to get an expression for $y(t)$, which will involve yet another unknown constant.

To solve for the constants, use the two pieces of information you are given:

The school has 1000 students in total. At 8 a.m., 107 students have heard the rumor

Now we have to decide how $t$ corresponds with hours of the day. You could say $t=0$ is 8 a.m., and time is measured in hours. Then the above says $y(0)=107/1000$.

and by noon, half the school has heard it

Since noon is 4 hours after 8 a.m., this says $y(4)=1/2$.

So you can plug in the two points $y(0)=107/1000$ and $y(4)=1/2$ into your solution for $y(t)$ to solve for the unknown constants. Once you've done that, you will have an expression for $y(t)$ where all the constants are known. Finally,

determine how much time passes before 90% of the students will have heard the rumor.

So we need to solve $y(t)=0.90$, which you can do since you now know $y(t)$.

kccu
  • 20,808
  • 1
  • 22
  • 41
  • Thank you very much but I still can't really grasp y(t); after integrating factors is it equal to r(y^2/2)-(y^3/3)+C ? @kccu – Alex.G Sep 24 '19 at 13:09
  • ${y'\over y}+{y'\over 1-y}=r$ – saulspatz Sep 24 '19 at 13:22
  • This is not a linear differential equation, I don't think you can use integrating factors. It is a separable differential equation though: $\frac{dy}{y(1-y)} = r\ dx$. As saulspatz points out, $\frac{1}{y(1-y)} = \frac{1}{y}+\frac{1}{1-y}$. – kccu Sep 24 '19 at 13:22
  • @kccu I don't understand how the t variable comes to be in the equation since it doesn't exist in the y'(t) equation you wrote. – Alex.G Sep 24 '19 at 15:38
  • @Alex.G Sorry, my comment should say $\frac{dy}{y(1-y)} = r \ dt$. $y$ is a function of $t$, and we have an expression for $\frac{dy}{dt} = r y(t)(1-y(t))$. – kccu Sep 24 '19 at 16:55
  • Thank you it worked! @kccu – Alex.G Sep 24 '19 at 17:03