# Differential Equation Help

1. Mar 16, 2005

### Jameson

I don't know why, but I am stuck on this seemingly easy question. Here's the question and the work I've done.

---------------------
A certain model for spread of rumors states that $$\frac{dy}{dt} = 3y(3-2y)$$ , where $$y$$ is the proportion of the population that has heard the rumor at time $$t$$. What proportion of the population has heard the rumor when it is spreading the fastest?

--------

Ok. You are given the derivative of the proportion function, so setting it equal to 0 will give you when it is changing the fastest/slowest. Solving the equation $$3y(3-2y) = 0$$ you get 0 and 1.5....

Next part is to find the original equation and evaluate it at 1.5. So I will need to separate the variables, and when I do I get:

$$\frac{1}{3y(3-2y)}dy = dt$$

This integral (I did it on my calculator) is $$\frac{-\ln{\frac{\mid2x-3\mid}{\mid{x}\mid}}}{9}$$

When I evaulate 1.5 I get $$\infty$$

Jameson

Last edited: Mar 16, 2005
2. Mar 16, 2005

### Data

Your procedure isn't quite right. Can you explain why the time when the proportion is changing fastest is when $$y^\prime = 0$$?

3. Mar 16, 2005

### Data

Also you need an initial condition.

4. Mar 16, 2005

### PBRMEASAP

You need to take the derivative one more time before you set it equal to zero. You are maximizing dy/dt, not y.

edit: sorry that wasn't very clear--I should say take the derivative with respect to y, since they have given you dy/dt as a function of y.

Last edited: Mar 16, 2005
5. Mar 16, 2005

### Data

Way to give away the answer to my question! :p

6. Mar 16, 2005

### Jameson

Alright then. So $$\frac{d^{2}y}{dx^{2}} = 9 - 12y$$ and setting it equal to zero you get $$y = \frac{3}{4}$$

Is my integral correct from my first post correct? So now I can just plug .75 in for y?

7. Mar 16, 2005

### Jameson

Can anyone give their thoughts to this question?

8. Mar 16, 2005

### Data

What is the initial condition?

9. Mar 17, 2005

### saltydog

Well, its been a day so I'd like to write this one up if you've moved on (it took me this long to figure it out so I couldn't help at the start). Can someone check it. I don't want to make mistakes.

The equation modeling rumor spread is:

$$\frac{dy}{dt} = 3y(3-2y)$$

Separating variables and integrating from $y_0$ to y yields:

$$ln|{\frac{2y}{3-2y}}|=9t+k$$

or:

$$y(t)=\frac{1}{2}[\frac{3e^{9t+k}}{1+e^{9t+k}}]$$

where:

$$k=ln|\frac{2y_0}{3-2y_0}|$$

Since y is a proportion: $0<y\leq1$, (assume y>0 since if no one knows the rumor at time 0 then it won't spread) the logarithm quantity is always positive and thus I can omit the absolute values.

A plot (for y(0)=0.1) is attached. Looking at the plot, one can see that the rate of y is fastest at the point of inflection, that is, when the second derivative is zero. But we know what the first derivative is:

$$y^{'}=9y-6y^2$$

Thus:

$$y^{''}=9y^{'}-12yy^{'}$$

or substituting in the first derivative:

$$y^{''}=72y^3-162y^2+81y=0$$

Solving this cubic equation, yields the roots:

0, 3/4, and 3/2.

Since y is a proportion between 0 and 1, we take the root 3/4 and conclude the rumor is spreading fastest when 75% of the group knows about it. Using the logarithm version of the solution, we can plug in 0.75 and determine, for a specific initial condition, the time when this occurs.

#### Attached Files:

• ###### rumor1.JPG
File size:
3.6 KB
Views:
47
Last edited: Mar 17, 2005
10. Mar 18, 2005

### saltydog

Ok, thanks to Daniel (from another post) I now understand why y cannot be 0 or 3/2: separating variables, one assumes that y can't be this since that would be dividing by zero. This is the reason we neglect absolute values and also why we must choose 3/4 as the root to the cubic. Yea, I know it's basic and in every Calculus text; I don't claim to be a wiz.