Differential Equation Help

[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.

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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?

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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

Help me please.
Jameson

[Data]

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

[Data]

Also you need an initial condition.

[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.

[Data]

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

[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?

[Jameson]

Can anyone give their thoughts to this question?

[Data]

What is the initial condition?

[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.

[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.