Logistics Growth Equation

**astuart** · November 5th 2012, 10:11 PM

I'm having an issue understanding what a question is asking me to do...Differential equations have taken a little bit to click when it comes to applying it with a real-world problem.

Anyway, I have a question that asks
The native Hawaiians lived for centuries in isolation from other peoples. When foreigners finally came to the islands they bought with them diseases such as measles, whooping cough and smallpox which decimated the population. Suppose such an island has a native population of 5000 and a sailor from a visiting ship introduces measles, which has an infection rate of 0.00005. Also suppose that the model for spread of an epidemic described in example three applies.
Write an equation for the number of natives who remain uninfected. Let t represent time in days.

Now, the equation used in Example three which it was referring to was the logistics equation which is $\frac {dy}{dt} = k(1 - \frac{y}{N})y$ of which the general solution of this equation is $\frac {N}{1 + (N-1)e^{-kt}}$ .

Now, based on the information I'm given, $N = 5000, \, k=0.25 (0.00005x5000) and t = time in days$ .

The correct equation is of the form $N - y = \frac {N(N-1)}{N-1 + e^{kt}}$ , which by subbing in the correct figures is...

$\frac {5000(4999)}{4999+e^{0.25t}}$

I have no idea however how I'm supposed to derive that formula above. Does anybody have any tips?

**astuart** · November 5th 2012, 10:44 PM

Originally Posted by astuart

I'm having an issue understanding what a question is asking me to do...Differential equations have taken a little bit to click when it comes to applying it with a real-world problem.

Anyway, I have a question that asks
The native Hawaiians lived for centuries in isolation from other peoples. When foreigners finally came to the islands they bought with them diseases such as measles, whooping cough and smallpox which decimated the population. Suppose such an island has a native population of 5000 and a sailor from a visiting ship introduces measles, which has an infection rate of 0.00005. Also suppose that the model for spread of an epidemic described in example three applies.
Write an equation for the number of natives who remain uninfected. Let t represent time in days.

Now, the equation used in Example three which it was referring to was the logistics equation which is $\frac {dy}{dt} = k(1 - \frac{y}{N})y$ of which the general solution of this equation is $\frac {N}{1 + (N-1)e^{-kt}}$ .

Now, based on the information I'm given, $N = 5000, \, k=0.25 (0.00005x5000) and t = time in days$ .

The correct equation is of the form $N - y = \frac {N(N-1)}{N-1 + e^{kt}}$ , which by subbing in the correct figures is...

$\frac {5000(4999)}{4999+e^{0.25t}}$

I have no idea however how I'm supposed to derive that formula above. Does anybody have any tips?

Hrmm, I seem to have already got the answer, though it's in a different form of what the textbook has given me...

I worked it out simply by figuring out N - y.

If $y = \frac {5000}{1 + be^{-kt}}$ where $b = \frac {5000-1}{1} = 4999$ .

Then $N - y = 5000 - \frac {5000}{1+4999e^{-0.25t}}$ [/TEX] which gets me the correct answer when I use t=30 and t=50.

So I guess what I'm asking now, how did the answer I had in the first question equal mine..

Or how does $\frac {5000(4999)}{4999+e^{0.25t}}$ $=$ $5000 - \frac {5000}{1+4999e^{-0.25t}}$

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