# Homework Help: Logistic equation problem

1. May 20, 2005

### shan

I just want to check my answer for this question:

The population P(t) of a city suburb is governed by the initial value problem

P' = P((10^-2) - (10^-6)P) P(0) = 2000

Time measured in months.
a) What is the limiting value of the population at t approaches infinity?
b) How long before the population is one-half of this limiting value?

a) I used the formula for the general solution
birth-rate/death-rate = 10^-2/10^-6 = 10^4

But I'm not sure if that is what I was supposed to do...

b) Solving the DE
P' = P((10^-2) - (10^-6)P)

$$\int \frac{dP}{P(10^{-2}-10^{-6})} = \int dt$$

Using partial fractions, the left can be seperated

$$\int \frac{10^2}{P} + \frac{10^{-4}}{10^{-2}-10^{-6}P} = \int dt$$

$$10^2 * ln(P-ln(10^{-2}-10^{-6}P) = t + c$$

Using the initial value P(0) = 2000, C = 250,000

Half of the limiting value = (10^4)/2 = 5000

Substituting into the equation, t = 139 months (3sf)

2. May 20, 2005

### OlderDan

Check your work. Looks like you have lost some terms. You are missing a P in the equation before partial fractions, but your partial fraction expansion looks OK. Your parentheses need some work in

$$10^2 * ln(P-ln(10^{-2}-10^{-6}P) = t + c$$

Consequently your C does not look right.

Part a) looks OK. P = $$10^4$$ is where the derivative goes to zero, so that is the limiting value of P.

3. May 20, 2005

### saltydog

So, what up Shan? What about when you correct your formula as Dan said, how about solving it "explicitly" for P(t). You can do that? Don't want to? I don't know. However if you did, it would clearly show what the limiting value is as t goes to infinity. Why not try and report it here.

Edit: Plot too but suppose I shouldn't push it.

Last edited: May 20, 2005
4. May 20, 2005

### shan

lol sorry, sorry, it was a typo. It's supposed to be

$$\int \frac{dP}{P(10^{-2}-10^{-6}P)} = \int dt$$

Another typo... V_V

$$10^2 * (lnP-ln(10^{-2}-10^{-6}P)) = t + c$$

so when I rearrange...

$$lnP-ln(10^{-2}-10^{-6}P)) = \frac{t + c}{10^2}$$

$$ln\frac{P}{10^{-2}-10^{-6}P} = \frac{t + c}{10^2}$$

$$\frac{P}{10^{-2}-10^{-6}P} = e^{\frac{t + c}{10^2}} = Ce^{\frac{t}{100}}$$

One thing I'm not sure about, can I solve for c using the second equation or the third one? Or does it not matter? Because I used the third equation so that I end up with

$$\frac{2000}{10^{-2}-10^{-6}*2000} = Ce^0$$

(P(0)=2000)

And c ends up being 250,000

lol, it's the second case. When I rearrange the equation up the top (presuming that it's right), I got

$$P = Ce^{\frac{t}{100}} * (10^{-2}-10^{-6}P)$$

And I don't know how to get rid of the P on the right... (yes, my algebra is bad lol)

5. May 20, 2005

### saltydog

Dude, you gotta' get those parenthesis right:

$$ln(P)-ln(10^{-2}-10^{-6}P) = \frac{t + c}{10^2}$$

We're good here. Actually leaving as:

$$10^2\{ln(P)-ln(10^{-2}-10^{-6}P)\} = t + c$$

makes it easy to calcualate c but leave it as 'c' until you need it.

Also, those quantities for logarithms, aren't they just:

$$\frac{10^{6}P}{10^4-P}$$

So we have:

$$ln\{\frac{10^{6}P}{10^4-P}\}=\frac{1}{100}(t+c)$$

So you take exponentials of both sides and get:

$$\frac{10^{6}P}{10^4-P}}=e^{1/100(t+c)}$$

This is where you're having problems right, isolating P.

Suppose I have:

$$\frac{ax}{1-x}=k$$

Well, multiplying by 1-x gives me:

$$ax=k(1-x)$$

soooooo . . . ax=k-kx
ax+xk=k
x(a+k)=k
You got it right?

6. May 22, 2005

### shan

Sorry for the late reply. So are you saying that

P = (10^4 * e^(1/100 (t+c))) / (10^6 + e^(1/100 (t+c)))

7. May 22, 2005

### saltydog

That's it. However, if you would divide the numerator and denominator by the numerator, then the limit as t goes to infinity is clear.

8. May 22, 2005

### shan

Um, is it 0? After I simplify, I get

P = 1 / (10^6 + e^(1/100 (t+c))) * (10^4 * e^(1/100 (t+c)))

Or did I mess up somewhere?