# Population Growth and Limiting Values: Solving the Logistic Equation

• shan
In summary: Um, is it 0? After I simplify, I getP = 1 / (10^6 + e^(1/100 (t+c))) * (10^4 * e^(1/100 (t+c)))Or did I mess up somewhere?No, you have it right. But when you take the limit as t goes to infinity, the 1 in the numerator disappears. What do you get?In summary, The population of a city suburb is described by the initial value problem P' = P((10^-2) - (10^-6)P), with P(0) = 2000. The limiting value of the population as t approaches infinity is 0, and it takes
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)

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.

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:
OlderDan said:
Check your work. Looks like you have lost some terms. You are missing a P in the equation before partial fractions,
lol sorry, sorry, it was a typo. It's supposed to be

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

OlderDan said:
Your parentheses need some work in

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

how about solving it "explicitly" for P(t). You can do that?
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)

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

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?

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

shan said:
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)))

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.

saltydog said:
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.
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?

## 1. What is the logistic equation problem?

The logistic equation problem is a mathematical model that describes the growth of a population over time. It takes into account factors such as limited resources and carrying capacity to predict the population's growth and eventual stabilization.

## 2. What are the variables in the logistic equation problem?

The variables in the logistic equation problem include the initial population size, growth rate, carrying capacity, and time. These variables are used to calculate the population size at any given time.

## 3. How is the logistic equation problem solved?

The logistic equation problem is solved using differential equations and numerical methods. The equation can be solved by hand, but more complex problems may require the use of computer software.

## 4. What are the applications of the logistic equation problem?

The logistic equation problem has various applications in fields such as biology, ecology, economics, and epidemiology. It can be used to study population dynamics, predict the spread of diseases, and analyze market trends.

## 5. What is the significance of the carrying capacity in the logistic equation problem?

The carrying capacity is a crucial factor in the logistic equation problem as it represents the maximum population size that an environment can support. It helps to predict when a population will reach its maximum size and stabilize, preventing overpopulation and resource depletion.

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