# Poplation Problem

1. Apr 19, 2005

### meggs521

I'm attempting to help someone with their diff eq homework and I'm having trouble remembeing how to do things and their book isn't the best.

So here's the problem:

"Suppose the number x(t) (with t in months) of alligators in a swamp satisfies the differential equation

dp/dt = 0.0001x^2 - 0.01x

(a) If there are initially 25 alligators in the swamp, solve this differential equation to determine what happens to the alligator population in the long run.

(b) Repeat for an initial population of 150 alligators."

The main part that's throwing me is the dp/dt = stuff with x, but they say that x is the number of alligators, which basically means population, right? So should I be setting this up as dx/dt = 0.0001x^2 - 0.01x instead?

I did that and I kept getting the ln of a negative number when x(0) was 25 [My actual equation was: 100(ln(x-100)-ln(x)) = t+C]. And lns of negative numbers don't work, so I assumed it's not supposed to be dx/dt = 0.0001x^2 - 0.01x.

But when I use dp/dt = 0.0001x^2 - 0.01x and integrate I get:

P = (0.0001x^2 - 0.01x)t + C

If this is the right way to do it then I know that if t is zero then x doesn't matter and I can solve for C, which gives me C = 25, but this doesn't seem right. If this is right then how do I find the population in the long run? I need to find when the population dies out, so P would be 0 and I would solve for t to find the time. But with my second version I still have x that I have no values for.

I know I'm making some really obvious mistake and I just can't see it yet, but I'm completely lost and any help would be super! All of the examples I've found online or in the book are dp/dt= stuff with p. I remember doing problems like the one in the HW problem, but I can't remember how I did them.

Thanks!

2. Apr 19, 2005

### dvs77

dx/dt=(x^2-10x)/10000 solve this equation u ll get corrct solution

3. Apr 19, 2005

### dvs77

dx/x(x-10)=dt/10000 use partial fractions

4. Apr 19, 2005

### meggs521

Thanks! I'll give it a try.

5. Apr 19, 2005

### meggs521

Alrighty, so I got the equation to be:

ln(x) + ln(x-10) = -(1/1000)t +C

And then if x(0) = 25, I get C= 5.92... or... (x)(x-10) = 5.92e^[-(1/1000)t]

Which works because as t approaches infinity x equals 0.

But when I use x(0) = 150 I get pretty much the same thing but C is now 9.9, so (x)(x-10) = 9.9e^[-(1/1000)t]. Then when I see when the population dies out I get that it happens at t= infinity as well... but the answer is supposed to be 9 years 2 months.

Do I not find the initial equation the same way with x(0) = 25 and x(0)= 150?

I'm sorry for being so slow tonite.

6. Apr 19, 2005

### dvs77

ln(x-10)-lnx=ln(x-10/x);
at t=150
ln140/150=150/1000+c

7. Apr 19, 2005

### saltydog

Well, when you separate variables and solve the ODE, you get:

$$x(t)=\frac{b}{a-Ke^{bt}}$$

with a=0.0001
and b=0.01

Look at the ODE (use x, the p just makes it confussing):

$$\frac{dx}{dt}=ax^2-bx$$

The derivative *slope", is given as a quadratic equation in x. If you plot this equation, you'll note that the curve is below the x-axis between x=0 and 100 and above it after 100. The slope then is negative for all initial conditions below 100 and positive for those above 100. Thus, any initial population below 100 will decrease down to 0, any initial population above 100 will increase without bounds. One right at 100 is an "equilibrium" point. The population won't change.

Anyway to me, when the population gets below 2 it dies off. You know, sexual reproduction and all. Thus, we can only study a decrease in the population when the initial number is below 100. So, assuming you want to know when there are only 2 individuals left starting with 25, When I substitute 2 in the solutions above and calculate what time that happens, I get 23.3 years. However, alligators are good eating (oh yea, just like Chicken) so you know, trapper Bienville, being hungry and all, may make the population drop quicker then this value, thus the 9 year 2 month value you're looking for may be due to him.

8. Apr 19, 2005

### dvs77

i didn't understand what u said

9. Apr 19, 2005

### saltydog

Hello dvs. Well, I'm sorry and disappointed in myself. Though I was good at explaining things. Please tell me exactly what you didn't understand. Please, not from the get-go is it?

10. Apr 19, 2005

### dvs77

i understand the equations but the theory ..........

11. Apr 19, 2005

### saltydog

Oh you mean the theory . . . well, I live in the swamp and really have eaten alligators and you see, the tail is . . . wait a minute, I think you mean the other theory . . . That it all about a "qualitative" analysis of differential equations. Devaney's book on the subject does a nice job of presenting ODEs this way. In the case above, it's telling you that the slope is a quadratic function of the population number. Since this function is negative for all values between 0 and 100, that means the slope is negative for all initial conditions between these values. So, if start with say 75 and the slope is negative, then at time t+deltat will have a value still less than 100 so slope is negative, so t+2deltat will still have a value less than 100 and so on down to 0 eventually. Same dif for values above 100.

12. Apr 19, 2005

### saltydog

You know, I've been think about Bienville. He has buddies you know. What influence would he and his buddies have to be in order for the alligator population to drop to 2 at 9 years, and 2 months assuming it started at 25?

Thus, we seek a solution of the following:

$$\frac{dx}{dt}=ax^2-bx+(Bienvielle factor)$$

with a=0.0001
and b=0.01;

with:
y(0)=25
y(110) just becomes below 2.

Keep in mind Bienville is "removing" gators so his effect on the rate that the gator population is changing is a negative one. Thus we have:

$$\frac{dx}{dt}=ax^2-(b+Bfact)x$$

You see, we can work the problem correctly AND arrive at the answer Meggs was looking for! Meggs, just tell them: "if we consider Bienville, then . . .
Now, just fill in the blanks.

Last edited: Apr 19, 2005