# Differential equation

1. Feb 19, 2008

### sara_87

1. The problem statement, all variables and given/known data

what is the meaning of the coefficients 8 and 80 in the following differential equation?

2. Relevant equations

dP/dt=2*(10^-3)(P-8)(80-P)

3. The attempt at a solution

I know that they are equilibrium solutions but i have to explain.

Thank you

2. Feb 19, 2008

### Dick

Figure out where dP/dt is positive, negative and zero as a function of P. If dP/dt=0 then you have an equilibrium value. Now explain why each equilibrium is stable or unstable.

3. Feb 19, 2008

### sara_87

yeah i did all that but i just have to explain in words the meaning of the coeff. 8 and 80.

4. Feb 19, 2008

### sutupidmath

Well, to do that i think we have to know what is the original problem, i mean to know what P stands for, is it something about a population model or?? If you want to explain the meaning of those constants in the context that the problem is stated!

5. Feb 19, 2008

### sara_87

oh right, yes it is a population model. P is the population and t is time

6. Feb 19, 2008

### sutupidmath

Well i think that in this case, an explanation would be like this:
I am not sure whether you got this part right, is it (P-8) or 8-P ?
When the initial size of the population you are dealing with is greater than 80 whatever the units are, it means that that population model will start decreasing, and it will reach, converge somehow to 80, it means that the most it can decrease is 80.
If 8<Po<80, it means that the most that population size can grow is about 80, so it means that the population starts grwoing but the largest it can get is 80(million, hundred, thousand or whatever). I am not sure about the last part, when the Po<8. It looks like if the initial population has a size of smaller than 8 for some reason it will start decreasing, but it looks wierd though!
I am sure Dick will give a much more accurate and decent explanation, or someone else, though i myself have just started to make a start on diff. eq.

7. Feb 19, 2008

### sara_87

thank you. that does help a lot. now, i can think outside the box.

8. Feb 21, 2008

### sara_87

if the equation is altered to:
dP/dt=2*(10^-3)(P-8)(80-P)-H
Where P is the population of fish, and
where H is rate per month. If the initial population is 16, then what would you recommend for a limit to the monthly catch of this fish population?
(we dont want the fish to be extinct and we dont want to starve).

9. Feb 21, 2008

### Dick

You want to fix the population at P=16? Put dP/dt=0 (since you want a stable population), P=16 and solve for H.

10. Feb 21, 2008

### sara_87

ah ok.
I found H to be:
less than or equal to 1.024
above this value would give below the solution.
is that it?
dont i have to consider that we dont want to starve?

11. Feb 21, 2008

### Dick

P=16, H=1.024, gives you dP/dt=0 So the solution is P(t)=16 for all t. It's not a stable solution. If the population is a little above P=16 it will start to increase and if it drops a little below it will keep dropping. Do you need a stable solution at some value of P besides 16?

12. Feb 21, 2008

### sara_87

i need a value for H so that the fish doesnt become extinct and at the same time...we dont starve