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Harvesting Model

  • #1
Here is the question:

Investigate the harvesting model in problem 5 both qualitatively and analytically in the case a=5, b=1, and h=25/4. Determine whether the population becomes extinct in finite time. If so, find that time.

The information from problem 5 is:

dP/dt= P(a-bP)-h , P(0)=P0


Attempt:
I subbed in the values for a,b, and h into the problem. Used variable separable and integrated both sides. My solution to the integration came out to be:

-2/(2P-5) = -t + C

I have absolutely no clue where to go from there. My roommate who took the class last semester couldn't figure out what to do either. Any help would be much appreciated.
 

Answers and Replies

  • #2
Redbelly98
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As a next step, I would:
  • Solve for P in terms of t
  • Take the derivitive of P, and verify that it is a solution to the differential equation
 
  • #3
by solving for P, I get: P= 1/t + 5/2. With what respect am I taking the derivative because if you do P you will get 1. Otherwise dP/dt= -1/t2. I'm still stuck either way because our professor doesn't explain topics very well. He just writes things out and doesn't actually say what he did so following it isn't easy.
 
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  • #4
Redbelly98
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by solving for P, I get: P= 1/t + 5/2.
Hmm, what happened to the C you had before? That should be in this equation somewhere. (Getting P(0)=P0=∞ at t=0 is a sign that something is wrong.)
With what respect am I taking the derivative because if you do P you will get 1. Otherwise dP/dt= -1/t2.
The differential equation you had was "dP/dt=....". So you need dP/dt in order to verify it.
After you account for the C term from before, compare dP/dt with the RHS of the differential equation you had.
I'm still stuck either way because our professor doesn't explain topics very well. He just writes things out and doesn't actually say what he did so following it isn't easy.
 
  • #5
I guess I forgot to write the C in. but I've come up with something else but I don't know if it's correct.

Taking P= 5/2 + 1/(C+t) I used P(0)=P0 = 1/C +5/2. then solved for C and got.
C=1/(P0-(5/2))
Then: used P(t1)=0 and plugged in my C value into the original P= 5/2 + 1/(C+t) getting:
5/2 +1/(1/P0-(5/2))+t1)=0. I then solved for t getting
t1= (2/5)(-5/(2P0+5)-2/5.

My next thought I had was to do: lim (t->infinity)P(t)= 5/2 + 1/(1/P0-(5/2))+t1). When I take the limit my answer for P(t)= 5/2. Im not sure what that answer actually gives me.
 
  • #6
cepheid
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Guys, I'm confused:

[tex]\frac{dP}{dt} = aP - bP^2 - h[/tex]

To me, this equation looks non-linear and non-separable. How did you solve it by straight integration???
 
  • #7
Redbelly98
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I guess I forgot to write the C in. but I've come up with something else but I don't know if it's correct.

Taking P= 5/2 + 1/(C+t) I used P(0)=P0 = 1/C +5/2. then solved for C and got.
C=1/(P0-(5/2))
Then: used P(t1)=0 and plugged in my C value into the original P= 5/2 + 1/(C+t) getting:
5/2 +1/(1/P0-(5/2))+t1)=0. I then solved for t getting
t1= (2/5)(-5/(2P0+5)-2/5.
Looks good. For a question like this, we're concerned only concerned with positive time ... i.e. does P ever become zero, after starting out at some initial value P0?

My next thought I had was to do: lim (t->infinity)P(t)= 5/2 + 1/(1/P0-(5/2))+t1). When I take the limit my answer for P(t)= 5/2. Im not sure what that answer actually gives me.
It tells you the solution to the equation at infinite times, but does not tell you what you need: does the solution ever become zero at any time?

Guys, I'm confused:

[tex]\frac{dP}{dt} = aP - bP^2 - h[/tex]

To me, this equation looks non-linear and non-separable. How did you solve it by straight integration???
But it is separable. Time appears only in the dt term, so it's trivial to separate.
 

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