Answer:Limiting Population: Solving Logistic Equation with P_0

[highlander2k5]

I'm stuck on what to do here. The question reads Consider a population P(t) satisfying the logistic equation dP/dt = aP-bP^2, where B = aP is the time rate at which births occur and D = bP^2 is the rate at which deaths occer. If the intial population is P(0) = P_0 (supposed to be P sub not), and B_0 births per month and D_0 deaths per month are occurring at time t=0, show that the limiting population is M = (B_0*P_0)/D_0.

My question is am I setting this up right? Where do I go from my last spot to get it to look like M = (B_0*P_0)/D_0?

Here's what I got:
1) dP/dt=(aP-bP^2)P
2) dP/dt=(a-bP)P^2
3) integral((1/P^2)+bP)dp = integral(a)dt
4) ? Please Help ?

 

[HallsofIvy]

I'm stuck on what to do here. The question reads Consider a population P(t) satisfying the logistic equation dP/dt = aP-bP^2, where B = aP is the time rate at which births occur and D = bP^2 is the rate at which deaths occer. If the intial population is P(0) = P_0 (supposed to be P sub not), and B_0 births per month and D_0 deaths per month are occurring at time t=0, show that the limiting population is M = (B_0*P_0)/D_0.

My question is am I setting this up right? Where do I go from my last spot to get it to look like M = (B_0*P_0)/D_0?

Here's what I got:
1) dP/dt=(aP-bP^2)P

Since you were told that dP/dt= (aP- bP^2) where did you get that additional P?

2) dP/dt=(a-bP)P^2
3) integral((1/P^2)+bP)dp = integral(a)dt
4) ? Please Help ?

I have no idea where you got (1/P^2 + bP)dP!

From dP/dt= (a-bP)P, you get dP/((a-bP)P)= dt and can integrate the left side using "partial fractions".

 

[dhris]

I think you are making it a little more complicated than it needs to be. If you take "limiting population" to mean "steady-state population", what condition does that place on the time derivative? Figuring this out will easily allow you to find M. Then to get the answer in the required form, you will need to figure out what a and b are in terms of P(0), B(0), and D(0).

 

[doubedylan]

I'm stuck on what to do here. The question reads Consider a population P(t) satisfying the logistic equation dP/dt = aP-bP^2, where B = aP is the time rate at which births occur and D = bP^2 is the rate at which deaths occer. If the intial population is P(0) = P_0 (supposed to be P sub not), and B_0 births per month and D_0 deaths per month are occurring at time t=0, show that the limiting population is M = (B_0*P_0)/D_0.

My question is am I setting this up right? Where do I go from my last spot to get it to look like M = (B_0*P_0)/D_0?

Here's what I got:
1) dP/dt=(aP-bP^2)P
2) dP/dt=(a-bP)P^2
3) integral((1/P^2)+bP)dp = integral(a)dt
4) ? Please Help ?

I am having the same problem (I have the same book, hoping bumping this will answer it). The error you had was you should START with dP/dt=aP-bP^2.

I started by separating the eqn into:
(1/aP-bP^2)dp=dt --> (1/(P(a-bP)))dP=dt
From there I tried partial fractions, but the answer seems way to complex.
Anyone know how to answer the initial question the other user typed?

 

[doubedylan]

I think I solved it. You don't integrate. M is known to be the limiting population, and you just need to prove B_o*P_o/D_o which if you substitute is (a/b)

Take P(a-bP) --> (b/b)(P(a-bP), which will can be simplified to bP(a/b-P)
If you compare to kP(M-P), they are equal. a/b is M, which is limiting capacity.