Modeling Mosquito Population Growth and Predation

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SUMMARY

The discussion focuses on modeling mosquito population growth and predation using differential equations. The initial population is 200,000 mosquitoes, which doubles weekly in the absence of predators, leading to the equation dP/dt = ln(2)P - 20,000. The correct solution involves understanding the exponential growth model and incorporating the predation factor, resulting in the population formula P = 201,977.31 - 1977.31e^(ln(2)t). The confusion arises from interpreting the proportionality factor and the impact of predators on the population dynamics.

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Ok, I have this modeling problem that I cannot figure out how to setup.
The problem is given:

The population of mosquitoes in a certain area increases at a rate proportional to the current population, and in the absence of other factors, the population doubles each week. There are 200,000 mosquitoes in the area initially, and predators (birds, bats, so forth) eat 20,000 mosquitoes/day. Determine the population of mosquitoes in the area at any time.


I don't understand what "absense of other factors" means. I'll put down what I have.

Let P(t) = the population of mosquitoes.
P(0)=200,000

dP/dt = 2*P(t) mosquitoes/week - 20,000 mosquitoes/day

(now simply convert week to days)
dP/dt = 2*P(t) mosquitoes/(7*day) - 20,000 mosquitoes/day

So yeah, that's the equation I have to model it. I have NO idea if that is right not, because like I said, I don't really understand what the question is asking.
Now, when I solve this problem I get a different answer then what the book has.
The book gives:
P=201,977.31-1977.31*e^(ln(2))*t, 0<=t<=t_f ~=6.6745 weeks

So, if anybody could point me in a direction of what I am doing wrong with the modeling... or maybe, just maybe my model is fine and I'm solving the DIFFEQ problem wrong...

Anyways... thanks in advance, I appreciate any help.
 
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Why do you think the proportionality factor is 2??
Let us consider the exponential growth scenario, that is, with no predators present.
Let time be measured in weeks.
We then have the diff.eq:
[tex]\frac{dp}{dt}=kp[/tex] where we regard k as momentarily an unkown constant.
This equation is easy to solve with an intial value [tex]p(0)=p_{0}\neq{0}[/tex]:
[tex]p(t)=p_{0}e^{kt}[/tex]
Now, we invoke the condition that after 1 week (t=1), the population is doubled ([tex]p(1)=2p_{0}[/tex]):
[tex]2p_{0}=p_{0}e^{k}\to{k}=ln(2)[/tex]

Thus, when we are to formulate the associated problem with predators, we get (with time still measured in weeks):
[tex]\frac{dp}{dt}=ln(2)p-7*a, a=20.000[/tex]
 
Ok, so the basic idea for problems like this is you hit condition A with some constant (say C from the integrating factor/solving process) and you hit the next condition with B (say with the constant K).


Yeah, that doubling process makes more sense. I was really just tossing variables and stuff in there, hoping it would work like magic without understanding it. That makes a hell of a lot more sense with what you are saying, and I'll play around with it when I get home tonight. Thanks man, I appreciate the help.
 

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