Population growth carrying capacity on a logistic model

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Homework Statement


Here is the equation I am given. I'm supposed to find the carrying capacity.

dp/dt=.05P-6.6666667e-5P^2

I know the general solution is rp-rp^2/k with k=carrying capacity, but the addition of the middle term has thrown me off.

The Attempt at a Solution



I tried ignoring the middle term and did r/p=5. Since r=.05, p would be .01 in this case, but that doesn't show up as the correct answer. What does the middle term change?
 

Answers and Replies

  • #2
lanedance
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can you explain which is the middle term you're talking about?
 
  • #3
lanedance
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guessing what you're asking, first consider the simple first order exponential DE
[tex]
\frac{dP}{dt}=0.05P
[/tex]

for any positive starting value P_0, [itex] \frac{dP}{dt}=0.05P[/itex] will always be positive and [itex] P(t) [/itex] will increase exponentially

now consider
[tex]
\frac{dP}{dt}=0.05P - \frac{2}{3} 10^{-5} P^2
[/tex]

for small [itex] P [/itex], this will behave the same as the first equation as [itex] P<1 \implies P^2<<1 [/itex]

however as [itex] P [/itex] increases, the [itex]P^2[/itex] term will get larger decreasing the growth rate, until at some point [itex] \frac{dP}{dt} = 0 [/itex], the population will assymtotically approach this point -

so can you find where [itex] \frac{dP}{dt} = 0 [/itex]?
 

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