# Population growth carrying capacity on a logistic model

## 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?

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lanedance
Homework Helper
can you explain which is the middle term you're talking about?

lanedance
Homework Helper
guessing what you're asking, first consider the simple first order exponential DE
$$\frac{dP}{dt}=0.05P$$

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

now consider
$$\frac{dP}{dt}=0.05P - \frac{2}{3} 10^{-5} P^2$$

for small $P$, this will behave the same as the first equation as $P<1 \implies P^2<<1$

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

so can you find where $\frac{dP}{dt} = 0$?