A model for population growth is given by:

$$\frac{dN}{dt} = f(N) = r N \left( 1-\frac{N}{K} \right) \left( \frac{N}{U}-1 \right) $$

where $r,\ U,$ and $K$ are positive parameters and $U < K$. (a) Find all equilibrium solutions of this equation and classify each as stable, semistable, or unstable using calculus.

Please see the picture for the question at hand. Currently working my way through some questions and I've got a bit stuck here. I seem to have a logistic model multiplied by the explosion extinction model. I can easily identify what the equilibrium solutions are but how would I go about grading the stability via calculus and not a directionfield chart?

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    $\begingroup$ Show what you have tried before asking for help here. $\endgroup$ – Auclair Apr 3 '19 at 16:14
  • $\begingroup$ So far, I have tried to seperate the variables but from what I've seen I'm supposed to then integrate what's left? I don't understand why this is and the only similar examples I could find were for first order not second order diff equations. $\endgroup$ – narutorulez Apr 3 '19 at 16:19
  • $\begingroup$ Hint: what does it mean to be in an equilibrium? $\endgroup$ – Ertxiem - reinstate Monica Apr 3 '19 at 22:13
  • $\begingroup$ I understand that I make f(N) = 0, and that at equilibrium N = 0,K or U - that's not the part I am struggling with, it's showing the stability of the solutions with calculus, but thank you for the hint, I should have made that clearer tbh $\endgroup$ – narutorulez Apr 3 '19 at 22:18
  • $\begingroup$ Another hint: the stability is related with $f'(x)$. $\endgroup$ – Ertxiem - reinstate Monica Apr 3 '19 at 22:20

It's a lot simpler than it looks. Since you've already known that $0, K, U$ are equilibria, where $0<K<U$. Also, since $N$ represents a population, we'll only consider $N \ge 0$.

For $0 < N < K$, $f(N) < 0$ meaning $N$ will decrease to $0$.

For $K < N < U$, $f(N) > 0$ meaning $N$ will increase.

For $N > U$, $f(N) < 0$ meaning $N$ will decrease to $U$.

This establishes stability for $0, K$ and $N$. This method actually reveals the meaning behind the values of $K$ and $N$.

Alternatively, you can take $f'(N)$ like one of the comments suggested. But this is such a nice expression, why make it ugly to study just the mathematics part?


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