We are given $\frac{dN}{dT}=2N(N-10)(1-\frac{N}{100}) - N(t)$ is population size at any given time $t$. We can find the equilibria by putting the rate of change of $N'(t)=0$. Now the question asked is - use the eigenvalue approach to determine the stability of the equilibria you found.
Need help in understanding, how $N''(t)$ will lead to the answer. I am not very clear about "Eigenvalues" either.
Being my first question, it may be trivial, but I am here to learn.
$\dfrac{dN}{dt} = f(N) $
The equilibria are the roots of $f(N) = 0$. An equilibrium will be stable if $f(N-\epsilon)$ is positive, and $f(N+\epsilon)$ is negative, where $\epsilon $ is a small positive scalar. Now using Taylor series expansion of $f(N-\epsilon)$, it is given by,
$f(N - \epsilon) = f(N) - f'(N) \epsilon = - f'(N) \epsilon $
and
$ f(N + \epsilon ) = f'(N) \epsilon $
Hence, an equilibrium will be stable iff $f'(N) $ is negative.
Now $f(N) = 2 N(N-10)(1 - \dfrac{N}{100}) $
So the critical points are $N = 0, N= 10, N = 100$
$f'(N) = 2 (N - 10)(1 - \dfrac{N}{100}) + 2 N (1 - \dfrac{N}{100}) - \dfrac{1}{50} N (N - 10) $
Hence, $f'(0) = 2 (-10) = -20 , f'(10) = 20 (1 - 0.1) = 18, f'(100) = -2(90) = -180$
Therefore, the stable equilibria are $N = 0$, and $N = 100$.
