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 $$N = 100$$.