stability of an equilibrium
Two equilibria can both be states of perfect balance and yet behave completely differently when you nudge them. A marble at the bottom of a bowl, pushed slightly, rolls back — that is stable. A marble balanced on top of a dome, pushed slightly, rolls away and never returns — that is unstable. Stability is the question of what happens to a solution that starts near an equilibrium: does it return, or does it run?
For an autonomous equation dy/dx = f(y) with equilibrium y* (where f(y*) = 0), stability is read straight off the phase line. If the arrows on both sides point toward y*, nearby solutions converge back to it — it is stable (an attractor). If they point away, nearby solutions flee — it is unstable (a repeller). There is a quick analytic test, the linearization test: look at the derivative f'(y*). If f'(y*) is negative the equilibrium is stable; if positive it is unstable. (The intuition: near y*, the equation behaves like dy/dx approximately f'(y*)*(y - y*), an exponential growth or decay in the displacement — a negative rate decays back, a positive rate amplifies away.) When f'(y*) = 0 the test is inconclusive and you must inspect the sign of f directly; such points can be semi-stable, attracting from one side and repelling from the other.
Stability is one of the most practically important ideas in all of applied mathematics, because real systems sit at their stable equilibria — a population at its carrying capacity, a chemical reaction at equilibrium concentration, a thermostat-controlled room at its set temperature. Unstable equilibria exist but are never observed for long, since the slightest perturbation drives the system away. This one-dimensional notion is the seed of the much larger stability theory for systems, where eigenvalues replace the single derivative f'(y*).
For dy/dt = y*(1 - y), f'(y) = 1 - 2*y. At y = 0, f'(0) = 1 > 0, so y = 0 is unstable; at y = 1, f'(1) = -1 < 0, so y = 1 is stable. A small population grows away from 0 and settles at the carrying capacity 1.
The sign of f'(y*) decides stability: negative means stable, positive means unstable.
The derivative test fails exactly when f'(y*) = 0 — it gives no verdict there. Do not conclude 'stable' from f'(y*) = 0; you must check the sign of f on each side directly, and the point may turn out semi-stable.