For small perturbations about an equilibrium, terms of higher order than linear may be considered negligible (see linearization). Small angles θ‎, that a simple pendulum makes about the vertical, satisfy $\ddot{\theta }=-(g/l)\theta$; this is SHM, which has small bounded trigonometric solutions, and so θ‎ = 0 is a stable equilibrium. For θ‎ = π‎ + ε‎, that is, small perturbations about the upward vertical, we have

which has unbounded exponential solutions, and so θ‎ = π‎ is an unstable equilibrium.

Linearizing about (0,0,0), the three equations governing the Lorenz attractor give

where α‎,β‎,γ‎ > 0. So r(t) = (x(t), y(t),z(t))^T satisfies

If the matrix has distinct real eigenvalues λ‎₁,λ‎₂,λ‎₃ with eigenvectors v₁,v₂,v₃, then the solution is

This is a small bounded solution if λ‎₁,λ‎₂,λ‎₃ are all negative; so the origin is stable when β‎<1.