Find the equilibrium solutions of the autonomous ordinary differential equation and determine whether they are stable or unstable.

dx/dt = (6-x)(x-2)

x≥ 0 and t≥0

Guest Aug 21, 2019

#1**+1 **

\(\text{Equilibrium solutions are found where $\dfrac{dx}{dt}=0$}\\ \text{Here these occur at $x=6,~x=2$}\\~\\ \text{To determine their stability we look at the second derivative}\\ \dfrac{d^2x}{dt^2} = (-1)(x-2) + 1(6-x) = 8-2x\\ \text{at $x=6,~\dfrac{d^2x}{dt^2}=8-12=-4$}\\ \text{A negative second derivative indicates an unstable equilibrium point}\\ \text{at $x=2,~\dfrac{d^2x}{dt^2}=8-4=4$}\\ \text{A positive second derivative indicates a stable equilibrium point}\\ \)

.Rom Aug 21, 2019