What is the long-term behavior of solutions for $dz/dt = 4z - z^3 + 4$?

The long-term behavior is determined by the phase line. Any initial condition strictly above the unstable equilibrium $z = 0$ will approach the stable equilibrium $z = 2$.

Conditions: The differential equation is $dz/dt = 4z - z^3$.; Initial conditions are not exactly at the unstable equilibrium $z = 0$.

A bifurcation diagram is a plot that shows how the equilibrium values of a dynamical system change as a parameter varies. It is constructed by plotting the equilibrium values (z) on the vertical axis against the parameter (C) on the horizontal axis.

Conditions: The system is described by a family of ODEs parameterized by C.; Equilibria are determined for a range of C values.

Equilibria are found by setting the time derivative to zero, which means finding the roots of the right-hand side function. By plotting $f(z) = 4z - z^3$, the equilibria correspond exactly to the points where the graph intersects the horizontal z-axis.

Conditions: The differential equation is autonomous, meaning the right-hand side depends only on z.; Equilibria are defined as constant solutions where $dz/dt = 0$.

Equilibria correspond to the roots of the right-hand side function $f(z) = 4z - z^3 + C$. As C increases, the graph of $f(z)$ shifts vertically upward.

Conditions: The differential equation is $dz/dt = 4z - z^3 + C$.; $C = 4$, which is greater than the critical value C* ≈ 3.079.

Increasing C from 0 to 1 translates the graph of the right-hand side function vertically upward by one unit. This shift causes the three equilibria to move: the two stable equilibria shift to the right, and the unstable equilibrium shifts to the left.

Conditions: The differential equation is $dz/dt = 4z - z^3 + C$.; C is changed from 0 to 1.