An equilibrium is a constant solution, found by setting the rate to zero and solving for y with no calculus. It is stable if nearby solutions move toward it, unstable if they move away, semi-stable if attracted from one side and repelled from the other. A phase line marks the equilibria and the sign of the rate between them, and long-run behavior is read from it.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Stable
A small disturbance dies away. A marble in a bowl, a cup of coffee at room temperature, a population at its carrying capacity. Push it and it returns.
Unstable
A pencil balanced on its point. dy/dt = y² − 4 has equilibria at ±2. At y = 3 the rate is 5, rising away from 2; at y = 0 the rate is −4, falling away from 2. Nothing returns to 2.
The other one
At y = −3 the rate is 5, rising toward −2; at y = 0 the rate is −4, falling toward −2. Both sides approach −2: stable. The phase line reads ↑ −2 ↓ 2 ↑, and every arrow is a sign test.
Semi-stable
dy/dt = y² has one equilibrium, y = 0. At y = 1 the rate is 1, rising away; at y = −1 the rate is also 1, rising toward. Attracted from below, repelled from above. Semi-stable.
A faster test
If f(y*) = 0 and f′(y*) < 0 the equilibrium is stable; f′(y*) > 0, unstable. For y(4 − y), f′ = 4 − 2y: at 4 it is −4, stable; at 0 it is 4, unstable. One derivative replaces two sign tests.
Why it matters
Long-run behavior is usually the real question. A carrying capacity, a terminal velocity and a steady temperature are all stable equilibria, and the phase line answers where a system ends up without solving it.
Step 2: Try It Yourself
Tap and try it out.
dy/dt = y² − 4. Find the equilibria, draw the phase line from sign tests, classify each, and predict the solutions starting at y = 1 and y = 3
- 1y² − 4 = (y − 2)(y + 2) = 0: equilibria −2 and 2set the rate to zero
Step 0 of 4
- The equationdy/dx = −a·y
- Through(-3, -3)
Every short line shows the slope the equation demands at that point. The curve simply follows them, and the marked point is the initial condition that picks it out of the family.
- The equationdy/dx = a·y
- Through(-3, 0.5)
Every short line shows the slope the equation demands at that point. The curve simply follows them, and the marked point is the initial condition that picks it out of the family.
Step 3: In Real Life
A thermostat and a pencil
Room temperature settles at the thermostat setting: a stable equilibrium. A pencil balanced on its tip is an equilibrium that runs from any nudge. Control engineers design for the first kind.
Step 4: Watch an Example
One step at a time.
Watch Priti Classify Two Equilibria
Priti analyses dy/dt = y(4 − y), the logistic equation.
- Step 1
She sets the rate to zero, finding equilibria at y = 0 and y = 4.
Step 5: Your Turn
Practice makes it stick.
The Capacity
Problem 1 of 2
dP/dt = P(500 − P). What is the non-zero equilibrium population?
The Cooling Cup
Problem 2 of 2
dT/dt = −0.2(T − 20). What is the equilibrium temperature?
Settle or Escape
1 of 8
dy/dt = y − 7. What is the equilibrium?
2 of 8
dy/dt = y − 7. Is that equilibrium stable?
3 of 8
dy/dt = 7 − y. Is the equilibrium stable?
4 of 8
dy/dt = y(6 − y). How many equilibria are there?
5 of 8
dy/dt = y(6 − y). What is the stable equilibrium?
6 of 8
dT/dt = −0.5(T − 15). What is the long-run temperature?
7 of 8
Sort each behavior by the kind of equilibrium it describes.
Tap something to move it.
- Empty
- Empty
8 of 8
A solution starting exactly at an unstable equilibrium. Does it move?
Step 6: Quick Check
Show what you know.
Question 1 of 2
dy/dt = y(9 − y). What is the stable equilibrium?
Question 2 of 2
What makes an equilibrium stable?
What You Learned
- An equilibrium is a constant solution, found by setting the rate to zero.
- It is stable when nearby solutions return, unstable when they run away, semi-stable when it depends on the side.
- Test one value in each interval between equilibria to draw the phase line, or check the sign of f′.
- Carrying capacities, terminal velocities and steady temperatures are all stable equilibria.