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: Picture It
Start with something you already know.
A marble and a bowl
- A marble rests at the very bottom of a bowl.
- Nudge it, and it rolls back down to the bottom.
- Now turn the bowl upside down and balance the marble on top.
- It can rest there too, as long as nothing touches it.
- The smallest nudge sends it rolling off, and it never comes back.
Both resting spots are equilibria, but only the bottom is stable: a nudge there dies away, while on top it grows.
Where the picture breaks: a real marble overshoots and rocks back and forth.
A solution in this lesson never overshoots; it creeps toward a stable value without ever passing it.
Step 2: Let's Learn
Read it, or press Listen and follow the words.
Try it first
A quantity y changes by the rule dy/dt = 3 − y. Right now y = 5. What is its rate of change at this moment?
Stuck is fine. Let’s Learn shows the way next. Got it first time? You can jump ahead to Watch an Example.
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 3: 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 4: 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 5: Watch an Example
One step at a time.
Watch Noor Build a Phase Line
A fresh equation, dy/dt = 9 − y². Noor finds its equilibria, tests one value in each interval, reads the arrows, and checks with the faster test.
- Step 1
Setting the rate to zero gives 9 − y² = 0, so y² = 9 and the equilibria are y = 3 and y = −3. These two constant solutions cut the line of y-values into three intervals.
Watch Ravi Follow a Solution Home
Ravi studies dy/dt = (y − 2)(y − 6). He finds the equilibria and tests the outer intervals; the middle test, one classification and where a solution from y = 5 ends up are yours.
- Step 1
A product is zero when one of its factors is, so the equilibria are y = 2 and y = 6.
Step 6: Your Turn
Practice makes it stick.
The Trout Pond
Problem 1 of 15
A pond is stocked with 900 trout. The number of trout P follows dP/dt = 0.001P(600 − P), with t in years. As the years go by, how many trout does the population approach?
trout
The Savings Account
Problem 2 of 15
An account earns 4% interest a year, and its owner withdraws $800 a year, spread evenly. The balance B, in dollars, follows dB/dt = 0.04B − 800. The balance today is $19,000. At what balance, in dollars, would the account hold exactly steady?
dollars
Settle or Escape
Set the rate to zero, test a value in each interval, and read the arrows. Nothing needs solving.
Problem 3 of 15
dy/dt = 12 − 3y. At what value of y is there an equilibrium?
Problem 4 of 15
dy/dt = y(y − 5). What is the rate at y = 2?
Problem 5 of 15
dy/dt = y(y − 5) has equilibria at 0 and 5. Which of them is stable?
Problem 6 of 15
dy/dt = f(y) with f(y) = y² − 6y + 8 has an equilibrium at y = 4. For the faster test, what is f′(4)?
Problem 7 of 15
dy/dt = (y − 3)². What kind of equilibrium is y = 3?
Whose Step?
One line on each page does not hold. Find it, then put the reading right.
Problem 8 of 15
Ana classified the equilibria of dy/dt = (y − 1)(y − 4). One line does not hold. Which is the first?
Ana’s classification for dy/dt = (y − 1)(y − 4)
- 1(y − 1)(y − 4) = 0 at y = 1 and y = 4
- 2y = 0: (−1)(−4) = 4 > 0, rises
- 3y = 2: (1)(−2) = −2 < 0, falls
- 4y = 5: (4)(1) = 4 > 0, rises
- Answer1 is unstable, 4 is stable
Problem 9 of 15
A solution of dy/dt = (y − 1)(y − 4) starts at y = 0. What value does it approach as t grows?
Problem 10 of 15
Leo built the phase line for dy/dt = (y − 4)(y + 2). One line does not hold. Which is the first?
Leo’s phase line for dy/dt = (y − 4)(y + 2)
- 1(y − 4)(y + 2) = 0 at y = 4 and y = −2
- 2y = −3: (−7)(−1) = 7 > 0, rises
- 3y = 0: (−4)(2) = 8 > 0, rises
- 4y = 5: (1)(7) = 7 > 0, rises
- Answer−2 and 4 are both semi-stable
Problem 11 of 15
For dy/dt = (y − 4)(y + 2), what kind of equilibrium is y = −2?
Pick the Move
Decide what to do next. Neither one needs the equation solved.
Problem 12 of 15
Kai found that dy/dt = y(10 − y) has equilibria at 0 and 10. To classify y = 0 by sign tests, where should he test?
Problem 13 of 15
Mia tries the faster test on dy/dt = (y + 1)² at its equilibrium y = −1, and finds f′(−1) = 0. What should she do next?
Same Phase Line?
Judge these from the equations. A few sign tests are all you need.
Problem 14 of 15
Select every equation whose phase line has a stable equilibrium at y = 2.
Problem 15 of 15
A phase line marks the equilibria and the sign of the rate between them. Select every pair of equations with exactly the same phase line.
Step 7: 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?
Ready for more?Optional. Past the lesson, for anyone who wants it.
Stretch 1 of 2
An equation of the form dy/dt = y² + by + c has a stable equilibrium at y = 1 and an unstable one at y = 6. What is c?
Stretch 2 of 2
A fishery holds P thousand fish, and P follows dP/dt = P(10 − P) − h, where h is the catch in thousands of fish a year. What is the largest catch h for which the population still has an equilibrium?
thousand fish a year
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.
For the grown-up
The misconception this lesson targets is reading stability from one side, or from the sign of the rate alone: "the rate is positive above it, so it is stable", when a positive rate above an equilibrium carries solutions away from it. Stability is about direction relative to the equilibrium, on both sides, so each sign test has to be turned into an arrow pointing toward or away from it before anything is concluded. A student who says "above 4 the solutions rise, which is away from 4" out loud will not make it. Two close cousins: a sign slip in a product of factors, which flips one arrow and turns a stable equilibrium into a semi-stable one on paper; and treating f′(y*) = 0 as a verdict, when the faster test then says nothing and only the sign tests decide.