An autonomous equation has the form dy/dt = f(y). The rate depends on where you are, not on what time it is.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
What follows
The slope field looks the same along every horizontal line. Shifting a solution left or right gives another solution.
The phase line
Because time does not matter, the whole picture compresses to one vertical line marked with the equilibria.
Reading the signs
Between equilibria, check the sign of f(y). Positive means solutions rise there; negative means they fall.
Predicting the future
A solution moves along the phase line in the direction of the arrows until it reaches an equilibrium.
Without solving
All of this needs no integration at all. The long-run answer comes from the signs alone.
The rate depends only on the state
An autonomous equation has no explicit t on the right-hand side. Time does not matter directly, only the current value, which makes the qualitative analysis much simpler.
The phase line
Plot the rate against the state, mark where it is zero, and note the sign in each interval. Arrows showing the direction of motion give the entire long-run behaviour on one line.
Solutions are time-translates
Because time does not appear explicitly, shifting a solution in time gives another solution. That symmetry is why one phase line describes every solution regardless of when it starts.
Solutions are monotone between equilibria
A solution cannot turn around between equilibria, because doing so would require passing through a point where the rate is zero. That constraint makes the qualitative picture rigorous rather than suggestive.
Step 2: Try It Yourself
Tap and try it out.
- The equationdy/dx = −a·y
- Through(-3, 2)
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: Watch an Example
One step at a time.
Watch Anita Predict Without Solving
Anita has dy/dt = (y − 1)(y − 5) and a starting value of y = 3.
- Step 1
She finds the equilibria at y = 1 and y = 5.
Step 4: Your Turn
Practice makes it stick.
The Sign
Problem 1 of 2
dy/dt = (y − 2)(y − 8) with y = 5. What is dy/dt?
The Destination
Problem 2 of 2
dy/dt = (y − 2)(y − 8) starting at y = 5. Which equilibrium does the solution approach?
Read the Phase Line
1 of 8
dy/dt = (y − 1)(y − 4). How many equilibria are there?
2 of 8
dy/dt = (y − 1)(y − 4) with y = 2. What is dy/dt?
3 of 8
dy/dt = (y − 1)(y − 4) with y = 6. What is dy/dt?
4 of 8
Is dy/dt = y + t autonomous? 1 yes, 0 no.
5 of 8
Is dy/dt = y² − 9 autonomous? 1 yes, 0 no.
6 of 8
dy/dt = y² − 9. What is the positive equilibrium?
7 of 8
Sort each equation by whether it is autonomous.
Tap something to move it.
- Empty
- Empty
8 of 8
A solution starting exactly at an equilibrium of an autonomous equation. Where does it end up? Enter the equilibrium value if it stays, using 0 for an equilibrium at zero.
Step 5: Quick Check
Show what you know.
Question 1 of 2
dy/dt = (y − 3)(y − 7) with y = 5. What is dy/dt?
Question 2 of 2
What makes an equation autonomous?
What You Learned
- An autonomous equation has dy/dt depending only on y, never on t directly.
- The whole picture compresses to a phase line marked with equilibria and arrows.
- Long-run behaviour follows from the signs alone, with no integration at all.