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Math101learn.math101.caAutonomous Equations
A phase-line approach to autonomous first-order equations, equilibria, stability, and exact solution curves.
Precise definition
An autonomous first-order differential equation has the form $y'=f(y)$: the rate depends on the current state $y$, not explicitly on the independent variable $t$. An equilibrium is a constant solution $y(t)=y_$ satisfying $f(y_)=0$. Non-equilibrium solutions can often be separated as $dy/f(y)=dt$ on intervals where $f(y)\ne0$.
Notation and mathematical language
Use $t$ for time, $y(t)$ for state, and $f(y)$ for the vector field. A phase line marks zeros of $f$ and arrows according to the sign of $f$: $f>0$ means $y$ increases and $f<0$ means $y$ decreases. Stability describes nearby solutions, not whether the equilibrium value itself changes.
Conceptual picture
Because every point at the same height has the same slope, horizontal translation of a solution curve produces another solution where defined. On the phase line, arrows pointing toward an equilibrium from both sides indicate asymptotic stability; arrows pointing away indicate instability; one-sided attraction gives semistability.
Conditions and key results
Sign analysis requires intervals on which $f$ is defined and continuous. Separating by dividing through $f(y)$ discards equilibrium solutions, so find them first. Uniqueness prevents distinct solution curves from crossing when the local hypotheses, such as continuity of $\partial f/\partial y$, hold.
A reliable strategy
- Solve $f(y)=0$ and record every constant equilibrium before any division.
- Partition the state axis at equilibria and singularities, then test the sign of $f$ on each interval.
- Classify stability from arrow directions and use separation if an explicit non-equilibrium solution is required.
- Apply the initial condition, state the maximal meaningful interval, and compare the formula with the phase-line prediction.
Fully worked example
Interpretation and application
Autonomous equations model populations, temperature feedback, chemical concentration, and one-dimensional control. The equilibrium classification predicts long-run behaviour without solving explicitly. That prediction remains conditional on the model: external time-dependent forcing would make the equation non-autonomous and can change the conclusion.
Common mistakes
Verification and reasonableness
- Differentiate the explicit formula and verify $y'=y(2-y)$.
- Check the initial value exactly and confirm the solution never crosses a uniqueness-protected equilibrium.
- Compare monotonicity and limiting values with the phase line.
Practice
- Find the equilibria of $y'=y(3-y)$.
- Classify $y=3$ for that equation.
- What was lost by dividing by $f(y)$?
Answers and brief solutions
- $y=0$ and $y=3$.
- Stable: the field is positive below 3 and negative above 3.
- All constant solutions at zeros of $f$; they must be restored separately.
Further deduction
Linearization sharpens the phase-line test. If $f$ is differentiable, $f(y_)=0$, and $f'(y_)<0$, then the local approximation $u'=f'(y_)u$ decays, so $y_$ is locally asymptotically stable; if $f'(y_)>0$, it is unstable. When $f'(y_)=0$, the test is inconclusive. For $y'=-y^3$, the origin is still stable and attracting even though the derivative test gives zero; direct sign analysis resolves the case.
Related topics
Explore the idea
Direction field and Euler step
Change one quantity at a time and connect what moves to Autonomous Equations.
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
For $y'=y(3-y)$, what is the long-run limit of a solution with $y(0)=1$?
- For $0<y<3$, $y(3-y)>0$, so the solution increases.
- Uniqueness prevents crossing $y=3$, and the stable equilibrium is the limit.
End of lesson
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