When the right side does not mention $t$, as in $y' = f(y)$, the equation is autonomous. The logistic equation is the standard example. You can learn almost everything about its solutions without solving it, just by looking at the sign of $f$.
🔑 Key idea
Equilibria are the constants $y^{*}$ with $f(y^{*}) = 0$. Between equilibria, solutions increase where $f > 0$ and decrease where $f < 0$, and they never cross an equilibrium. An equilibrium is stable if solutions on both sides move toward it, unstable if both move away, and semi-stable if one side approaches and the other leaves.
💡 Why it works
A constant $y = y^{*}$ with $f(y^{*}) = 0$ solves the equation. By uniqueness, no other solution can touch it, so a solution that starts between two equilibria stays between them. Its slope $f(y)$ keeps one sign there, so it is monotone and must level off at an equilibrium (or run off to $\pm\infty$). Near $y^{*}$, $f(y) \approx f'(y^{*})(y - y^{*})$, so the gap behaves like $e^{f'(y^{*})t}$: $f'(y^{*}) < 0$ means stable, $f'(y^{*}) > 0$ unstable.
Drawing the phase line
Mark the equilibria on a vertical line, test the sign of $f$ in each gap, and draw an arrow up where $f > 0$ and down where $f < 0$. The arrows tell you each equilibrium's type at a glance.
✏️ Worked example: Two equilibria
Classify the equilibria of $y' = y^{2} - 4y + 3$.
Factor: $f(y) = (y - 1)(y - 3)$, so the equilibria are $y = 1$ and $y = 3$.
Signs: on $(-\infty, 1)$, $f > 0$ (up); on $(1, 3)$, $f < 0$ (down); on $(3, \infty)$, $f > 0$ (up).
At $y = 1$ the arrows point toward it from both sides; at $y = 3$ they point away.
Classify the equilibria of $y' = (y - 2)^{2}(y + 1)$.
Equilibria: $y = -1$ and $y = 2$.
Below $-1$: $f < 0$ (down). Between $-1$ and $2$: $f > 0$ (up). Above 2: $f > 0$ (up).
At $y = 2$, solutions below rise toward it, but solutions above rise away. Here $f'(2) = 0$, so the derivative test is silent.
Answer: $y = -1$ is unstable and $y = 2$ is semi-stable.
✏️ Worked example: Long-run behavior
For $y' = y(4 - y)$, describe $y(t)$ if $y(0) = 1$ and if $y(0) = 6$.
Equilibria $y = 0$ and $y = 4$; $f > 0$ on $(0, 4)$ and $f < 0$ above 4.
From $y(0) = 1$ the solution rises and cannot pass 4, so it levels off at 4.
From $y(0) = 6$ it falls toward 4.
Answer: In both cases $y(t) \to 4$, the stable equilibrium.
Left: the graph of $f(y)$ turned on its side, with the phase line and its arrows. Right: solutions $y(t)$. Dots: filled = stable, hollow = unstable, half-filled = semi-stable.
⚠️ Common mistake
Calling an equilibrium stable because solutions approach it from one side. Stability needs every nearby solution, on both sides, to approach. At $y = 2$ above, a solution that starts at $2.01$ moves away, so $y = 2$ is not stable. Check both neighbors of every equilibrium.
The direction field of $y' = (y - 1)(y - 3)$: the slopes are the same along every horizontal line, which is what autonomous means.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.