Differential Equations · step 12 of 14

Slope fields, equilibria & qualitative behaviour

A slope field draws the direction dy/dx = f(x,y) at each point, so solution curves follow the arrows without any formula being solved. Equilibria are the constant solutions where f = 0, and their stability follows from the sign of f nearby: solutions approaching from both sides means stable, moving away means unstable, and mixed means semi-stable.

Practise slope fields, equilibria & qualitative behaviour free → All Differential Equations subtopics

✓ Unlimited questions · marked criterion by criterion · no card needed

Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Find the equilibriaSolve f(x,y) = 0 for y. These horizontal lines are constant solutions.
Sign-test between the equilibriaPick a test y in each interval and evaluate f. Positive means solutions rise; negative means they fall.
Classify each equilibriumArrows pointing in from both sides means stable; out on both sides means unstable; one of each means semi-stable.
Sketch representative curvesThey follow the arrows and can never cross an equilibrium, by uniqueness.

Worked example

For dy/dt = y(1 − y), find the equilibria and classify them.

Equilibria: y(1 − y) = 0, so y = 0 and y = 1.
For 0 < y < 1, take y = 0.5: f = 0.5(0.5) = 0.25 > 0, so solutions increase towards 1.
For y > 1, take y = 2: f = 2(−1) = −2 < 0, so solutions decrease towards 1.
For y < 0, take y = −1: f = −1(2) = −2 < 0, so solutions decrease away from 0.

Answer. y = 0 is unstable and y = 1 is stable — the logistic equation, where any positive starting population tends to the carrying capacity 1.

Where marks get dropped

These are the specific errors that cost credit on slope fields, equilibria & qualitative behaviour questions — QED's rubric penalises each of them separately.

Drawing solution curves that cross an equilibrium line. Uniqueness forbids it, so curves approach asymptotically instead.
Testing only one side of an equilibrium. Stability depends on the behaviour on BOTH sides.
Confusing the sign of f with the sign of y. The arrows are determined by dy/dt, not by the value of y.

Practise this until it is automatic

Unlimited fresh questions

QED generates new slope fields, equilibria & qualitative behaviour problems on demand at warm-up, exam and challenge level, so you can drill this one skill until it stops costing you marks.

Marked like an examiner

Every answer is scored against a point-by-point rubric with partial credit, so you see exactly which step of the method broke down — not just a tick or a cross.

Answer in real notation

A one-tap symbol palette, a visual equation editor and a truth-table builder — or photograph your handwritten working and QED converts it to LaTeX.

Saved to your library

Every question you generate is kept and re-takeable as a timed exam, and your Differential Equations mastery is tracked so you know when this is exam-ready.

Slope fields, equilibria & qualitative behaviour — frequently asked questions

What is an autonomous equation?

One where f depends only on y, not on t. Its slope field looks the same in every vertical strip, which is why the phase-line analysis works.

How do I classify stability algebraically?

For an autonomous equation, an equilibrium y* is stable when f′(y*) < 0 and unstable when f′(y*) > 0. For the logistic f′ = 1 − 2y, giving f′(0) = 1 > 0 and f′(1) = −1 < 0 ✓.

Why bother when I can solve the equation?

Because most equations cannot be solved in closed form. Qualitative analysis gives the long-run behaviour regardless.

The rest of Differential Equations

Solving and modelling with ODEs, from separable to systems. Each subtopic below has its own method, worked example and mark-losing traps.

Ready to make slope fields, equilibria & qualitative behaviour exam-proof?

Generate your first questions free — no card, no setup, no personal data stored. Practise until the method is second nature.

Start practising free →