Differential Equations · step 5 of 14

Substitutions: homogeneous & Bernoulli

Two standard substitutions convert awkward first-order equations into solvable ones. If dy/dx can be written as a function of y/x alone, the substitution y = vx makes it separable. If the equation has the Bernoulli form y′ + P(x)y = Q(x)yⁿ, then v = y^(1−n) makes it linear — solvable by the integrating factor.

Practise substitutions: homogeneous & bernoulli 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.

Recognise the homogeneous formEvery term has the same total degree, or the right side depends only on y/x.
Substitute y = vxThen dy/dx = v + x dv/dx, and the equation becomes separable in v and x.
Recognise the Bernoulli formy′ + P(x)y = Q(x)yⁿ with n ≠ 0, 1 (those cases are already linear or separable).
Substitute v = y^(1−n)Then v′ = (1−n)y^(−n)y′, and the equation becomes v′ + (1−n)Pv = (1−n)Q — linear in v.

Worked example

Solve the Bernoulli equation y′ + y = xy² .

Here n = 2, so substitute v = y^(1−2) = y⁻¹, giving v′ = −y⁻²y′.
Divide the equation by y²: y⁻²y′ + y⁻¹ = x, i.e. −v′ + v = x.
Rearrange to linear form: v′ − v = −x, with integrating factor e^(−x).
Then (e^(−x)v)′ = −xe^(−x); integrating by parts gives e^(−x)v = xe^(−x) + e^(−x) + C.

Answer. v = x + 1 + Ce^x, so y = 1/(x + 1 + Ce^x).

Where marks get dropped

These are the specific errors that cost credit on substitutions: homogeneous & bernoulli questions — QED's rubric penalises each of them separately.

Forgetting the product rule in dy/dx = v + x dv/dx for the homogeneous substitution. Writing just x dv/dx loses a term.
Applying the Bernoulli substitution when n = 0 or 1, where it degenerates — those cases are already linear.
Not converting back at the end. The answer must be in terms of y, not v.

Practise this until it is automatic

Unlimited fresh questions

QED generates new substitutions: homogeneous & bernoulli 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.

Substitutions: homogeneous & Bernoulli — frequently asked questions

How do I spot a homogeneous equation?

Replace x by tx and y by ty. If the equation is unchanged, it is homogeneous and y = vx will work.

Why does v = y^(1−n) linearise Bernoulli?

Because dividing by yⁿ turns the y-terms into y^(1−n) and its derivative, which is exactly v and v′ up to the constant 1−n.

Are there other useful substitutions?

Yes — for y′ = f(ax + by + c), substitute u = ax + by + c to get a separable equation.

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 substitutions: homogeneous & bernoulli 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 →