# Homogeneous Differential Equations

> Homogeneous differential equations for JEE Mathematics: y = vx substitution, separable exit routine, reducible forms and full worked solutions.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/homogeneous-differential-equations
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Homogeneous Differential Equations", PrepElephant, https://prepelephant.com/topics/jee/mathematics/homogeneous-differential-equations

## Direct answer

Homogeneity means every term of the differential equation carries the same total degree in x and y, so the right side of dy/dx = F(x, y) collapses into a function of the single ratio y/x — and the substitution y = vx, with dy/dx = v + x(dv/dx), turns the equation separable on the spot. The exit routine never varies: isolate x(dv/dx), separate, integrate both sides, then replace v by y/x. The reducible variants extend the reach: dy/dx = (ax + by + c)/(a′x + b′y + c′) with proportional coefficient rows becomes homogeneous after one linear substitution ax + by = v, while non-proportional constants demand shifting the origin to the intersection of the two lines first.

## What you must remember

- **The test:** divide the right side by the highest power of x; if only y/x survives (every term same total degree), the equation is homogeneous.
- **The substitution:** y = vx gives dy/dx = v + x dv/dx — forgetting the v term and writing just dv/dx is the chapter's top error.
- **Separation shape:** after substitution you always reach g(v) dv = dx/x; the right side integrates to ln|x| + C, never ln x alone.
- **Reverse gear:** equations homogeneous in x and y but arranged as dx/dy respond to x = vy with dx/dy = v + y(dv/dy).
- **Proportional-constants case:** if (a, b) and (a′, b′) rows are proportional, put ax + by = v and the equation separates immediately.
- **Non-proportional case:** shift the origin to the intersection of ax + by + c = 0 and a′x + b′y + c′ = 0, then proceed as homogeneous in the new variables.
- **Standard integrals on the exit:** arctan v and ln(1 + v²) appear whenever (1 + v²) sits below; recognise their derivatives on sight.

## A full homogeneous solve

Solve dy/dx = (x + y)/(x − y). Divide numerator and denominator by x: the right side becomes (1 + v)/(1 − v) with v = y/x, so v + x dv/dx = (1 + v)/(1 − v). Subtract v: x dv/dx = (1 + v)/(1 − v) − v = (1 + v − v + v²)/(1 − v) = (1 + v²)/(1 − v). Separate: (1 − v)/(1 + v²) dv = dx/x. The left side splits into 1/(1 + v²) − v/(1 + v²), integrating to arctan v − (1/2)ln(1 + v²), while the right side is ln|x| + C. Back-substitute v = y/x for the implicit solution arctan(y/x) − (1/2)ln(1 + y²/x²) = ln|x| + C, cleanable by noting ln(1 + y²/x²) = ln(x² + y²) − 2ln|x|. Every homogeneous item in the syllabus is this same five-move sequence: substitute, subtract v, separate, integrate, restore — and the integrals that appear are drawn from a short, learnable list.

## Recognising before solving

JEE Main tests recognition as hard as technique: presented with (x² + y²) dx = 2xy dy or dy/dx = (y² − x²)/(xy)? — both are homogeneous, both surrender to y = vx, and the paper's numerical answer usually comes from evaluating C at a given initial point. The planted non-homogeneous distractor is dy/dx = (x + y + 1)/(x − y + 3): the moment constants ride along, plain y = vx fails, and the coefficients of (1, 1) and (1, −1) are not proportional, so the origin must shift to the intersection of x + y + 1 = 0 and x − y + 3 = 0 — solving, that point is (−2, 1) — before the homogeneous machinery starts. JEE Advanced adds the dx/dy arrangement, where x² dy/dx type equations flip to x = vy faster than forcing y = vx. Marks leak in three places: the dropped v in v + x dv/dx, the missing modulus and constant on ln|x|, and leaving the answer in v instead of restoring y/x. Finish by differentiating the answer mentally — thirty seconds that catch every one of those errors at once.

## Frequently asked questions

### How do you test whether an equation is homogeneous?

Replace x by tx and y by ty throughout; if every term carries the same power of t, the equation is homogeneous and depends only on y/x.

### What substitution solves a homogeneous equation?

y = vx, converting dy/dx to v + x dv/dx and making the variables separable.

### What if constants appear alongside x and y in the numerator?

If the coefficient rows are proportional, substitute ax + by = v; otherwise shift the origin to the lines' intersection point first.

### When should x = vy be used instead?

When the equation arranges more naturally as dx/dy — same machinery, roles of x and y exchanged.

### What is the commonest error in these solutions?

Writing dv/dx instead of v + x dv/dx after the substitution — the chain rule contribution is the whole point of the method.