# Bernoulli Differential Equations

> Bernoulli differential equations for JEE Mathematics: dy/dx + Py = Qy^n reduced to linear by v = y^(1-n), with integrating factor and the lost y = 0 solution.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/bernoulli-differential-equation
- 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: "Bernoulli Differential Equations", PrepElephant, https://prepelephant.com/topics/jee/mathematics/bernoulli-differential-equation

## Direct answer

Bernoulli equations look nonlinear but solve like linear ones. The form dy/dx + P(x) y = Q(x) y^n, with n neither 0 nor 1, converts by dividing through by y^n and substituting v = y^(1-n): since dv/dx = (1 - n) y^(-n) dy/dx, the equation becomes dv/dx + (1 - n) P(x) v = (1 - n) Q(x), a first-order linear equation solved with the integrating factor e^∫(1-n)P dx. The two excluded values are degenerate: n = 0 is already linear, and n = 1 makes it separable (homogeneous linear). One caution survives every such solution: dividing by y^n discards the solution y ≡ 0 whenever n > 0, and a complete answer either notes it or checks whether the constant-zero solution satisfies the original equation.

## What you must remember

- **Standard form:** dy/dx + P y = Q y^n with n ≠ 0, 1; identify P and Q as functions of x before anything else.
- **The substitution:** divide by y^n, then set v = y^(1-n); the equation turns into dv/dx + (1 - n) P v = (1 - n) Q, linear in v.
- **Integrating factor:** IF = e^∫(1-n)P dx, and the solution is v × IF = ∫(1 - n) Q × IF dx + C.
- **Lost solution:** y ≡ 0 solves the original equation for n > 0 but is destroyed by the division — mention it or verify it separately.
- **Verify n first:** an equation like dy/dx + y = xy^2 is Bernoulli with P = 1, Q = x, n = 2; recognising the pattern is the entire diagnostic step.
- **Linear special case n = 0** gives the ordinary linear equation with IF e^∫P dx; n = 1 gives dy/dx = (Q - P) y, separable with exponential solutions.
- **Physical cameo:** n = 2 produces logistic-shaped solutions y = 1/(C e^{...} ...) whenever Q and P are constants — the Verhulst population model in disguise.

## Solving one end to end

Take dy/dx + y = y^2. Here P = 1, Q = 1, n = 2. Divide by y^2: y^(-2) dy/dx + y^(-1) = 1. Substitute v = y^(-1) (so 1 - n = -1 and dv/dx = -y^(-2) dy/dx): the equation reads -dv/dx + v = 1, or dv/dx - v = -1. The integrating factor is e^∫(-1)dx = e^(-x), so d/dx (v e^(-x)) = -e^(-x). Integrate: v e^(-x) = e^(-x) + C, hence v = 1 + C e^x. Return to y through v = 1/y: y = 1/(1 + C e^x). Check the constant C = 0 branch — y = 1 — by substitution: dy/dx = 0 and the equation demands 0 + 1 = 1, satisfied. And the lost solution y ≡ 0 also works in the original but cannot be recovered for any finite C, exactly the caveat promised. Notice the three signatures of a clean Bernoulli solution: the (1 - n) factor appears early, the IF carries it, and the final answer is checked at one point before submission.

## How the exam frames it

JEE Main presents the equation either overtly (solve dy/dx + y/x = x^2 y^2... style) or lightly disguised with a preliminary rearrangement — collecting y-terms on one side exposes the y^n structure. Advanced rarely announces the name; it hides the form inside substitution questions (an equation in x(y) that is Bernoulli in y as the independent variable) or asks for particular solutions through a given point where the constant matters. The predictable losses: forgetting the (1 - n) multiplier on the right side (answers off by a factor of 1 - n survive the algebra and look plausible); dividing by y^n and never revisiting y = 0; and misintegrating the IF when P involves 1/x — the classic P = 1/x gives IF = x^k shapes, and sign slips there are endemic. Bernoulli equations sit inside the differential-equations unit of both syllabi (NCERT Class 12 treats the linear case; Bernoulli is its standard JEE extension) and are among the few nonlinear first-order equations a JEE student is expected to finish reliably.

## Frequently asked questions

### What makes an equation Bernoulli?

The structure dy/dx + P(x) y = Q(x) y^n with n ≠ 0, 1 — a linear left side corrupted by a power of y on the right.

### Which substitution linearises a Bernoulli equation?

Divide by y^n and set v = y^(1-n); then dv/dx = (1 - n) y^(-n) dy/dx converts the whole equation to dv/dx + (1 - n)Pv = (1 - n)Q.

### Why is n = 1 excluded from the Bernoulli method?

Because y^(1-n) = y^0 = 1 makes v a constant; but the equation then reads dy/dx = (Q - P)y, which is separable and solves directly by exponentials.

### Which solution is at risk of being lost?

y ≡ 0, discarded by the division by y^n when n > 0; it always satisfies the original equation and should be reported or checked.

### What is the integrating factor after the substitution?

e^∫(1-n)P(x) dx, multiplying the linear equation in v = y^(1-n); the (1 - n) inside the exponent is the step most often fumbled.