Bernoulli Differential Equations
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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.