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The Bernoulli Equation and Linearizing Substitutions

A nonlinear first-order equation can sometimes be tamed by changing what you call the unknown. The Bernoulli substitution turns a stubborn y^n into a clean linear equation — and the same trick reappears in disguise everywhere.

The trick behind every substitution

By now you have a small arsenal: you can solve a separable equation by splitting variables, a linear first-order equation with an integrating factor, an exact equation by recovering a potential, and — just last guide — a homogeneous-degree equation by setting v = y/x. Notice what the homogeneous trick really did: it did not solve the equation directly, it *changed the unknown* from y to v, and in the new variable the equation became separable. That is the whole spirit of substitution.

A substitution is a bet: you guess a new variable v = (some combination of x and y) such that, after rewriting, the equation lands in a form you already know how to solve. The art is choosing v so the messy piece collapses. There is no universal recipe — but for a few famous patterns the right substitution is known in advance, and the most important of these is the Bernoulli equation. It looks nonlinear and frightening, yet one clever change of variable straightens it into a linear equation you can finish on autopilot.

Spotting a Bernoulli equation

A Bernoulli equation is almost a linear equation, spoiled by a single power of y on the right. Its shape is y' + p(x) y = q(x) y^n, where n is any real number. Compare it to the linear form y' + p(x) y = q(x): the only difference is that lonely y^n factor multiplying the right-hand side. When n = 0 the right side is just q(x) and the equation is already linear; when n = 1 you can absorb the y back into the left and it is linear too. So the genuinely Bernoulli, genuinely nonlinear cases are every other n — n = 2, n = 1/2, n = -3, and so on.

linear:     y' + p(x) y = q(x)
Bernoulli:  y' + p(x) y = q(x) y^n      (n != 0, 1)

  the only wrinkle is the  y^n  on the right
A Bernoulli equation is the linear template with one nonlinear power of y bolted on.

You have already met one without being told its name. The logistic equation y' = r y (1 - y) expands to y' = r y - r y^2, which is y' - r y = -r y^2 — exactly Bernoulli with n = 2. So population growth with a carrying cap is a Bernoulli equation in disguise, and the substitution below gives you a second, completely different route to its solution besides separating variables.

The substitution that linearizes it

Here is the magic move, the Bernoulli substitution: let v = y^(1-n). Why that strange power? Because differentiating it produces exactly the combination we need. By the chain rule, v' = (1-n) y^(-n) y'. The factor y^(-n) is the perfect counterweight to the y^n bothering us. If you divide the whole Bernoulli equation by y^n first, the right side becomes plain q(x), the left becomes y^(-n) y' + p(x) y^(1-n) — and both of those pieces are exactly what v and v' are made of.

  1. Confirm the form: rearrange to y' + p(x) y = q(x) y^n and read off n. (If n is 0 or 1, stop — it is already linear, no substitution needed.)
  2. Divide every term by y^n, giving y^(-n) y' + p(x) y^(1-n) = q(x).
  3. Substitute v = y^(1-n), so v' = (1-n) y^(-n) y'. Replace y^(1-n) by v and y^(-n) y' by v'/(1-n).
  4. You now have v'/(1-n) + p(x) v = q(x) — a linear equation in v. Multiply through by (1-n) and solve it with the integrating-factor recipe.
  5. Convert back: once you have v(x), recover y from y = v^(1/(1-n)). Don't forget to undo the substitution at the end.

Watch it work on the logistic case y' - r y = -r y^2, where n = 2 so 1 - n = -1 and v = y^(-1) = 1/y. Dividing by y^2 and substituting turns it into the linear equation v' + r v = r, which the integrating factor e^(rt) cracks instantly. The nonlinear logistic curve becomes a one-line linear computation — the substitution did the hard part. Notice that v = 1/y is exactly the reciprocal you may already have used to solve the logistic equation by hand; the Bernoulli method explains *why* that particular reciprocal was the right thing to try.

The same idea, many disguises

Once you see substitution as 'rename the unknown until the equation simplifies', a whole zoo of tricks becomes one idea. The v = y/x substitution from last guide is the same move for homogeneous equations. When an equation depends only on the combination ax + by, the substitution v = ax + by often makes it reducible to separable. And for an equation that is *almost* linear but has an extra quadratic term — the Riccati equation y' = p(x) + q(x) y + r(x) y^2 — there is no general elementary solution at all, but if you can guess one particular solution, a substitution turns the remainder into a Bernoulli equation you *can* solve.

Be honest about the limits here. Bernoulli always works for its template, mechanically and reliably. The others are opportunistic: Riccati needs a lucky guess, the v = ax + by trick needs the equation to depend on that exact combination, and the Clairaut equation (next guide) brings a genuinely new phenomenon — solutions the general formula cannot reach. Substitution is powerful, but it is a collection of pattern-matched gifts, not a single algorithm that conquers every first-order equation. The overwhelming majority of first-order ODEs still have no closed-form solution; these named cases are the lucky, solvable islands.

Why linearizing is worth the trouble

You might ask: why bend over backwards to reach a *linear* equation specifically? Because linear is the one corner of the subject where we have complete, guaranteed machinery. A first-order linear equation always has an explicit solution by the integrating factor — no guessing, no luck, no special functions required. Nonlinear equations have none of these guarantees. So when a substitution converts a nonlinear equation into a linear one, it is not cosmetic; it moves the problem from 'maybe unsolvable' into 'definitely solvable'.