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Bernoulli Differential Equations

Solve Bernoulli first-order differential equations by reducing them to a linear ODE via a substitution, with a worked example.

  • University Physics Year 1
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A Bernoulli equation is a first-order ODE of the form

dy/dx + P(x) y = Q(x) yⁿ,

where n ≠ 0,1.

This is one of the main “nonlinear, but still solvable” first-order types. In physics it can appear whenever the rate depends on a power of the state variable (a common modelling move for nonlinear loss, nonlinear absorption, saturating effects, etc.).

Quick links (big picture + applications)
Why it’s nice

After a substitution, a Bernoulli equation becomes a standard first-order linear ODE.

Method (reduce to linear)

Start with

y' + P(x) y = Q(x) yⁿ.
Template (the one you want to memorize)
  1. Divide by yⁿ.
  2. Substitute u = y¹⁻ⁿ.
  3. Solve the resulting linear ODE in u.
  4. Substitute back to get y.

  1. Divide by yⁿ:

    y⁻ⁿy' + P(x) y¹⁻ⁿ = Q(x).

  2. Substitute

    u = y¹⁻ⁿ.

    Then

    u' = (1-n)y⁻ⁿy'.

  3. Multiply the divided equation by (1-n) to get a linear ODE in u:

    u' + (1-n)P(x) u = (1-n)Q(x).

  4. Solve for u(x) using the integrating factor method, then substitute back y = u^(1/(1-n)).

Special cases

If n = 0 or n = 1, the equation is already linear (not “Bernoulli” in the usual sense).

Pitfalls
  • The factor (1-n) matters: u' = (1-n)y⁻ⁿy'; dropping (1-n) will break the algebra.
  • Don’t divide by yⁿ blindly: if y = 0 is a possible solution, handle it separately before dividing.
  • Check the final answer: differentiate your y(x) and verify it satisfies the original ODE.

Worked example

Solve

y' + y = x y².

This is Bernoulli with n = 2, P(x) = 1, Q(x) = x.

Divide by y²:

y⁻²y' + y⁻¹ = x.

Let u = y¹⁻² = y⁻¹ = 1/y. Since u' = -y⁻²y', the equation becomes:

-u' + u = x ⇒ u' - u = -x.

Solve the linear ODE u'-u = -x. The integrating factor is μ = e^(∫ (-1) dx) = e^(-x), so:

(d/dx)(u e^(-x)) = -x e^(-x).

Integrate:

u e^(-x) = ∫ -x e^(-x) dx + C = (x + 1)e^(-x) + C.

So

u = x + 1 + Ce^x ⇒ y = 1/(x + 1 + Ce^x).

Worked example (physics-flavoured): linear loss + quadratic loss

In some physics contexts you meet “linear + quadratic loss” models. A common mathematical form is:

dI/dx + α I = -β I²,

where I(x) is an intensity/flux-like quantity and α,β are constants. This is Bernoulli with n = 2.

Divide by I²:

I⁻²I' + α I⁻¹ = -β.

Let u = I⁻¹ so u' = -I⁻²I'. Then:

-u' + α u = -β ⇒ u' - α u = β.

Solve the linear ODE:

u(x) = -β/α + Ce^(α x).

So:

I(x) = 1/(Ce^(α x)-β/α).

If I(0) = I₀, then 1/I₀ = C-β/α, so C = 1/I₀ + β/α and

I(x) = 1/((1/I₀ + β/α)e^(α x)-β/α).

Practice (with hints + answers)

1) Identify n and solve: y' + 2y = 2y^2

Hint: This is Bernoulli with n = 2. Use u = y¹⁻ⁿ = 1/y.

Answer: Divide by y²:

y⁻²y' + 2y⁻¹ = 2.

Let u = 1/y so u' = -y⁻²y', giving

-u' + 2u = 2 ⇒ u' - 2u = -2.

Solve: u = 1 + Ce^2x. Therefore

y(x) = 1/(1 + Ce^2x).
2) Classify (no solving needed): y' + (1/x) y = x^2 y^3

Hint: Compare with y' + P(x)y = Q(x)yⁿ.

Answer: Bernoulli with n = 3, P(x) = 1/x, Q(x) = x².

3) Show the first move: y' - y = e^x y^4

Hint: Divide by y⁴ and set u = y¹⁻⁴ = y⁻³.

Answer:

y⁻⁴y' - y⁻³ = e^x

then with u = y⁻³ you get a linear ODE in u.

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