Bernoulli Differential Equation

What is a Bernoulli equation?

A Bernoulli equation is any first-order ODE of the form $\frac{dy}{dx} + P(x) \, y = Q(x) \, y^{n}, \qquad n \ne 0, 1.$

The $y^n$ on the right makes it non-linear in general, so the linear-ODE recipe (integrating factor) doesn't apply directly. But there's a slick trick: the substitution $v = y^{\,1 - n}$ transforms it into a linear ODE in $v$, which you can then solve with an integrating factor. This is one of the most satisfying dimensional reductions in the first-order toolbox.

The edge cases $n = 0$ and $n = 1$ are excluded because they're already linear (or separable) with no substitution needed.

The given equation

$\frac{dy}{dx} + y = y^{3}$

Standard Bernoulli form with $P(x) = 1$, $Q(x) = 1$, $n = 3$.

Step-by-step solution

Step 1 — Substitution.

With $n = 3$, let $v = y^{1 - 3} = y^{-2}$.

Differentiate (chain rule): $\frac{dv}{dx} = -2 y^{-3} \, \frac{dy}{dx} \quad\Longleftrightarrow\quad \frac{dy}{dx} = -\tfrac{1}{2} y^{3} \, \frac{dv}{dx}$

Step 2 — Rewrite the ODE in terms of $v$.

Divide the original equation through by $y^{3}$ (assuming $y \ne 0$): $y^{-3} \, \frac{dy}{dx} + y^{-2} = 1$

Use $v = y^{-2}$ so $v' = -2 y^{-3} y'$ $\implies$ $y^{-3} y' = -\tfrac{1}{2} v'$: $-\tfrac{1}{2} v' + v = 1$

Multiply by $-2$: $\frac{dv}{dx} - 2 v = -2$

That is a linear ODE in $v$. Non-linearity absorbed.

Step 3 — Solve the linear ODE.

Standard form: $P = -2$, $Q = -2$. Integrating factor $\mu(x) = e^{\int -2 \, dx} = e^{-2x}$. $\frac{d}{dx}\!\left[ e^{-2x} v \right] = -2 e^{-2x}$ $e^{-2x} v = \int -2 e^{-2x} \, dx = e^{-2x} + C$ $v(x) = 1 + C e^{2x}$

Step 4 — Back-substitute $v = y^{-2}$. $y^{-2} = 1 + C e^{2x}$ $\boxed{\, y(x) = \pm \frac{1}{\sqrt{\, 1 + C e^{2x} \,}} \,}$

The sign is chosen by the initial condition.

The singular solution we almost lost

Dividing by $y^{3}$ required $y \ne 0$. Check directly: $y \equiv 0$ satisfies $0 + 0 = 0$, so it is also a solution. This one is not recovered for any finite $C$ in our family — it sits outside and is a genuine singular solution.

Verification (for a specific $C$)

Take $C = 0$: $y = 1$ (constant). Plug into the ODE: $0 + 1 = 1 = 1^{3}$ ✓.

Take $C = -e^{-2}$, so $y(0) = 1/\sqrt{1 - e^{-2}}$ (some concrete non-trivial value). Then $y' = \ldots$ you can confirm $y' + y - y^3 = 0$ holds; the algebra is straightforward but tedious — the structural check above (constant solution) plus the derivation itself is usually enough.

Equilibria of the original equation

Set $y' = 0$ in $y^{3} - y = 0$: $y (y^{2} - 1) = 0 \implies y = 0, \pm 1$

• $y = 0$ is a stable equilibrium (small perturbation pulls back toward it from the $|y| < 1$ side).
• $y = \pm 1$ are unstable — solutions that start with $|y| > 1$ blow up in finite time.

Look at our solution family: $y = 1 / \sqrt{1 + C e^{2x}}$. If $C > 0$ the denominator grows, so $y \to 0$ as $x \to \infty$. If $C < 0$ the denominator can hit zero in finite $x$ — that's the blow-up time.

Bernoulli recipe (memorize)

1. Identify $n$ in $y' + P y = Q y^{n}$.
2. Substitute $v = y^{1 - n}$; compute $v' = (1-n) y^{-n} y'$.
3. Divide original equation by $y^{n}$, then rewrite in $v$: you get a linear ODE $v' + (1 - n) P(x) \, v = (1 - n) Q(x).$
4. Solve with integrating factor.
5. Back-substitute $y = v^{\,1/(1-n)}$.
6. Check for singular solutions where you divided.

Examples of $n$:

• $n = 2$ → logistic-like (actually the logistic equation is Bernoulli with $P = -r$, $Q = -r/K$).
• $n = 3$ → today's equation.
• $n = -1$ → $v = y^{2}$; shows up in drag problems.
• $n = 1/2$ → $v = y^{1/2}$.

Common mistakes

• Using $v = y^{n}$ instead of $v = y^{1-n}$. The exponent that linearizes is $1 - n$, derived by forcing the substitution to eliminate $y^{n}$.
• Forgetting the factor $(1 - n)$ on the linear ODE. Dividing by $y^{n}$ and computing $v'$ together produces that factor naturally — show your work.
• Missing $y = 0$ as a singular solution. Division by $y^{n}$ loses it; check explicitly.
• Dropping the $\pm$ on the back-substitution. $y^{-2} = v$ gives two branches $y = \pm v^{-1/2}$; the initial condition picks one.

Try it in the visualization

Slide $n$ (the exponent), watch the Bernoulli equation shape change, and see the substitution $v = y^{1-n}$ "unbend" the non-linearity into a linear problem in $v$. Overlay the stable and unstable equilibria and watch solution curves flow toward / away from them.