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Bernoulli Differential Equation — Definition, Formula & Examples

A Bernoulli differential equation is a first-order ordinary differential equation of the form y′+P(x)y=Q(x)yny' + P(x)y = Q(x)y^n, where nn is any real number other than 0 or 1. It is nonlinear when n≠0,1n \neq 0,1, but a clever substitution transforms it into a linear equation that you can solve with standard methods.

A Bernoulli equation is a first-order ODE dydx+P(x)y=Q(x)yn\frac{dy}{dx} + P(x)y = Q(x)y^n with n∈R∖{0,1}n \in \mathbb{R} \setminus \{0,1\}. The substitution v=y1−nv = y^{1-n} reduces it to the linear first-order equation dvdx+(1−n)P(x)v=(1−n)Q(x)\frac{dv}{dx} + (1-n)P(x)v = (1-n)Q(x), which can be solved using an integrating factor.

Key Formula

dydx+P(x) y=Q(x) yn\frac{dy}{dx} + P(x)\,y = Q(x)\,y^n
Where:
  • yy = Unknown function of $x$
  • P(x)P(x) = Coefficient function multiplying $y$
  • Q(x)Q(x) = Coefficient function on the right side
  • nn = Real exponent, with $n \neq 0$ and $n \neq 1$

How It Works

Start by dividing both sides by yny^n to isolate the nonlinear term. Then substitute v=y1−nv = y^{1-n}, so that dvdx=(1−n)y−ndydx\frac{dv}{dx} = (1-n)y^{-n}\frac{dy}{dx}. This converts the equation into a linear first-order ODE in vv. Solve that linear equation using an integrating factor, then back-substitute v=y1−nv = y^{1-n} to recover yy.

Worked Example

Problem: Solve the Bernoulli equation y′+y=y3y' + y = y^3.
Identify parameters: Here P(x)=1P(x) = 1, Q(x)=1Q(x) = 1, and n=3n = 3.
Substitute: Let v=y1−3=y−2v = y^{1-3} = y^{-2}. Then dvdx=−2y−3dydx\frac{dv}{dx} = -2y^{-3}\frac{dy}{dx}. Divide the original equation by y3y^3 to get y−3y′+y−2=1y^{-3}y' + y^{-2} = 1. Multiply both sides by −2-2:
dvdx−2v=−2\frac{dv}{dx} - 2v = -2
Solve the linear ODE: The integrating factor is μ=e∫−2 dx=e−2x\mu = e^{\int -2\,dx} = e^{-2x}. Multiply through: ddx(ve−2x)=−2e−2x\frac{d}{dx}\bigl(ve^{-2x}\bigr) = -2e^{-2x}. Integrate both sides:
ve−2x=e−2x+Cve^{-2x} = e^{-2x} + C
Back-substitute: Divide by e−2xe^{-2x} to get v=1+Ce2xv = 1 + Ce^{2x}. Since v=y−2v = y^{-2}:
y−2=1+Ce2x⟹y=±11+Ce2xy^{-2} = 1 + Ce^{2x} \quad\Longrightarrow\quad y = \pm\frac{1}{\sqrt{1 + Ce^{2x}}}
Answer: y=±11+Ce2xy = \pm\dfrac{1}{\sqrt{1 + Ce^{2x}}}

Why It Matters

Bernoulli equations appear in population dynamics (logistic growth), fluid mechanics, and circuit analysis. Mastering the substitution technique is essential in a first course on differential equations, as it illustrates how nonlinear problems can be linearized through a change of variable.

Common Mistakes

Mistake: Forgetting to multiply by the factor (1−n)(1-n) after substituting v=y1−nv = y^{1-n}.
Correction: The chain rule gives dvdx=(1−n)y−ndydx\frac{dv}{dx} = (1-n)y^{-n}\frac{dy}{dx}, so after dividing by yny^n you must multiply the entire equation by (1−n)(1-n) to correctly express it in terms of vv.