Bernoulli Equations and Substitutions

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The idea

A Bernoulli equation has the form $y'+p(t)y=q(t)y^n$ . For $n\ne0,1$ , the substitution $v=y^{1-n}$ makes it linear on a branch where the powers and division are valid. The cases $n=0,1$ are already linear.

Visual guide

VISUAL GUIDEA reciprocal substitution changes the equation

Nonlinear variable

Original y

Linear variable

Transformed v = 1/y

For y′ + y = y² with y(0) = 1/2, y = 1/(1 + eᵗ). Taking v = 1/y produces v = 1 + eᵗ, which solves the linear equation v′ − v = −1. The paired graphs describe the same solution in different dependent variables.

Method and assumptions

Divide by $y^n$ and use $v'=(1-n)y^{-n}y'$ . The transformed equation is $v'+(1-n)pv=(1-n)q$ . Solve by an integrating factor, then transform back. Check $y=0$ separately whenever the original equation is defined there.

Worked example: logistic form

For $y'+y=y^2$ , let $v=1/y$ . Then $v'-v=-1$ , so $v=1+Ce^t$ and $y=1/(1+Ce^t)$ . Initial value $y(0)=1/2$ gives $C=1$ . The excluded equilibrium $y=0$ also solves the original equation.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

For the Bernoulli equation y′+p(t)y=q(t)yⁿ with n≠0,1 and y≠0, which substitution linearizes it?

v=y¹⁻ⁿ

v=yⁿ

v=ln t

Hint 1 · Find a starting point

Divide by yⁿ and look for a derivative.

Hint 2 · Take the next step

d(y¹⁻ⁿ)/dt=(1−n)y⁻ⁿy′.

Show the reasoning

Answer: v=y¹⁻ⁿ

v=y¹⁻ⁿ gives v′+(1−n)pv=(1−n)q. Excluded zero solutions require a separate check.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: a cubic term

For $y'+y=y^3$ with positive $y$ , set $v=y^{-2}$ . Then $v'-2v=-2$ and $v=1+Ce^{2t}$ . With $y(0)=2$ , $C=-3/4$ . The positive solution is $[1-(3/4)e^{2t}]^{-1/2}$ while its bracket remains positive.

Interpreting the result

The inverse substitution can introduce branch restrictions or a finite-time singularity. A linear transformed solution need not correspond to a real original solution on its entire interval.

Practice

What substitution handles exponent $n=3$ ?
What happens when $n=1$ ?
List equilibria of $y'+y=y^3$ .

Show worked solutions

$v=y^{-2}$ on a nonzero branch.
The equation becomes $y'+(p-q)y=0$ , already linear.
Solve $y=y^3$ : $y=0,1,-1$ .

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

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