Recipe

Solve a Bernoulli equation

Use when: the equation has the form $y'+p(x)y=q(x)y^n$ with $n\ne0,1$

Goal: a linear equation in $v$ you already know how to solve, then the answer converted back to $y$

Taught in The Bernoulli equation

Quick recipe

Identify $n$ and divide by $y^n$
Substitute $v=y^{1-n}$
Write the resulting linear equation in $v$
Solve for $v$ using the integrating-factor method
Substitute back to recover $y$

1
Match the equation against $y'+p(x)y=q(x)y^n$ to read off $n$ , then divide the whole equation by $y^n$ .

Example

Take $\dfrac{dy}{dx}+y=xy^2$ : matching against $y'+p(x)y=q(x)y^n$ gives $p(x)=1$ , $q(x)=x$ , $n=2$ . Dividing by $y^2$ : $y^{-2}\dfrac{dy}{dx}+y^{-1}=x$

Why: isolates the exact combination, $y^{-n}y'$ , that the substitution below turns into $\frac{dv}{dx}$ .
2
Let $v=y^{1-n}$ , so $\dfrac{dv}{dx}=(1-n)y^{-n}\dfrac{dy}{dx}$ .

Example

With $n=2$ : $v=y^{-1}$ , and $\dfrac{dv}{dx}=-y^{-2}\dfrac{dy}{dx}$ , so $y^{-2}\dfrac{dy}{dx}=-\dfrac{dv}{dx}$ .

Why: this change of variable is chosen for exactly one reason — it makes $\frac{dv}{dx}$ come out proportional to the $y^{-n}y'$ combination Step 1 already isolated, which is what cancels the nonlinearity.
3
Substitute to get $\dfrac{dv}{dx}+(1-n)p(x)v=(1-n)q(x)$ , an ordinary first-order linear equation in $v$ .

Example

Substituting $y^{-2}\dfrac{dy}{dx}=-\dfrac{dv}{dx}$ and $y^{-1}=v$ into Step 1's equation: $-\dfrac{dv}{dx}+v=x$ , so $\dfrac{dv}{dx}-v=-x$ matching the general form with $(1-n)p(x)=(1-2)(1)=-1$ and $(1-n)q(x)=(1-2)(x)=-x$ .

Why: dividing by $y^n$ and substituting $v$ together strip out every power of $y$ , leaving a genuinely linear equation — the nonlinearity was in the substitution, not in the underlying structure.
4
Solve the linear equation in $v$ exactly as in "Solve a first-order linear equation" — build $\mu(x)$ , multiply through, integrate.

Example

For $\dfrac{dv}{dx}-v=-x$ : $\mu(x)=e^{\int(-1)\,dx}=e^{-x}$ . Multiplying through: $\dfrac{d}{dx}[e^{-x}v]=-xe^{-x}$ . Integrating (by parts): $e^{-x}v=xe^{-x}+e^{-x}+C$ , so $v=x+1+Ce^{x}$

Why: this is now first-order linear with no trace of the original $y^n$ term, the same technique this chapter already covers.
5
Once $v$ is solved, substitute $y=v^{1/(1-n)}$ to write the answer in terms of the original variable.

Example

Since $v=y^{-1}$ , $y=v^{-1}$ : $\boxed{y=\dfrac{1}{x+1+Ce^{x}}}$

Why: $v$ was never the actual unknown — it was a device to make the equation solvable. Leaving the final answer in terms of $v$ is an unfinished substitution, not a different valid answer.

Common mistakes

Dividing by $y^n$ using the wrong sign of $n$ , or forgetting to divide at all before substituting.
Writing $v=y^{n-1}$ instead of $v=y^{1-n}$ — the exponent's sign is easy to flip.
Solving the linear equation in $v$ correctly, then reporting $v$ itself as the final answer instead of converting back to $y$ .
Forgetting the $(1-n)$ factor picked up when substituting $\frac{dv}{dx}$ back into the equation.