# Solution to Bernoulli's Equation

## Theorem

$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$

where:

$n \ne 0, n \ne 1$

has the general solution:

$\ds \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$

where:

$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$

## Proof

Make the substitution:

$z = y^{1 - n}$

in $(1)$.

Then we have:

 $\ds \frac {\d z} {\d y}$ $=$ $\ds \paren {1 - n} y^{-n}$ Power Rule for Derivatives $\ds \leadsto \ \$ $\ds \frac {\d z} {\d y} \frac {\d y} {\d x} + \map P x y \paren {1 - n} y^{-n}$ $=$ $\ds \map Q x y^n \paren {1 - n} y^{-n}$ $\ds \leadsto \ \$ $\ds \frac {\d z} {\d x} + \paren {1 - n} \map P x y^{1 - n}$ $=$ $\ds \paren {1 - n} \map Q x$ Chain Rule for Derivatives $\ds \leadsto \ \$ $\ds \frac {\d z} {\d x} + \paren {1 - n} \map P x z$ $=$ $\ds \paren {1 - n} \map Q x$

This is now a linear first order ordinary differential equation in $z$.

It has an integrating factor:

 $\ds \map \mu x$ $=$ $\ds e^{\int \paren {1 - n} \map P x \rd x}$ $\ds$ $=$ $\ds e^{\paren {1 - n} \int \map P x \rd x}$

and this can be used to obtain:

$\ds \map \mu x z = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$

Substituting $z = y^{1 - n} = \dfrac 1 {y^{n - 1} }$ finishes the proof.

$\blacksquare$

## Also see

When $n = 0$ or $n = 1$ the equation is linear, and Solution to Linear First Order Ordinary Differential Equation can be used.