# Bernoulli differential equation

## Definition

In normalized form, this first-order first-degree differential equation looks like:

$y' + p(x)y = q(x)y^n$

where $n \ne 0,1$. (Note that the cases $n = 0,1$ give first-order linear differential equations).

### Solution method and formula

Divide both sides by $y^n$. If $n > 0$, this means that we may be potentially discarding the stationary solution $y = 0$, and must remember to add that back to the solution family at the end.

We get:

$\frac{y'}{y^n} + \frac{p(x)}{y^{n-1}} = q(x)$

Now put $w = 1/y^{n-1}$ to get:

$\frac{w'}{1 - n} + p(x)w = q(x)$

Multiply by $1 - n$ to get:

$w' + (1 - n)p(x)w = (1 - n)q(x)$

This is now a first-order linear differential equation in $w$, and can be solved to get a family of functional solutions for $w$ in terms of $x$. Plugging back $w = 1/y^{n-1}$ gives a family of functional solutions for $y$ in terms of $x$. We can now add back $y = 0$.