## Bernoullis Equation

The differential equation
$frac{dy}{dx}+P(x)y=Q(x)y^n$
is called Bernoulli's equation. It can be transformed using the substitution
$t=y^{1-n}$
. Then
$\frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}=1/ \frac{dy}{dt} frac{dt}{dx} = \frac{1}{(1-n)y^{-n}} \frac{dt}{dx}$
.
The equation becomes
$\frac{1}{(1-n)y^{-n}} \frac{dt}{dx}+ P(x)y=Q(x)y^n$
.
Multiplying each term by
$(1-n)y^{-n}$
gives
$\frac{dt}{dx}+ (1-n)P(x)y^{1-n}=(1-n)Q(x) \rightarrow \frac{dt}{dx} +P(x)t=Q(x) \rightarrow \frac{dt}{dx}=P(x)t=Q(x)$
.
The transformed equation can be solved by The Integrating Factor Method.
Example: Solve
$\frac{dy}{dx} - \frac{2}{x}y=xy^2$
> After the transformation
$t=y^{1-2}=y^{-1}$
the equation becomes
$\frac{dt}{dx}+ \frac{2}{x}y=-x$
.
The integrating factor is
$e^{\int \frac{2}{x} dx}= e^{2 lnx}=e^{lnx^2}=x^2$
then
$\frac{d}{dx}(tx^2)=-x^3 \rightarrow tx^2=-\frac{x^4+4A}{4} \rightarrow t=- \frac{x^4+A}{4x^2}$
.
$t=\frac{1}{y} \rightarrow y=\frac{1}{t} = - \frac{4x^2}{x^4+A}$
.