\[frac{dy}{dx}+P(x)y=Q(x)y^n\]
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 \[\frac{d}{dx}(tx^2)=-x^3 \rightarrow tx^2=-\frac{x^4+4A}{4} \rightarrow t=- \frac{x^4+A}{4x^2}\]
.\[t=\frac{1}{y} \rightarrow y=\frac{1}{t} = - \frac{4x^2}{x^4+A}\]
.