Answer
$\frac{du}{dx}+(1-n)~P(x)~u = (1-n)~Q(x)$
Work Step by Step
$u = y^{1-n}$
$\frac{du}{dx} = (1-n)y^{-n}~\frac{dy}{dx}$
We can consider the given equation:
$\frac{dy}{dx}+P(x)y = Q(x)y^n$
$(1-n)y^{-n}\frac{dy}{dx}+(1-n)y^{-n}~P(x)y = (1-n)y^{-n}~Q(x)y^n$
$(1-n)y^{-n}\frac{dy}{dx}+(1-n)~P(x)y^{1-n} = (1-n)~Q(x)$
$\frac{du}{dx}+(1-n)~P(x)~u = (1-n)~Q(x)$