Chapter 9 - Section 9.5 - Linear Equations - 9.5 Exercises - Page 646: 27

$\frac{du}{dx}+(1-n)~P(x)~u = (1-n)~Q(x)$

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