Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Section 9.5 - Linear Equations - 9.5 Exercises - Page 625: 23

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