Answer
$y=\frac{x^{3}-3x+C}{3(x-1)}$
Work Step by Step
$(x-1)y'+y=x^{2}-1$
$y'+(\frac{1}{x-1})y=x+1$
Integrating factor: $e^{\int[\frac{1}{x-1}]dx}=e^{ln|x-1|}=x-1$
$y(x-1)=\int(x^{2}-1)dx=\frac{1}{3}x^{3}-x+C_{1}$
$y=\frac{x^{3}-3x+C}{3(x-1)}$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 6 - Differential Equations - 6.4 Exercises - Page 428: 11
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