Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.4 Exercises: 11

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: $e^{\int[\frac{1}{x-1}]dx}=e^{ln|x-1|}=x-1$ $y(x-1)=\int(x^{2}-1)dx=\frac{1}{3}x^{3}-x+C_{1}$ $y=\frac{x^{3}-3x+C}{3(x-1)}$
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