Answer
y=$\frac{Ce^{x}}{(2+x)^{2}}$
Work Step by Step
(2+x)y'-xy+0
$\frac{1}{y}$dy=$\int$$\frac{x}{2+x}$dx
$\frac{1}{y}$dy=$\int$(1-$\frac{x}{2+x}$)dx
y=-3-$\frac{1}{x+C}$
y=$\frac{Ce^{x}}{(2+x)^{2}}$
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 - Review Exercises - Page 431: 19
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