## Calculus 10th Edition

Published by Brooks Cole

# Chapter 6 - Differential Equations - Review Exercises: 19

#### 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=$\frac{Ce^{x}}{(2+x)^{2}}$

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