## Calculus 10th Edition

Published by Brooks Cole

# Chapter 6 - Differential Equations - Review Exercises: 17

#### Answer

y=-3-$\frac{1}{x+C}$

#### Work Step by Step

$\frac{dy}{dx}=(3+y)^{2}$ $\int$$(3+y)^{-2}dy$=$\int$dx $-(3+y)^{-2}$=x+C $\frac{4}{15}(x-y)^{\frac{3}{2}}$(3x+14)+C y=-3-$\frac{1}{x+C}$

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