Calculus 10th Edition

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

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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