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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