## anonymous 5 years ago 3y^3 - 1 + (2y + 3xy^2)y' = 0 help

Solve for y' $y' = -{{3 y^3 - 1}\over{3 x y^2+2 y}}$ then integrate both sides. $y = {{x(1-3y^3)}\over{3 x y^2+2y}}+c$