Question

3y^3 - 1 + (2y + 3xy^2)y' = 0 help

Answer 1

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