Cubic logistic differential equation ($2\dot{y}=y-y^3$)

The substitution $z=y^{-2}$ turns the nonzero solutions into $\dot{z}=1-z$. Thus $y=\pm (1+K{e}^{-t}{)}^{-1/2}$ where the denominator is positive, together with $y=0$. The nonzero equilibrium points $\pm 1$ are attracting and $0$ is unstable. The sign of nonzero initial data is preserved.