Consider the solution of the ODE $y'(t) = -y^3+y^2+2y$ subject to $y(0) = y_0 \in (0, 2)$. Then $\lim_{t \to \infty}y(t)$ belongs to
- $\{-1, 0\}$
- $\{-1, 2\}$
- $\{0, 2\}$
- $\{0, \infty\}$
We have $$\frac{dy}{y(y^2-y-2)} = \frac{dy}{y(y+1)(y-2)} = - \frac{dt}{1}$$ Further, we can write $$\bigg[ -\frac{1}{2y}+\frac{1}{3\left(y+1\right)}+\frac{1}{6\left(y-2\right)} \bigg] dy= - \frac{dt}{1}$$Integrating both sides we get $$ \frac{(y-2)^{\frac{1}{6}}(y+1)^{\frac{1}{3}}}{y^{\frac{1}{2}}} = ce^{-t} $$The initial condition $y(0) = y_0$ implies $$\frac{(y-2)^{\frac{1}{6}}(y+1)^{\frac{1}{3}}}{y^{\frac{1}{2}}} = \frac{(y_0-2)^{\frac{1}{6}}(y_0+1)^{\frac{1}{3}}}{y_0^{\frac{1}{2}}} e^{-t}$$ Similarly, Now on taking the limit, we get $y=2$ or $y=-1$. Is my solution correct? Is $(3)$ one more solution?