Solution of the ordinary differential equation $y'(t)=-y^3+y^2+2y$ Consider the solution of the ordinary differential equation $y'(t)=-y^3+y^2+2y$ subject to $y(0)=y_0 \in (0,2). $ then $\lim \limits_{t \to \infty}y(t)$ belongs to 


*

*{-1,0} 

*{-1,2}

*{0,2}

*{0, $+\infty $}
My Attempt:  $y'(t)=-y^3+y^2+2y \\ \Rightarrow y'(t)=-y(y^2-y-2) \\ \Rightarrow y'(t)=-y(y-2)(y+1) \\ \Rightarrow \frac{dy}{dt}=-y(y-2)(y+1) \\ \Rightarrow \frac{dy}{y(y-2)(y+1)}=-dt \\ \Rightarrow \int\frac{dy}{y(y-2)(y+1)}=\int-dt \\by \;\;partial \;fraction \; we \; get \\ \int [\frac{-1}{2y}+\frac{1}{6(y-2)}+\frac{1}{3(y+1)}]dy =-\int dt \\ \Rightarrow -\frac{1}{2} log|y| +\frac{1}{6} log|y-2| + \frac{1}{3} log|y+1|=-t + logc_1 \\ \Rightarrow \frac{-3log|y|+log|y-2|+2log|y+1|}{6}=-t + logc_1 \\ \Rightarrow log\frac{(y-2)(y+1)^2}{y^3}=-6t+6logc_1 \\ \Rightarrow log\frac{(y-2)(y+1)^2}{y^3}-logc_1^6=-6t \\ \Rightarrow log\frac{(y-2)(y+1)^2}{cy^3}=-6t \;\; where \;c_1^6=c \\ \Rightarrow \frac{(y-2)(y+1)^2}{cy^3}=e^{-6t}..........(1) \\ when \;\;t=0 \;\; \frac{(y_0-2)(y_0+1)^2}{y_0^3}=c \\ (1) \Rightarrow \frac{(y-2)(y+1)^2}{y^3}=\frac{(y_0-2)(y_0+1)^2}{y_0^3} e^{-6t}$
Now I am not getting how to apply the limit for this and how to find $\lim \limits_{t \to \infty} y(t)$ 

you got $\frac{(y-2)(y+1)^2}{y^3}=\frac{(y_0-2)(y_0+1)^2}{y_0^3} e^{-6t}$ take the limit and you get $$\lim\limits_{t\to\infty}\frac{(y-2)(y+1)^2}{y^3}=0\implies \lim\limits_{t\to\infty}(y-2)(y+1)^2=0 \implies \lim\limits_{t\to\infty}y(t)=\begin{cases}2\\-1\end{cases}$$ hence option (2) is true
