Let $y(t)$ be solution to the boundary value problem $$\frac{dy}{dt}=e^yy^2(y-1)(y-2)^3(y-3)^5,\qquad y(3)=\frac{1}{2}$$ Find $$\lim_{t\to+\infty}y(t)$$
My Try: For $y\lt0$, $\frac{dy}{dt}\lt0$. For $0\lt y\lt1$, $\frac{dy}{dt}\lt0$. For $1\lt y\lt2$, $\frac{dy}{dt}\gt0$. For $2\lt y\lt3$, $\frac{dy}{dt}\lt0$. For $y\gt3$, $\frac{dy}{dt}\gt0$. Hence for $y=\frac{1}{2}$, $\frac{dy}{dt}\lt0$, then $y(t)$ decrease as $t$ increase. But I don't know how to determine if $y(t)$ will decrease below $0$. Also if $y(t) $ go below $0$, then still have $\frac{dy}{dt}\lt0$. Now whether $y(t) $ will go to negative infinity.
$y=0$ is a solution and the right side is differentiable, so the uniqueness theorem applies. Thus crossing the value zero is not possible. The same goes for the other fixed points $y=1,2,3$.