Given a differential equation
$$y'=y(y-3)^2\cos y $$ $$y(0)=a $$
Find an $$ a, a\neq 3 $$
such that $$\lim_{t\to\infty} y(t)= 3$$
I'm not too sure what to do for this. I know the equilibrium solutions to the DE are $$y=0,y=3,y=n\pi/2$$ and then you would check whether these are stable or unstable. However, I'm not sure what to do from there. What should I do next?
Thanks in advance!
The equilibrium solutions are $y=0$, $y=3$ and $y=\pi/2+k\,\pi$, $k\in\mathbb{Z}$ are. They are ordered as $$ \dots<-\frac\pi2<0<\frac\pi2<3<\frac\pi2+\pi<\dots $$ Let's call $y_a$ the solution such that $y_a(0)=a$.
If $\pi/2<a<3$, then $\pi/2<y_a(t)<3$ for all $t\in\mathbb{R}$. Moreover $y(y-3)^2\cos y<0$ if $\pi/2<y<3$. This implies that $y_a$ is decreasing and $\lim_{t\to\infty}y_a(t)=\pi/2$.
If $3<a<\pi/2+\pi$, then $3<y_a(t)<\pi/2+\pi$ for all $t\in\mathbb{R}$. Moreover $y(y-3)^2\cos y<0$ if $3<y<\pi/2+\pi$. This implies that $y_a$ is decreasing and $\lim_{t\to\infty}y_a(t)=3$.