Let a solution $y\colon I\to \mathbb{R}$, where $I$ is a subset of the real numbers, be defined as follows:
$$y'=\frac{\cos(t)}{4-(\sin(t))^2}(4-y^2),\quad y(0)=1$$
Is it possible to have a solution $y$ of the given IVP such that $y(a)=2$ for some $a$ in $I$?
Question: Can someone explain me how this $y(a)=2$ for some $a\in I$ contradicts with the uniqueness and existence theorem?
Set $y(a)=b$ as initial value
get the solution$$y(t)=\frac{\sin (a) (b \sin (t)+4)-4 (b+\sin (t))}{\sin (a) (b+\sin (t))-b \sin (t)-4}$$
look for the maximum $$y'(t)=-\frac{2 \left(b^2-4\right) (\cos (2 a)+7) \cos (t)}{(-2 \sin (a) (b+\sin (t))+2 b \sin (t)+8)^2}$$ $$y'(t)=0\to t=\pm\frac{\pi}{2}+2k\pi$$ $$f(\pi/2)=\frac{(b+4) \sin (a)-4 (b+1)}{(b+1) \sin (a)-b-4}$$ $$f(\pi/2)=2\to \sin a=-2\to\text{ impossible in }\mathbb{R}$$ if the max is at $-\pi/2$ this leads to $\sin a=2$ which is impossible.
Therefore the function is strictly less than $2$ for any initial value.
Maybe there is a shortcut, tho.