Show that the solution of the initial value problem $$y'=-2\sin(x)\sqrt{y}, \quad y(0)=1, \quad y\in[0,2]$$ that are defined for all $x\in\mathbb{R}$ are not unique.
I have found one such solution: $$\dfrac{dy}{dx}=-2\sin(x)\sqrt{y}$$ $$y=(\cos(x)+C)^2$$ combining with the initial value yields: $$y=(\cos(x))^2$$
Is it possible for me to find more? Is there a certain pattern here?
Given he IVP $$y'=-2\sin (x)\>\sqrt{y},\qquad y(0)=1,$$ there is no problem with the "micro solution" defined in a small neighborhood of $x=0$: There is the unique solution $$y(x)=\cos^2(x)\qquad\left(-{\pi\over2}<x<{\pi\over2}\right)$$ you have found. It is even possible to extend this solution to all of ${\mathbb R}$. But this extension is not unique, since the basic "technical assumption" of the existence and uniqueness theorem for ODEs is not fulfilled at the points of the $x$-axis. The function $$y_1(x)=\left\{\eqalign{\cos^2 x\qquad&\bigl(|x|\leq{\pi\over2}\bigr) \cr 0\qquad\quad&\bigl(|x|\geq{\pi\over2}\bigr)}\right.$$ is a global solution of the given IVP as well. Since you actually have two choices for each interval $\left(n\pi-{\pi\over2}, n\pi+{\pi\over2}\right)$, $\>n\ne0$, there are in fact uncountably many very different solutions.