I have the following initial value problems:
$$ \begin{cases} y' = y^{\frac{1}{2}}\cos(x) \\[6pt] y(-2)= 1 \end{cases} $$
$$ \begin{cases} y' = y^{\frac{1}{2}}\cos(x) \\[6pt] y(5)=-8 \end{cases} $$
$$ \begin{cases} y' = y^{\frac{1}{2}} \cos(x) \\[6pt] y(0)= 0 \end{cases} $$
I want to determine if they have a unique solution.
I'm using the uniqueness theorem that says if $f(x, y)$ and $\partial f/\partial y$ are continuous on a rectangular region enclosing $(x_0, y_0)$, then there is a unique solution on an interval $I$.
$$\frac{\partial f}{\partial y} = \frac{1}{2} \cdot \frac{\cos(x)}{y^{\frac{1}{2}}}$$
So I know that I can only have unique solutions for $y > 0$. For the first IVP, it's easy to find a rectangular region containing $(-2, 1)$, so it must have a unique solution. But the next one lies below the line $y = 0$ and the last one on the line. The textbook says that if the hypothesis of the theorem doesn't hold, then it's inconclusive how many solutions there are.
My question is, how can I determine if there is a unique solution if I can't use the theorem? Should I try to solve the IVPs and just see what happens?