Consider the ordinary differential equation $$y'=y^2, y(0) =0$$
Clearly $y=0$ is a solution of it. Can i say that this is only solution? I tried as follows
Choose rectangle $|x|\leq a, |y|\leq b$, then $|f(x, y) |\leq b^2$ . Now using existence theorem we have $$h=min\{a, \frac{b}{b^2}\} =\frac{1}{b}\to\infty$$ So by Existence theorem we can say that $y=0$ is a global solution of given differential equation . But I don't know about uniquness of solution. Did I correctly apply Existence theorem? Please suggest. Thank you.
Suppose there exist a different solution $y\ne 0$. Then divide both sides of the equation by $y^2$:$$\frac{y'}{y^2}=1$$ Integrating, you get $$-\frac 1y=x+C$$or $$y=-\frac 1{x+C}$$But this can't verify the condition $y(0)=0$ for any finite $C$.