Suppose $y=\phi(x)$ is the solution of DE $\frac{dy}{dx}=y^2$ which satisfy $y(x_0)=y_0$. If $y=\phi(x)$ is bounded in $(-\infty,+\infty)$ , then find $\phi(x)$.
My try: First suppose $y_0=0$, then $\phi(x)=0$. If $y_0\neq0$, then $$y^{-2}dy=dx\Rightarrow-\frac{1}{y}=x+c$$ since $y(x_0)=y_0$. Then we have $c=-x_0-\frac{1}{y_0}$. Hence $$y=-\frac{1}{x-x_0-\frac{1}{y_0}}$$ So when $x\to x_0+\frac{1}{y_0}$ , then $y=\phi(x)\to \infty$. Now that $y=\phi(x)$ is bounded . We have $y=\phi(x)=0$.
Is this proof correct?