For the initial condition problem $x'(t)=f(x(t)), \ x(t_0)=x_0$ there is exactly one solution if $f$ is lipschitz continuous. Using this statement, but without referring to the explicitly known solution of the problem, show that the solutions of $$x'(t)=x(t), \ x(t_0)=\alpha $$ are strictly positive for all choices of $\alpha>0.$
I know the solution to the problem is supposed to be the exponential function but I'm having trouble seeing how the first statement can be used to get the second one, especially unless I'm not mistaken, when $e^x$ is not even lipschitz on $\mathbb{R}$.
Let $\tau=\min\{t \ge t_0: x(t) \le \alpha/2\}$ if this closed set is nonempty.
(If the set is empty, we are done.) If this set was indeed nonempty, then the mean value theorem would yield some $s \in (t_0,\tau)$ where $$(\tau-t_0)\cdot x(s)=(\tau-t_0) \cdot x'(s)=x(\tau)-x(t_0)=-\alpha/2 \,,$$ a contradiction, since the LHS is positive.