Non-vanishing Solution in Differencial Equation

Say we wish to solve $y' = y$ on $(-\infty,\infty)$ *.

1) $y$ is non-vanishing i.e. $y(\xi) \not = 0$ for any $\xi \in \mathbb{R}$ .
Then we can write $y'/y = 1$ and thus $(\log |y|)' = 1$ .
This leads to $\log |y| = x + k_1 \implies y = ke^x$ with $k\not = 0$

2) $y$ is the zero function but that is definitely a solution.

3) $y$ vanishes at some point i.e. $y(\xi) = 0$ for some $\xi \in \mathbb{R}$ .
This is were the difficultly is. If it vanishes at $\xi$ and in the open interval $I$ containing $\xi$ then I think this leads to $y$ being the zero function**. However, if $y$ does not vanish on any open interval containing it then it means there is a point $\xi_1$ such that $y(\xi_1) \not = 0$ . But by continuity it means there is an interval $J$ containing $\xi_1$ such that $y$ is non-vanishing on $J$ . By applying #1 to this restricted function, $y|J$ , we find that $y|J$ must be of the form $ke^x,k\not = 0$ . And this perhaps shows that $y$ is $ke^x$ everywhere.*** Therefore $y$ is non-vanishing everywhere. A contradiction if $y$ is a non-zero function. Thus, this case does not exist for non-zero functions.

How do we complete the proof?
And is there a general theorem that tells us we can divide the functions we are solving for without worrying it can vanish.

*)This can be solved without seperation of variables i.e. with integrating fractor.
But with this something interesting happens.

**)Though I did not find the complete proof. I guess the proof comes down to a analogue of the identity theorem from complex analysis applied to real functions. This is because the solution to the differencial equation is $\mathcal{C}^{\infty}$ and we can possibly even proof its analyticity. With analyticity we can have an analogue of the identity theorem.

***)Again by identity theorem for real function - by using power series.