uniqueness differential equation It is know that one solution for $\frac{dy}{dx}=y$ is $Ce^x, C \ne0$ (This is valid if y is not zero anywhere, we divide by y when finding this). I want to prove that if y takes the value 0 for one $x_0$it must be zero on the entire line.
So I assume that $y(x_0)=0$, but for contradiction that y is not zero on the entire line. Then there is a z, such that $y(z) \ne 0$. Because of continuity, there has to be an interval around z, where y is not zero, and on this interval y must be on the form $Ce^x$.
But how do I extend this argument to the entire line, so that I show that y can not be zero anywhere?
I tried looking at $x^*=\sup\{x: f([x_0,x])=\{0\}\}$, if $z>x_0$. Ofcourse I can assume that $z>x_0$ without loss of generality. But I wasnt able to force a contradiction or something at $x^*$.
 A: If 
$$
\frac{dy}{dx}=y,
$$
then
$$
\mathrm{e}^{-x}\left(y'(x)-y(x)\right)=0,
$$
or equivalently
$$
\left(\mathrm{e}^{-x}y(x)\right)'=0
$$
which is equivalent to
$$
\mathrm{e}^{-x}y(x)=c,
$$
for some $c$ constant. Or $y(x)=c\mathrm{e}^{x}$. 
So if $y(x_0)=0$, then clearly $c=0$, and thus $y\equiv 0$.
A: Let's $y = f(x)$ be such that it is differentiable everywhere and $f'(x) = f(x)$ for all $x$. Also let it vanish at some point say $a$ so that $f(a) = 0$. I will show that this will lead to $f(x) = 0$ everywhere. If we set $F(x) = f(x + a)$ then we see that $F'(x) = F(x)$ and $F(0) = 0$. Hence it is sufficient to consider that the functions $f(x)$ vanishes at point $a = 0$.
Let us assume that there is some point $b$ for which $f(b) \neq 0$ and consider $g(x) = f(b + x)f(b - x)$ so that $$g'(x) = f'(b + x)f(b - x) - f(b + x)f'(b - x) = f(b + x)f(b - x) - f(b + x)f(b - x) = 0$$ so that $g(x)$ is a constant and hence $g(x) = g(0) = f(b)^{2} > 0$. Then we have $f(b + x)f(b - x) > 0$ for all $x$. Putting $x = b$ we get $f(2b)f(0) > 0$ which is a contradiction because $f(0) = 0$. Hence there is no point $b$ such that $f(b) \neq 0$. Thus $f(x) = 0$ for all $x$.
This proof does not involve $e$.
