Showing solution to $y'=y$ is never zero Is there a simple Calc I level (i.e., IVT and MVT) way to show that if $y'=y$ on $\mathbb{R}$, then $y$ is either identically zero or nowhere zero? 
One way is to use the fact that $e^x$ is a solution; then if $g$ is another solution, $(ge^{-x})'\equiv 0$, so $g=Ce^x.$
Is there a simple way to show this using only the relation $y'=y$, the MVT, and the IVT?
As stated in a comment, my issue is how to rule out something which is zero then transitions to an increasing/decreasing function.  I feel like there should be a simple way to rule this out without using the fact that $e^x$  is a solution.
 A: If there is some $x_0$ for which $y(x)>0$ then as $y'(x_0)>0$, $y(x)>0$ for all $x>x_0$. On the other hand, if in addition, there is some $x_1<x_0$ for which $y(x_1)<0$ then $y'(x_1)<0$ and forever after $y$ is decreasing, so it could never increase to $y(x_0)>0$ a contradiction.
To rule out a case where $y=0$ for a long time then suddenly goes up or down, let's WLOG assume it suddenly increases and becomes positive (the other case is symmetric). Then there is some maximum $x_0$ for which $y(x)=0$ for $x\le x_0$ and positive ever after. But then let $x_0<x_1$ and consider ${y(x_1)-y(x_0)\over x_2-x_1}={y(x_1)\over x_1-x_0} =y'(c) =y(c) $ for some $x_0<c<x_1$ by assumption. Choose $0<x_1-x_0<1$ so that $y(x_1)<y(c)$ a contradiction to monotonicity.

Also, an alternative proof I thought up while working on the technical one:
Consider $f(x) = \log y(x)$ so that $f'(x) ={y'\over y}\equiv 1$. But then if $x_0$ is the largest real number for which $y(x) = 0$ we have that $\displaystyle\lim_{x\to x_0}f(x) = -\infty$, however $f'(x) \equiv 1$ verifies that $f(x) = x+c$ to the right of $x_0$, that is $\log y(x) = ce^{x}$ (without assuming it a priori!). Then the limits and derivatives agree to the right of $x_0$, implying that
$$\lim_{x\to x_0^+} y = \lim_{x\to x_0}ce^x= ce^{x_0}>0$$
showing it's impossible for $y(x_0)=0$ because continuity (given because the function is differentiable) implies the limit is equal to the value.
A: If $y(x_0)>0$ then $y(x)>0$ for all $x>x_0$. Take $[y=0]$ the zero set of $y$, and pick its maximum element $z$(it exists since $[y=0]$ is closed and bounded from above). From continuity $y$ is positive on $(z,\infty)$, and is also increasing there.
Now use the IVT on $(z,z+1/2)$ and obtain
$$y(z+1/2)=\frac{1}{2}y'(\xi)=\frac{1}{2}y(\xi), \, \xi \in (z,z+1/2).$$
A contradiction since the motonicity implies $y(z+1/2)>y(\xi)$.
A: Suppose $y(x_0)=0.$ For any $x\ne x_0$ let $$M_x=\sup \{|y(t)|: |t-x_0|\leq |x-x_0|\}.$$ We have $M_x<\infty$ because $y$ is continuous and $\{t:|t-x_0|\leq |x-x_0|\}$ is a closed bounded interval.
For any $r>0$ take $n\in \mathbb N$ where $n$ is large enough that $$M_x|x-x_0|^{n+1}/(n+1)!<r.$$ We have $$|y(x)|=|y(x_0)+\left(\sum_{j=1}^n(x-x_0)^jy^{(j)}(x_0)\right) +y^{(n+1)}(t)(x-x_0)^{n+1}/(n+1)!|$$ for some $t$ such that $|t-x_0|<|x-x_0|.$
But $y(x_0)=0,$ and $y^{(j)}(x_0)=y(x_0)=0 $ for $1\leq j\leq n,$ so  $$|y(x)|=|y^{(n+1)}(t)(x-x_0)^{n+1}/(n+1)!.$$  But $|y^{(n+1)}(t)|=|y(t)|\leq M_x$ so $$|y(x)|\leq M_x|x-x_0|^{n+1}/(n+1)!<r.$$ Since $|y(x)|<r$ for every $r>0,$ therefore $y(x)=0.$
A: If $y' = y$,
then, differentiating,
$y'' = y' = y$;
by induction,
$y^{(n)} = y$
for all $n$.
If
$y(a) = 0$,
then
$y^{(n)}(a) = 0$
for all $n$,
so $y$ is constant zero.
