Properties of the Solutions of $y' = y$ In this section of my notes we take $E(x)$ as a solution of the initial value problem $y'=y , y(0)= 1 $
We show that $E(x+r)$ is also a solution to this problem and show $E(x+r)=E(x)E(r)$.
The notes then go on to prove that $E(x)>0$. It is done as the following:
Suppose $E(s) = 0$ for some
$s  \in \mathbb{R}$. Then, for arbitrary $x$, we have $E(x + s) = E(x)E(s) = 0$, so $E(x) ≡ 0$.
This contradicts $E(0) = 1$.
My question is, surely $E(x)$ doesn't have to be $0$, surely it could be anything because for whatever value of $E(x)$, since we've said $E(s)=0$ then $E(x)E(s)$ will always be $0$ no matter what the value of $E(x)$. Also I'm a little confused why $E(0)=1$. How do we know this?
Thanks in advance!
 A: I think you are correct that the inference doesn't follow as stated.
However the notes can be fixed by adding an extra step at the end:
Since $x$ was arbitrary, we can choose $x=-s$, then we get:
\begin{align}
E(x+s)&=E(-s+s)
\\&=E(-s)E(s)
\\&=E(-s).0
\\&=0
\end{align}
But $-s+s=0$, so $E(-s+s)=E(0)=1$ Contradiction.
Notice how the value of $E(-s)$ is irrelevant.
I hope this helps.
A: Not an answer but too long for a comment and semi useful
Here's a fun argument. Assume that $f$ is a real differentiable function that satisfies the equation $f'=f$. Assume $g$ is another such function. Then $(\frac{f}{g})'=\frac{f'g-g'f}{g^2}=\frac{fg-gf}{g^2}=0$ and thus there exists a $c\in \mathbb{R}$ such that $f=cg$.
This shows that up to real multiples, there is only one real differentiable function $f$ satisfying $f'=f$. As you know that the function $e^x$ satisfies this equation, you know that all solutions are of the form $ce^x$. Only one of these functions further satisfies $f(0)=1$, namely $1\cdot e^x$.
