Why can we substitute the exponential function while deriving the characteristic equation I'm studying the way characteristic equation works. The derivation, according to Wikipedia, follows:
We have a function $y(x)$ and an equation
$$a_n y^{(n)} + \cdots + a_1 y' + a_0 y = 0.$$
Then we substitute $$y = e^{rx}.$$
Why can we make such a substitution? I suspect we can't represent any function as $e^{rx}$. Besides it is said that $\forall z: e^{z} \neq 0$ (can you point me to a proof too), so if $\exists x_0: y(x_0) = 0$ then we can't represent this value in this substitution.
 A: We only bravely try$^*$ to substitute $y=e^{rx}$ in 
$$a_n y^{(n)} + ... + a_1 y' + a_0 y = 0. $$
We get then
$$a_n r^ne^{rx} + ... + a_1 re^{rx} + a_0 e^{rx} = 0.\tag 1$$
If we substitute $x=0$ (or if we divide both sides by $e^{rx}>0$) then we get the famous characteristic equation:
$$a_n r^n + ... + a_1 r + a_0  = 0.$$
We realize now that having solved this characteristic equation in $r$ and having substituted the solutions in $(1)$ for $r$ then we get a series of solutions of the differential equation.
Why?
Just imagine that $r_i$ is a solution of the characteristic equation and we substitute it for $r$ in $(1)$. Now, we have
$$a_n r_i^ne^{r_ix} + ... + a_1 r_ie^{r_ix} + a_0 e^{r_ix}=0 $$
since $e^{r_ix}>0$ for all $x$ we can divide both sides of this equation by $e^{r_ix}$. Then we do get zero exactly because $r_i$ is a solution. And , of course, this is the case for all $i$.
Note that we haven't yet determined all of the solutions. The result of our brave action is only a set of solutions.
$^*$ See the comment to the OP by  Martín-Blas Pérez Pinilla.
Edited because the last question remained unanswered
I've just realized that the OP asked the following question at the end of his/her post:
"Why can we make such a substitution? [$e^rx$] I suspect we can't represent any function as $e^x$. Besides it is said that $\forall z \ e^z\not =0$ (can you point me to a proof too), so if $\exists \ x_0 : y(x_0)=0$ then we can't represent this value in this substitution."


*

*Of courese we cannot represent any functions as $e^x$. (To be honest I do not understand this sentence.)

*Why is it true that $\forall z :\ e^z\not =0$? The most general case is when $z=a+ib$, a complex number. Then $e^z=e^{a+ib}=e^ae^{ib}.$ Here $a$ is real so $e^a$ is never zero. The question reamins: is there a real $b$ for which $e^{ib}=0$? That is impossible because the absolute value of $e^{ib}$ is $1$ independently of the argument $b$.

*Even if $\exists \ x_0 : y(x_0)=0$ there are functions $y$ that can be represent by a homogenuous linear differential equation. Remember, the linear combinations of the solutions of such a differential equations are solutions to it as well. Let's see an example. 


Solve the following differential equation:
$$y''+y'-2y=0.$$
The characteristic equation is $r^2+r-2=0$ whose solutions are: $r_1=-2$ and $r_2=1$. That is,  $y_1(x)=e^{-2x}$ and $y_2(x)=e^x$ are certainly solutions. But the linear combinations are also solutions! So
$$y(x)=e^{-2x}-e^x$$
is a solution again. Substituting $0$ for $x$ we get that $y(0)=0$ when $y$ is a solution. 
So, even if we cannot describe all kinds of functions as solutions of certain differential equations, the existence of an $x_0$ for which $y(x_0)=0$ is not an obstacle.
A: I would refer to that (setting $y= e^{rx}$) as "an educated guess".
