If $f$ is even and $y'=f(y)$ then $y$ is odd 
Let $f\in C^1(\mathbb{R}, \mathbb{R})$ be an even function.
Consider the maximal solution $y\colon\left]\alpha ,\beta\right[\to \mathbb{R}$ of the IVP $$y'=f(y),\ y(0)=0$$
Prove that $y$ is an odd function and $\beta =-\alpha$.

To be able to prove that $y$ is odd, I first need its domain to be symmetric ($x\in \left]\alpha ,\beta\right[\implies -x\in \left]\alpha ,\beta\right[$), from here I can conclude that $\alpha=-\beta$. But how to prove the domain is symmetric?
And how to prove that $y(-x)=-y(x)$ for all $x\in \left]\alpha,\beta\right[$? I suspect it has something to do with the fact that the derivative of an even function is odd and $y'=f(y)$, but I can't see how to get the desired result.
 A: Clearly, $\alpha<0<\beta$.
Let $r=\min\{-\alpha,\beta\}$ so that we at least have a solution on $\left]-r,r\right[$.
Then show: If $y$ is a solution on $]\alpha,\beta[$, then $x\mapsto-y(-x)$ is a solution on $\left]-\beta,-\alpha\right[$.
By uniqueness, these coincide on $\left]-r,r\right[$.
Then you can combine $y$ with $x\mapsto- y(-x)$ to find a solution on $\left]-\max\{-\alpha,\beta\},\max\{-\alpha,\beta\}\right[$. By maximality, we conclude $\alpha=-\max\{\alpha,\beta\}=-\beta$.
A: Define $h(x)=-y(-x)$. Then 
$$
h'(x)=y'(-x)=f(y(-x))=f(-h(x))=f(h(x)),
$$
where we have used the eveness of $f$. Now, notice that $h(0)=-y(0)=0=y(0)$. As the solution should be unique you obtain that $y$ is odd.
A: There is a little problem in the formulation of this question. 
In order to even wonder whether $\alpha=\beta$, a further piece of information is necessary: That the solution is defined in a maximum possible domain.
Clearly, if the flux function (here $f$) satisfies conditions which guarantee uniqueness (which here it does as $f$ is $C^1$, and thus locally uniformly Lipschitz continuous), then it is possible to talk about a "solution defined in a maximum open interval". This is obtained as the set-theoretic union of all solutions of this particular IVP; these solutions do not have the same domain, but each two of them agree in their common domain due to uniqueness. (Note that a function, in Set Theory, is defined as a set on ordered pairs.) If uniqueness is not enjoyed by an IVP, then we simply talk about a "solution defined in a maximal interval", and apparently its definition requires the use of Zorn's  lemma.
A: I'll leave this here, for someone that has a similar question.
Since f is odd, y' = f(y)=f(-y) — This means that not only the function is odd, but f[y(t)]=f[-y(t)] (we'll get to this later)
From this, we come to the point that y(t) has a known point (0,0) , from the inicial condition. 
Let us consider a point α : 
f[y(α)] = y'(α) = f[-y(α)]
Considering now α and -α=β, we come to : 
f[y(α)] = y' = f[-y(-α)] =f[-y(β)]
This means that f[y(α)] = f[-y(β)] = f[-y(-α)]
This implies that y(α) = -y(-α) , condition for a odd function f(x) = -f(-x) , so y(t) is odd, and it's done.
