Two variable function in this ODE? (Peano theorem) In the following ODE, the function $f$ is a one variable function, $f:\mathbb{R}\rightarrow \mathbb{R}$, right?
\begin{align}
y'(x)=f(x,y(x))
\end{align}
But in Peano existence theorem, the function $f$ is a two variable function, $f:\mathbb{R}^2\rightarrow \mathbb{R}$:
\begin{align}
y'(x)&=f(x,y(x))\\
y(x_0)&=y_0
\end{align}
It's an ODE, so why is the function a two variable function?
For ODE:s, aren't we always consider $f:\mathbb{R}\rightarrow \mathbb{R}$ or $f:\mathbb{R}\rightarrow \mathbb{R}^n$, i.e. only functions of one variable?
 A: $\underline{\text{More general case}}$ :
$f(x,y)\quad$ is a two variables function : The variables are $x$ and $y$.
In the case of those variables are functions of another common variable, say $t$ , this generates a new function of only one variable $t$ :
$f\left(x(t),y(t)\right)= F(t)$
$F(t)$ and $f(x,y)$ are not the same function since they are functions of different variables and the form of those functions are different.
The functions are different but, of course, the values taken by $\quad F(t)\quad$ and by $\quad f(x,y)\quad$ are equal when $\quad x=x(t)\quad$ and when $\quad y=y(t)$.
$\underline{\text{This is the same in the case considered in the raised question}}$ :
$f(x,y)\quad$ is a function of two variables. If $y$ is function of $x$ this generates a new function of only one variable $x$ :
$f\left(x,y(x)\right)= F(x)$
$F(x)$ and $f(x,y)$ are not the same function since they are functions of different variables and the form of those functions are different.
The functions are different but, of course, the values taken by $\quad F(x)\quad$ and by $\quad f(x,y)\quad$ are equal when $\quad y=y(x)$.
