A doubt regarding Picard's theorem. Picard's theorem states:

If $f(x,y)$ and $\frac{\partial f}{\partial x}$ are continuous functions on a closed rectangle $R$, then through each point $(x_0,y_0)\in R$ there passes a unique integral curve of the equation $\frac{dy}{dx}=f(x,y)$. 

Here, should we assume that $x$ and $y$ are independent? If they were independent, $\frac{dy}{dx}$ would simply equal $0$, which would imply $f(x,y)=0$ passes through every point inside $R$. Why could the theorem not have simply stated $f(x,y)=0$ passes through every point in $R$, instead of saying $\frac{dy}{dx}=f(x,y)$ passes through every such point? Is there some subtle difference between these two statements that I'm missing?
Thanks in advance!
 A: Good question. I think you are confused about the order in which things are taking place. Let me say the same theorem in a more wordy manner.
We start life out in the $x, y$ plane (or actually some small rectangle in it, parallel to the axes), where it doesn't make sense to say $x$ and $y$ are dependent or independent. (They are just two variables. I guess if someone put a gun to my head I would say they are "independent," but I don't think this vocabulary is helpful.)
Then someone hands us a differential equation, of the shape
$$
\frac{dy}{dx} = f(x, y).
$$
We don't have much control over this differential equation, but we do know two things:


*

*The right hand side is continuous (it is a function of two variables, so "continuous" means with respect to them both. Note that the right hand side is just some function of two variables, it doesn't care about any physical interpretation where $y$ depends on $x$ or not. The left hand side you can think of as just a formal symbol for now; we'll interpret it later.)

*We can take a partial of the RHS with respect to the variable $x$, and what we get is still continuous (same caveats as in (1) above).


(Actually we just know these two things inside the rectangle)
Then the theorem assures us that if you give me a point $(a, b)$ in the rectangle, then 
(1) I can find a solution to the differential equation, whose graph passes through your point, which is to say a differentiable function $g(x)$ such that
$$
g'(x) = f(x, g(x))
$$
such that $g(a) = b$ (this is how I am "interpreting the left hand side later").
(2) That solution is unique.
I think when you have a doubt regarding the case where $y$ depends on $x$, you are worried that somehow $f(x, y)$ is constant? That's OK -- in that case the solution curves will be lines of constant slope.
A: "Here, should we assume that x and y are independent? "
What is "independent" for you? The $x$ and $y$ are related by the equation.
If all the solutions verify $y'(x)=0$ $\forall x$, then we are in the case $f(x,y)=0$ $\forall (x,y)$, and the differential equation is... $y'=0$. Surprising? In this case, the solutions (integral curves) are constant functions $y(x)=y_0$.
