What does it mean by 'implicitly depends on x' ?

[KFC]

Hi there,
I am reading an introduction on solving ODE, there the exact equation is mentioned. Suppose a ODE is in the following form

M(x, y) + N(x, y)\frac{dy}{dx} = 0

assume there is a function F(x, y) such that

\frac{\partial F(x, y)}{\partial x} = M(x, y) + N(x, y)\frac{dy}{dx}

Hence,
\frac{\partial F(x, y)}{\partial x} = 0

The text says that the above equation imply that F(x, y) = \text{Const}

But here are my doubts

1) Let's assume the above equation is true, so does it mean F(x, y) has no way to be a function of x if the original equation is exact?

2) What about if F(x,y) = g(y), in this case, we don't know the exact form of y, but we know that y is depending on x, so can we say F(x,y)=g(y) is implicitly depending on x? If it is true, how can we conclude that F(x, y) = \text{Const} instead of some functions of y?

Well, you might find what I am asking is vague. What I actually means is if F(x,y)=g(y), so can we safely say that

\frac{\partial F(x, y)}{\partial x} = \frac{\partial g(y)}{\partial x} = 0

It is quite confusing to use the term 'implicitly'! Because we do know that y=y(x), so how come we put
\frac{\partial g(y)}{\partial x} = 0 instead of \frac{\partial g(y)}{\partial y}\frac{dy}{dx} = 0 ?

 

[owlpride]

I agree with the conclusion that

<br /> \frac{\partial F(x, y)}{\partial x} = 0<br />

implies that F(x,y) = const under two assumptions:

(1) y is a function of x, not an independent variable

(2) \frac{\partial F(x, y)}{\partial x} means "take the partial derivative of F with respect to x" and not "take the partial derivative of F with respect to the first variable."

With these two assumptions, I can write F(x,y(x)) = G(x) and dF(x,y)/dx = dG(x)/dx = 0, so G(x) = const = F(x,y).