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

1. Aug 6, 2010

### 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 $$\frac{\partial g(y)}{\partial y}\frac{dy}{dx} = 0$$ ???

Last edited: Aug 7, 2010
2. Aug 7, 2010

### owlpride

I agree with the conclusion that

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

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).