Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Aug 6, 2010 #1


    User Avatar

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

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

    assume there is a function [tex]F(x, y)[/tex] such that

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

    [tex]\frac{\partial F(x, y)}{\partial x} = 0[/tex]

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

    But here are my doubts

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

    2) What about if [tex]F(x,y) = g(y)[/tex], 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 [tex]F(x,y)=g(y)[/tex] is implicitly depending on x? If it is true, how can we conclude that [tex]F(x, y) = \text{Const}[/tex] instead of some functions of y?

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

    [tex]\frac{\partial F(x, y)}{\partial x} = \frac{\partial g(y)}{\partial x} = 0[/tex]

    It is quite confusing to use the term 'implicitly'! Because we do know that y=y(x), so how come we put
    [tex]\frac{\partial g(y)}{\partial x} = 0[/tex] instead of [tex]\frac{\partial g(y)}{\partial y}\frac{dy}{dx} = 0[/tex] ???
    Last edited: Aug 7, 2010
  2. jcsd
  3. Aug 7, 2010 #2
    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) [tex]\frac{\partial F(x, y)}{\partial x} [/tex] 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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook