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

  1. Aug 6, 2010 #1

    KFC

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    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]

    Hence,
    [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

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

    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).
     
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