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Homework Help: Exact Equation - theory

  1. Jul 18, 2005 #1
    An exact equation has the form

    [tex]M(x,y) + N(x,y) \: y^{\prime} = 0[/tex]

    where

    [tex]M(x,y) = \frac{\partial \psi}{\partial x} (x,y)[/tex]

    and

    [tex]N(x,y) = \frac{\partial \psi}{\partial y} (x,y) \mbox{.}[/tex]

    If [tex]y=\phi (x)[/tex] and [tex]\psi (x,y) = c[/tex], then

    [tex]M(x,y) + N(x,y) \: y^{\prime} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right] = 0 \mbox{.}[/tex]

    I can't understand this:

    [tex]\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}[/tex]

    Any help is highly appreciated.
     
  2. jcsd
  3. Jul 18, 2005 #2

    OlderDan

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    From the definition of an "exact differential" or "total differential" for a function f(x,y) whose integral is path independent

    [tex] df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy [/tex]

    http://mathworld.wolfram.com/ExactDifferential.html

    If you divide by dx you have

    [tex] \frac{df}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{dy}{dx} [/tex]

    and since you have y as a function of x, there is a derivative of y wrt x
     
  4. Jul 18, 2005 #3

    James R

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    [tex]\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = \frac{d}{dx} \psi \left[ x, \phi (x) \right]\mbox{.}[/tex]

    Well, [itex]\psi[/itex] is a function of x and y, so

    [tex]\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y}\frac{\partial y}{\partial x}[/tex]

    Now put [itex]y=\phi (x)[/itex], and that's all there is to it.
     
  5. Jul 19, 2005 #4
    Thank you, guys! Thus, we have

    [tex]M(x,y) + N(x,y) \: y^{\prime} = 0[/tex]

    [tex]d\psi = M(x,y) \: dx + N(x,y) \: dy = 0[/tex]

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

    [tex]\frac{d\psi}{dx} = \frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{dy}{dx} = 0 \mbox{.}[/tex]
     
  6. Jul 19, 2005 #5
    I'm not exactly sure of how pragmatic it is to convert an ordinary differential equation into a partial differential equation. Maybe I miss something?
     
  7. Jul 19, 2005 #6

    HallsofIvy

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    No one changed an ordinary differential equation into a partial differential equation. They just used partial derivatives to show show that the sum of two differentials can, in some circumstances, be written as a single differential- which makes the problem very easy.
     
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