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Homework Help: Looking for the name of a class of ODE:

  1. Aug 16, 2007 #1
    1. The problem statement, all variables and given/known data

    [tex]M(x,y)y^{'}(x) + N(x,y) = 0[/tex]
    There exists:
    [tex]\phi(x,y)[/tex]
    Such that
    [tex]\frac{\partial\phi(x,y)}{\partial x}=N(x,y)[/tex]

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

    I'm not looking for a solution to anything particular to this but I can't find the type in my notes and I can't google it unless I know the name.
     
  2. jcsd
  3. Aug 16, 2007 #2

    HallsofIvy

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    That is an exact equation since
    [tex]d\phi= M(x,y)dy+ NIx,y)dx[/itex]
    is an exact differential.

    Of course, since the differential equation says [itex]d\phi= 0[/itex], [itex]\phi(x,y)= 0[/itex] is the general solution.
     
  4. Aug 16, 2007 #3

    HallsofIvy

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    By the way, in physics, such a differential would correspond to a "conservative force field" and the function [itex]\phi[/itex] would be the "potential function".
     
  5. Aug 17, 2007 #4
    Thanks.
    Just a question: The I after the N is a typo yes? And how did you get [itex]d\phi= 0[/itex]?
     
    Last edited: Aug 17, 2007
  6. Aug 17, 2007 #5

    HallsofIvy

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    Yex, that was a typo- my finger was aiming at "("!

    Since [itex]\phi[/itex] is a function of both x and y, [itex]d\phi /dx[/itex] would make no sense. By the chain rule, if x and y are functions of some third variable, t,
    [tex]\frac{d\phi}{dt}= \frac{\partial \phi}{\partial x}\frac{dx}{dt}+ \frac{\partial \phi}{\partial y}\frac{dy}{dt}[/tex]
    or, in differential notation,
    [tex]d\phi= \frac{\partial \phi}{\partial x}dx+ \frac{\partial \phi}{\partial y}dy[/tex]
     
  7. Aug 17, 2007 #6
    :rofl:

    Thanks I forgot that this was dealing with partials. It makes sense now.
     
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