1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?