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Differential Equations : Integrating factors

  1. Sep 18, 2011 #1
    If u(x,y) and v(x,y) are two integrating factors of a diff eqn M(x,y)dx + N(x,y)dy,
    u/v is not a constant, then
    u(x,y) = cv(x,y) is a solution to the differential eqn, for every constant c.
     
  2. jcsd
  3. Sep 18, 2011 #2
    What does an integrating factor mean?
     
  4. Sep 19, 2011 #3
    an "integrating factor" is a function F ( for example ) so that when you multiply the differential equation out by F, the result is easier to integrate ( i.e. with a product rule or something )
     
  5. Sep 19, 2011 #4
    What is the strict condition that has to be satisfied?
     
  6. Sep 19, 2011 #5
    the integrating factors are more of a technique than something that is explicitly defined; certain types of equations will have different conditions for the integrating factors ( as the thing the OP wants to show is a condition about the integrating factors ).. or for an equation of the form y ' + p( t )y = g( t ), an integrating factor can be of the form e^integral ( p( t ) ) [ when you multiply the equation out by this, you can easily integrate because the expression will be in the "product rule form" ]
     
  7. Sep 19, 2011 #6
    But, we are considering a different kind of equation here, namely the so called exact differential equation:

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

    What does it mean in this context?
     
  8. Sep 20, 2011 #7

    HallsofIvy

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    Theoretically, every first order differential equation has an "integrating factor". It is only for linear equations that it is possible to give a general formula for that integrating factor.

    Saying that u(x,y) is an integrating factor for the diffrerential equation M(x,y)dx+ N(x,y)dy= 0 means that u(x,y)M(x,y)dx+ u(x,y)N(x,y)dy= d(f(x,y)) for some differentiable function f(x,y). That, of course, means that u(x,y)M(x,y)dx+ u(x,y)N(x,y)dy is "exact". In particular that
    [tex]\frac{\partial f}{\partial x}= u(x,y)M(x,y)[/tex]
    and that
    [tex]\frac{\partial f}{\partial y}= u(x,y)N(x,y)[/tex]
     
  9. Sep 20, 2011 #8
    So, if those are the first partial derivatives, what are the second mixed derivatives?
     
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