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Exact differential

  1. Sep 2, 2007 #1
    how would i show that this is the exact differential

    z=xy-y+lnx+2
    dz= (y+ 0)dx + (x-1)dy (i hope the total differential is right)

    so how would i show that dz is the exact differential ???
     
  2. jcsd
  3. Sep 2, 2007 #2

    quasar987

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    The definition of the total differential of a function f(x,y) is the expression

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

    Now with the z you're given, is the expression for dz correct? That is to say, is

    [tex]\frac{\partial z}{\partial x}=(y+0)[/tex]

    and

    [tex]\frac{\partial z}{\partial y}=(x-1)[/tex]

    ??
     
  4. Sep 2, 2007 #3
    i believe dz is correct because i used partial differentials to solve it, and i hope its correct
     
  5. Sep 2, 2007 #4

    quasar987

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    what is the derivative of ln(x)?
     
  6. Sep 3, 2007 #5

    HallsofIvy

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    Once you fixed the derivative of ln x thing, the fact that dz is the total derivative of z pretty much means it is an "exact" differential. That's the definition of "exact" differential!
     
  7. Sep 3, 2007 #6
    thx and the derivative of lnx is just 1/x
     
  8. Sep 3, 2007 #7

    HallsofIvy

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    Yes.
    z=xy-y+lnx+2 so
    dz= (y+ 1/x)dx + (x-1)dy

    And that is an "exact" differential precisely because it is the differential of z.

    Now suppose you were given the differential form (y+ 1/x)dx+ (x- 1)dy without having been given z. How would you determine whether it was an exact differential?
    If
    [tex](y+ 1/x)= \frac{\partial z}{\partial x}[/tex]
    for some function z and
    [tex]x-1= \frac{\partial z}{\partial y}[/tex]
    then
    [tex]\frac{\partial(y+ 1/x)}{\partial y}= \frac{\partial^2 z}{\partial x\partial y}[/tex]
    and
    [tex]\frac{\partial(x-1)}{\partial x}= \frac{\partial^2 z}{\partial y\partial x}[/tex]
    and so must be equal. Fortunately, it is easy to see that each second derivative is 1 and so they are in fact equal.
    Okay, now, how would we find z? Knowing that
    [tex]x-1= \frac{\partial z}{\partial y}[/tex]
    integrating with respect to y (treating x as a constant) we get z= xy- y+ C, except that, since we are treating x as a constant, that "constant of integration", C, may depend on x: z= xy- y+ C(x). Differentiating that with respect to x,
    [tex]\frac{\partial z}{\partial x}= y+ C'(x)= y+ 1/x[/tex]
    Notice that the "y" terms cancel (it was the "cross condition" above that guarenteed that) and so we have C'(x)= 1/x. Then C(x)= ln(x)+ C where "C" now really is a constant.
    Any z(x,y)= xy- y+ ln(x)+ C satisfies dz= (y+ 1/x)dx+ (x-1)dy.
     
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