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Question of integration

  1. Feb 23, 2010 #1
    So in doing a homework problem I have convinced my self that [tex]\int\frac{V_{x}}{V}=ln(V)[/tex] which I vaguely remember learning in class, but I'm having trouble deriving it. Can someone help me out?
     
  2. jcsd
  3. Feb 23, 2010 #2

    CompuChip

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    You mean, you want to show that
    [tex]\frac{d}{dx} \ln(V(x)) = \frac{dV/dx}{x}[/tex] ?

    Well, this follows from the chain rule and the fact that [itex]\frac{d}{du} \ln(u) = \frac{1}{u}[/itex].
    The latter can be derived, for example by differentiating [tex]x = e^{\ln(x)}[/tex] w.r.t. x using the chain rule, and solving for d ln(x)/dx.
     
  4. Feb 23, 2010 #3
    To give a bit more context.

    I was trying to solve the partial differential equation: [tex]3U_{y}+U_{xy}=0[/tex] with the hint, let [tex]V=U_y[/tex]

    substituting we have [tex] 3V+V_{x}=0[/tex]
    then [tex]-3=\frac{V_{x}}{V}[/tex]

    I didn't really know how to continue from here so I just played around and figured out that [tex]V=e^{3x}[/tex] and [tex]U(x,y)=ye^{-3x}[/tex]

    I'm trying to figure out what to do when I come to [tex]-3=\frac{V_{x}}{V}[/tex].
     
  5. Feb 23, 2010 #4

    CompuChip

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    Ah, then indeed you can use the integral you gave.

    An alternative is to note that 3V + Vx = 0 is a linear ordinary differential equation. In that case, you can usually stick in [itex]V(x) = e^{\lambda x}[/itex] as a "trial" solution. The differential equation then turns into a polynomial equation for [itex]\lambda[/itex] with solutions [itex]\lambda_1, \cdots, \lambda_n[/itex], where n is the order of the ODE (in this case, n = 1). The general solution to the ODE is then
    [tex]V(x) = A_1 e^{\lambda_1 x} + \cdots + A_n e^{\lambda_n x}[/itex]
    where the Ai[/i] are to be determined (for example from initial or boundary values).
     
  6. Feb 23, 2010 #5

    arildno

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    Note:

    You can rewrite this as:
    [tex]\frac{\partial}{\partial{y}}(3U+U_{x})=0\to{3}U+U_{x}=F(x)[/tex]
    where F is some arbitrary function of x.

    This can be re-written as:
    [tex]\frac{\partial}{\partial{x}}(e^{3x}U)=e^{3x}F(x)[/tex]

    whereby the solution is given as:
    [tex]U(x,y)=e^{-3x}\int(e^{3x}F(x))dx+G(y)e^{-3x}[/tex]

    where G(y) is some arbitrary function in y.

    Obviously, the first term can be replaced by some arbitrary function H(x).
     
    Last edited: Feb 23, 2010
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