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Change of variables in DE

  1. Apr 7, 2006 #1
    Am I insane or is this a typo:

    Consider the nonlinear DE

    dy/dx = (y-x)^2 + 1

    Show that the change of variables, u=x, v = y-x transforms this DE into the seperable DE: dv/du = v^2

    dv = dy
    du=dx
    dv/du = dy/dx = (y-x)^2 + 1 = v^2 + 1

    not v^2

    am i wrong ???
     
  2. jcsd
  3. Apr 7, 2006 #2
    In your expression for dv, where did dx go??
    it should be dv = dy - dx


    or you could have just differentiated v wrt x
    taht is dv/dx which is the same as dv/du because of the way u is defined.
     
  4. Apr 7, 2006 #3
    ok so dv/dx = -1, so dv/du = -1. so dv/du = -1(dy/dx), which is:

    -1 - (y-x)^2 or -1 - v^2

    i got that dv/du = dy/dx = (y-x)^2 + 1, but i cant get that dv/du = v^2 by itself.
     
    Last edited: Apr 7, 2006
  5. Apr 8, 2006 #4
    ok this time you went and eliminated the y

    this is what it should be
    [tex] \frac{dv}{du} = \frac{dv}{dx} = \frac{dy}{dx} - \frac{dx}{dx} = \frac{dy}{dx} - 1 [/tex]
    v and y are both funcions of x, so when you differentiate eitehr you get dv/dx and dy/dx.
     
  6. Apr 8, 2006 #5

    arildno

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    I can't see the point of doing the u-substitution.

    Just define [tex]v(x)=y(x)-x\to{y}(x)=v(x)+x\to\frac{dy}{dx}=\frac{dv}{dx}+1[/tex]

    A simple substitution then yields:
    [tex]\frac{dv}{dx}=v^{2}[/tex]
     
  7. Apr 9, 2006 #6
    ahhhh!!!!!!!!!

    thank you guys so much!
     
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