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Calculation question

  1. Sep 1, 2005 #1
    for this equation, y=vx, it's the same as v=y/x
    however, if i differenate the 1st equation, dy/dx=v+x*dv/dx
    and if i differenate the 2nd equation, dv/dx=(1/x)*dy/dx + [-1/(x^2)]y
    but both differenated equations don't seem the same...
    does somebody know where my calculuations went wrong?
  2. jcsd
  3. Sep 1, 2005 #2


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    Staff Emeritus
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    dv/dx=(1/x)*dy/dx + [-1/(x^2)]y

    becomes x (dv/dx) = dy/dx - y/x, but v = y/x

    so x (dv/dx) = dy/dx - v


    x (dv/dx) + v = dy/dx, which is same as other equation.
  4. Sep 2, 2005 #3
    thank you very much!!! :)
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