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Solving an IC

  1. Mar 7, 2007 #1
    Its funny.. because I got this right on a test.. but.. I'm looking back at it.. and it doesn't make sense how I figured it out...

    I had to find the IC of [tex]d(\frac{x^2}{y})[/tex]

    this was my work...

    [tex]d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2[/tex]
    [tex]d(\frac{x^2}{y}) = \frac{2xy - x^2y}{y^2}[/tex]

    I'm just confused because I would of thought that it would of been something like this...

    [tex]d(\frac{x^2}{y}) = 2xy^-^1 + x^2y^-^2[/tex]

    hmm...?
     
  2. jcsd
  3. Mar 8, 2007 #2

    HallsofIvy

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    What do you mean by "IC"? When I saw "Solving an IC" in the differential equation forum, I thought you mean an "initial condition" problem but I see no differential equation at all here!

    Certainly, the derivative of x2/y is, by the quotient rule,
    [tex]\frac{2xy- x^2}{y^2}[/tex]
    There is no "y" in the second term in the numerator. (I surely don't see where you get your first line
    [tex]d(\frac{x^2}{y}) = 2xy - x^2y^2 - y^2[/tex]
    that makes no sense to me.)

    Of course, you could also write x2/y as x2y-1 and use the product rule. In that case, the dervative is
    [tex]2xy^{-1}+ (-1)x^2y^{-2}[/tex]
    Just rewrite that as
    [tex]\frac{2x}{y}- \frac{x^2}{y^2}[/tex]
    multiply the numerator and denominator of the first fraction by y and combine and you get the same answer as with the quotient rule.

    Since this has nothing to do with differential equations, I moving it to calculus.
     
  4. Mar 8, 2007 #3
    sorry, I didn't know the difference, this was on a DE test

    but thanks for explaining it
     
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