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Leibnitz notation

  1. Jul 17, 2009 #1
    Hi,

    I was looking at a solution to a differential equation problem. And there are some parts that i don't understand.

    [tex]\frac{dy}{dx} - \frac{2y}{x} = x^2[/tex]

    And the solution looks like this:

    [tex]e^\int{ \frac{-2}{x}} = \frac{1}{x^2}[/tex]

    [tex]\frac{1}{x^2}\frac{dy}{dx} -\frac{2}{x^3}y=1[/tex]

    [tex]\frac{d}{dx}\frac{y}{x^2}=1[/tex]

    This is the part i don't understand, what does [tex]\frac{d}{dx}\frac{y}{x^2}[/tex] mean, and why is it equal to 1 ?

    To be honest i find this notation pretty confusing.
     
  2. jcsd
  3. Jul 17, 2009 #2

    tiny-tim

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    Hi James889! :smile:

    It just means d/dx of y/x2

    use the product rule, and that's (dy/dx)(1/x2) + y(-2/x3) …

    which is the LHS of the previous line! :wink:
     
  4. Jul 17, 2009 #3

    rock.freak667

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    Because

    [tex] \frac{d}{dx}( \frac{y}{x^2}) = \frac{1}{x^2}\frac{dy}{dx}- \frac{2}{x^3}y[/tex]
     
  5. Jul 17, 2009 #4

    HallsofIvy

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    What that means is that you differentiate [itex]yx^2[/itex] with respect to x! That's pretty basic terminology. If you going to study calculus, you had better learn it:
    [tex]\frac{df}{dx}[/tex] means "differentiate f with respect to x".

    In this case, y is a function of x so we need to use both the product rule and the chain rule.
    [tex]\frac{d yx^2}{dx}= \frac{dy}{dx}(x^2)+ y\frac{dx^2}{dx}[/tex]
    exactly what is on the left side of your equation above. It is equal to one because it is equal to the left side of the equation whose right side is 1: If A= B and B= C then A= C.
     
  6. Jul 17, 2009 #5

    Mark44

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    James,
    You're not asking the right question. The question is not why that derivative is 1, but what does it mean that that derivative is 1. The equation is
    [tex]\frac{d}{dx} \frac{y}{x^2} = 1[/tex]

    This is saying that the derivative wrt x of y/x2 = 1, which can only be true if y/x2 = x. This means that y/x2 = x ==> y = x3. Notice that this is the solution of the given differential equation.
     
  7. Jul 18, 2009 #6

    tiny-tim

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    plus a constant! :wink:
     
  8. Jul 18, 2009 #7
    Hello,

    What is the difference in meaning between these notations?
    [tex]\frac{dx}{e^x}[/tex] and [tex]e^x dx[/tex]
     
  9. Jul 18, 2009 #8

    tiny-tim

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    Hello James889! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    There's nothing special about dx "over" something …

    (unless the "something" also starts with "d", of course!)

    dx/ex is the same as (1/ex)dx :wink:
     
  10. Jul 18, 2009 #9

    HallsofIvy

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    That's a peculiar question!

    [itex]e^x dx[/itex] is dx times [itex]e^x[/itex] and [itex]dx/e^x[/itex] is dx divided by [itex]e^x[/itex]

    You could also write [itex]dx/e^x[/itex] as [itex](1/e^x)dx[/itex] or [itex]e^{-x}dx[/itex].
     
  11. Jul 18, 2009 #10
    Yes yes, im not the sharpest knife in the box :rolleyes:
     
  12. Jul 18, 2009 #11

    Mark44

    Staff: Mentor

    Right, but in the DE the constant was zero, so I omitted it in that step.
     
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