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Solving linear differential equation

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm trying to make sense of the steps for solving a linear differential equation and I'm stuck at the part where
    [tex] \mu (x) \frac{\mathrm{d}y}{\mathrm{d}x} + \mu (x) P(x)y=\frac{\mathrm{d}}{\mathrm{d}x} [\mu (x)y] [/tex]

    I guess I'm not sure what [itex] \frac{\mathrm{d}}{\mathrm{d}x} [/itex] even means. Doesn't it mean the derivative with respect to x?

    Here's an example, perhaps that will help to clarify the trouble.
    [tex] x^{-2} \frac{\mathrm{d}y}{\mathrm{d}x} - 2x^{-3}y [/tex] is supposedly equivalent to
    [tex] \frac{\mathrm{d}}{\mathrm{d}x} (x^{-2}y) [/tex]

    Now how does that equate?

    2. Relevant equations



    3. The attempt at a solution

    The derivative of [tex] (x^{-2}y) [/tex] is [tex] - 2x^{-3}y [/tex].
    So... why doesn't [tex] \frac{\mathrm{d}}{\mathrm{d}x} (x^{-2}y) = - 2x^{-3}y [/tex] ?
     
  2. jcsd
  3. Jan 15, 2012 #2

    rock.freak667

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    d/dx here is the total differential.

    So for d/dx(x-2y), x and y are both variables. So you'd apply the product law here i.e. u = x^-2 and v=y.
     
  4. Jan 15, 2012 #3

    I like Serena

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    Happy new year, ArcanaNoir! :smile:

    In your equation you have ##{dy \over dx}##.
    This confirms that y is a function of x. that is, y=y(x).

    When you determine the derivative ##\frac{\mathrm{d}}{\mathrm{d}x} (x^{-2}y)##, you need to apply the product rule for derivatives.
     
  5. Jan 15, 2012 #4
    What is the difference between d/dx and dy/dx?
     
  6. Jan 15, 2012 #5

    I like Serena

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    d/dx is the derivative with respect to x of the function that comes after.

    dy/dx is the derivative of y=y(x) with respect to x.
     
  7. Jan 15, 2012 #6
    How come the derivative of the d/dx expression pops out with a dy/dx?
     
  8. Jan 15, 2012 #7

    I like Serena

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    Suppose you had to differentiate ##x^{-2}g(x)##, what would the derivative be?
     
  9. Jan 15, 2012 #8
    -2x^(-3)g(x)+x^(-2)g'(x)

    Ah, so since y is a function of x, aha. Thank you a lot for helping me dissect this.
     
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