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How to know here on what variable its a derivative of(diff)

  1. Aug 16, 2009 #1
    [tex](2x^2ylny-x)y'=y[/tex]
    [tex](2x^2ylny-x)dy=ydx[/tex]
    then i divide both sides by dy
    [tex](2x^2ylny-x)=yx'[/tex]
    then i divide both sides by y
    [tex](2x^2lny-\frac{x}{y})=x'[/tex]
    [tex]x'+\frac{x}{y}=2x^2lny[/tex]
    so i have here a bernuly foruma
    i divide both sides by [tex]x^2[/tex]
    [tex]\frac{x'}{x^2}+\frac{1}{xy}=2lny[/tex]
    [tex]z=x^{-1}[/tex]
    [tex]z'=-1x^{-2}x'[/tex]
    [tex]-z'+\frac{z}{y}=2ln y[/tex]

    z is defined to be a function of x
    so [tex]z'=\frac{dz}{dx}[/tex]
    why the book interprets [tex]z'=\frac{dz}{dy}[/tex]
    ??

    z is linked to y not in a direct way .
    but z linked to x in a direct way
    z and x are more close to each other.

    i cant see a mathematical way of figuring it out
    its all intuition.and i my intuition is very bad
     
  2. jcsd
  3. Aug 16, 2009 #2
    [tex]\frac{dz}{dx} = -\frac{1}{x^2}[/tex]
    [tex]dz = -\frac{1}{x^2} dx[/tex]
    [tex]\frac{dz}{dy} = -\frac{1}{x^2} \frac{dx}{dy}[/tex]
     
  4. Aug 16, 2009 #3
    z=x'
    z'=-x^(-2)x'

    i cant undestand how you cameup with the first step
    [tex]
    \frac{dz}{dx} = -\frac{1}{x^2}
    [/tex]

    i cant understand what are you doing here?
     
  5. Aug 16, 2009 #4
    and your conclution
    doesnt show that z is a derivative by y
    it shows that z' is a derivative by x
     
  6. Aug 16, 2009 #5
    What I was doing up there was proving mathematically that [tex]\frac{x'}{x^2}[/tex] necessary equals [tex]-\frac{dz}{dy}[/tex]. The 1st step was obtained from your substitution variable [tex]z = x^{-1}[/tex].

    On a more intuitive basis, what are you attempting to do when performing the substitution? It's to simplify the DE via substituting x with a new variable z, and thus why would there be a dx lying around after you have completed your substitution? If [tex]z' = \frac{dz}{dx}[/tex], then how are you going to solve [tex]-z'+\frac{z}{y}=2ln y[/tex]? You now have what, three variables?
     
  7. Aug 16, 2009 #6

    D H

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    Staff Emeritus
    Science Advisor

    Because you defined it that way, right here:
    In the first line you defined z to be a function of x. In the second, you defined z' to be in a derivative of the same variable as used for x' -- which was y.
     
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