1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Aug 16, 2009 #1
    then i divide both sides by dy
    then i divide both sides by y
    so i have here a bernuly foruma
    i divide both sides by [tex]x^2[/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

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

    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

    User Avatar
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook