Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Require help

  1. Dec 12, 2006 #1
    In an investigation of a physics problem, I ran into the following equation:

    d^2(y)/(dt)^2 = k * y * (y^2 + c)^-1.5

    I know how to solve separable first order differential equations but this one seems to be beyond me. Assistance?
  2. jcsd
  3. Dec 12, 2006 #2
    hmm I don't think that one can be solved analytically, can you settle for a numeric answer?
  4. Dec 12, 2006 #3
    Well, one thing you can do is multiply by y prime

    [tex]y^{\prime} y^{\prime \prime} = \frac{k y y^{\prime}}{(y^2 + c)^\frac{3}{2}} [/tex]

    and then integrate to get

    [tex] \frac{1}{2} y^{\prime 2} = - \frac{k}{\sqrt{y^2 + c}} + A [/tex]

    where A is a constant of integration.

    You can then square root the y prime square, pull over all the y stuff on one side (and integrate again) to get x as some horrendous integral in y.


    [tex]x = \int{\frac{dy}{\sqrt{2(A- \frac{k}{\sqrt{y^2 + c}})}}} [/tex]

    or rather

    [tex]x = \frac{1}{\sqrt{2}} \int{\sqrt{\frac{\sqrt{y^2 +c}}{A \sqrt{y^2 +c} - k }} dy} [/tex]

    Other than that, I dunno.
    Last edited: Dec 12, 2006
  5. Dec 13, 2006 #4
    That looks intractable. I expected there to be a "clean" or closed (or whatever you call it) solution. This equation arose from me trying to plot the position of a point mass in a field generated by another point mass. The y is the vertical position (the reference point mass is at the origin and is stationary).
  6. Dec 13, 2006 #5
    With the use of a clever substitution, it may yet be soluble. You never know.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook