Physics Forums (http://www.physicsforums.com/index.php)
-   Differential Equations (http://www.physicsforums.com/forumdisplay.php?f=74)

 Atomos Dec12-06 11:25 AM

Require help

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?

 CPL.Luke Dec12-06 01:30 PM

hmm I don't think that one can be solved analytically, can you settle for a numeric answer?

 Matthew Rodman Dec12-06 03:57 PM

Well, one thing you can do is multiply by y prime

$$y^{\prime} y^{\prime \prime} = \frac{k y y^{\prime}}{(y^2 + c)^\frac{3}{2}}$$

and then integrate to get

$$\frac{1}{2} y^{\prime 2} = - \frac{k}{\sqrt{y^2 + c}} + A$$

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.

i.e.,

$$x = \int{\frac{dy}{\sqrt{2(A- \frac{k}{\sqrt{y^2 + c}})}}}$$

or rather

$$x = \frac{1}{\sqrt{2}} \int{\sqrt{\frac{\sqrt{y^2 +c}}{A \sqrt{y^2 +c} - k }} dy}$$

Other than that, I dunno.

 Atomos Dec13-06 09:56 PM

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).

 Matthew Rodman Dec13-06 10:42 PM

With the use of a clever substitution, it may yet be soluble. You never know.

 All times are GMT -5. The time now is 05:25 PM.