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

2nd Order Elliptic Equation

  1. Jul 12, 2015 #1
    I've been searching for exact solution of d2y/dx2 = C/y, where C is some constant, such equation take place when deriving equation of motion in gravitationnal field, I'm more interested in how to solve it, yet I only managed to express it as power series using taylor's theorem at x = 0, just pick y(0) = c1, y'(0) = c2, so y''= C/c1, y'''(x) = C*c2/c12 and so on, until y ≈ c1 + c2*x + C/2c1 * x2+ C*c2/6c12 *x3 + ......., Is there's any special function that solves the equation or any other approximisation ?, I want to hear some thoughts !,
  2. jcsd
  3. Jul 12, 2015 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If we let ##v = dy/dx##, then

    $$\frac{C}{y} = \frac{d^2y}{dx^2} = \frac{dv}{dx} = \frac{dy}{dx} \frac{dv}{dy} = v \frac{dv}{dy}.$$

    We can integrate this for ##v(y)## (it's the square root of log). Solving ##v = dy/dx## for ##y## in closed form is probably not possible.
  4. Jul 12, 2015 #3


    User Avatar
    Science Advisor

    There is a trick for solving second order equations of this type, where the independent variable x is missing. Using the notation y' = dy/dx, write d2y/dx2 as follows:
    [tex] \frac{d^2 y}{dx^2} = \frac{dy'}{dx} = \frac{dy'}{dy} \frac{dy}{dx} = y' \frac{dy'}{dy}[/tex]

    Now you have eliminated x from the equation and turned it into a first order equation in y and y'. For your equation:

    [tex] y' \frac{dy'}{dy} = \frac{C}{y} [/tex]


    [tex] y' dy'= \frac{C}{y} dy[/tex]

    Now you can integrate both sides, and solve for y' in terms of y. Then you replace y' with dy/dx and integrate a second time. So you should be able to solve your equation analytically without needing power series.
  5. Jul 13, 2015 #4
    Thanks for quick replies, as far as I tried this seems "Damn" hard to solve but can get other approximisation from it , thanks :p
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook