How can the 2nd Order Elliptic Equation be solved for exact solutions?

[Noctisdark]

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 !,

 

[fzero]

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.

 

[phyzguy]

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:
\frac{d^2 y}{dx^2} = \frac{dy'}{dx} = \frac{dy'}{dy} \frac{dy}{dx} = y' \frac{dy'}{dy}

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

y' \frac{dy'}{dy} = \frac{C}{y}

or:

y' dy'= \frac{C}{y} dy

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.

 

[Noctisdark]

Thanks for quick replies, as far as I tried this seems "Damn" hard to solve but can get other approximisation from it , thanks :p