I'm trying to find a general solution to the following system of ODEs:
$\ddot{x}=\frac{-x}{(x^2+y^2)^{3/2}} \\ \ddot{y}=\frac{-y}{(x^2+y^2)^{3/2}} \\ x(0)=0, \dot{x}(0)=-.25,y(0)=1,\dot{y}(0)=.5 \\ \text{where }\ddot{x}=\frac{d^2x}{dt^2},\ddot{y}=\frac{d^2y}{dt^2}$
I've been trying to work out an analytic solution but cannot find one. I have worked out a numerical solution using Mathematica. Is there a technique I should be using for this, I am new to solving systems of ODE's.
HINT : Change to polar coordinates.
The general solution is given below. Continue for computing the parameters $c_1,c_2,c_3,c_4$ according to the given initial conditions.
Note : some parameters can be real or complex, depending on the elliptic or hyperbolic trajectory.
The integrals obtained below can be written on close forms with elementary functions.
Addition to the first answer, with conditions $x(0)=0,y(0)=1,x'(0)=-1/4,y'(0)=1/2$ :
Obviously, $\rho(0)=1$ and $\theta(0)=\pi/2$. Putting these values into the above equations $x'(\rho,\theta)$ and $y'(\rho,\theta)$ leads to $\rho'(0)=1/2$ and $\theta'(0)=1/4$
$$\theta'(0)=\frac{c_1}{\rho(0)^2}=\frac{1}{4}=\frac{c_1}{1^2} \quad\to\quad c_1=\frac{1}{4}$$
$$\rho'(0)=\frac{1}{\rho(0)}\sqrt{c_2\rho(0)^2+2\rho(0)-c_1^2}=\frac{1}{2}=\frac{1}{1}\sqrt{c_2(1)^2+2(1)-(1/4)^2} \quad\to\quad c_2=-\frac{27}{16}$$
With the lower bound $=1$ of the integrals, one can set $c_3=0$ and $c_4=\pi/2$ so that : $\rho(t=0)=1$ and $\theta(t=0)=\theta(\rho=1)=\pi/2$