Find bounded solutions of second order ODE system Resulting from a Classical Mechanics problem, I have the following system of second order ODEs
$$
\begin{align}
\ddot{x} &= \frac{-kx}{(x^2+y^2)^{3/2}} \\
\ddot{y} &= \frac{-ky}{(x^2+y^2)^{3/2}}
\end{align}
$$
With $k>0$ a real constant. I want to find the initial conditions for which the solutions of the system are bounded, and the initial conditions for which the system passes through the origin.
I know the origin is an unstable equilibrium point of the system, but I have no idea how to continue from there. I guess converting into polar coordinates would help with it, but I have only seen examples of converting systems of first order equations, not second order.
 A: Hint.
Calling $p = (x(t),y(t))$ and also
$\cases{
T = \frac 12 ||\dot p||^2\\
U = -\frac{k}{||p||}}
$
we have the lagrangian
$$
L = T - U = \frac 12 ||\dot p||^2+\frac{k}{||p||}
$$
with the movement equations
$$
\cases{
\ddot x(t) = -\frac{k x(t)}{\left(x(t)^2+y(t)^2\right)^{3/2}}\\
\ddot y(t) = -\frac{k y(t)}{\left(x(t)^2+y(t)^2\right)^{3/2}}}
$$
The total energy is given by
$$
E = T + U = \frac 12 ||\dot p||^2-\frac{k}{||p||}
$$
NOTE
Making a change of coordinates
$$
\cases{
x(t) = r(t)\cos(\theta(t))\\
y(t) = r(t)\sin(\theta(t))
}
$$
on the former lagrangian we arrive at
$$
L = \frac{1}{2} \left(r'(t)^2+r(t)^2 \theta '(t)^2-\frac{2 k}{r(t)}\right)
$$
and now deriving the movement equations we arrive at
$$
\left\{
\begin{array}{c}
r''(t) = r(t) \theta '(t)^2-\frac{k}{r(t)^2} \\
 2 r(t)r'(t) \theta '(t)+r(t)^2 \theta ''(t)= (r(t)^2\theta'(t))' = 0 \\
\end{array}
\right.
$$
so we have
$$
r(t)^2\theta'(t) = h_0
$$
and substituting into the first movement equation we get
$$
r''(t) = \frac{h_0^2}{r(t)^3}-\frac{k}{r(t)^2} 
$$
etc.
