Solution of two second order differential equations Hi guys I have a second order time dependent differential equation. The solution is linked to the orbit of a body around a slightly non-symmetric gravitational field. I am unable to find the solution myself. 
The equations are:
$\ddot{x}=\frac{1}{x^2+y^2}[1-\frac{x}{(x^2+y^2)^{0.5}}-\frac{2x^2}{x^2+y^2}]$
and
$\ddot{y}=\frac{1}{x^2+y^2}[-\frac{2xy}{(x^2+y^2)}-\frac{y}{(x^2+y^2)^2}]$
If it is any help I  am considering a perturbation to a symmetric field which has solution x=cos(t) and y=sin(t). The initial conditions are x(0)=1, y(0)=0, $\dot{x(0)}=0$ and $\dot{y(0)}=1$. I am aware that this is easy to solve numerically but I require an analytic solution. Any help would be greatly appreciated.
 A: If you started from the potential $V(x,y)=-\frac1r-ϵ\frac{x}{r^2}$, then the force terms are 
$$\begin{align}
\ddot x=-\frac{\partial V}{\partial x}&=-\frac{x}{r^3}+ϵ\frac1{r^2}-ϵ\frac{2x^2}{r^4}\\
\ddot y=-\frac{\partial V}{\partial y}&=-\frac{y}{r^3}-ϵ\frac{2xy}{r^4}\\
\end{align}$$
or in polar coordinates $x+iy=re^{i\phi}$
$$
\ddot r+2i\dot r\dot \phi-r\dot \phi^2+ir\ddot \phi=-\frac1{r^2}-ϵ\frac{e^{i\phi}}{r^2}
$$
Now insert $r=1+ϵr_1$, $ϕ=t+ϵϕ_1$ to get in first order
$$
\ddot r_1+2i\dot r_1-r_1-2\dot \phi_1+i\ddot \phi_1=2r_1- e^{it} 
$$
Collecting real and imaginary terms this gives the IVP $r_1(0)=\dot r_1(0)=0$, $\phi_1(0)=\dot\phi_1(0)=0$,
\begin{align}
\ddot r_1-3r_1-2\dot \phi_1&=-\cos t 
\\
\ddot\phi_1+2\dot r_1&=-\sin t 
\\
\implies \dot\phi_1+2r_1&=\cos t -1
\\
\ddot r_1+r_1&=\cos t - 2\\
r_1&=2(\cos t -1)+\frac12t\sin t
\end{align}
Now all that is left is the integration of $\phi_1$. By perturbation theory, this approximation should be good for $t\ll \frac1ϵ$. As the phase space is 4-dimensional and only the energy conserved, the motion on the 3-dimensional energy hypersurface does not enforce periodicity nor boundedness, so there is no expectation that some other approach, such as multiple time scales, will change the nature of the approximation. However, the resonance might be resolved for some longer lasting (in terms of accuracy) terms.
