I have the following coupled differential equations: $$m\ddot x=-\frac{x}{(x^2+y^2)^{3/2}}$$ $$m\ddot y=-\frac{y}{(x^2+y^2)^{3/2}}$$ I've tried defining $\eta=x+iy$ and multiplying the equation of $m\ddot y$ by $i$ but this didn't work due to the $x^2+y^2$ on the bottom splitting into $(x+iy)(x-iy)$. The $(x^2+y^2)^{3/2}$ on the bottom makes it seem like there isn't a nice analytic solution to this, however it just describes a radial field, so with intial velocity of $0$ it should come out with a straight line.
Any help would be appreciated, thanks.
This is the motion of a mass $m$ in the gravitational potential $-\frac1{\sqrt{x^2+y^2}}$. Thus a constant of motion is the energy $$ E(x,\dot x)=\frac m2 (\dot x^2+\dot y^2)-\frac1{\sqrt{x^2+y^2}}=const. $$ A second constant is obtained from $$ x\ddot y-y\ddot x=0\implies x\dot y-y\dot x=const. $$
As is classical, the solutions are the Kepler ellipses.
Inserting polar coordinates leads to $$ r^2\dot \phi=L $$ and \begin{align} E&=\frac{m}2(\dot r^2+r^2\dot\phi^2)-\frac1r =\frac{m}2\left(\dot r^2+\frac{L^2}{r^2}-\frac2{mr}\right) \\&=\frac{m}2\left(\dot r^2+\left(\frac{L}{r}-\frac1{mL}\right)^2\right)-\frac1{2mL^2} \end{align} which allows to reduce by one dimension by inserting a circle parametrization and so on.