This is a long one, so thanks in advance if you actually read the whole thing...
I'm reviewing Lee's Riemannian Manifolds and am trying to discover the geometric implications of torsion in a connection by just coming up with a sample connection on $\mathbb{R}^2$ and trying to solve e.g. the geodesic equations and parallel transport equations. I'm going to pick a connection $\nabla$ with Christoffel symbols $\Gamma_{ij}^k$ gives as
$$\Gamma_{11}^1= 0,\ \Gamma_{11}^2= 1,\ \Gamma_{12}^1= -1,\ \Gamma_{12}^2= 0,\\ \Gamma_{21}^1= 0,\ \Gamma_{21}^2=1 ,\ \Gamma_{22}^1= -1,\ \Gamma_{22}^2= 0.$$
You can check that this connection is compatible with the Riemannian connection, but clearly it's not the Euclidean connection. So I'd expect its geodesics would look strange. According to Lee, a curve $x(t)=(x^i(t))$ is a geodesic if and only if its components satisfy the equations
$$\ddot{x}^k + \dot{x}^i\dot{x}^j\Gamma_{ij}^k = 0.$$
Using my Christoffel symbols, I get eight equations:
$$\ddot{x}^1 + \dot{x}^1\dot{x}^1\Gamma_{11}^1 = 0,\ \ddot{x}^2 + \dot{x}^1\dot{x}^1\Gamma_{11}^2 = 0,\ \ddot{x}^1 + \dot{x}^1\dot{x}^2\Gamma_{12}^1 = 0,\ \ddot{x}^2 + \dot{x}^1\dot{x}^2\Gamma_{12}^2 = 0,\\ \ddot{x}^1 + \dot{x}^2\dot{x}^1\Gamma_{21}^1 = 0,\ \ddot{x}^2 + \dot{x}^2\dot{x}^1\Gamma_{21}^2 = 0,\ \ddot{x}^1 + \dot{x}^2\dot{x}^2\Gamma_{22}^1 = 0,\ \ddot{x}^2 + \dot{x}^2\dot{x}^2\Gamma_{22}^2 = 0,$$
which simplify to
$$\ddot{x}^1 = 0,\ \ddot{x}^2 + \dot{x}^1\dot{x}^1= 0,\ \ddot{x}^1 - \dot{x}^1\dot{x}^2 = 0,\ \ddot{x}^2 = 0,\\ \ddot{x}^1 = 0,\ \ddot{x}^2 + \dot{x}^2\dot{x}^1 = 0,\ \ddot{x}^1 - \dot{x}^2\dot{x}^2 = 0,\ \ddot{x}^2= 0.$$
But now you can easily see that both components of your curve $x^1(t)$ and $x^2(t)$ must be constant, which seems to say that the only geodesics in this connection are constant paths with zero velocity. But doesn't the existence theorem say that we should be able to find geodesics with any initial position and velocity? Have I broken ODE's?