Suppose we are given a certain Riemannian metric $$\frac{(dx)^2 + (dy)^2}{y^2}$$
How do I calculate geodesics?
I know that we have the geodesic equation $$\ddot{x}^k(t) + \dot{x}^i(t)\dot{x}^j\Gamma_{ij}^k(x(t)) = 0$$
and that we have only two components, hence a geodesic $\gamma$ should be of the form $\gamma(t) = (x^1(t),x^2(t))$. Writing out the case $k = 1$ we get $$\ddot{x}^1(t) + (\dot{x}^1(t))^2\Gamma_{11}^1(x(t)) + 2\dot{x}^1(t)\dot{x}^2(t)\Gamma_{12}^1(x(t)) + (\dot{x}^2(t))^2\Gamma_{22}^1(x(t)) = 0$$
I also calculated $\Gamma^1_{11} = \Gamma^1_{22} = 0$ and $\Gamma^1_{12} = -1/y$. Now my problem is that I do not really know how to plug in the Christoffel symbols since they are evaluated in $x(t)$.
I know that there is a similar problem here, but I do know how to solve the resulting ODE. My problem is, that I would like to get to the initial equation.
Edit. The geodesic equation is taken from Riemannian Manifolds by John M. Lee, p. 58.
$x(t)$ means the same thing as $\gamma(t)$ in our situation, where $\gamma(t)=(x^1(t),\ldots,x^n(t))$, since $x(t)$ is usually short for $(x^1(t),\ldots,x^n(t))$. So we only need to plug $(x^1(t),\ldots,x^n(t))$ in the Christoffel symbol $\Gamma_{ij}^k(x^1,\ldots,x^n)$. For example, in this particular problem we have $$\Gamma_{11}^1=\Gamma_{22}^1=\Gamma_{12}^2=\Gamma_{21}^2=0,~\Gamma_{11}^2=1/y,~\Gamma_{22}^2=\Gamma_{12}^1=\Gamma_{21}^1=-1/y$$ Then plug in $x=x(t),y=y(t)$ to see that along the geodesic $\gamma$, we have $$\Gamma_{11}^1=\Gamma_{22}^1=\Gamma_{12}^2=\Gamma_{21}^2=0,~\Gamma_{11}^2=1/y(t),~\Gamma_{22}^2=\Gamma_{12}^1=\Gamma_{21}^1=-1/y(t)$$ The geodesic equation becomes \begin{align} \ddot x(t)-2\dot x(t)\dot y(t)/y(t)=0\\ \ddot y(t)+(\dot x(t))^2/y(t)-(\dot y(t))^2/y(t)=0 \end{align} The first equation can be integrated easily to $$\dot x(t)=c(y(t))^2$$ While the second equation is more complicated to deal with. However, it can be checked that $$\dfrac d{dt}\frac{\dot x^2+\dot y^2}{y^2}=\frac{2}{y^3}((\dot x\ddot x+\dot y\ddot y)y-(\dot x^2+\dot y^2)\dot y)=0$$ which leads to another integration. Now it's not hard to solve those equations. The complete solution can be found here, which is a semicircle centered on the real axis.