$$x''=\frac{2x'y'}{y},$$ $$y''=\frac{(y')^2 -(x')^2}{y}.$$
It is also well known that the geodesics are semi-circles lying on the x-axis. However, I am encountering difficulties when I attempt to verify this fact.
I am trying to follow the solution to part (b) of problem 5 on paper 1 here.
The path $\tilde \gamma(t)=(\cos(t),\sin(t))$, when re-parameterized, should be a geodesic. At the point $(\cos(t),\sin(t))$, the unit tangent vector is $$(-\sin^2(t), \sin(t)\cos(t)).$$ So we need $$x''(t)=\frac{2(-\sin^2(t))(\sin(t)\cos(t)}{\sin(t)},$$ or $$x''(t)=-2\sin^2(t)\cos(t).$$ However, $(-\sin^2(t))'=-2\cos(t)\sin(t)$, which is not quite what we want. I'm confused why this does not line up. I suppose because the velocity field does not arise from a curve.
How can this be fixed up?
Recall that $$ 0=E_1(E_j,E_k)= \frac{1}{y^2} \{ \Gamma_{1j}^k + \Gamma_{1k}^j \} \Rightarrow \Gamma_{1j}^k =-\Gamma_{1k}^j $$
$$ -2y^{-3}= E_2(E_2,E_2)= \frac{1}{y^2} (2\Gamma_{22}^2) \Rightarrow \Gamma_{22}^2= - \frac{1}{y} $$
By definition,
$$ \Gamma_{12}^1 = -\frac{1}{y}$$
If $$ v= (-\sin^2 t, \sin\ t\cos\ t) :=fE_1+ g E_2$$ then $$ v (F(t))= y \frac{d}{dt} F(t) $$ and $$ \nabla_vv = \nabla_v (fE_1) + \nabla_v (gE_2) $$ $$= v(f) E_1+ v(g)E_2 + f\nabla_vE_1 + g\nabla_vE_2 $$ $$ = y(f',g') + f ( f \Gamma_{11}^i E_i + g \Gamma_{21}^i E_i) + g( f \Gamma_{12}^i E_i + g \Gamma_{22}^i E_i ) $$
$$ = y(f',g') + f^2 \Gamma_{11}^2E_2 +2f g \Gamma_{12}^1 E_1 + g^2 \Gamma_{22}^2 E_2 =0 $$