Why is $\gamma(t)=(0,t)$ a geodesic in the hyperbolic plane? I'm having trouble to understand a very simple fact of the book of DoCarmo "Riemannian Geometry". In the page 73 he calculates the geodesics of the hyperbolic plane:
$$ \mathbb{R}^+_2 = \{ (x,y) \in \mathbb{R}^² :y>0 \} \qquad g_{ij}= y^{-2} \delta_{ij} $$
To do so he considers the curve $\gamma(t)=(0,t)$ which is a geodesic. But when I consider the equations of the geodesics I don't get this conclusion. 
The Cristoffel symbols for the hyperbolic plane with this metric are:
$$\Gamma^x_{xx}=\Gamma^y_{xy}=\Gamma^x_{yy}=0 \quad \Gamma^y_{xx}=\frac{1}{y} \quad \Gamma^x_{xy} = \Gamma^y_{yy} = -\frac{1}{y}$$
And then the equations of the geodesics:
$$ \frac{d^2x}{dt^2} - \frac{1}{y}\frac{dx}{dt} \frac{dy}{dt} = 0 $$
$$ \frac{d^2y}{dt^2} + \frac{1}{y} \left( \frac{dx}{dt} \right)^2 - \frac{1}{y} \left( \frac{dy}{dt} \right)^2 = 0$$
Then when I consider the given curve it satisfies the first equation but not the second. What I am missing here?

As he says, it's the image of a geodesic. It's not parametrized by arclength.
