Why is $\gamma(t)=(0,t)$ a geodesic in the hyperbolic plane?

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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?

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The geodesics parametized by arc length are, with constants $A,B$ and $B > 0,$ $$ ( A, \; e^t), $$ $$ ( A + B \, \tanh t, \; \; B \, \operatorname{sech} t) $$ In both cases we can decide on some favorite constant $t_0$ and replace $t$ by $t - t_0.$

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