Geodesic Equation Problem I am reading through the book "Geometry, Topology and Physics" by Prof. Mikio Nakahara, page 285, an example of finding the geodesics of Poincare metric.
$$g=\frac{dx \otimes dx + dy \otimes dy}{y^2}$$
The geodesics equations given are:
$$x'' - \frac{2}{y}x'y'=0 ......(1)$$
$$y''-\frac{1}{y}(x'^{2} + 3y'^{2}) = 0 ......(2)$$
$x'=\frac{dx}{ds}$.
I managed to obtain equation (1) but I obtain a different equation for equation (2).
I have Christoffel symbols $$ \Gamma^{2}_{11} = \frac{1}{y}$$
$$ \Gamma^{2}_{22}=-\frac{1}{y}$$
$$ \Gamma^{2}_{12}= 0 $$
$$ \Gamma^{2}_{21}= 0 $$
and I obtained instead the following equation(2):
$$y''+ \frac{1}{y}x'^{2} - \frac{1}{y}y'^{2}=0$$
May someone help to verify where did I go wrong ?
Thank you.
 A: The length of a curve in this surface is
$\ell=\int_I\sqrt{\dfrac{x'^2+y'^2}{y^2}}ds$,
where $x'=\frac{dx}{ds}$ and $y'=\frac{dy}{ds}$.
But, since we are seeking a curve in a surface we can eliminate the $s$ parameter and do the supposition that $x=x(y)$.
Now our functional takes the form
$\ell=\int_I\sqrt{\dfrac{x'^2+1}{y^2}}dy$,
where $x'=\dfrac{dx}{dy}$.
So the Lagrangian is ${\cal L}=\sqrt{\dfrac{x'^2+1}{y^2}}$
and the Euler-Lagrange equation (which is
$\dfrac{\partial{\cal L}}{\partial x}-\dfrac{d}{dy}
\dfrac{\partial{\cal L}}{\partial x'}=0$ in general)
now is assembled with
$$\frac{\partial{\cal L}}{\partial x}=0\quad \mbox{and}\quad 
\frac{\partial{\cal L}}{\partial x'}=\frac{x'}{y\sqrt{1+x'^2}}.$$
This gives us
$\dfrac{d}{dy}\left(\dfrac{x'}{y\sqrt{1+x'^2}}\right)=0$ and
implies that
$\dfrac{x'}{y\sqrt{1+x'^2}}=K$.
The differential equation is transformed into
$x'=\dfrac{Ky}{\sqrt{1-K^2y^2}}$
that has solution
$$x=C-\frac{1}{K}\sqrt{1-K^2y^2},$$
where $K,C$ are two integration constants.
