Geodesics in Poincare Disk I would like to find the geodesics in the Poincare disk.
I know that the metric is  $$\frac{dx^2+dy^2}{(1-x^2-y^2)^2}$$ so $$s=\int \frac{\sqrt{1+y'^2}}{1-x^2-y^2}\, dx$$ Then I try to find y(x) using the Euler-Lagrange equations. The ODE in which I end up is: $$yy'^2+xy'+y=0$$ I can't solve the above ODE.
I also tried using polar coordinates: $$s= \int \frac{\sqrt{r^2+r'^2}}{1-r^2}\,d\theta$$ but the expression I get is also not easy to handle with. 
 A: One method is to use symmetry:

*

*First prove (using your ODE, if you like) that the only geodesic with a vertical tangent vector is a vertical geodesic $x(t)=$constant, $y(t) = e^{\pm t + C}$.


*Next prove that the metric is invariant under the action of the group $SL(2,\mathbb{R})$ acting by fractional linear transformations. For the most pedestrian was to do this, consider a fractional linear transformation on the upper half plane of the form
$$w = \frac{az+b}{cz+d} \quad\text{(with $a,b,c,d \in \mathbb R$, $ad-bc=1$)}
$$
Assign variables for the real and complex parts $z=x+iy$ and $w=x'+iy'$. Rewrite the above equation as a system of two equations in two unknowns
$$x' = f(x,y), \qquad y' = g(x,y)
$$
and prove that
$$\frac{(dx')^2 + (dy')^2}{(y')^2} = \frac{dx^2 + dy^2}{y^2}
$$
There are probably slicker ways to do this.


*One should also prove that the fractional linear action of $SL(2,\mathbb R)$ acts transitively on points and on unit tangent vectors in the upper half plane. Transitivity on unit tangent vectors at $z=i$ is proved using the matrix $$\begin{pmatrix}a & b \\ c & d \end{pmatrix} = \begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$$
which represents a rotation of the unit tangent circle at $z=i$ by  the angle $\theta/2$. Transitivity on points is proved by setting $z=i$, choosing $w$ arbitrarily, and finding $a,b,c,d$.


*Finally, transform the equations for vertical geodesics in part 1 using the factional linear transformation in part 2. This will give you the general equations for geodesics.
A: $$y = \pm i \sqrt{2} x $$ 
Is a solution. It can be motivated (without the use of any assumptions) through a lot of work, but I'm trying to first see if using it I can find a general form, Namely I'm trying to find a general $f$ such that if 
$$ r(r')^2 + xr' + r = 0$$
then
$$ f(r)(f(r'))^2 + xf(r)r' + f(r) = 0 $$
Your equation is also equivalent to 
$$ (w')^2 + 2xw' + w = 0$$
where $w = y^2$
