Geodesic Upper Half Plane help

[mmmboh]

The metric is $$ds^2=\frac{dx^2+dy^2}{y^2}.$$ I have used the Euler-Lagrange equations to find the geodesics, and my equations are $$\dot{x}=Ay^2,$$ $$\ddot{y}+\frac{\dot{x}^2-\dot{y}^2}{y}=0.$$ I cannot seem to find the first integral for the second equation. I know it is $$\dot{y}=y\sqrt{1-Ay^2},$$ but I can't seem to derive it. The only trick I currently know for doing these type of things is to multiple by $$\dot{y}$$ and then integrate, but that doesn't work here. Can anyone offer some guidance?

I tried it a slightly different way, but it doesn't seem to work for some reason:
Instead of parametrizing, x=x(t), y=y(t) and minimizing, I just minimized $$\frac{1+y'(x)^2}{y^2}.$$ Using the Euler-Lagrange equations, I get $$y''y-y'^2+1=0,$$ and $$y(x)=sinh(x)$$ is a solution to this...but the geodesics are suppose to be half circles, and this doesn't give me a half circle..I am quite confused.

 

[lippyka]

The issue here is that you are not solving the correct equation. The Euler-Lagrange equation you wrote down is \ddot{y}+\frac{\dot{x}^2-\dot{y}^2}{y}=0, which is not the same as the equation you wrote down for minimizing \frac{1+y'(x)^2}{y^2}. The correct equation for minimizing \frac{1+y'(x)^2}{y^2} is \ddot{y}+\frac{1}{y}=0.Once you have the correct equation, you can solve it by integrating twice and using the appropriate boundary conditions to obtain the solution y(x)=sinh(x). From this, you can then determine the first integral of the equation by multiplying by \dot{y} and integrating. This gives you the result \dot{y}=y\sqrt{1-Ay^2}, where A is a constant of integration.