I'm trying to solve the geodesic problem on a plane using the parametrization $x=x(t), y=y(t), x'=\frac{dx}{dt}, y'=\frac{dy}{dt}$.
But when using the Euler-Lagrange equation, I'm obtaining that
$$ -\frac{d}{dt}\left[\frac{x'}{\sqrt{x'^2 +y'^2}} \right]-\frac{d}{dt}\left[\frac{y'}{\sqrt{x'^2 +y'^2}} \right]=0$$
where I've been stuck since I can't find the solution to obtain the straight line equation.
The Euler-Lagrange equations tell you in this case that
$$\frac{\partial L}{\partial x}=\frac{d}{dt}(\frac{\partial L}{\partial x'})$$ $$\frac{\partial L}{\partial y}=\frac{d}{dt}(\frac{\partial L}{\partial y'})$$
For the Lagrangian $L(x', y')=\sqrt{x'^2+y'^2}$, the derivatives on the left hand side of the equations are zero. Integrating both sides with respect to $t$ then yields
$$\frac{\partial L}{\partial x'} = A$$ $$\frac{\partial L}{\partial y'} = B$$
Where $A, B$ are constants. Now, we can plug in our Lagrangian, take the derivatives, and get
$$\frac{2x'}{2\sqrt{x'^2+y'^2}} = A \implies 2\sqrt{x'^2+y'^2} = \frac{2x'}{A}$$ $$\frac{2y'}{2\sqrt{x'^2+y'^2}} = B \implies 2\sqrt{x'^2+y'^2} = \frac{2y'}{B}$$
Combining the equations gives
$$\frac{2x'}{A}=\frac{2y'}{B} \implies y'=\frac{B}{A}x'$$
Finally, we can integrate both sides with respect to $t$, and conclude
$$y = \frac{B}{A}x +C$$
Which is obviously a line (it is of the form $y = mx+b$).