geodesics in a Liouville's surface A surface $S$ is a Liouville's surface if its coefficients of the first fundamental form satistifes $E=G=U+V$ and $F=0,$ where $U=U(u)$ and $V=V(v).$
I need to prove that the geodesics in a Liouville's surfaces  can be obtained by primitivation of the form $$\int \frac{\textrm{d}u}{\sqrt{U(u)-c}}=\pm\int\frac{\textrm{d}v}{\sqrt{V(v)+c}}. $$
My first try is to use the parallelism equations for a geodesic curve. If $\gamma$ is a geodesic, $\gamma(t)=\mathbb{x}(u(t),v(t)),$ where $\mathbb{x}$ is a local parametrization of $S,$ so 
$$
\begin{cases}
\displaystyle u''+\Gamma_{11}^{1}(u')^{2}+2\Gamma_{12}^{1}u'v'+\Gamma_{22}^{1}(v')^{2}=0, \\
\displaystyle v''+\Gamma_{11}^{2}(u')^{2}+2\Gamma_{12}^{2}u'v'+\Gamma_{22}^{2}(v')^{2}=0.
\end{cases}
 $$
Computing the Christofell symbols, I find
$$\Gamma_{11}^{1}=\frac{U_{u}}{2(U+V)},\;\Gamma_{11}^{2}=-\frac{V_{v}}{2(U+V)},\; \Gamma_{12}^{1}=\frac{V_v}{2(U+V)\;}, \\
\Gamma_{12}^{2}=-\frac{U_{u}}{2(U+V)},\;\Gamma_{22}^{1}=-\frac{U_{u}}{2(U+V)},\;\Gamma_{22}^{2}=\frac{V_v}{2(U+V)}.
$$
So, applying this to the ODE,
$$
\begin{cases}
\displaystyle u''+\frac{U_{u}(u)}{2(U(u)+V(v))}(u')^{2}+\frac{V_{v}(v)}{U(u)+V(v)}u'v'-\frac{U_{u}(u)}{2(U(u)+V(v))}(v')^{2}=0, \\
\displaystyle v''-\frac{V_{v}(v)}{2(U(u)+V(v))}(u')^{2}+\frac{U_{u}(u)}{U(u)+V(v)}u'v'+\frac{V_{v}(v)}{2(U(u)+V(v))}(v')^{2}=0.
\end{cases}
$$
But... well, I don't know if I can do something from here.
I tried to manipulate the equations, and I got
$$
\begin{cases}
2\left[u'\left(U(u)+V(v)\right)\right]'-\left((u')^{2}+(v')^{2}\right)U_{u}(u)=0 \\
2\left[v'\left(U(u)+V(v)\right)\right]'-\left((u')^{2}+(v')^{2}\right)V_{v}(v)=0
\end{cases}
 $$
Since, both equations equals to zero, one is equal to the other, so, manipulating,
$$2\left[\left(U(u)+V(v)\right)\left(u'-v'\right)\right]'+\left((u')^{2}+(v')^{2}\right)\left(V_{v}(v)-U_{u}(u)\right)=0, $$
and now I don't know what to do.
 A: Starting from the equation Ted Shifrin told you, $$ \frac{d}{dt}\left[ (U+ V) \dot{u} \right] = \frac{1}{2(U + V)} \frac{dU}{du}. $$ 
Therefore, 
$$ \frac{dU}{du} = 2(U + V) \frac{d}{dt} \left[ (U +V) \dot{u} \right] $$
$$ \therefore \frac{dU}{du} = 2 (U + V) \left[ \frac{d(U + V)}{dt} \dot{u} + (U + V) \ddot{u}  \right] $$
$$ = 2 (U + V) \frac{d(U + V)}{dt} \dot{u} + 2 (U + V)^2 \ddot{u}$$
$$  =\frac{d (U + V)^2}{dt} \dot{u} + 2 (U + V)^2 \frac{d \dot{u}}{dt}. $$
However, notice that 
$$ \frac{d}{dt} \left[ (U + V)^2 \dot{u}^2 \right] = 2 (U + V) \frac{d(U+V)}{dt} \dot{u}^2 + (U + V)^2 2 \dot{u} \ddot{u}$$
$$ = \dot{u} \left[ \frac{d(U + V)^2}{dt}\dot{u} + (U + V)^2 2 \ddot{u} \right] = \dot{u} \frac{dU}{du}. $$
Since $\frac{dU}{du} \frac{du}{dt} = \frac{dU}{dt} $,
integrating you obtain for some constant A
$$ (U + V)^2 \dot{u}^2 = U - A. $$
The geodesic is parameterized by arclength, which implies
$$ (U + V)(\dot{u}^2 + \dot{v}^2) = 1 \implies $$
$$ (U + V)^2(\dot{u}^2 + \dot{v}^2) = U + V. $$
Substitute what you found for $(U + V)^2 \dot{u}^2$, reaching that $ (U + V)^2 \dot{v}^2 = V + A $ . 
Finally, take square roots (here the sign possibilities appear) and "equaling the $ dt $s":
$$ \frac{(U + V)}{\sqrt{U - A}} du = \pm \frac{(U + V)}{\sqrt{V + A}} dv $$
Cancelling $ U + V $ and integrating, you reach, for some constant $ B $:
$$ \int \frac{du}{\sqrt{U - A}} = \pm \int \frac{dv}{\sqrt{V + A}} + B .$$
