Show that the function $y$ which minimizes $\int_C F(y)\,ds$ satisfies $\frac{F(y)}{\sqrt{1+\left(y'\right)^2}}=C$ Say we want to minimize
$$\int_C F(y)\, ds$$
for some function $F(y)>0$. Using some physics, you can derive that if $y$ is a function for which the above functional is stationary, then
$$\frac{F(y)}{\sqrt{1+\left(y'\right)^2}}=C$$
The derivation of this can be found here. 
I wanted to prove this in a more rigorous manner using Euler-Lagrange equations. First, I rewrote the integral:
$$\int_CF(y)ds=\int_a^bF(y)\sqrt{1+\left(y'\right)^2}\,dx$$
Assuming that $F(y)$ is solely a function of $y$, we can write the Euler-Lagrange equation for the above functional. 
$$\frac{dF(y)}{dy}\sqrt{1+\left(y'\right)^2}-\frac{d}{dx}\left[\frac{F(y)y'}{\sqrt{1+\left(y'\right)^2}}\right]=0$$
I can rewrite this equation as follows: 
\begin{align*}
\frac{d}{dx}\left[\frac{F(y)y'}{\sqrt{1+\left(y'\right)^2}}\right]&=\frac{dF(y)}{dy}\sqrt{1+\left(y'\right)^2}\\
\frac{d}{dx}\left[\frac{F(y)}{\sqrt{1+\left(y'\right)^2}}\right]y'+\frac{F(y)y''}{\sqrt{1+\left(y'\right)^2}}&=\frac{dF(y)}{dy}\sqrt{1+\left(y'\right)^2}\\
\frac{d}{dx}\left[\frac{F(y)}{\sqrt{1+\left(y'\right)^2}}\right]&=\frac{\frac{dF(y)}{dy}\left[1+\left(y'\right)^2\right]-F(y)y''}{y'\sqrt{1+\left(y'\right)^2}}
\end{align*}
From here it is sufficient to prove that 
$$\frac{dF(y)}{dy}\left[1+\left(y'\right)^2\right]-F(y)y''=0$$
Unfortunately, I don't know how to proceed.
 A: This is easiest to prove using the Beltrami identity.  Suppose we want to extremize the quantity 
$$
S = \int_a^b f(y, y') \, dx,
$$
for which $f$ does not explicitly depend on $x$, i.e., $\partial f/\partial x = 0$.  This implies that
$$
\frac{df}{dx} = \frac{\partial f}{\partial y} \frac{dy}{dx} +  \frac{\partial f}{\partial y'} \frac{dy'}{dx}. 
$$
According to the Euler-Lagrange equation, we will have
$$
\frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) = \frac{\partial f}{\partial y},
$$
and so
$$
\frac{df}{dx} = \frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) y' + \frac{\partial f}{\partial y'} \frac{dy'}{dx} = \frac{d}{dx} \left( \frac{\partial f}{\partial y'} y' \right)
$$
This then implies that
$$
\frac{d}{dx} \left( f - \frac{\partial f}{\partial y'} y' \right) = 0 \quad \Rightarrow \quad f - \frac{\partial f}{\partial y'} y' = C,
$$
where $C$ is a constant.
In your case, you have $f(y,y') = F(y) \sqrt{ 1 + (y')^2}$, and so
$$
f - \frac{\partial f}{\partial y'} y' = F(y) \sqrt{ 1 + (y')^2} - \frac{F(y) y
'}{\sqrt{ 1 + (y')^2}} y' = \frac{F(y)}{ \sqrt{ 1 + (y')^2}} \left( 1 + (y')^2 - (y')^2 \right) \\=  \frac{F(y)}{ \sqrt{ 1 + (y')^2}}, 
$$
and the fact that this is a constant follows from the Beltrami identity.

Alternately:  if you wish to proceed from the last step of your derivation, note that
$$
\frac{d}{dx} \left[ \frac{F(y)}{\sqrt{1 + (y')^2}} \right] = \left( 1 + (y')^2 \right)^{-3/2} \left[ \frac{\partial F}{\partial y} \left( 1 + (y')^2 \right) - F(y) y'' \right] y'.
$$
The quantity in square brackets is what you want to vanish.  Combining this equation with the last equation in your derivation, you can obtain
$$
\frac{d}{dx} \left[ \frac{F(y)}{\sqrt{1 + (y')^2}} \right] = \frac{\left( 1 + (y')^2 \right)}{(y')^2} \frac{d}{dx} \left[ \frac{F(y)}{\sqrt{1 + (y')^2}} \right],
$$
which (since the pre-factor is never zero) implies the result sought.  (But the Beltrami identity is so useful in calculus of variations and in physics that I much prefer the derivation that uses it.)
