Q:

Making Integral Stationary

We want to write and solve the Euler equation to make the integral stationary.
let $ I = \int_{x_1}^{x_2} e^x\sqrt {1+y'^2} $.
so $ F(x,y,y') = e^x\sqrt {1+y'^2} $.
we find $ F_{y'} =  \frac{e^xy'}{2\sqrt{1+y'^2}}$ and $ F_{y}= 0 $.
Now we use the Euler-Lagrange equation to find y.
$$ \frac{d}{dx}\frac{\partial F}{\partial y'}-\frac{\partial F}{\partial y} = 0 $$
This gives $$\frac{d}{dx}\frac{\partial F}{\partial y'} = 0 $$
$$\implies\frac{d}{dx}{\frac{e^xy'}{2\sqrt{1+y'^2}}} = 0.$$
Now we integrate to get 
$$\frac{e^xy'}{2\sqrt{1+y'^2}} = c_1$$ where $c_1$ is a constant of integration.
Now we wish to solve for $y'$ to get to $y$. How do we do this?

A:

Your last equation is reduced to
$${{\rm e}^{x}}{\frac {d}{dx}}y \left( x \right) =2\,c_{{1}}\sqrt {1+
 \left( {\frac {d}{dx}}y \left( x \right)  \right) ^{2}}
$$
Then you have
$$\left( {\frac {d}{dx}}y \left( x \right)  \right) ^{2}{{\rm e}^{2\,x}}=4\,{c_{{1}}}^{2} \left( 1+ \left( {\frac {d}{dx}}y \left( x \right) 
 \right) ^{2} \right) 
$$
From this last equation you derive that
$$\left( {\frac {d}{dx}}y \left( x \right)  \right) ^{2}={\frac {4{c_{{1}}}^{2}}{{{\rm e}^{2\,x}}-4\,{c_{{1}}}^{2}}}
$$
It is to say
$${\frac {d}{dx}}y \left( x \right) ={\frac {2c_{{1}}}{\sqrt {{{\rm e}
^{2\,x}}-4\,{c_{{1}}}^{2}}}}
$$
Making the integration you obtain
$$y \left( x \right) =\arctan \left( {\frac {\sqrt {{{\rm e}^{2\,x}
}-4\,{c_{{1}}}^{2}}}{2c_{{1}}}} \right) +{\it c_2}
$$
Do you agree?

