Solving the Euler Lagrange equation for the functional $L(y)=\int_{-1}^7\sqrt{1+y'^2} dx$ I want to solve this functional
\begin{equation}
L(y)=\int_{-1}^7\sqrt{1+y'^2} \ dx
\end{equation}
with IC:  $y(0)=1,\ y(1)=2$
I start using the Euler Lagrange equation
\begin{equation}
\frac{d}{dx}\frac{\partial F}{\partial y'}-\frac{\partial F}{\partial y}=0
\end{equation}
But I encounter immediately a problem, namely that
$\frac{d}{dx}\frac{\partial F}{\partial y'}=0$, since $\frac{\partial F}{\partial y'}=\frac{y'}{\sqrt{1+y'^2}}$ which has no x-variable. Also $\frac{\partial F}{\partial y}=0$, hence it does not make sense.
Alterantively, I set this equal to a constant instead, and get:
\begin{equation}
\begin{array}
f\frac{d}{dx}\frac{y'}{\sqrt{1+y'^2}}=C \\
\frac{y'}{\sqrt{1+y'^2}}=C_1x + C_3\\
\frac{y'}{\sqrt{1+y'^2}}=C_2x+C_4
\end{array}
\end{equation}
How do I solve this?
Thanks
 A: Method I:
Obviously, the functional
$$
L(y)=\int_{a}^b \ \sqrt{1+y'^2} \ dx
$$
represents the arc-length of a curve, and the variational problem reduces to finding a curve of shortest arc-length connecting two points on the plane. This is one of the classical problems in the Calculus of Variations.
Geometrically, the well-known answer is a straight line, $$y = m x + c.$$
The constants $m$ and $c$ can be found from the given boundary conditions. (we need two conditions basically)
You have given the conditions in the problem as
$$
y(0) = 1, \ \ y(1) = 2
$$
Substituting, we get
$$
1 = m (0) + c \ \ \mbox{or} \ \ c = 1
$$
$$
2 = m(1) + c \ \ \mbox{or} \ \ m = 2 - c = 2 - 1 = 1
$$
Hence, the integral of the given variational problem is
$$
y = x + 1
$$
(Clearly, it satisfies the conditions $y(0) = 1, y(1) = 2$.
Method II:
Using the Euler-Lagrange equation
$$
{\partial f \over \partial y} - 
{d \over dx}\left({\partial f \over \partial y'}
\right) = 0 \tag{1}
$$
Since $f$ is purely a function of $y'$, we have
$$
{\partial f \over \partial y} = 0
$$
Thus, (1) simplifies to
$$
{d \over dx}\left({\partial f \over \partial y'}
\right) = 0 
$$
or
$$
{\partial f \over \partial y'} = \mbox{constant}
$$
Thus, we get
$$
{1 \over 2 \sqrt{1 + y'^2}} (2 y') = k_1
$$
or
$$
{y' \over \sqrt{1 + y'^2}} = k_1
$$
Simplifying, we get
$$
1 + y'^2 = k_1 y'^2
$$
where $k_1 = {1 \over c_1^2} = $constant
Simplifying again, we get
$$
y' = \mbox{constant} = m (\mbox{say})
$$
Integrating, we get
$$
y = m x + c
$$
which is a straight line.
The constants $m$ and $c$ can be found as in Method I.
Finally, we arrive at the optimal solution as
$$
y = x + 1
$$
A: I found a second method:
Using the formula
\begin{equation}
F-y'F_{y'}=C
\end{equation}
I get:
\begin{equation}
\begin{array}
a\sqrt{1+y'^2}-y'\frac{y'}{\sqrt{1+y'^2}}=C \\
-y'^2+1+y'^2=C\sqrt{1+y'^2}\\
1=C(1+y'^2)\\
Cy'^2+C=1\\
C(y'^2+1)=1\\
y'=\sqrt{C-1} \ set\ \sqrt{C-1}=A\\
\int y'=\int A \\
y(x)=Ax+B
\end{array}
\end{equation}
With initial conditions, this gives $y(x)=x+1$
