Variation Problem with Euler-Lagrange Differential Equation I'm just trying to understand that type of equations, and I can't solve this, kind of a simple minimization problem. Maybe someone can help me ? Here is my equation
$$\eqalign{
  & A(y(x)) = \int_{ - 1}^1 {{{\sqrt {1 + {{\left[ {y'(x)} \right]}^2}} } \over {y(x)}}dx}   \cr 
  & y( - 1) = 1  \cr 
  & y(1) = 1 \cr} $$
Thanks!
 A: Here the Lagrangian $L(t,y,v)=\frac{\sqrt{1+v^2}}{y}$ does not depend directly on the variable $t$, in this case is useful to apply the second formulation of Euler-Lagrange equation:
$$L(y(t),y'(t))-y'L_v(y(t),y'(t))=cost.$$
In your case it produces the following differential equation
$$y(t)\sqrt{1+y'(t)^2}=cost.=C$$
We can rewrite the previous equation in the form:
$$\pm \frac{y'(t)y(t)}{\sqrt{C^2-y(t)^2}}=1$$
and this is equivalent to
$$\pm\big(\sqrt{C^2-y(t)^2}\big)'=1.$$
Hence
$$\pm \sqrt{C^2-y(t)^2}=t+D$$
and finally squaring we get
$$C^2-y(t)^2=(t+D)^2.$$
Now just plug in the conditions $y(1)=y(-1)=1$ to get $D=0$ and $C^2=2$.
A: Just to make a complementing remark to @mrprottolo's solution. 
It happens in some variational problems that the Beltrami identity gives more complicated equations to solve than the canonical E-L equation. This problem is one of the examples. If we take the classical E-L we get after elementary simplifications that 
$$
y'(1+y'^2+yy'')=0,\quad y(-1)=y(1)=1.
$$
It gives one possibility $y'=0$ and another to be
$$
1+y'^2+yy''=1+(yy')'=1+\frac{1}{2}(y^2)''=0\qquad\Leftrightarrow\qquad (y^2)''=-2.
$$
Both DE's are very elementary (without square roots etc).
