Solution of a first order non linear differential equation I am trying to find the extrema of the integral below 
$$
I= \int_0^1 y^2 \mathrm dx
$$
under the conditions
$$
\int_0^1 \left(\frac{dy}{dx}\right)^2 \mathrm dx =1
$$
and y(0)=y(1)=0
Using Lagrange multiplier λ, equivalently, I can find the extrema of the below quantity 
$$
I*= \int_0^1 y^2 \mathrm dx+λ\int_0^1 \left(\frac{dy}{dx}\right)^2 \mathrm dx
$$
Since there is not explicit dependence on x, by using beltrami identity
$$
F-y'\frac{\partial F}{\partial y'} = C
$$
the problem above reduced to the solution of the 2nd order differential equation
$$
2λ(y’)^2y’’-λ(y')^{2}- y^{2}-c=0.
$$
And after the substitution u=dy/dx 
The second order differential equation above become the first order one below
$$
2λu^{3}u’-λ(u)^{2}- y^{2}-c=0.
$$
where the differentiation refers to y.
And here is where I got stuck. 
Can someone give me some hint on how I can procced with that equation?
Many thanks in advance!
 A: Taking
$$
I = \int_0^1\left(y^2+\lambda\left(y'^2-1\right)\right)dx
$$
we have the Euler-Lagrange equations
$$
\cases{
y-\lambda y'' = 0\\
\int_0^1 y'^2 = 1
}
$$
Solving we have
$$
y = c_1 e^{\frac{x}{\sqrt{\lambda}}}+c_2 e^{-\frac{x}{\sqrt{\lambda}}}\\
\frac{c_1^2 \left(e^{\frac{2}{\sqrt{\lambda }}}-1\right) \sqrt{\lambda }+c_2^2
   \left(1-e^{-\frac{2}{\sqrt{\lambda }}}\right) \sqrt{\lambda }-4 c_2 c_1}{2 \lambda }=1
$$
now depending on your initial/boundary conditions, after determining $c_1,c_2$ we proceed with the second equation to determine the feasible value(s) for $\lambda$ and then it is done. 
Example
Supposing $y(0) = 1, y(1) = 1$ we have 
$$
c_1 = 0.0914543\\
c_2 = 0.908546\\
\lambda = 0.189694
$$
Follows a MATHEMATICA script making the necessary calculations
L = y[x]^2 + lambda (y'[x]^2 - 1)
EL = D[L, y[x]] - D[D[L, y'[x]], x]
yx = y[x] /. DSolve[EL == 0, y, x][[1]]
cond = Integrate[D[yx, x]^2, {x, 0, 1}] - 1
equ1 = (yx /. x -> 0) - 1;
equ2 = (yx /. x -> 1) - 1;
equs0 = {equ1, equ2, cond} /. {C[1] -> c1, C[2] -> c2};
obj = equs0.equs0
sol = NMinimize[{obj, lambda >= 0}, {c1, c2, lambda}, Reals]

NOTE
The boundary conditions $y(0) = y(1) = 0$ require a little different setup 
L = y[x]^2 - lambda^2 (y'[x]^2 - 1)
EL = D[L, y[x]] - D[D[L, y'[x]], x]
yx = y[x] /. DSolve[EL == 0, y, x][[1]]
cond = Integrate[D[yx, x]^2, {x, 0, 1}] - 1
equ1 = (yx /. x -> 0);
equ2 = (yx /. x -> 1);
equs0 = {equ1, equ2, cond} /. {C[1] -> c1, C[2] -> c2};
obj = equs0.equs0
sol = NMinimize[{obj}, {c1, c2, lambda}, Reals]

this is necessary to avoid complex numbers in the calculation procedure. This last case gives
$$
\cases{
c_1 = 0\\
c_2 = -0.450158\\
\lambda = -0.31831
}
$$
A: The Beltrami equation can not contain a second derivative, where should it come from? It should evaluate to
$$
y^2-λy'^2=C
$$
which easily leads to solutions in terms of trigonometric or hyperpolic functions.
