For the functional $J$ defined by $$J(y)=\int_0^1y'\sqrt{1+(y'')^2}dx,$$Find an extremal satisfying the conditions $y(0)=0,~y'(0)=0,~y(1)=1$ and $y'(1)=2$?
My attempt:
Let $F(x,y,y',y'')=y'\sqrt{1+(y'')^2}$, then second order Euler's equation $$F_y-\frac{d}{dx}F_{y'}+\frac{d^2}{dx^2}F_{y''}=0,$$ gives $$0-\frac{d}{dx}\sqrt{1+(y'')^2}+\frac{d^2}{dx^2}\left(\frac{y' \cdot y''}{\sqrt{1+(y'')^2}}\right)=0,$$ How to proceed now with simple steps? It seem very tedious to solve the equation.
As I point out in the comment, we can rephrase the problem in terms of $z= y'$. The initial conditions $y(0) = 0$, $y(1) = 1$ then became the constraint $\int_0^1 z\text dx = 1$. OP's variational problem is therefore equivalent to the following one : $$\tilde J[z] = \int_0^1 (z\sqrt{1+z'^2} - Cz)\text dx $$ where $C$ is a Lagrange multiplier and with boundary conditions $z(1)=2$, $z(0) = 0$
In other words, we have the Lagrangian $L(z,z') = z\sqrt{1+z'^2} - Cz$.
Since this is independent of $x$, we know that : $$H = pz' - L$$ is conserved along the solutions of the Euler-Lagrange equation, where $p = \frac{\partial L}{\partial z'}$. Explicitly, we have : \begin{align} p &= \frac{zz'}{\sqrt{1+z'^2}} \\ H &= Cz - \frac{z}{\sqrt{1+z'^2}} = E \tag 1 \end{align} for some constant $E$.
This is then rewritten : $$1+z'^2 = \frac{z^2}{(Cz-E)^2} \tag 2$$
Edit @Diger provided the end of the solution in the comments below. I reproduced them here to provide a complete answer.
Evaluating $(1)$ at $x= 0$ and using $z(0) = 0$, we get $E= 0$. Then, $(2)$ becomes : $$1+z'^2 = \frac{1}{C^2}$$
ie $z'$ is constant and $z(x) = ax+b$. The boundary conditions then fix $a=2, b=0$. Then $\int_0^1 z(x) \text dx$ is automatically satisfied.
Therefore, $y(x) = \int_0^x z(x')\text dx' = x^2$ is the solution.