Am I solving dido's isoperimetric (variational) problem correctly? I am trying to solve dido's isoperimetric problem, more specifically, the version where we have to maximize the area under a curve, given that the two endpoints are on the x-axis, and given a fixed arclength:



That is, we have to maximize
$$J(y)=\int_a^by(x)dx$$
subject to constraint
$$C(y)=\int_a^b\sqrt{1+(y'(x))^2}dx$$
Where we assume $a$ is fixed, but $b$ is allowed to vary. 

Am I solving this correctly?

Besides the boundary constraint (because of variable boundary $b$) $L_{y'}(b)=0$, we of course formulate the Euler-Lagrange equation $L_y=\frac d {dx}L_{y'}$, based on $J$ subject to the constraint $C$, so that the Lagrangian becomes:

$$L(y,y')=y(x)+\lambda\sqrt{1+(y'(x))^2}$$

Therefore: 
$$L_y=1$$
$$L_{y'}=\lambda \frac {y'(x)} {\sqrt{1+(y'(x))^2}}$$
$$\frac d {dx} L_{y'}=\lambda y''(x)\left(\frac {1} {\sqrt{1+(y'(x))^2}}-\frac {(y'(x))^2}{\left (1+(y'(x))^2\right)^{\frac 3 2}}\right)$$ 

which is equal to

$$\frac d {dx} L_{y'}=\lambda y''(x)\left(\frac {1}{\left (1+(y'(x))^2\right)^{\frac 3 2}}\right)$$ 


  This gives the Euler-Lagrange equation $$\lambda
y''(x)=\left (1+(y'(x))^2\right)^{\frac 3 2}$$


I have no idea how to solve this ODE, and moreover, it doesn't seem like this is what I should be getting.

Did I derive this result correctly? If so, How do I solve it?

note: I know that it is also possible to solve by parameterizing $x=x(t), y(x)=y(x(t))$. I want to do this as well, later, but I would first like to understand the approach I'm taking here.
First, integrate both sides with regards to $x$. Next, we can get an explicit function $x=f(y')+C_1$, $C_1$ is a constant. Here we have a fact : 
                        $y'(x)=0$ if and only if $x=C_1$
Geometrically, curve $y$ reaches the maximum value when $x=C_1$. Hence, $x$ increases for $x&lt;C_1$ and x decrease for $x&gt;C_1$. So, $y'(x)$ will never be zero when $x$ in $(0,C_1) \cup (C_1,1)$. Finally, with some algebraic manipulations, we can get a circle equation (with radius $\lambda$).
First, integrate both sides with regards to $x$. Next, we can get an explicit function $x=f(y')+C_1$, $C_1$ is a constant. Here we have a fact : 
                        $y'(x)=0$ if and only if $x=C_1$
Geometrically, curve $y$ reaches the maximum value when $x=C_1$. Hence, $x$ increases for $x&lt;C_1$ and x decrease for $x&gt;C_1$. So, $y'(x)$ will never be zero when $x$ in $(0,C_1) \cup (C_1,1)$. Finally, with some algebraic manipulations, we can get a circle equation (with radius $\lambda$).