Vtyjkyuk

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
$$C(y)=int_a^bsqrt1+(y'(x))^2dx$$
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 $L_y=frac d dxL_y'$, based on $J$ subject to the constraint $C$, so that the Lagrangian becomes:

$$L(y,y')=y(x)+lambdasqrt1+(y'(x))^2$$

Therefore:
$$L_y=1$$
$$L_y'=lambda frac y'(x) sqrt1+(y'(x))^2$$
$$frac d dx L_y'=lambda y''(x)left(frac 1 sqrt1+(y'(x))^2-frac (y'(x))^2left (1+(y'(x))^2right)^frac 3 2right)$$

which is equal to

$$frac d dx L_y'=lambda y''(x)left(frac 1left (1+(y'(x))^2right)^frac 3 2right)$$

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

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 $x<C_1$ and x decrease for $x>C_1$. 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 $lambda$).