TITLE: How to solve following differential equation?
QUESTION [5 upvotes]: $$
\int \limits_{0}^{\infty}\sqrt{1 + y'^{2}(x)}dx = 2 \sqrt{x} + y \qquad (.1) 
$$
The solution is
$$
3y = x\sqrt{x} - 3\sqrt{x} .
$$
I don't know how to solve this type of equations. Also I don't understand how the right-hand side of the equation (.1) may depend on the variable x. 
Can you help me?

REPLY [2 votes]: There is something I don't understand here.
If $y$ is a solution of this equation, note that the lhs is a constant.
So
$$
y=-2\sqrt{x}+C.
$$
Now plugging this back into the integral yields $+\infty$.
So I find that there are no solutions.