$$ \int \limits_{0}^{\infty}\sqrt{1 + y'^{2}(x)}dx = 2 \sqrt{x} + y \qquad (.1) $$ 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?
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.