This comes up in an exercise of a physics book for finding the Brachistrochrone with calculus of variation. I should get that $x=\int\sqrt{\frac{y}{2a-y}}\,dy$ for an appropriate value of the constant $C$. However, I get:
$$\begin{aligned}\sqrt{\frac{y'^2+1}{y}} - y'\frac{y'}{\sqrt{\frac{y'^2+1}{y}}} &= \frac {1}{2a}\qquad\text{I'm guessing that }C=\frac{1}{2a}\\ \frac{y'^2+1}{y}-2y'^2+\frac{y'^2y}{y'^2+1}&=\frac {1}{4a^2}\\ (y'^2+1)^2+(y'y)^2-2y'^2(y'^2+1) &= \frac{y(y'^2+1)}{4a^2}\\ (y'^2+1-y'^2)^2+y'^4+(y'y)^2&=\frac {y(y^2+1)}{4a^2}\\ y'^2(y'^2+y^2)&=\frac {y(y^2+1)}{4a^2}\\ z&:=y'\\ z^2+zy-\frac{y(y^2+1)}{4a^2}&=0\end{aligned}$$
So $$z=\left(\frac{dy}{dx}\right)^2=-\frac y 2\pm\sqrt{\left(\frac{y}{2}\right)^2-\frac{y(y^2+1)}{a^2}}$$
And $$x=\int\frac{dy}{\sqrt{-\frac y 2\pm\sqrt{\left(\frac{y}{2}\right)^2-\frac{y(y^2+1)}{a^2}}}}$$
The Lagrangian for the Brachistrochrone is $$f(y,y') = \sqrt{\frac{{1+(y')^2}}{y}}$$ and thus $$\frac{\partial f}{\partial y'} = \frac{ y'}{\sqrt{ y \, \Big(1+(y')^2\Big)}}$$ and so your Hamiltonian, i.e. conserved energy function in $y, y'$ variables, is $$H(y,y') = f(y,y') - \frac{\partial f}{\partial y'} y' = C$$ which should yield the first order ordinary differential equation $$\sqrt{\frac{{1+(y')^2}}{y}} - \frac{ (y')^2}{\sqrt{ y \, \Big(1+(y')^2\Big)}} = C$$ where $C$ is a constant.