Differential equation (Brachistochrone problem) I'm really only supposed to solve the differential equation $(1+(y')^2)y=k^2.$

I haven't encountered any problem with $(y')^2$. How do you start with a problem like this, I did try googling it but all I got was basic differential equation, should I substitute for something?

Grateful for any help at this point!
$$\left(1+\left(\frac{dy}{dx}\right)^2\right)y(x)=k^2$$
$$\frac{dy}{dx}=\pm \sqrt{\frac{k^2}{y}-1}$$
ODE of the separable kind :
$$dx=\pm \sqrt{\frac{k^2-y^2}{y}}\:dy \quad\to\quad x=\pm\int \sqrt{\frac{k^2-y^2}{y}}\:dy $$
$$\pm x=k^2\tan^{-1}\left(\sqrt{\frac{y}{k^2-y}}\right)-\sqrt{(k^2-y)y}\:+c$$
The solution is on the form of $x$ as a function of $y$. There is no simple closed form for the inverse function $y(x)$.

The result can be presented on parametric form , with $\quad\tan(\theta)=\sqrt{\frac{y}{k^2-y}}$ :

$$\begin{cases}
x=k^2\left(\theta -\sin(\theta)\cos(\theta)\right)+c \\
y=k^2\sin^2(\theta)
\end{cases}$$
which is a parametric form of equation of cycloid.
There is an algebraic mistake in the above answer:  In the third line,
dx should be sqrt[ y/(k^2 -y^2) ] dy
(The argument of the sqrt has the numerator and denominator
interchanged).