ODE problem shooting Please help me spot my mistake:
I have an equation
$$(u(x)^{-2} + 4u'(x)^2)^{\frac{1}{2}} - u'(x)\frac{d}{du'}(u(x)^{-2} + 4u'(x)^2)^{\frac{1}{2}} = k$$ where $k$ is  a constant.
I am quite sure that if I take $u(x) = \sqrt{y(x)}$ I would have the brachistochrone equation, hence I am expecting a cycloid equation if I let $u(x) = \sqrt{y(x)}$ in the result, but I don't seem to get it :(
My workings are as follows:
$$u(x)^{-2} + 4u'(x)^2- 4u'(x)^2 = k \times (u(x)^{-2} + 4u'(x)^2)^{\frac{1}{2}}$$
$$\implies u(x)^{-4} = k^2 \times (u(x)^{-2} + 4u'(x)^2)$$
$$\implies u'(x)= \frac{1}{2k}\sqrt{u(x)^{-4} - k^2u(x)^{-2}}$$
$$\implies \int \frac{1}{u \sqrt{u^2 - k^2}} du = \int \frac{1}{2k} dx$$
Change variable: let $v = \frac{u}{k}$
$$\implies \int \frac{1}{v \sqrt{v^2 - 1}} dv = \frac{x+a}{2}$$, where $a$ is a constant 
$$\implies \operatorname{arcsec}(v) =  \frac{x+a}{2} $$
$$\implies \operatorname{arcsec}\left(\frac{\sqrt{y}}{k}\right) = \frac{x+a}{2}$$
which does not seem to describe a cycloid...
Help would be very much appreciated! Thanks.

In the line after 
$$ 2k u' = \sqrt{ u^{-4} - k^2 u^{-2}} $$
(the fourth line of your computations) when you divided, you divided wrong. The integrand in the LHS should be
$$ \int \frac{u^2}{\sqrt{1-k^2 u^2}} \mathrm{d}u $$ 
