How to simplify this differential equation of orbital motion While solving for expression of orbit of a body under central force , this method used some substitution to simplify given problem 
But I cannot understand the simplification of $\left(\frac{\mathrm{d} r}{\mathrm{~d} \theta}\right)^{2}=\left(\frac{L^{2}}{k m}\right)^{2} \frac{1}{u^{4}}\left(\frac{\mathrm{d} u}{\mathrm{~d} \theta}\right)^{2}$       into $u^{\prime}(\theta)^{2}=-u^{2}+2 u+\frac{2 E L^{2}}{m k^{2}}$          so, anyone can please describe it in bit more detailed steps
 A: Starting from here
$$
\left(\frac{L}{mr^2}\frac{dr}{d\theta}\right)^2 = \frac{2E}{m}-\frac{L^2}{m^2r^2} + \frac{2k}{mr}
$$
and applying the substitutions $1/r\rightarrow kmu/L^2$ and $dr/d\theta = -L^2/(kmu^2)(du/d\theta)$, we get
$$
\left[\frac{L}{m}\left(\frac{kmu}{L^2}\right)^2\left(-\frac{L^2}{kmu^2}\right)\frac{du}{d\theta}\right]^2 = \frac{2E}{m}-\frac{L^2}{m^2}\left(\frac{kmu}{L^2}\right)^2 + \frac{2k}{m}\left(\frac{kmu}{L^2}\right).
$$
Now do the cancellations to get
$$
\left[\frac{k}{L}\frac{du}{d\theta}\right]^2 = \frac{2E}{m}-\frac{k^2u^2}{L^2} + \frac{2k^2}{L^2}u,
$$
then multiply through by $L^2/k^2$ and rearrange to get
$$
\left(\frac{du}{d\theta}\right)^2 = -u^2 + 2u + \frac{2EL^2}{mk^2}.
$$

That's the raw algebra version. If you want a somewhat more intuitive version that explains why this substitution is done, it comes from
$$
\frac{1}{r^2}\frac{dr}{d\theta} = -\frac{d}{d\theta}\left(\frac{1}{r}\right)
$$
If we just let $r = 1/s$, we'd get
$$
\left(\frac{L}{m}\frac{ds}{d\theta}\right)^2 = \frac{2E}{m}-\frac{L^2}{m^2}s^2 + \frac{2k}{m}s
$$
which has no reciprocals, but still has all those constants running around. $s = (m/L)t$ is the obvious choice to get rid of a bunch, leaving
$$
\left(\frac{dt}{d\theta}\right)^2 = -t^2 + \frac{2k}{L}t +\frac{2E}{m} 
$$
but there's still that constant in front of the $t$ term. However, if we had $t = (k/L)u$, then all the $t$ terms would have the same coefficient, so it can be divided out to get
$$
\left(\frac{du}{d\theta}\right)^2 = -u^2 + 2u + \frac{2EL^2}{mk^2}.
$$
And the net substitution we did was $u = (L/k)t = L^2/(mk)s = L^2/(mkr)$, as expected
