Second order non- linear differential equation solution EDIT1:
Please suggest a substitution for solving or any method of solving:
$$y'' (x)\cot( y(x) ) = y'(x){^2} +c  $$
Thanks in advance.
 A: Substituting $u=\sin(y)$ one gets
$$
u''=(\sin(y))''=(\cos(y)y')'=\cos(y)y''-\sin(y)y'^2=c\sin(y)=cu
$$
which is, depending on the sign of $c$, an oscillation equation or an exponential function which is easily solvable for $u$ and thus for $y$.
$$
u=\begin{cases}
a_1\cos(\sqrt{|c|}x)+a_2\sin(\sqrt{|c|}x)&\text{ for }c<0,\\
b_1+b_2x&\text{ for }c=0,\\
c_1e^{\sqrt{c}x}+c_2e^{\sqrt{c}x}&\text{ for }c>0.
\end{cases}
$$
A: $$\cot y \, y^{''} = y^{'}{^2} +c$$
Substitute $y'=p \implies y''=p\frac {dp}{dy}$
$$pp'\cot y = p^2 +c$$
$$\int \frac {2p}{ p^2 +c}dp=2\int \tan(y) dy$$
$$\ln|p^2 +c|=-2\ln |\cos(y)|+K_1$$
$$p^2 =\frac {K_1} {\cos^2(y)}-c$$
$$y'=\sqrt{\frac {K_1} {\cos^2(y)}-c}$$
$$\int \frac {dy}{\sqrt{\frac {K_1} {\cos^2(y)}-c}}=x+K_2$$
With $u=\sin(y)$
$$\int \frac {du}{\sqrt{K_1+cu^2}}=x+K_2$$
A: HINTS :
If $y$ is a function of $x$, namingly $y(x)$ and $c$ is just a constant such that $c \in \mathbb R$, then a (very complicated) solution can be yielded after a string of operations using the substitution 
$$v(y) = \frac{\mathrm{d}y(x)}{\mathrm{d}y}$$
which gives 
$$\frac{\mathrm{d}^2y(x)}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{\mathrm{d}y(x)}{\mathrm{d}x}\bigg)=\frac{\mathrm{d}v(y)}{\mathrm{d}x}=\frac{\mathrm{d}v(y)}{\mathrm{d}y}\frac{\mathrm{d}y}{\mathrm{d}x}=v(y)\frac{\mathrm{d}v(y)}{\mathrm{d}x}$$
and thus integrating solving for $v(y)$ and substituting back to find an expression in terms only of $y(x)$ and $x$.

If $y$ is a function of $c$, then there isn't a solution in terms of standard functions. Specifically, the solution is 
$$y(c) = \arcsin\big[c_1\big(c_2\mathrm{Bi}(c) + \pi \mathrm{Ai}(c)\big)\big]$$
which gives some rather interesting plots-properties while sampling an initial $y(0)$, where $\mathrm{Ai}(x)$ is the Airy function and $\mathrm{Bi}(x)$ is the Airy Bi function.  
$\qquad\qquad\qquad\qquad\qquad$
