How to solve this second order DE question? How to solve this DE?
$(\cos y+p)y''+q{ (y') }^{ 2 }+r=0$
where $p, q, r$ are constants. 
I know that this a type of second order DE whose solution is given as $y=CF+PI$. Here the $PI$ is $0$.
Generally it can be represented as, $$a{ D }^{ 2 }+bD+c=0$$
Now we can solve this as a quadratic equation and write the solution. But I'm stuck at this because of the power on $y'$.
 A: Polyanin's EqWorld proposed the nice substitution $\;w(y):=(y')^2\,$ to rewrite $\,(\cos y+p)\,y''+q(y')^2+r=0\;$ as :
$$\frac{\cos y+p}2\,w'(y)+q\;w(y)+r=0$$
(since $\,\displaystyle\;2\;y'\,y''=\frac d{dx}w(y)=\frac d{dy}w(y)\frac {dy}{dx}=y'\,w'(y)\,$ and ignoring the case $\,\frac {dy}{dx}=0\,$ here...)
To solve the resulting first order linear ODE begin with the homogenous ODE :
$$\frac {dw}w=-\frac{2q\,dy}{\cos y+p}$$
to get :
$$\log(w)=C-2q\int \frac {dy}{\cos y+p}$$
that you may solve with the $\,t:=\tan\left(\frac y2\right)$ substitution 
to get Alpha's result.
Adding a trivial solution of the complete linear ODE should give you Alpha's complete result :
$$(y')^2=w(y)=C_1\; \exp\left[{\frac{4\,q\;\operatorname{atanh\left(\tan\left(\dfrac y2\right)\dfrac{p-1}{\sqrt{1-p^2}}\right)}}{\sqrt{1-p^2}}}\right]-\dfrac rq$$
With all this work we still only have an expression of the square of $y'$ as a function of $y$.
This should be helpful to study the phase space of the solution but far from an explicit solution.
We may obtain $x$ in function of $y$ by rewriting this as (up to a constant) :
$$x=\int \frac {dy}{\sqrt{C_1\,\exp\left[\dfrac{4\,q\;\operatorname{atanh\left(\tan(y/2)\;(p-1)/\sqrt{1-p^2}\right)}}{\sqrt{1-p^2}}\right]-\frac rq}}$$
(or $\sqrt{-\cdots}$ in the denominator) but I am not sure that a closed form may be found even if we ignore the additional constants...
