Solve: $y=x+2\tan^{-1} p,$ where $p=\frac{dy}{dx}.$ (My actual question is in the body below.) Given equation is:
$y=x+2\tan^{-1} p\tag{1}.$
Well, the equation $(1)$ is of the form:
$y=xF(p) +f(p),$ which is basically the Lagrange's equation for ODEs of first order but not of first degree. So I proceeded as follows:
Differentiating the given equation w.r.t. $x$ and writing $p$ for $\frac{dy}{dx},$ we get:
$p=1+\frac{1}{1+p^2}\cdot\frac{dp}{dx},$ which after some simplifications and on integration gives:
$x=\ln\left(\frac{c|p-1|}{\sqrt{1+p^2}}\right)-\tan^{-1} p\tag{2}.$
Noting that elimination of $p$ is not possible between $(1)$ and $(2) $, as the method suggests I substituted the value of $x$ given by $(2)$ in $(1)$ and obtained:
$y=\ln\left(\frac{c|p-1|}{\sqrt{1+p^2}}\right) +\tan^{-1} p\tag{3}.$
As we can see elimination of $p$ between $(2)$ and $(3)$ isn't possible either so we declare that Eq.s $(2)$ and $(3)$ together form the required general solution in parametric form, $p$ being the parameter. (This matches with the answer given in the book.)
Now my actual question is:
Can't we substitute the value of $p$ from Eq. $(1),$ which is $\tan\left(\frac{y-x}{2}\right),$ in $(3)$ and thus obtain a solution which is free from $p$(i.e., to say elimination of $p$ is possible!)? 
 A: $$y=x+2\tan^{-1} p\tag{1}.$$
$$x=\ln\left(c\frac{|p-1|}{\sqrt{1+p^2}}\right)-\tan^{-1} p\tag{2}.$$
There is a slight missunderstanding in thinking that the elimination of $p$ is not possible between $(1)$ and $(2).$
It is possible to eliminate $p$ between $(1)$ and $(2) $ : This leads to the solution expressed on the form of an implicit equation (As shown below).
What is impossible is to express the solution on explicit form $y(x)$ with a finite number of standard functions.
From $(1)$ :
$$p=\tan\left(\frac{y-x}{2} \right)$$
Then, putting $p$ into $(2)$  :
$$x=\ln\left|\tan\left(\frac{y-x}{2} \right)-1\right| -\frac12\ln\left(1+\tan^2\left(\frac{y-x}{2} \right)\right) -\frac{y-x}{2} +C$$
After simplification :
$$\boxed{y+x-\ln\left|\sin\left(\frac{y-x}{2} \right)-\cos\left(\frac{y-x}{2} \right)\right|=C'}$$
This is the solution of the EDO on the form of implicit equation.
Solving it for the explicit form $y(x)$ is not possible with a finite number of the available elementary and special functions.
A: Hoping that I am not mistaken
Let $y=x+2z$ to make
$$\tan ^{-1}\left(2 z'+1\right)=z\implies 2z'+1=\tan(z)\implies\frac 2 {x'}+1=z  \implies x'=\frac{2}{z-1}$$
