The differential equation: $ \arctan (y) = \arctan(x)+C .$ I solved the equation and stalled. Help with decision please.
$$(1+y^2)\,dx=(1+y^2)\,dy \iff \int \frac{dx}{1+x^2} = \int\frac{dy}{1+y^2}  $$ 
Transformed expression for the table of integrals.
$$ \arctan (y) = \arctan(x)+C $$
Prompt how to further transform expression.(Find the general solution)
 A: The differential set
$$
(1+y^{2}) \,dx  = (1+x^{2})\, dy 
$$
can be integrated as seen by
\begin{align}
\int \frac{dy}{1+y^{2}} = \int \frac{dx}{1+x^{2}}
\end{align}
and leads to
\begin{align}
\tan^{-1}(y) = \tan^{-1}(x) + c
\end{align}
or
\begin{align}
y = \tan(\tan^{-1}(x) + c).
\end{align}
Now using 
\begin{align}
\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a) \tan(b)}
\end{align}
the general result becomes
\begin{align}
y(x) = \frac{x+c_{1}}{1 - c_{1} x}
\end{align}
since $\tan(\tan^{-1}(x)) = x$ and $c_{1} = \tan(c)$ is a constant. 
A: Take the tangent of each side of the equation$$\tan(\arctan(\theta)) = \theta$$
$$\tan\Big(\arctan(y)\Big) =\tan\Big(\arctan(x) + C\Big) \iff y = \tan\Big(\arctan(x) + C\Big)$$
As Lucian notes, you can use the following double-angle formula for $\tan$ to expand the right-hand side of the result:
$$\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a) \tan(b)}$$
Here, put $a = \arctan(x)$, and $b = C$. 
A: Assume that an initial point $(x_0,x_0)\in{\mathbb R}^2$ is given. The formula
$$(1+y^2)\,dx=(1+x^2)\,dy\tag{1}$$
should then be interpreted as follows: $x$ and $y$ are functions of a "hidden variable" $t$, and we are told to solve
$${\dot x\over 1+x^2}={\dot y\over 1+y^2}\tag{2}$$
with the initial condition $x(0)=x_0$, $\>y(0)=y_0$. Integrating $(2)$ with respect to $t$ from $t=0$ to $t=T$ gives
$$\arctan x(T)-\arctan x_0=\arctan y(T)-\arctan y_0\ .$$
Taking the $\tan$ on both sides (note that $\tan(\arctan x)\equiv x$) we obtain
$${x(T)-x_0\over 1+x(T)x_0}={y(T)-y_0\over 1+y(T)y_0}\ .$$
This means that the solution curve of $(1)$ through the point $(x_0,y_0)$ satisfies the equation
$${x-x_0\over 1+xx_0}={y-y_0\over 1+yy_0}\ ,$$
or
$$y={x-{x_0-y_0\over 1+x_0y_0}\over {x_0-y_0\over 1+x_0y_0}x+1}\ .$$
