Equations solvable for y Solve the following equation:
$$y=x+a\tan^{-1}p$$
$$\text{where p}=\frac{dy}{dx}$$
Differentiating both side w.r.t. x,
$$\frac{dy}{dx}=1+\frac{a}{1+p^2}\frac{dp}{dx}\\
\implies p=1+\frac{a}{1+p^2}\frac{dp}{dx}$$
I have tried till this...but what to do next?..please help..
 A: $$\frac{dy}{dx}=1+\frac{a}{1+p^2}\frac{dp}{dx}$$
$$\implies p=1+\frac{a}{1+p^2}\frac{dp}{dx}$$
It's separable
$$ \frac {dp}{(p-1)({1+p^2})}=\frac {dx} a$$
Use  fraction decomposition. And integrate.

Edit
It's better to keep the original equation 
$$y=x+a\arctan (y')$$
$$y'=\tan \left (\frac {y-x}{a} \right )$$
Substitute $y-x=u \implies u'=y'-1$
The equation becomes:
$$u'+1= \tan \left (\frac {u}{a} \right )$$
This last De is separable:
$$\int \frac {du}{\tan \left (\frac {u}{a} \right )-1} =\int dx$$
A: The equation can be written as
$$\frac{dy}{dx}=\tan \frac{1}{a}(y-x).$$
Now make the substitution $u=y-x$ to get 
$$
\frac{du}{dx}+1=\tan\frac{u}{a}.
$$
This equation is now separable.
A: $$y=x+a\tan^{-1}p\\
\begin{align}
&\implies\frac{dy}{dx}=1+\frac{a}{1+p^2}\frac{dp}{dx}\\
&\implies p=1+\frac{a}{1+p^2}\frac{dp}{dx}\\
&\implies (p-1)(1+p^2)=a\frac{dp}{dx}\\
&\implies\int{\frac{adp}{(p^2+1)(p-1)}}=\int{dx}\\
&\implies\frac a2\int{\frac{1}{p-1}-\frac{p+1}{p^2+1}dp}=\int{dx}\\
&\implies\frac a2\left[\int{\frac{dp}{p-1}}-\int{\frac{pdp}{p^2+1}}-\int{\frac{dp}{p^2+1}}\right]=x+c\\
&\implies\frac a2\left[\log {(p-1)}-\frac 12\log{(p^2+1)}-\tan^{-1}p\right]=x+c\\
\end{align}$$
