Solving $y=2xy'+\tan^{-1}(xy'^2)$ $$y=2xy'+\tan^{-1}(xy'^2)$$
Using the substitution $y'=p$ gives $ y=2xp+\tan^{-1}(xp^2)$. Now taking differentiate from both sides,
$$dy=2(xdp+pdx)+\frac{p^2dx+2px\;dp}{1+x^2p^4}$$
Dividing by $dx$,
$$p=(2p+\frac{p^2}{1+x^2p^4})+(2x+\frac{2px}{1+x^2p^4})\frac{dp}{dx}$$
$$(p+\frac{p^2}{1+x^2p^4})dx+(2x+\frac{2px}{1+x^2p^4})dp=0$$
$$(pdx+xdp)+xdp+\frac{p^2dx+2px\;dp}{1+x^2p^4}=0$$
$$d(px)+d(\tan^{-1}(xp^2))+xdp=0$$If we integrate the last equation we get
$$px+\tan^{-1}(xp^2)+\int xdp=C$$
But I don't know how to deal with $\displaystyle \int xdp$.
 A: $$y=2xy'+\tan^{-1}(xy'^2)$$
Change the variable $u=\sqrt x$:
$$y=uy'+\tan^{-1}\left(\dfrac {y'^2}4 \right)$$
This is Clairaut's differential equation.
$$y=uy'+f(y')$$
You can read the wiki page on Clairaut equation here
A: Try to make $\sqrt{x}y'$ a derivative. That is achieved by setting $v(u)=y(u^2)$, that is, $x=u^2$, so that $v'(u)=2y'(u^2)u$ and $v'(u)^2=4xy'(x)^2)$. Inserted into the equation this gives
$$
v(u)=uv'(u)+\tan^{-1}(v'(u)^2/4),
$$
which is now in the form of a Clairaut equation with its regular solutions
$$
v(u)=Cu+\tan^{-1}(C^2/4)
$$
and the singular solutions that solve
$$
0=u+\frac{v'(u)/2}{1+v'(u)^4/16}.
$$
A: Using @mkcpz's suggestion on the deleted answer, from $(p+\frac{p^2}{1+x^2p^4})dx+(2x+\frac{2px}{1+x^2p^4})dp=0$ we have,
$$(p\;dx+2x\;dp)+\frac{d(xp^2)}{1+(xp^2)^2}=0$$
$$\frac{d(xp^2)}{p}+\frac{d(xp^2)}{1+(xp^2)^2}=0$$
$$1+x^2p^4+p=0$$
Which we can solve for $x^2$. Hence the final answer is $${ \begin{cases}{x^2=\dfrac{-(p+1)}{p^4}} \\ {y=2xp+\tan^{-1}(xp^2)}\end{cases} }$$
