First order differential equation with $y,y',$ and $\sqrt y$ I have been struggling with this equation:
$(x^2+1)y'-2xy=4\sqrt{(x^2+1)y}\arctan x$
I have tried with $y=z^m$ to make homogeneous equation, but I didn't get anything anything useful. 
Left side also looks a lot like quotient rule, so I tried solving in that direction, but again it didn't work. Whatever I do I can't seem to get it to any standard form.
Thanks for help. 
 A: Let $y = z^2$. We have: 
$$ 2(x^2 + 1) z z' - 2x z^2 = 4 z \sqrt{x^2+1} \arctan(x)$$
$$ z' - \frac{xz}{x^2 + 1} = 2 \frac{1}{\sqrt{x^2+1}}\arctan(x)$$
Homogeneous equation:
$$z' = \frac{xz}{x^2+1} $$
Therefore, $z = c \sqrt{x^2+1}$
Variation of constant: 
$$c' \sqrt{x^2+1} + c \frac{x}{\sqrt{x^2+1}} - c \frac{x}{\sqrt{x^2+1}} = 2 \frac{1}{\sqrt{x^2+1}}\arctan(x)  $$
$$c' = \frac{2}{x^2+1}\arctan(x)$$
$$c = \arctan(x)^2 + a$$
$$z = \arctan(x)^2 \sqrt{x^2+1} + a\sqrt{x^2+1}$$
A: As suggested by metamorphy in the comments above, we set $y = z^2$, in which case $y' = 2 \, z \, z'$, and this DE becomes linear:
$$
(1+x^2) z' - x \, z \;=\; 2\sqrt{1+x^2}\; \arctan(x)
$$
Next, we define $z = \sqrt{1+x^2}\; w$, so that $w = \sqrt{y\, /\, (1+x^2)}$. Then
$$
z' \;=\; \frac{x}{\sqrt{1+x^2}}\, w \;+\; \sqrt{1+x^2}\, w'\, ,
$$
and the DE becomes:
$$
w' \;=\; \frac{dw}{dx} \;=\; 2 \frac{\arctan(x)}{1+x^2}
\;=\;
\frac{d}{dx}\, \arctan^2(x)\, .
$$
This in turn implies that
$$
w(x) \;=\; 
\sqrt{\frac{y(x)}{1+x^2}} \;=\;  \arctan^2(x) \;+\; c
\quad\rightarrow\quad
y(x) \;=\; (1+x^2)\, {\left[\arctan^2(x) \;+\; c\right]}^2\, .
$$
Here $c \;=\; \sqrt{y(0)}$ is a constant.
Edited to add:
Note that (1) $y = 0$ is also a solution, and (2) I have implicitly assumed here that $y(0) \ge 0$.
