<s>[INST] 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. 
 [/INST] 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: 
$$c' \sqrt{x^2+1} + c \frac{x}{\sqrt{x^2+1}} - c \frac{x}{\sqrt{x^2+1}} = 2 \frac{1}{\sqrt{x^2+1}}\arctan(x)  $$
$$c' = \frac{2}{x^2+1}\arctan(x)$$
$$c = \arctan(x)^2 + a$$
$$z = \arctan(x)^2 \sqrt{x^2+1} + a\sqrt{x^2+1}$$
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