<s>[INST] Nonlinear first order ODE In trying to compute an integral curve for a certain fiber of the Hopf fibration in $\mathbb R^3$ under stereographic projection, I've come across the following IVP:
$$
2y' - y^2 - 1=0, \quad y(0) = 0.
$$
I'm at a loss on how to solve this simple looking thing. Any help would be appreciated!
 [/INST] This is Riccati equation
Rewrite $$y'-\frac{1}{2}y^2=\frac{1}{2}$$
Let $y=-2\frac{w'}{w}$
then $$y'=-2\frac{w''(x)}{w(x)}+2\frac{w'(x)^2}{w(x)^2}=-2\frac{w''(x)}{w(x)}+\frac{1}{2}y^2$$
or
$$y'-\frac{1}{2}y^2=-2\frac{w''(x)}{w(x)}$$
thus we got
$$-2\frac{w''(x)}{w(x)}=\frac{1}{2}$$ or
$${w''(x)}+\frac{1}{4}w(x)=0$$ 
which has a general solution $$w=a\sin\frac{1}{2}x+b\cos\frac{1}{2}x$$
therefore $$y=-2\frac{w'}{w}=-2\frac{a\frac{1}{2}\cos\frac{1}{2}x-b\frac{1}{2}\sin\frac{1}{2}x}{a\sin\frac{1}{2}x+b\cos\frac{1}{2}x}$$
Since $y(0)=0$ we got that $$a\frac{1}{2}\cos\frac{1}{2}\cdot0-b\frac{1}{2}\sin\frac{1}{2}\cdot0=a\frac{1}{2} + 1=0$$
that is $a=0$, and we arrived at $$y=\tan\frac{1}{2}x$$
Another option
Thanks to @Evgeny
This is separable DE
Rewrite as $$2y'=1+y^2$$
then
$$\frac{2y'}{1+y^2}=1$$
so we get 
$$2\,\mathrm{arctan}\,y=x+C$$
using initial condition one get $C=0$ and therefore 
$$y=\tan\frac{1}{2}x$$
</s>