Non linear Differential Equation Let $\Omega:=\{(x_1,x_2) \subset \mathbb{R}^2 | x_2>0\}$. I want to solve the differential equation  $$\begin{pmatrix} \dot{x_1} \\\dot{x_2} \end{pmatrix}=\begin{pmatrix}x_2^2-x_1^2 \\-2x_1x_2\end{pmatrix}$$I have no idea how to do that, since I have only seen linear differential equations so far. I'd appreaciate any starting point here. I have the hint to look at $z=x_1+ix_2$ but that confuses me even more. Thank you!
 A: If $z(t)=x_1 (t) + \Bbb i x_2 (t)$ then $$- z^2 = (x_2 ^2 - x_1 ^2) - 2 x_1 x_2 \Bbb i$$ Therefore, your equation can be rewritten as $\dot z = - z ^2$ or, equivalently, $$\frac{\dot z}{z^2} = -1$$ Integrate this with respect to $t$, getting $1 /z = t + C$, or $z(t) = 1/(t+C)$, with $C=a + b \Bbb i$ an integration constant. To get back to $x_1$ and $x_2$, write $$x_1 + x_2 \Bbb i = \frac1{t+a + b \Bbb i} = \frac{t+a - b \Bbb i}{(t+a)^2 + b^2}$$ Equating the real and imaginary parts, $$x_1 = \frac{t+a}{(t+a)^2 + b^2}\qquad x_2 = \frac{-b}{(t+a)^2 + b^2}$$ To have $x_2 > 0$ (as in the definition of $\Omega$), you will have to take $b<0$.
A: Observe how the right hand side looks like the real and the imaginary term of a squared complex number. In fact, if you write
$$z^2=x_1^2-x_2^2+i (2x_1x_2)$$
you can see, that if you multiply the second equation by $i$ and add it to the first one, you will get
$$\dot{z}=-z^2$$
which is trivially solvable.
This is a very handy trick that can be used in physics quite frequently when there is some intrinsic rotational motion going on (or something like that).
