I'm reading a book of differential geometry that states that it's easy to get the integral curves of this vector field:
$$X(x,y)=(x^2-y^2,2xy)$$
But proceeding the way the book says, first I have to take that equation as
$$\frac{dx}{dt}=x^2-y^2$$
$$\frac{dy}{dt}=2xy$$
But from that I don't know what more to do to proceed. I know that the integral curves must be circles of center $(0,t)$, and the vector field seems as this
Best Answer
Indentifying $(x,y)$ with $z=x + y \, \mathrm{i}$ the differential equation reads $$\frac{\partial z}{\partial t}=z^2.$$ Rewriting this we find $$\frac{-\partial z^{-1}}{\partial t} = \frac{\partial z}{\partial t} z^{-2} = 1$$ and so $$-z^{-1}= t+z_0 \textrm{ or } z = -\frac1{t+z_0}$$ for some fixed initial value $z_0 \in \mathbb{C}$. For $t \in \mathbb{R}$ these describe circular trajectories as expected.
Alternative 1: transform the vector field under the inversion
$$\sigma(x, y) = \left(\frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right).$$
This map is an involution ($\sigma^2 = \mathrm{id}$) and its Jacobian is $$D\sigma_{(x,y)}=\frac{1}{(x^2+y^2)^2}\begin{pmatrix} y^2-x^2 & -2xy \\ -2xy & x^2-y^2 \end{pmatrix}.$$
Therefore it maps the element $$\begin{pmatrix}x^2 - y^2 \\ 2xy \end{pmatrix}$$ at point $(x, y)$ to the element
$$\frac{1}{(x^2+y^2)^2}\begin{pmatrix} y^2-x^2 & -2xy \\ -2xy & x^2-y^2 \end{pmatrix} \begin{pmatrix}x^2 - y^2 \\ 2xy \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \end{pmatrix}$$ at point $\sigma(x,y)$. The trajectories of the transformed vector field are easy to find. Then apply the inversion $\sigma$ to find the trajectories of the original vector field.
Alternative 2: Since you know that the trajectories are circular start with the ansatz $$\begin{cases} x = -R \sin \alpha \\ y = R + R \cos \alpha \end{cases}$$ for some function $\alpha$ of $t$. Then the given differential equations are consistent and both lead to $$\frac{\partial \alpha}{\partial t} = 2R \, (1 + \cos\alpha).$$ One solution to this equation is $\alpha(t) = 2 \arctan(2R \, t)$. Substituting this in the expressions for $x$ and $y$ gives the trajectory
$$\left(\frac{-t}{t^2 + (2R)^{-2}}, \frac{(2R)^{-1}}{t^2+(2R)^{-2}}\right).$$