Given the 2D velocity field $v_x = -2y$, $\, v_y = 2x$ find the path line equations. The book said the path lines are circles, i.e. $x^2 + y^2 = \text{constant} = X^2 + Y^2$. Where
\begin{cases} x = x(X,Y,t) \\ y = y(X,Y,t) \\ X = x(X,Y,t_0) \\Y = y(X,Y,t_0) \end{cases}
I'm approaching it like this \begin{align} v_x &= \frac{dx}{dt} = -2y \\ \\ v_x &= \frac{d\left[ x(X,Y,t) \right]}{dt} = -2\left[ y(X,Y,t) \right] \\ \\ \int_{t_0}^{t} &\frac{dx}{dt} dt = -2 \int_{t_0}^{t} y \, dt \\ \\ x(&X,Y,t) - x(X,Y,t_0) = -2\int_{t_0}^{t} y \, dt \\ \\ x - &X = -2\int_{t_0}^{t} y \, dt \end{align}
A similar equation can be made for $y$ and $Y$. But this is where I hit a wall. I'm thinking there's a much simpler way to do this.