Finding pathline

Question

I've been trying to find the pathline of a particle dropped in a steady flow defined by the following vector components: $$ u= \frac{-2x}{(x^2+y^2+1)^2} \hat i + \frac{-2y}{(x^2+y^2+1)^2}\hat j $$ in such a way that when I solve the ODE $$ \frac {dx}{dt}=u_x =\frac{-2x}{(x^2+y^2+1)^2} $$ $$ \frac{dy}{dt}=u_y=\frac{-2y}{(x^2+y^2+1)^2}$$ I get an $ x(t) $ and an $ y(t) $ that I could then turn into a parametric plot?

Answer 1

The pathlines are curves in space, and parametric functions of $t$. So if you eliminate t, you get an expression for $\frac{dy}{dx}$.

$$ \frac{dy}{dx} = \frac{y}{x} $$

On integrating this, you have $y=cx$, so the pathlines are straight lines through the origin.