= Solution
The divergence is $\partial_xu_x+\partial_yu_y=1-1=0$, while the scalar vorticity is $\partial_xu_y-\partial_yu_x=0$. The flow is therefore <incompressible flow> and <irrotational flow>. Integrating the <streamfunction> equations gives
$$
\boxed{\psi(x,y,t)=x(y-t)+C(t).}
$$
The time dependence must be retained: no time-independent function $\psi(x,y)$ represents this unsteady field for every $t$. At $t=0$ the <streamlines> are $xy=\mathrm{constant}$, with the coordinate axes as separatrices. Their arrows point towards increasing $|x|$ and decreasing $|y|$, as shown below.
A <pathline> solves the material ODEs, giving
$$
x(t)=x_0e^t,\qquad y(t)=t-1+(y_0+1)e^{-t}.
$$
For $x_0\ne0$, elimination of time yields
$$
\boxed{y=\log(x/x_0)-1+(y_0+1)\frac{x_0}{x},\qquad x/x_0>0.}
$$
If $x_0=0$, the <pathline> stays on the $y$ axis and the parametric expression remains valid; it cannot in general be written as a single-valued $y=f(x)$ along that vertical trajectory. For $x_0=y_0=1$, the path is $y=\log x-1+2/x$: it initially descends, reaches its minimum $(2,\log2)$, then rises. Unlike a frozen <streamline>, it records the time-varying flow.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ib/paper-1-flow-trajectories.png]
{title=Frozen saddle streamlines at time zero and the material path through (1,1)}
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