= Solution
{c}
Adopt the <streamfunction> convention $u=\psi_y$, $v=-\psi_x$. Integrating the first component of the <velocity field> in $y$ gives $\psi=-xy+\tfrac12\Omega e^{-2t}y^2+f(x,t)$. The second component then requires $f_x=\Omega e^{2t}x$. Thus, omitting an arbitrary function of time,
$$
\boxed{\psi(x,y,t)=\frac\Omega2(e^{2t}x^2+e^{-2t}y^2)-xy}.
$$
At a fixed instant, the <streamlines> are the connected curves $\psi=K$. To classify this <quadratic form>, introduce the invertible scaled and rotated coordinates
$$
\xi=e^t x,\quad \upsilon=e^{-t}y,\qquad
s=\frac{\xi+\upsilon}{\sqrt2},\quad d=\frac{\xi-\upsilon}{\sqrt2}.
$$
Then
$$
\boxed{2\psi=(\Omega-1)s^2+(\Omega+1)d^2}.
$$
For $\Omega>1$, both coefficients are positive. Every $K>0$ therefore gives an <ellipse>, and $K=0$ is the <stagnation point> at the origin. For $0<\Omega<1$, the <quadratic form> is an <indefinite quadratic form>: each $K\ne0$ gives a <hyperbola>. The level $K=0$ consists of the straight <separatrices>
$$
d=\pm\sqrt{\frac{1-\Omega}{1+\Omega}}\,s,
$$
whose rays approach or leave the origin; the origin itself is a <stagnation point>. At the transition $\Omega=1$, $\psi=d^2$, so the nonstationary <streamlines> are parallel straight lines and $d=0$ is a line of <stagnation points>. These are instantaneous <streamlines>; the time dependence means that they need not coincide with <pathlines>.
\Image[/past-exam-of-the-fluids-and-waves-course-of-the-university-of-oxford/2021-streamlines.png]
{title=Instantaneous elliptical and hyperbolic streamlines at the same positive time}
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