partial differential equation for ruled surfaces
Here is a test for when a surface of the form $z = f(x,y)$, where $f$ is a sufficiently smooth function of two variables, is ruled.
To begin, set $I\!I = f_{xx} dx^2 + 2f_{xy}dxdy + f_{yy}dy^2$. If $I\!I$ vanishes identically, then the surface is a plane, so it is ruled.
Suppose that $I\!I$ is nonzero. The discriminant of $I\!I$ is defined (up to a factor of $(\mathrm{d}x\wedge\mathrm{d}y)^{\otimes 2}$) to be $$ \Delta(I\!I) = f_{xx}f_{yy}- {f_{xy}}^2. $$
If $\Delta(I\!I) >0$, then the surface is locally strictly convex and so cannot be ruled.
If $\Delta(I\!I) = 0$, then the surface is ruled. In fact, it has vanishing Gauss curvature. Moreover, $I\!I = \pm \alpha^2$ for some nonzero $1$-form $\alpha$ on the domain of $f$, and the curves in this domain defined by $\alpha = 0$ (which turn out to be straight lines) lift to the graph $z = f(x,y)$ to be straight lines.
If $\Delta(I\!I) < 0$, set $I\!I\!I = f_{xxx}\ dx^3+3f_{xxy}\ dx^2dy+3f_{xyy}\ dxdy^2+f_{yyy}\ dy^3$ and let $I\!I\!I_0$ be the $I\!I$-trace-free part of $I\!I\!I$. Then the surface $z = f(x,y)$ is ruled if and only if the discriminant of $I\!I\!I_0$ vanishes.
(Added later: This latter condition (i.e., the vanishing of the discriminant of $I\!I\!I_0$) turns out to be equivalent to the condition that $I\!I$ and $I\!I\!I$ have a common linear factor, say, $\alpha$ (which will necessarily be real when $\Delta(I\!I) < 0$), and hence is equivalent to the vanishing of the resultant of $I\!I$ and $I\!I\!I$, i.e., $\textrm{Reslt}(I\!I,I\!I\!I) = 0$. When such an $\alpha$ exists, the leaves of $\alpha=0$ are lines on the surface.)
Notes:
The discriminant of a cubic form $C = p\ dx^3 + 3q\ dx^2dy + 3r\ dxdy^2 + s\ dy^3$ is, by definition, $$ \Delta(C) = s^2p^2 + 4r^3p + 4 q^3s - 3 r^2q^2 - 6 sqrp. $$ It is, up to a multiple, the unique polynomial of degree $4$ in the coefficients that vanishes if and only if $C$ has a multiple factor.
Given a quadratic form $Q = a\ dx^2 + 2b\ dxdy + c\ dy^2$ with nonvanishing discriminant $D$, the $Q$-trace of a form $C$ of degree $3$ is the linear form $$ tr_Q(C ) = \frac{(ar-2bq+cp)\ dx + (as-2br+cq)\ dy}{D}. $$ Any cubic form $C$ can be uniquely written in the form $$ C = C_0 + L\cdot Q $$ where $L$ is a linear form and $tr_Q(C_0) = 0$. (In fact, $L = \tfrac34 tr_Q(C)$.) The term $C_0$ is called the $Q$-trace-free part of $C$.
The resultant $\textrm{Reslt}(Q,C)$ of a quadratic form $Q$ and a cubic form $C$ is the (unique up to nonzero multiples) polynomial that is cubic in the coefficients of $Q$, quadratic in the coefficients of $C$, and vanishes exactly when they have a common linear divisor.
From the paper to which 'lowerbound' linked, "Symmetry groups and Lagrangians associated with Tzitzeica surfaces," by Nicoleta Bila (also arXiv:math/9910138v1), here is Theorem 1 and its preamble:
Consider $D \subset \mathbb{R}^2$ and let
I take no credit (or blame!) for this; just posting as a community service.
Does anybody know if there is a partial differential equation whose solutions are all ruled surfaces and only them?
An example is theorem 1 of http://www.kurims.kyoto-u.ac.jp/EMIS/journals/BJGA/10.1/bt-bil.pdf
Edit: I have made this a community wiki answer so that if anyone would care to type in the theorem then they are free to do so.