orthogonal trajectories
Given a family of curves filling the plane — say all the circles centred at the origin — there is a second family that crosses every one of them at a right angle. Those crossing curves are the orthogonal trajectories. The everyday picture is a map: the contour lines of constant height and the paths of steepest descent (the lines water would run down) always meet at right angles, and one family is the orthogonal trajectories of the other.
Finding them is a clean three-step application of first-order methods. First, take the given one-parameter family F(x, y) = C and differentiate to get its differential equation dy/dx = f(x, y), eliminating the parameter C along the way. Second, at each point the orthogonal curve must have slope equal to the negative reciprocal, so replace dy/dx by -1/(dy/dx): the new differential equation is dy/dx = -1/f(x, y). Third, solve that equation; its solution family is the set of orthogonal trajectories.
This is one of the most physical uses of first-order theory. In electrostatics the equipotential lines and the electric field lines are orthogonal trajectories of each other; in fluid flow the streamlines and equipotentials; in heat conduction the isotherms and heat-flow lines. The negative-reciprocal step is the whole idea — perpendicular slopes multiply to minus one — packaged as a differential equation.
Find the orthogonal trajectories of the family of circles x^2 + y^2 = C. Differentiating gives 2x + 2y y' = 0, so y' = -x/y. The orthogonal slope is the negative reciprocal y' = y/x, a separable (and homogeneous) equation whose solution is y = k x — straight lines through the origin, which indeed cross every circle at a right angle.
Replace the slope by its negative reciprocal, then solve the new equation.
You must eliminate the parameter C before flipping the slope: the orthogonality condition is about the slope FIELD, not any single curve, so the differential equation that defines the field must be free of C.