Page 54 - Advanced engineering mathematics

34 CHAPTER 1 First-Order Differential Equations

4

2

y
0
–4 –2 2 4

–2

–4

FIGURE 1.9 Two families of orthogonal trajectories.

Orthogonal Trajectories Two curves intersecting at a point P are orthogonal if their tan-
gents are perpendicular at P. This occurs when the slopes of these tangents at P are negative
reciprocals of each other.
Suppose we have two sets (or families) of curves, F and G. We say that F is a set of orthogo-
nal trajectories of G if, whenever a curve of F intersects a curve of G, these curves are orthogonal
at the point of intersection. When F is a family of orthogonal trajectories of G, then G is also a
family of orthogonal trajectories of F.
For example, let F consist of all circles about the origin and G of all straight lines through
the origin. Figure 1.9 shows some curves of these families, which are orthogonal trajectories of
each other. Wherever one of the lines intersects one of the circles, the line is orthogonal to the
tangent to the circle there.
Given a family F of curves, suppose we want to find the family G of orthogonal trajectories
of F. Here is a strategy to do this. The curves of F are assumed to be graphs of an equation
F(x, y,k)=0 with different choices of k giving different curves. Think of these curves as integral
curves (graphs of solutions) of some differential equation y = f (x, y). The curves in the set of

orthogonal trajectories are then integral curves of the differential equation y =−1/f (x, y) with

the negative reciprocal ensuring that curves of one family are orthogonal to curves of the other
family at points of intersection. The idea is to produce the differential equation y = f (x, y) from

F; then solve the equation y =−1/f (x, y) for the orthogonal trajectories.

EXAMPLE 1.18
Let F consist of curves that are graphs of

2
F(x, y,k) = y − kx = 0.
These are parabolas through the origin. We want the family of orthogonal trajectories. First obtain
2
2
2
the differential equation of F.From y −kx =0 we can write k = y/x . Differentiate y −kx =0
to get y = 2kx. Substitute for k in this derivative to get

y

y − 2kx = 0 = y − 2 x = 0.

x 2
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