Page 84 - Schaum's Outline of Differential Equations

CHAP. 7] APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS 67

Here/(X y) = -x/y, so that (7.15) becomes

This equation is linear (and, in differential form, separable); its solution is

which represents the orthogonal trajectories.
In Fig. 7-7 some members of the family of circles are shown in solid lines and some members of the family
(1), which are straight lines through the origin, are shown in dashed lines. Observe that each straight line intersects
each circle at right angles.

Fig. 7-7

2
7.24. Find the orthogonal trajectories of the family of curves y = ex .
2
The family, which is given by (7.12) with F(x, y, c)=y-cx , consists of parabolas symmetric about the
y-axis with vertices at the origin. Differentiating the given equation with respect to x, we obtain dyldx = 2cx.
2
To eliminate c, we observe, from the given equation, that c = y/x ; hence, dyldx = 2y/x. Hetef(x, y) = 2y/x, so (7.15)
becomes

2
The solution of this separable equation is ^x 1 + y =k. These orthogonal trajectories are ellipses. Some members
of this family, along with some members of the original family of parabolas, are shown in Fig. 7-8. Note that each
ellipse intersects each parabola at right angles.

1
2
7.25. Find the orthogonal trajectories of the family of curves x + y = ex.
2
1
Here, F(x, y, c) = x + y — ex. Implicitly differentiating the given equation with respect to x, we obtain