possible. iii) Pick several typical starting points and
sketch typical trajectories in the phase plane. iv)
Interpret the outcomes predicted by your graphical
analysis in terms of the constants a, b, m, n, and
Note: When you get to part (iii), you should
gravity encounters an air resistance proportional to
the square of velocity, then the bodys velocity t
seconds into the fall satisfies the equation where k
is a constant that depends on the bodys
aerodynamic properties and the density of the air.
(We assu
Sociologists recognize a phenomenon called social
diffusion, which is the spreading of a piece of
information, technological innovation, or cultural
fad among a population. The members of the
population can be divided into two classes: those
who have the
gives four distinct regions in the plane A, B, C, D,
with their respective trajectory directions shown in
Figure 16.30. Next, we examine what happens near
the two equilibrium points. The trajectories near
point away from it, upward and to the right. The
b
not a constant multiple of (see Exercise 61). From
Theorem 2 we conclude the following result. er1 x
er2 x y2 = er2 x y1 = er1 x r1 r2 b2 4ac>0. b2 - 4ac.
r2 = -b - 2b2 - 4ac 2a r1 = . -b + 2b2 - 4ac 2a ay +
by + cy = 0 THEOREM 3 If and are two real and
u
resistance is a constant R ohms and whose selfinductance, shown as a coil, is L henries, also a
constant. There is a switch whose terminals at a and
b can be closed to connect a constant electrical
source of V volts. From Section 16.2, we have where
i is
rates of change of these populations. As time
passes, each species breeds, so we assume its
population increases proportionally to its size. Taken
by itself, this would lead to exponential growth in
each of the two populations. However, there is a
counter
constants and , the function is also a solution to
Equation (2). y(x) = c1y1(x) + c2 y2(x) c1 c2 y1(x)
y2(x) Proof Substituting y into Equation (2), we have
144442444443 144442444443 is a solution 0, is a
solution Therefore, is a solution of Equation (2).
infinitely many solutions to the boundary value
problem 66. Show that if a, b, and c are positive
constants, then all solutions of the homogeneous
differential equation approach zero as x : q. ay +
by + cy = 0 y + 4y = 0, y(0) = 0, y(p) = 0. y + 4y =
0, y
2. For the system (2a and 2b), show that any
trajectory starting on the unit circle will traverse the
unit circle in a periodic solution. First introduce
polar coordinates and rewrite the system as and 3.
Develop a model for the growth of trout and bass
a
particular solution of the form We need to
determine the unknown coefficients A, B, and C.
When we substitute the polynomial and its
derivatives into the given nonhomogeneous
equation, we obtain or, collecting terms with like
powers of x, This last equati
16.20. b. Using y(t) for the pearls velocity as a
function of time t, write a differential equation
modeling the velocity of the pearl as a falling body.
c. Construct a phase line displaying the signs of and
d. Sketch typical solution curves. e. What is t
trajectory starts inside the unit circle, it spirals
outward, asymptotically approaching the circle as .
If a trajectory starts outside the unit circle, it spirals
inward, again asymptotically approaching the circle
as . The circle is called a limit cycle
0, (3) P, Q, R where a, b, and c are constants. To
solve Equation (3), we seek a function which when
multiplied by a constant and added to a constant
times its first derivative plus a constant times its
second derivative sums identically to zero. One
func
16-38 Chapter 16: First-Order Differential Equations
In setting up our competitive-hunter model, precise
values of the constants a, b, m, n will not generally
be known. Nonetheless, we can analyze the system
of Equations (1) to learn the nature of its sol
y = 2 sin 2x. c1 = 0 c2 = 2 y a p 12 b = c1 cos a p 6 b +
c2 sin a p 6 b = 1 y(0) = c1 # 1 + c2 # 0 = 0 y = c1 cos
2x + c2 sin 2x r r = ;2i 2 + 4 = 0 y + 4y = 0, y(0) = 0, y
a p 12 b = 1 x1 x2 y1 y2 y(x1) = y1 y(x2) = y2 y = ex 2xex . c1 = 1 c2 = -2. 1 =
general solution to Equation (6) is . This situation of
motion is called critical damping and is not
oscillatory. Figure 17.6a shows an example of this
kind of damped motion. Case 2: . The roots of the
auxiliary equation are real and unequal, given by
and
the auxiliary equation . Its roots are and . Therefore,
the complementary solution to the associated
homogeneous equation is . We now seek a
particular solution . That is, we seek a function that
will produce when substituted into the left-hand
side of th
+ 2 = (r - 1)(r - 2) = 0, y - 3y + 2y = 0 y = ex 0 = 5ex
Aex - 3Aex + 2Aex = 5ex yp = Aex y - 3y + 2y = 5ex
yp = cos x - sin x A = -1 B = 1 B - A = 2 A + B = 0 (B A) sin x - (A + B) cos x = 2 sin x -A sin x - B cos x - (A
cos x - B sin x) = 2 sin x yp = A
. The general solution to the differential equation is .
You may find the following table helpful in solving
the problems at the end of this section. + 5xex - 1 10
cos 2x + 1 5 c1e sin 2x x y = yc + yp = + c2 yp = 5xex 1 10 cos 2x + 1 5 sin 2x A = 5, B =
survive over time, and mutual coexistence of the
species is highly improbable. Limitations of the
Phase-Plane Analysis Method Unlike the situation
for the competitive-hunter model, it is not always
possible to determine the behavior of trajectories
near a
the initial conditions we have Thus, and The unique
solution satisfying the initial conditions is The
solution curve is shown in Figure 17.1. Another
approach to determine the values of the two
arbitrary constants in the general solution to a
second-order
value satisfying this last inequality. That is, for each
there is a unique value of x satisfying the equation
in part (d). Thus there can exist only one trajectory
solution approaching from below, as shown in
Figure 16.37. f. Use a similar argument to sho
equation (1) where a, b, and c are constants and G is
continuous over some open interval I. Let be the
general solution to the associated complementary
equation (2) (We learned how to find in Section
17.1.) Now suppose we could somehow come up
with a part
interval (finite or infinite). We cannot look at just
one of these equations in isolation to find solutions
or since each derivative depends on both x and y. To
gain insight into the solutions, we look at both
dependent variables together by plotting the