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CHAPTER 4 Systems of ODEs. Phase Plane.

Qualitative Methods
Major Changes
This chapter was completely rewritten in the eighth edition, on the basis of suggestions
by instructors who have taught from it and my own recent experience. The main reason
for rewriting was the increasing emphasis on linear algebra in our standard curricula, so
that we can expect that students taking material from Chap. 4 have at least some working
knowledge of 2 ⫻ 2 matrices.
Accordingly, Chap. 4 makes modest use of 2 ⫻ 2 matrices. n ⫻ n matrices are mentioned
only in passing and are immediately followed by illustrative examples of systems of two
ODEs in two unknown functions, involving 2 ⫻ 2 matrices only. Section 4.2 and the
beginning of Sec. 4.3 are intended to give the student the impression that, for first-order
systems, one can develop a theory that is conceptually and structurally similar to that in
Chap. 2 for a single ODE. Hence if the instructor feels that the class may be disturbed by
n ⫻ n matrices, omission of the latter and explanation of the material in terms of two
ODEs in two unknown functions will entail no disadvantage and will leave no gaps of
understanding or skill.
To be completely on the safe side, Sec. 4.0 is included for reference, so that the student
will have no need to search through Chap. 7 or 8 for a concept or fact needed in Chap. 4.
Basic throughout Chap. 4 is the eigenvalue problem (for 2 ⫻ 2 matrices), consisting
first of the determination of the eigenvalues l1, l2 (not necessarily numerically distinct)
as solutions of the characteristic equation, that is, the quadratic equation

a11 ⫺ l a12
2 2 ⫽ (a11 ⫺ l)(a22 ⫺ l) ⫺ a12a21
a21 a22 ⫺ l
⫽ l2 ⫺ (a11 ⫹ a22)l ⫹ a11a22 ⫺ a12a21 ⫽ 0,

and then an eigenvector corresponding to l1 with components x 1, x 2 from

(a11 ⫺ l1)x 1 ⫹ a12x 2 ⫽ 0

and an eigenvector corresponding to l2 from

(a11 ⫺ l2)x 1 ⫹ a12x 2 ⫽ 0.

It may be useful to emphasize early that eigenvectors are determined only up to a nonzero
factor and that, in the present context, normalization (to obtain unit vectors) is hardly
of any advantage.
If there are students in the class who have not seen eigenvalues before (although the
elementary theory of these problems does occur in every up-to-date introductory text on
beginning linear algebra), they should not have difficulties in readily grasping the meaning
of these problems and their role in this chapter, simply because of the numerous examples
and applications in Sec. 4.3 and in later sections.
Section 4.5 includes three famous applications, namely, the pendulum and van der
Pol equations and the Lotka–Volterra predator–prey population model.

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SECTION 4.0. For Reference: Basics of Matrices and Vectors, page 124
Purpose. This section is for reference and review only, the material being restricted to
what is actually needed in this chapter, to make it self-contained.
Main Content
Matrices, vectors
Algebraic matrix operations
Differentiation of vectors
Eigenvalue problems for 2 ⫻ 2 matrices
Important Concepts and Facts
Matrix, column vector and row vector, multiplication
Linear independence
Eigenvalue, eigenvector, characteristic equation
Some Details in Content
Most of the material is explained in terms of 2 ⫻ 2 matrices, which play the major role
in Chap. 4; indeed, n ⫻ n matrices for general n occur only briefly in Sec. 4.2 and at the
beginning in Sec. 4.3. Hence the demand of linear algebra on the student in Chap. 4 will
be very modest, and Sec. 4.0 is written accordingly.
In particular, eigenvalue problems lead to quadratic equations only, so that nothing
needs to be said about difficulties encountered with 3 ⫻ 3 or larger matrices.
Example 1. Although the later sections include many eigenvalue problems, the complete
solution of such a problem (the determination of the eigenvalues and corresponding
eigenvectors) is given in Sec. 4.0.
Emphasize to your students that the eigenvalues of a given square matrix are uniquely
determined (and some of them can very well be 0), whereas eigenvectors must not be zero
vectors and are determined only up to a nonzero multiplicative constant.

SECTION 4.1. Systems of ODEs as Models in Engineering Applications,

page 130
Purpose. In this section the student will gain a first impression of the importance of
systems of ODEs in physics and engineering and will learn why they occur and why they
lead to eigenvalue problems.
Main Content
Mixing problem
Electrical network
Conversion of single equations to system (Theorem 1)
The possibility of switching back and forth between systems and single ODEs is practically
quite important because, depending on the situation, the system or the single ODE will
be the better source for obtaining the information sought in a specific case.
Background Material. Secs. 2.4, 2.8.
Short Courses. Take a quick look at Sec. 4.1, skip Sec. 4.2 and the beginning of Sec. 4.3,
and proceed directly to solution methods in terms of the examples in Sec. 4.3.
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Some Details on Content

Example 1 extends the physical system in Sec. 1.3, consisting of a single tank, to a system
of two tanks. The principle of modeling remains the same. The problem leads to a typical
eigenvalue problem, and the solutions show typical exponential increases and decreases
to a constant value.
Problem Set 4.1
The mixing problems (Probs. 1–6) should lead to an understanding of the physical
parameters involved (tank size, flow rate, amount of fertilizer), similarly in the networks
in Probs. 7–9.
Problems 10–13 show conversions from ODEs to first-order systems.
Problem 14 shows the principle of extending the physical system in Sec. 2.5 to a system
of more than one mass and spring, with (11) probably best understood by looking at Fig. 81.

SOLUTIONS TO PROBLEM SET 4.1, page 136

2. The system is
y1r ⫽ 0.02y2 ⫺ 0.01y1
y r2 ⫽ 0.01y1 ⫺ 0.02y2
where 0.01 appears because we divide by the content of the tank T1, which is twice
the old value. In proper order, the system becomes
y 1r ⫽ ⫺0.01y1 ⫹ 0.02y2
y r2 ⫽ 0.01y1 ⫺ 0.02y2.
As a single vector equation,

d.c
⫺0.01 0.02
y r ⫽ Ay, where A⫽
0.01 ⫺0.02
A has the eigenvalues l1 ⫽ 0 and l2 ⫽ ⫺0.03 and corresponding eigenvectors

c d, c d,
1 1
x (1) ⫽ x (2) ⫽
0.5 ⫺1
respectively. The corresponding general solution is
y ⫽ c1x (1) ⫹ c2x (2)eⴚ0.03t.
From the initial values,

y(0) ⫽ c1 c d ⫹ c2 c d c d.
1 1 0
⫽
0.5 ⫺1 150
In components this is c1 ⫹ c2 ⫽ 0, 0.5c1 ⫺ c2 ⫽ 150. Hence c1 ⫽ 100, c2 ⫽ ⫺100.
This gives the solution

y ⫽ 100 c d ⫺ 100 c d e ⴚ0.03t.

1 1

0.5 ⫺1
In components,
y1 ⫽ 100(1 ⫺ e ⴚ0.03t)
y2 ⫽ 100(12 ⫹ eⴚ0.03t).
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4. With
Flow rate
a⫽
Tank size
we can write the system that models the process in the following form:
y1r ⫽ ay2 ⫺ ay1
y r2 ⫽ ay1 ⫺ ay2,
ordered as needed for the proper vector form
y1r ⫽ ⫺ay1 ⫹ ay2
y r2 ⫽ ay1 ⫺ ay2.
In vector form,

c d.
⫺a a
y r ⫽ Ay, where A⫽
a ⫺a
The characteristic equation is
(l ⫹ a)2 ⫺ a 2 ⫽ l2 ⫹ 2al ⫽ 0.
Hence the eigenvalues are 0 and ⫺2a. Corresponding eigenvectors are

c d c d
1 1
and
1 ⫺1

respectively. The corresponding “general solution” is

y ⫽ c1 c d ⫹ c2 c d eⴚ2at.
1 1

1 ⫺1
This result is interesting. It shows that the solution depends only on the ratio a, not
on the tank size or the flow rate alone. Furthermore, the larger a is, the more rapidly
y1 and y2 approach their limit.
The term “general solution” is in quotation marks because this term has not yet been
defined formally, although it is clear what is meant.
6. The matrix of the system is

⫺0.02 0.02 0

A ⫽ D 0.02 ⫺0.04 0.02 T .

0 0.02 ⫺0.02

The characteristic polynomial is

l3 ⫹ 0.08l2 ⫹ 0.0012l ⫽ l(l ⫹ 0.02)(l ⫹ 0.06).
This gives the eigenvalues and corresponding eigenvectors

1 1 1

l1 ⫽ 0, x (1) ⫽ D 1 T , l2 ⫽ ⫺0.02, x (2) ⫽ D 0 T , l3 ⫽ ⫺0.06, x (3) ⫽ D ⫺2 T .

1 ⫺1 1
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Hence a “general solution” is

1 1 1

y ⫽ c1 D 1 T ⫹ c2 D 0 T e ⴚ0.02t ⫹ c3 D ⫺2 T e ⴚ0.06t.

1 ⫺1 1
We use quotation marks since the concept of a general solution has not yet been
defined formally, although it is clear what is meant.
8. The first ODE remains as before. The second ODE is obviously changed to

I r2 ⫽ 0.4I 1r ⫺ 0.54I 2.
Substitution of the first ODE into the new second one, as in the text, gives

I 2r ⫽ ⫺1.6I1 ⫹ 1.06I2 ⫹ 4.8.

Hence the matrix of the new system is

c d.
⫺4 4
A⫽
⫺1.6 1.06
Its eigenvalues are ⫺1.5 and ⫺1.44. The corresponding eigenvectors are x (1) ⫽
[1 0.625]T and x (2) ⫽ [1 0.64]T, respectively. The corresponding general solution
is

y ⫽ c1x (1)e ⴚ1.5t ⫹ c2x (2)e ⴚ1.44t.

10. The system is

y 1r ⫽ y2
y r2 ⫽ ⫺3y1 ⫺ 4y2
The matrix has eigenvalues ⫺1 and ⫺3 and corresponding eigenvectors [1, ⫺1]T and
[1, ⫺3]T, respectively. From this y ⫽ c1eⴚt ⫹ c2eⴚ3t.
11. y r1 ⫽ y2, y r2 ⫽ y1 ⫹ 32y2, y ⫽ c1[1/2,1]Te2t ⫹ [⫺2,1]Teⴚt/2
12. The system is

y r1 ⫽ y2
y r2 ⫽ y3
y3 ⫽ ⫺2y1 ⫹ y2 ⫹ 2y3
The eigenvalues of its matrix are ⫺1, 1, 2. Eigenvalues are [1, ⫺1, 1]T, [1, 1, 1]T,
[1, 2, 4]T, respectively. The corresponding general solution is

y ⫽ c1[1, ⫺1, 1]Teⴚt ⫹ c2[1, 1, 1]Tet ⫹ c3[1, 2, 4]Te2t.

13. y r1 ⫽ y2, y r2 ⫽ 12y1 ⫺ y2, y1 ⫽ c1eⴚ4t ⫹ c2e3t, y2 ⫽ y r1

14. TEAM PROJECT. (a) From Sec. 2.5 we know that the undamped motions of a mass
on an elastic spring are governed by my s ⫹ ky ⫽ 0 or

my s ⫽ ⫺ky
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where y ⫽ y(t) is the displacement of the mass. By the same arguments, for the two
masses on the two springs in Fig. 81 we obtain the linear homogeneous system
m 1y s1 ⫽ ⫺k 1y1 ⫹ k 2( y2 ⫺ y1)
(11)
m 2y s2 ⫽ ⫺k 2( y2 ⫺ y1)
for the unknown displacements y1 ⫽ y1(t) of the first mass m 1 and y2 ⫽ y2(t) of the
second mass m 2. The forces acting on the first mass give the first equation, and the
forces acting on the second mass give the second ODE. Now m 1 ⫽ m 2 ⫽ 1, k 1 ⫽ 3,
and k 2 ⫽ 2 in Fig. 81 so that by ordering (11) we obtain
y1s ⫽ ⫺5y1 ⫹ 2y2
y s2 ⫽ 2y1 ⫺ 2y2
or, written as a single vector equation,

c d c d c d.
y s1 ⫺5 2 y1
ys ⫽ ⫽
y s2 2 ⫺2 y2

(b) As for a single equation, we try an exponential function of t,

y ⫽ xevt. Then y s ⫽ v2xevt ⫽ Axevt.
Then, writing v2 ⫽ l and dividing by evt, we get
Ax ⫽ lx.
Eigenvalues and eigenvectors are

c d c d.
1 2
l1 ⫽ ⫺1, x (1) ⫽ and l2 ⫽ ⫺6, x (2) ⫽
2 ⫺1

Since v ⫽ ⫾ 2l and 2⫺1 ⫽ i and 2⫺6 ⫽ i 26, we get

y ⫽ x (1)(c1eit ⫹ c2e ⴚit ) ⫹ x (2)(c3ei26t ⫹ c4e ⴚi26t )

or, by (7) in Sec. 2.3,

y ⫽ a1x (1) cos t ⫹ b1x (1) sin t ⫹ a2x (2) cos 26t ⫹ b2x (2) sin 26t

where a1 ⫽ c1 ⫹ c2, b1 ⫽ i(c1 ⫺ c2), a2 ⫽ c3 ⫹ c4, b2 ⫽ i(c3 ⫺ c4). These four

arbitrary constants can be specified by four initial conditions. In components, this
solution is
y1 ⫽ a1 cos t ⫹ b1 sin t ⫹ 2a2 cos 26t ⫹ 2b2 sin 26t
y2 ⫽ 2a1 cos t ⫹ 2b1 sin t ⫺ a2 cos 26t ⫺ b2 sin 26t.
(c) The first two terms in y1 and y2 give a slow harmonic motion, and the last two a
fast harmonic motion. The slow motion occurs if, at some instant, both masses are
moving downward or both upward. For instance, if a1 ⫽ 1 and the three other con-
stants are zero, we get y1 ⫽ cos t, y2 ⫽ 2 cos t; this is an example of such a motion.
The fast motion occurs if, at each instant, the two masses are moving in opposite
directions, so that one of the two springs is extended, whereas the other is simul-
taneously compressed. For instance, if a2 ⫽ 1 and the other constants are zero, we
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have y1 ⫽ 2 cos 26t, y2 ⫽ ⫺cos 26t; this is a fast motion of the indicated type.
Depending on the initial conditions, one or the other motion will occur or a super-
position of both.

SECTION 4.2 Basic Theory of Systems of ODEs. Wronskian, page 137

Purpose. This survey of some basic concepts and facts on nonlinear and linear systems
is intended to give the student an impression of the conceptual and structural similarity
of the theory of systems to that of single ODEs.
Content, Important Concepts
Standard form (1) of first-order systems, nonlinear or linear. The point is that each
equation contains only one derivative, which appears on the left. For instance, Kirchhoff’s
Voltage Law (KVL) for electric systems gives a system of ODEs that can be transformed
into the form (1) by algebra and differentiation.
Form of corresponding initial value problems (1), (2). If an ODE is converted to a
system, using Theorem 1 of Sec. 4.1, a corresponding IVP written as in Chaps. 2 and 3
converts to the form (1), (2) and conversely.
Existence and Uniqueness Theorem 1 for solutions of IVPs (1), (2). Note that the
(sufficient) conditions correspond to those for single ODEs.
Standard form and notations (3) for linear systems of ODEs. These notations agree
with the usual notations for matrices involving double subscripts.
Homogeneous and nonhomogeneous linear systems of ODEs.
Basis, general solution (5), Wronskian (7), which is the determinant of the fundamental
matrix Y (see (6)) such that a general solution can be written (8) y ⫽ Yc, c the vector
with the arbitrary constants as components.
Background Material. Sec. 2.6 contains the analogous theory for single equations. See
also Sec. 1.7.
Short Courses. This section may be skipped, as mentioned before.

SECTION 4.3. Constant-Coefficient Systems. Phase Plane Method,

page 140
Purpose. Typical examples show the student the rich variety of pattern of solution
curves (trajectories) near critical points in the phase plane, along with the process of
actually solving homogeneous linear systems. This will also prepare the student for a
good understanding of the systematic discussion of critical points in the phase plane in
Sec. 4.4.
Main Content
Solution method for homogeneous linear systems
Examples illustrating types of critical points
Solution when no basis of eigenvectors is available (Example 6)
Important Concepts and Facts
Trajectories as solution curves in the phase plane
Phase plane as a means for the simultaneous (qualitative) discussion of a large number
of solutions
Basis of solutions obtained from basis of eigenvectors
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Background Material. Short review of eigenvalue problems from Sec. 4.0, if needed.
Short Courses. Omit Example 6.
Some Details on Content
In addition to developing skill in solving homogeneous linear systems, the student is
supposed to become aware that it is the kind of eigenvalues that determines the type of
critical point. The examples show important cases. (A systematic discussion of all cases
follows in the next section.)
Example 1. Two negative eigenvalues give a node.
Example 2. A real double eigenvalue gives a node.
Example 3. Real eigenvalues of opposite sign give a saddle point.
Example 4. Pure imaginary eigenvalues give a center, and working in complex is
avoided by a standard trick, which can also be useful in other contexts.
Example 5. Genuinely complex eigenvalues give a spiral point. Some work in complex
can be avoided, if desired, by differentiation and elimination. The first ODE is
(a) y2 ⫽ y r1 ⫹ y1.
By differentiation and from the second ODE as well as from (a),
y s1 ⫽ ⫺y r1 ⫹ y r2 ⫽ ⫺y r1 ⫺ y1 ⫺ (y r1 ⫹ y1) ⫽ ⫺2y r1 ⫺ 2y1.
Complex solutions e(ⴚ1⫾i)t give the real solution
y1 ⫽ e ⴚt(A cos t ⫹ B sin t).
From this and (a) there follows the expression for y2 given in the text.
Example 6 shows that the present method can be extended to include cases when A
does not provide a basis of eigenvectors, but then becomes substantially more involved.
In this way the student will recognize the importance of bases of eigenvectors, which also
play a role in many other contexts.
A further illustration of the situation is given in Probs. 8 and 16.

SOLUTIONS TO PROBLEM SET 4.3, page 147

1. y1 ⫽ c1eⴚt ⫹ c2et, y2 ⫽ 3c1eⴚt ⫹ c2et
2. The eigenvalues are 3 and 9. Eigenvectors are [3 ⫺1]T and [3 1]T, respectively.
The corresponding general solution is
y1 ⫽ 3c1e3t ⫹ 3c2e9t
y2 ⫽⫺c1e3t ⫹ c2e9t.
3. y1 ⫽ c1 ⫹ c2et, y2 ⫽ 2c2et ⫹ 43c1
4. The matrix has the double eigenvalue ⫺6. An eigenvector is [1 ⫺1]T. Hence the
vector u needed is obtained from

c du⫽ c d.
⫺2 ⫺2 1
(A ⫹ 6I) u ⫽
2 2 ⫺1
We can take u ⫽ 30, ⫺124. With this we obtain as a general solution
y1 ⫽ c1e ⴚ6t ⫹ c2te ⴚ6t
y2 ⫽ ⫺c1e ⴚ6t ⫺ c2(t ⫹ 12)e ⴚ6t.
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6. The eigenvalues are complex, 2 ⫹ 2i and 2 ⫺ 2i. Corresponding complex eigenvectors

are [1 ⫺i]T and [1 i]T, respectively. Hence a complex general solution is
y1 ⫽ c1e(2⫹2i)t ⫹ c2e(2–2i)t
y2 ⫽ ⫺ic1e(2⫹2i)t ⫹ ic2e(2–2i)t.
From this and the Euler formula we obtain a real general solution
y1 ⫽ e2t3(c1 ⫹ c2) cos 2t ⫹ i(c1 ⫺ c2) sin 2t4
⫽ e2t (A cos 2t ⫹ B sin 2t),
y2 ⫽ e2t3(⫺ic1 ⫹ ic2) cos 2t ⫹ i(⫺ic1 ⫺ ic2) sin 2t4
⫽ e2t(⫺B cos 2t ⫹ A sin 2t)
where A ⫽ c1 ⫹ c2 and B ⫽ i(c1 ⫺ c2).
7. y1 ⫽ ⫺(1/2)c2 22 cos(22ax) ⫹ (1/2)c3 sin(22ax)22 ⫹ c1,
y2 ⫽ c2 sin(22ax) ⫹ c3 cos(22ax),
y3 ⫽ (1/2)c2 22 cos(22ax) ⫺ (1/2)c3 sin(22ax)22 ⫹ c1.
8. The eigenvalue of algebraic multiplicity 2 is 9. An eigenvector is [1 ⫺1]T. There
is no basis of eigenvectors. A first solution is

c d e9t.
1
y (1) ⫽
⫺1

A second linearly independent solution is (see Example 6 in the text)

c d te9t ⫹ c d e9t
1 u1
y (2) ⫽
⫺1 u2

with 3u 1 u 24T determined from

c dc d c d.
⫺1 ⫺1 u1 1
(A ⫺ 9I) u ⫽ ⫽
1 1 u2 ⫺1

Thus u 1 ⫹ u 2 ⫽ ⫺1. We can take u 1 ⫽ 0, u 2 ⫽ ⫺1. This gives the general solution

c d e9t ⫹ c2 c d e9t.
1 0
y ⫽ c1y (1) ⫹ c2y (2) ⫽ (c1 ⫹ c2t)
⫺1 ⫺1

10. The eigenvalues are 1 and ⫺3. The corresponding general solution is
y1 ⫽ c1et ⫹ 5c2eⴚ3t and y2 ⫽ c1et ⫹ c2eⴚ3t
Using the initial conditions we obtain
y1(0) ⫽ c1 ⫹ 5c2 ⫽ 0 and y2(0) ⫽ c1 ⫹ c2 ⫽ 4
Hence the solution of the IVP is
y1 ⫽ 5et ⫺ 5eⴚ3t
y2 ⫽ 5et ⫺ eⴚ3t
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ⴚt/4
11. y1 ⫽ 85et ⫺ 18
5e and y2 ⫽ 85et ⫺ 85eⴚt/4
12. The eigenvalues are 0 and 2. The corresponding general solution is
y1 ⫽ c1 ⫹ c2e2t
y2 ⫽ ⫺13 c1 ⫹ 13 c2e2t.
From this and the initial values we obtain
y1(0) ⫽ c1 ⫹ c2 ⫽ 12
1
y2(0) ⫽ ⫺3 c1 ⫹ 13 c2 ⫽ 2.
Hence the solution of the IVP is
y1 ⫽ 3 ⫹ 9e2t
y2 ⫽ ⫺1 ⫹ 3e2t.

13. y1 ⫽ sinh 2x, y2 ⫽ cosh 2x

14. The eigenvalues are ⫺1 ⫹ i and ⫺1 ⫺ i. The corresponding real general solution is
y1 ⫽ e ⴚt (c1 cos t ⫹ c2 sin t)
y2 ⫽ e ⴚt (⫺c2 cos t ⫹ c1 sin t).
From this and the initial conditions we obtain the solutions of the IVP
y1 ⫽ eⴚt cos t
y2 ⫽ eⴚt sin t.

15. y1 ⫽ 0.25eⴚt, y2 ⫽ ⫺0.25eⴚt

16. The system is
(a) y 1r ⫽ 8y1 ⫺ y2
(b) y r2 ⫽ y1 ⫹ 10y2.
10(a) ⫹ (b) gives 10y r1 ⫹ y r2 ⫽ 81y1; hence
(c) y r2 ⫽ ⫺10y r1 ⫹ 81y1.
Differentiating (a) and using (c) gives
0 ⫽ y s1 ⫺ 8y r1 ⫹ y r2
⫽ y s1 ⫺ 8y r1 ⫺ 10y r1 ⫹ 81y1
⫽ y s1 ⫺ 18y r1 ⫹ 81y1.
A general solution is
y1 ⫽ (c1 ⫹ c2t)e9t.
From this and (a) we obtain
y2 ⫽ ⫺y r1 ⫹ 8y1 ⫽ 3⫺c2 ⫺ 9(c1 ⫹ c2t) ⫹ 8(c1 ⫹ c2t)4e9t
⫽ (⫺c2 ⫺ c1 ⫺ c2t)e9t,
in agreement with the answer to Prob. 8.
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18. The restriction of the inflow from outside to pure water is necessary to obtain a
homogeneous system. The principle involved in setting up the model is
Time rate of change ⫽ Inflow ⫺ Outflow.
For Tank T1 this is (see Fig. 88)

y r1 ⫽ a12 ⴢ 0 ⫹ y2 b ⫺
4 16
y1.
200 200
For Tank T2 it is
16 4 ⫹ 12
y r2 ⫽ y1 ⫺ y2.
200 200
Performing the divisions and ordering terms, we have
y r1 ⫽ ⫺0.08y1 ⫹ 0.02y2
y r2 ⫽ 0.08y1 ⫺ 0.08y2.
The eigenvalues of the matrix of this system are ⫺0.04 and ⫺0.12. Eigenvectors are
[1 2]T and [1 ⫺2]T, respectively. The corresponding general solution is

y ⫽ c1 c d e ⴚ0.04t ⫹ c2 c d e ⴚ0.12t.
1 1

2 ⫺2
The initial conditions are y1(0) ⫽ 100, y2(0) ⫽ 200. This gives c1 ⫽ 100, c2 ⫽ 0. In
components the answer is
y1 ⫽ 100e ⴚ0.04t
y2 ⫽ 200e ⴚ0.04t.
Both functions approach zero as t : ⬁, a reasonable result because pure water flows
in and mixture flows out.

SECTION 4.4. Criteria for Critical Points. Stability, page 148

Purpose. Systematic discussion of critical points in the phase plane from the standpoints
of both geometrical shapes of trajectories and stability.
Main Content
Table 4.1 for the types of critical points
Table 4.1 for the stability behavior
Stability chart (Fig. 92), giving Tables 4.1 and 4.2 graphically
Important Concepts
Node, saddle point, center, spiral point
Stable and attractive, stable, unstable
Background Material. Sec. 2.4 (needed in Example 2).
Short Courses. Since all these types of critical points already occurred in the previous
section, one may perhaps present just a short discussion of stability.
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Some Details on Content

The types of critical points in Sec. 4.3 now recur, and the discussion shows that they
exhaust all possibilities. With the examples of Sec. 4.3 fresh in mind, the student will
acquire a deeper understanding by discussing the stability chart and by reconsidering
those examples from the viewpoint of stability. This gives the instructor an opportunity
to emphasize that the general importance of stability in engineering can hardly be
overestimated.
Example 2, relating to the familiar free vibrations in Sec. 2.4, gives a good illustration
of stability behavior, namely, depending on c, attractive stability, stability (and instability
if one includes “negative damping,” with c ⬍ 0, as it will recur in the next section in
connection with the famous van der Pol equation).

Problem Set 4.4

Problems 1–10 give straightforward applications of the criteria in this section.
Problems 11–12 are related to previous material, for instance, to oscillations of a mass
on a spring.
For Prob. 13 let the student reconsider the discussion of (10)–(13) in Sec. 4.3 as
discussed there and then, in addition, from the new viewpoint of stability.
Problems 15–17 concern the basic concept of perturbation, which, in practice, may be
due to inaccuracies of measurement.
The purpose of Probs. 18–20 is fairly obvious.

SOLUTIONS TO PROBLEM SET 4.4, page 151

1
1. Unstable improper node, y1 ⫽ c1et, y2 ⫽ e2t.
2. p ⫽ ⫺7, q ⫽ 12 ⬎ 0, ¢ ⫽ 49 ⫺ 48 ⬎ 0, stable and attractive improper node,
y1 ⫽ c1e⫺4t, y2 ⫽ c2eⴚ3t
3. Center, always stable, y1 ⫽ c1 sin 2t ⫹ c2 cos 2t, y2 ⫽ 2c1 cos 2t ⫺ 2c2 sin 2t
4. p ⫽ 0, q ⫽ ⫺9, saddle point, always unstable. A general solution is

y1 ⫽ c1eⴚ3t ⫹ c2e3t
y2 ⫽ ⫺5c1eⴚ3t ⫹ c2e3t.

5. Stable spiral, y1 ⫽ eⴚt(c1 sin t ⫹ c2 cos t), y2 ⫽ ⫺eⴚt(⫺c1 cos t ⫹ c2 sin t)

6. p ⫽ ⫺12, q ⫽ 27, ¢ ⫽ 144 ⫺108 ⬎ 0, stable and arrtactive node. A general
solution is

y1 ⫽ c1eⴚ3t ⫹ c2eⴚ9t
y2 ⫽ ⫺3c1eⴚ3t ⫹ 3c2eⴚ9t.

7. Saddle point, always unstable, y1 ⫽ c1eⴚ2t ⫹ c2e3t, y2 ⫽ c1eⴚ2t ⫺ 4c2e3t.

8. p ⫽ ⫺3, q ⫽ ⫺10, saddle point, always unstable. A general solution is

y1 ⫽ c1eⴚ5t ⫹ 4c2e2t
y2 ⫽ ⫺c1eⴚ5t ⫹ 3c2e2t.

9. Unstable node, y1 ⫽ c1e2t ⫹ c2e3t, y2 ⫽ ⫺43c1e2t ⫺ c2e3t.

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10. p ⫽ ⫺2, q ⫽ 5, ¢ ⫽ 4 ⫺ 20 ⬍ 0, stable and attractive spiral. The components

of a general solution are

y1 ⫽ c1e(ⴚ1⫹2i)t ⫹ c2e(ⴚ1ⴚ2i)t
⫽ eⴚt((c1 ⫹ c2) cos 2t ⫹ i(c1 ⫺ c2) sin 2t)
⫽ eⴚt(A cos 2t ⫹ B sin 2t),
y2 ⫽ (⫺1 ⫹ 2i)c1e(ⴚ1⫹2i)t ⫹ (⫺1 ⫺ 2i)c2e(ⴚ1ⴚ2i)t
⫽ eⴚt((⫺c1 ⫺ c2 ⫹ 2ic1 ⫺ 2ic2) cos 2t ⫹ i(2ic1 ⫹ 2ic2 ⫺ c1 ⫹ c2) sin 2t)
⫽ eⴚt((⫺A ⫹ 2B) cos 2t ⫺ (2A ⫹ B) sin 2t)

where A ⫽ c1 ⫹ c2 and B ⫽ i(c1 ⫺ c2).

11. y ⫽ eⴚt(c1 sin 2t ⫹ c2 cos 2t). Stable and attractive spirals.
12. y ⫽ A cos 13t ⫹ B sin 13t. The trajectories are the ellipses
1 2
9 y1 ⫹ y 22 ⫽ const.

This is obtained as in Example 4 in Sec. 4.3.

14. y r1 ⫽ ⫺dy1>dt, y r2 ⫽ ⫺dy2>dt, reversal of the direction of motion; to get the usual
form, we have to multiply the transformed system by ⫺1, which amounts to
multiplying the matrix by ⫺1, changing p into ⫺p, but leaving q and ¢ unchanged.
In the example, we get an unstable node.
16. At a center, p ⫽ a11 ⫹ a22 ⫽ 0, q ⫽ det A ⬎ 0, hence ¢ ⬍ 0. Under the change, p
changes into a11 ⫹ k ⫹ a22 ⫹ k ⫽ 2k ⫽ 0; q remains positive because

(a11 ⫹ k)(a22 ⫹ k) ⫺ a12a21 ⫽ q ⫹ k 2 ⬎ 0.

Finally, ¢ remains unchanged because

(p ⫹ 2k)2 ⫺ 4(q ⫹ k 2) ⫽ (2k)2 ⫺ 4(q ⫹ k 2) ⫽ ⫺4q ⬍ 0.

Hence we obtain a spiral point, which is unstable if k ⬎ 0 and stable and attractive
if k ⬍ 0.
We can reason more simply as follows. For a center, the eigenvalues are pure
imaginary (to have closed trajectories). An eigenvalue l of A gives an eigenvalue
l ⫹ k of A, causing a damped oscillation (when k ⬍ 0) or an increasing one (when
k ⬎ 0), thus a spiral.

SECTION 4.5 Qualitative Methods for Nonlinear Systems, page 152

Purpose. As a most important step, in this section we extend phase plane methods from
linear to nonlinear systems and nonlinear ODEs.
The particular importance of phase plane methods for nonlinear ODEs and systems
results from the difficulty (or impossibility) of solving them, explicitly as explained in
the text.
Main Content
Critical points of nonlinear systems
Their discussion by linearization
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Transformation of single autonomous ODEs

Applications of linearization and transformation techniques
Important Concepts and Facts
Linearized system (3), condition for applicability
Linearization of pendulum equations
Self-sustained oscillations, van der Pol equation
Short Courses. Linearization at different critical points seems the main issue that the student
is supposed to understand and handle in practice. Examples 1 and 2 may help students to
gain skill in that technique. The other material can be skipped without loss of continuity.
Some Details on Content
This section is very important, because from it the student should learn not only techniques
(linearization, etc.) but also the fact that phase plane methods are particularly powerful
and important in application to systems or single ODEs that cannot be solved explicitly.
The student should also recognize that it is quite surprising how much information these
methods can give. This is demonstrated by the pendulum equation (Examples 1 and 2)
for a relatively simple system, and by the famous van der Pol equation for a single ODE,
which has become a prototype for self-sustained oscillations of electrical systems of
various kinds.
We also discuss the famous Lotka–Volterra predator–prey model.
For the Rayleigh and Duffing equations, see the problem set.
Problem Set 4.5
Problems 1–3 concern the undamped pendulum and limit cycles and their deformation
in the famous van der Pol equation.
Problems 4–8 are formal and concern linearization of systems in the case of several
limit points, usually of a different type.
Problems 9–13 concern linearization of single nonlinear second-order ODEs, which are
first converted to systems.
Team Project 14 discusses further self-sustained oscillations.

SOLUTIONS TO PROBLEM SET 4.5, page 159

2. A limit cycle is approached by trajectories (from inside and outside). No such approach
takes place for a closed trajectory.
4. Writing the system in the form
y r1 ⫽ y1(4 ⫺ y1)
y r2 ⫽ y2
we see that the critical points are (0, 0) and (4, 0). For (0, 0) the linearized system is
y r1 ⫽ 4y1
y r2 ⫽ y2.
The matrix is

c d.
4 0

0 1
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Hence p ⫽ 5, q ⫽ 4, ¢ ⫽ 25 ⫺ 16 ⫽ 9. This shows that the critical point at (0, 0) is

an unstable node.
For (4, 0) the translation to the origin is
y1 ⫽ 4 ⫹ 苲 y1, y2 ⫽ 苲
y2.
This gives the transformed system
苲
y 1r ⫽ (4 ⫹ 苲
y1)(⫺y苲1)
苲
y 2r ⫽ 苲
y2
and the corresponding linearized system
苲
y 1r ⫽ ⫺4y 苲1
苲 苲
y r2 ⫽ y 2.
Its matrix is

c d.
⫺4 0

0 1
Hence 苲p ⫽ ⫺3, q苲 ⫽ ⫺4. Thus the critical point at (4, 0) is a saddle point, which is
always unstable.
5. Center at (0, 0). Saddle point at (4, 0). At (4, 0) set y1 ⫽ 4 ⫹ ~y 1. Then ~y r2 ⫽ ~y 1.
6. At (0, 0), y 1r ⫽ y2, y r2 ⫽ ⫺y1, p ⫽ 0, q ⫽ 1, ¢ ⫽ ⫺4, center. The other critical point
is at (⫺1, 0). We set y1 ⫽ ⫺1 ⫹ 苲 y 2. Then ⫺y1 – y 21 ⬇ 苲
y1, y2 ⫽ 苲 y 1r ⫽ 苲
y1. Hence 苲 y 2,
苲 苲
y 2r ⫽ y 1. This gives a saddle point.
7. The point (0, 0), y 1r ⫽ ⫺y1 ⫹ y2, y 2r ⫽ ⫺y1 ⫺ 12y2, is stable and an attractive spiral
point. The point (⫺1, 2) is a saddle point, y1 ⫽ ⫺1 ⫹ ~y 1, y2 ⫽ 2 ⫹ ~y 2, ~y 1r ⫽
⫺2y~1 ⫺ 3y~2, ~y r2 ⫽ ⫺y~1 ⫺ 12~y 2.
8. The system may be written
y 1r ⫽ y2(1 ⫺ y2)
y r2 ⫽ y1(1 ⫺ y1).
From this we see immediatly that there are four critical points, at (0, 0), (0, 1),
(1, 0), (1, 1).
At (0, 0) the linearized system is
y 1r ⫽ y2
y r2 ⫽ y1.
The matrix is

c d.
0 1

1 0

Hence p ⫽ 0, q ⫽ ⫺1, so that we have a saddle point.

At (0, 1) the transformation is y1 ⫽ 苲
y1, y2 ⫽ 1 ⫹ 苲
y 2. This gives the transformed
system
苲
y r1 ⫽ (1 ⫹ 苲y 2)(⫺y苲2)
y r2 ⫽ 苲
苲 y 1(1 ⫺ 苲y 1).
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Linearization gives
苲
y 1r ⫽ ⫺y苲2
c d
0 ⫺1
with matrix
苲
y 2r ⫽ 苲
y1 1 0
苲
for which p苲 ⫽ 0, q苲 ⫽ 1, ¢ ⫽ ⫺4, and we have a center.
At (1, 0) the transformation is y1 ⫽ 1 ⫹ 苲
y1, y2 ⫽ 苲
y 2. The transformed system is
苲
y 1r ⫽ 苲
y 2(1 ⫺ 苲
y 2)
苲
y r2 ⫽ (1 ⫹ 苲y1)(⫺y苲1).
Its linearization is
苲 苲
c d
y 1r ⫽ y2 0 1
with matrix
苲
y 2r ⫽ ⫺y苲1 ⫺1 0
苲
for which 苲p ⫽ 0, 苲 q ⫽ 1, ¢ ⫽ ⫺4, so that we get another center.
At (1, 1) the transformation is
y1 ⫽ 1 ⫹ 苲
y1, y2 ⫽ 1 ⫹ 苲
y2.
The transformed system is
y 1r ⫽ (1 ⫹ 苲
苲 苲2)
y 2)(⫺y
苲
y 2r ⫽ (1 ⫹ 苲
y1)(⫺y苲1).

Linearization gives
苲
y 1r ⫽ ⫺y苲2
c d
0 ⫺1
with matrix
y 2r ⫽ ⫺y苲1
苲 ⫺1 0

for which p苲 ⫽ 0, 苲
q ⫽ ⫺1, and we have another saddle point.
9. (0, 0) saddle point, (⫺2, 0) and (2, 0) centres.
10. y s ⫹ y ⫺ y 3 ⫽ 0 written as a system is
y1r ⫽ y2
y 2r ⫽ ⫺y1 ⫹ y 31.
Now ⫺y1 ⫹ y 31 ⫽ y1(⫺1 ⫹ y 21) ⫽ 0 shows that there are three critical points, at
(y1, y2) ⫽ (0, 0), (⫺1, 0), and (1, 0).
The linearized system at (0, 0) is

c d.
y 1r ⫽ y2 0 1
Matrix:
y 2r ⫽ ⫺y1. ⫺1 0

From the matrix we see that p ⫽ a11 ⫹ a22 ⫽ 0, q ⫽ 1. Hence (0, 0) is a center
(see Sec. 3.4).
For the next critical point we have to linearize at (⫺1, 0) by setting
y1 ⫽ ⫺1 ⫹ 苲
y1, y2 ⫽ 苲
y 2.
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Then
y1(⫺1 ⫹ y 21 ⫽ (⫺1 ⫹ 苲
y1)3⫺1 ⫹ (⫺1 ⫹ 苲 y1)24
⫽ (⫺1 ⫹ 苲 y 14 ⬇ 2y苲1.
y1)3⫺2y苲1 ⫹ 苲 2

Hence the linearized system is

苲
y 1r ⫽ 苲
c d.
y2 0 1
Matrix:
苲
y 2r ⫽ 2y苲1. 2 0
Hence q ⫽ det A ⫽ ⫺2 ⬍ 0, that is, the critical point at (⫺1, 0) is a saddle point.
Similarly, to linearize at (1, 0), set
y1 ⫽ 1 ⫹ 苲
y1, y2 ⫽ 苲
y 2.
Then
⫺y1 ⫹ y 31 ⬇ 2y苲1
and we obtain another saddle point, as just before.
11. (p p p
4 ⫾ np, 0) saddle points; (⫺ 4 ⫾ np, 0) centres. For the critical point ( 4 , 0), after
p ~ ~
linearizing and using the substitutions y 1r ⫽ 4 ⫹ y 1, and y r2 ⫽ y 2, we get

y r1 ⫽ ~y r1
y r2 ⫽ ⫺cos 2(p ~
4 ⫹ y r1)
⫽ sin 2y~1
⫽ 2y~1

where we have used the fact 2y ⫽ 2y ⫺ 43y 3 ⫹ Á .

12. y 1r ⫽ y2, y 2r ⫽ y1(⫺9 ⫺ y1). (0, 0) is a critical point. The linearized system at (0, 0) is

c d
y 1r ⫽ y2 0 1
with matrix
y r2 ⫽ ⫺9y1 ⫺9 0
for which p ⫽ 0 and q ⫽ 9 ⬎ 0, so that we have a center.
A second critical point is at (⫺9, 0). The transformation is

y1 ⫽ ⫺9 ⫹ 苲
y 1, y2 ⫽ 苲
y 2.
This gives the transformed system
苲
y 1r ⫽ 苲
y2
苲
y 2r ⫽ (⫺9 ⫹ 苲
y1)(⫺y苲1).
Its linearization is
苲
y 1r ⫽ 苲
c d
y2 0 1
with matrix
苲
y r2 ⫽ 9y苲1 9 0
for which 苲
q ⫽ ⫺9 ⬍ 0, so that we have a saddle point.
13. (⫾2np, 0) centres; y1 ⫽ (2n ⫹ 1)p ⫹ ~y 1, (p ⫾ 2np, 0) saddle points.
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14. TEAM PROJECT. (a) Unstable node if ␮ ⭌ 2, unstable spiral point if 2 ⬎ ␮ ⬎ 0,

center if ␮ ⫽ 0, stable and attractive spiral point if 0 ⬎ ␮ ⬎ ⫺2, stable and attractive
node if ␮ ⬉ ⫺2.
(c) As a system we obtain
(A) y r1 ⫽ y2
(B) y r2 ⫽ ⫺(v20y1 ⫹ by 31).
The product of the left side of (A) and the right side of (B) equals the product of the
right side of (A) and the left side of (B):

y r2y2 ⫽ ⫺(v20y1 ⫹ by 31)y r1.

Integration on both sides and multiplication by 2 gives

y 22 ⫹ v20 y 21 ⫹ 12 by 41 ⫽ const.

For positive b these curves are closed because then by 3 is a proper restoring term,
adding to the restoring due to the y-term. If b is negative, the term by 3 has the opposite
effect, and this explains why then some of the trajectories are no longer closed but
extend to infinity in the phase plane.
For generalized van der Pol equations, see e.g., K. Klotter and E. Kreyzig, On a
class of self-sustained oscillations. J. Appl. Math. 27(1960), 568–574.

SECTION 4.6. Nonhomogeneous Linear Systems of ODEs, page 160

Purpose. We now turn from homogeneous linear systems considered so far to solution
methods for nonhomogeneous systems.
Main Content
Method of undetermined coefficients
Modification for special right sides
Method of variation of parameters
Short Courses. Select just one of the preceding methods.
Some Details on Content
In addition to understanding the solution methods as such, the student should observe the
conceptual and technical similarities to the handling of nonhomogeneous linear ODEs in
Secs. 2.7–2.10 and understand the reason for this, namely, that systems can be converted
to single equations and conversely. For instance, in connection with Example 1 in this
section, one may point to the Modification Rule in Sec. 2.7, or, if time permits, establish
an even more definite relation by differentiation and elimination of y2,
y1s ⫽ ⫺3y 1r ⫹ y 2r ⫹ 12eⴚ2t
⫽ ⫺3y 1r ⫹ ( y1 ⫺ 3y2 ⫹ 2eⴚ2t) ⫹ 12eⴚ2t
⫽ ⫺3y 1r ⫹ y1 ⫺ 3( y 1r ⫹ 3y1 ⫹ 6eⴚ2t) ⫹ 14eⴚ2t
⫽ ⫺6y r1 ⫺ 8y1 ⫺ 4eⴚ2t,

solving this for y1 and then getting y2 from the solution.

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Problem Set 4.6

Problems 2–7 concern general solutions of nonhomogeneous systems, where the particular
solution needed is obtained by the method of undetermined coefficients.
Problems 8 and 9 are general questions on the method of undetermined coefficients.
Problems 10–15 concern initial value problems; here it is important to emphasize that
the initial conditions are to be used only after a general solution of the nonhomogeneous
system has been obtained.
Typical applications to nonhomogeneous systems are shown in Probs. 17–20.

SOLUTIONS TO PROBLEM SET 4.6, page 163

2. The matrix of the system has the eigenvalues 2 and ⫺2. Eigenvectors are [1 1]T
and [1 ⫺3]T, respectively. Hence a general solution of the homogeneous system is

c d e2t ⫹ c d eⴚ2t.
1 1
y (h) ⫽ c1 c2
1 ⫺3
We determine y (p) by the method of undetermined coefficients, starting from

c d.
a1 cos t ⫹ b1 sin t
y (p) ⫽
a2 cos t ⫹ b2 sin t
Substituting this and its derivative into the given nonhomogeneous system, we obtain,
in terms of components,

⫺a1 sin t ⫹ b1 cos t ⫽ (a1 ⫹ a2) cos t ⫹ (b1 ⫹ b2) sin t ⫹ 10 cos t
⫺a2 sin t ⫹ b2 cos t ⫽ (3a1 ⫺ a2) cos t ⫹ (3b1 ⫺ b2) sin t ⫺ 10 sin t.
By equating the coefficients of the cosine and of the sine in the first of these two
equations we obtain

b1 ⫽ a1 ⫹ a2 ⫹ 10, ⫺a1 ⫽ b1 ⫹ b2.

Similarly from the second equation,

b2 ⫽ 3a1 ⫺ a2, ⫺a2 ⫽ 3b1 ⫺ b2 ⫺ 10.

The solution is a1 ⫽ ⫺2, b1 ⫽ 4, a2 ⫽ ⫺4, b2 ⫽ ⫺2. This gives the answer

y1 ⫽ c1e2t ⫹ c2eⴚ2t ⫺ 2 cos t ⫹ 4 sin t

y2 ⫽ c1e2t ⫺ 3c2eⴚ2t ⫺ 4 cos t ⫺ 2 sin t.
3. y1 ⫽ c1et ⫹ c2eⴚt, y2 ⫽ c1et ⫺ c2eⴚt ⫺ 53e2t
4. From the characteristic equation,

c d, c d.
4 1
l1 ⫽ 2, x (1) ⫽ l2 ⫽ ⫺4, x (2) ⫽
1 1

y (p) can be obtained by the method of undetermined coefficients, starting from

y (p) ⫽ a cosh t ⫹ b sinh t.
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Substitution gives
y (p) r ⫽ a sinh t ⫹ b cosh t

⫽ c d (a cosh t ⫹ b sinh t) ⫹ c d cosh t ⫹ c d sinh t.

4 ⫺8 2 0

2 ⫺6 1 2
Comparing sinh terms and cosh terms (componentwise), we get from this
a1 ⫽ 4b1 ⫺ 8b2
f sinh terms
a2 ⫽ 2b1 ⫺ 6b2 ⫹ 2
b1 ⫽ 4a1 ⫺ 8a2 ⫹ 2
f cosh terms.
b2 ⫽ 2a1 ⫺ 6a2 ⫹ 1
To solve this, one can substitute the first two equations into the last two, solve for
b1 ⫽ 2, b2 ⫽ 1, and then get from the first two equations a1 ⫽ a2 ⫽ 0. This gives
the general solution
y1 ⫽ 4c1e2t ⫹ c2eⴚ4t ⫹ 2 sinh t, y2 ⫽ c1e2t ⫹ c2eⴚ4t ⫹ sinh t.
5. y1 ⫽ ⫺13 5
4 ⫺ 2 t ⫹ c1e
2t
⫹ c2et, y2 ⫽ 72 ⫺ 23c1e2t ⫺ c2et ⫹ 3t
6. From the characteristic equation we obtain

c d, c d.
1 1
l1 ⫽ 4, x (1) ⫽ l2 ⫽ ⫺4, x (2) ⫽
1 ⫺1
By the method of undetermined coefficients, we set
y (p) ⫽ u ⫹ vt ⫹ wt 2.
By substitution,

c d (u ⫹ vt ⫹ wt 2) ⫹ c d c d t 2.
0 4 0 0
y (p) r ⫽ v ⫹ 2wt ⫽ ⫹
4 0 2 ⫺16
We now compare componentwise the constant terms, linear terms, and quadratic
terms:
v1 ⫽ 4u 2
f constant terms
v2 ⫽ 4u 1 ⫹ 2
2w1 ⫽ 4v2
f linear terms
2w2 ⫽ 4v1
0 ⫽ 4w2
f quadratic terms.
0 ⫽ 4w1 ⫺ 16
We then obtain, in this order,
w1 ⫽ 4, w2 ⫽ 0, v1 ⫽ 0, v2 ⫽ 2, u 1 ⫽ 14 (v2 ⫺ 2) ⫽ 0, u 2 ⫽ 0.
The corresponding general solution is

y ⫽ c1x (1)e4t ⫹ c2x (2)eⴚ4t ⫹ vt ⫹ wt 2;

in components,
y1 ⫽ c1e4t ⫹ c2eⴚ4t ⫹ 4t 2, y2 ⫽ c1e4t ⫺ c2eⴚ4t ⫹ 2t.
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7. y1 ⫽ c1eⴚ2t ⫹ c2e3t ⫹ 34eⴚt ⫹ 16t ⫺ 36

1

ⴚ2t 47 175
y2 ⫽ c1e ⫺ 4c2e ⫹ 6 t ⫹ 36
3t

c d , x (2) ⫽ c
d . Now et on the right is a
1 4
10. y ⫽ c1x (1) et ⫹ c2x (2)e2t, x (1) ⫽
⫺1 ⫺5
solution of the homogeneous system. Hence, to find y (p), we have to proceed as in
Example 1, setting
y (p) ⫽ utet ⫹ vet.
Substitution gives

c d (utet ⫹ vet) ⫹ c d et.

⫺3 ⫺4 5
y (p) r ⫽ u(t ⫹ 1)et ⫹ vet ⫽
5 6 ⫺6

Equating the terms in et (componentwise) gives

u 1 ⫹ v1 ⫽ ⫺3v1 ⫺ 4v2 ⫹ 5
u 2 ⫹ v2 ⫽ 5v1 ⫹ 6v2 ⫺ 6
and the terms in tet give
u 1 ⫽ ⫺3u 1 ⫺ 4u 2
u 2 ⫽ 5u 1 ⫹ 6u 2.
Hence u 1 ⫽ 1, u 2 ⫽ ⫺1, v1 ⫽ 1, v2 ⫽ 0. This gives the general solution

y1 ⫽ c1et ⫹ 4c2e2t ⫹ tet ⫹ et, y2 ⫽ ⫺c1et ⫺ 5c2e2t ⫺ tet.

From the initial conditions we obtain c1 ⫽ ⫺2, c2 ⫽ 5. Answer:

y1 ⫽ ⫺2et ⫹ 20e2t ⫹ tet ⫹ et, y2 ⫽ 2et ⫺ 25e2t ⫺ tet.

11. y1 ⫺ cosh t ⫹ 2 sinh t ⫹ et, y2 ⫽ 2 cosh t ⫺ sinh t ⫺ eⴚt

12. A general solution of the homogeneous system is

c d, c d.
2 2
y (h) ⫽ c1x (1)e3t ⫹ c2x (2)eⴚt, x (1) ⫽ x (2) ⫽
1 ⫺1
Answer:
y1 ⫽ 2eⴚt ⫹ t 2, y2 ⫽ ⫺eⴚt ⫺ t.
13. y1: ⫽ ⫺sin 2t ⫺ cos 2t ⫹ 2 sin t, y2: ⫽ 2 cos 2t ⫺ 2 sin 2t

14. The matrix of the homogeneous system

c d
0 4

⫺1 0
has the eigenvalues 2i and ⫺2i and eigenvectors [2 i]T and [2 ⫺i]T, respectively.
Hence a complex general solution is

c d e2it ⫹ c2 c d eⴚ2it.
2 2
y (h) ⫽ c1
i ⫺i
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By Euler’s formula this becomes, in components,

y 1(h) ⫽ (2c1 ⫹ 2c2) cos 2t ⫹ i(2c1 ⫺ 2c2) sin 2t
y 2(h) ⫽ (ic1 ⫺ ic2) cos 2t ⫹ (⫺c1 ⫺ c2) sin 2t.
Setting A ⫽ c1 ⫹ c2 and B ⫽ i(c1 ⫺ c2), we can write
y 1(h) ⫽ 2A cos 2t ⫹ 2B sin 2t
y 2(h) ⫽ B cos 2t ⫺ A sin 2t.
Before we can consider the initial conditions, we must determine a particular solution
y (p) of the given system. We do this by the method of undetermined coefficients, setting
y 1(p) ⫽ a1et ⫹ b1eⴚt
y 2(p) ⫽ a2et ⫹ b2eⴚt.
Differentiation and substitution gives
a1et ⫺ b1eⴚt ⫽ 4a2et ⫹ 4b2eⴚt ⫹ 5et
a2et ⫺ b 2eⴚt ⫽ ⫺a1et ⫺ b1eⴚt ⫺ 20eⴚt.
Equating the coefficients of et on both sides, we get
a1 ⫽ 4a2 ⫹ 5, a2 ⫽ ⫺a1, hence a1 ⫽ 1, a2 ⫽ ⫺1.
Equating the coefficients of e⫺t, we similarly obtain
⫺b1 ⫽ 4b2, ⫺b2 ⫽ ⫺b1 ⫺ 20, hence b1 ⫽ ⫺16, b2 ⫽ 4.
Hence a general solution of the given nonhomogeneous system is
y1 ⫽ 2A cos 2t ⫹ 2B sin 2t ⫹ et ⫺ 16eⴚt
y2 ⫽ B cos 2t ⫺ A sin 2t ⫺ et ⫹ 4eⴚt.
From this and the initial conditions we obtain
y1(0) ⫽ 2A ⫹ 1 ⫺16 ⫽ 1, y2(0) ⫽ B ⫺ 1 ⫹ 4 ⫽ 0.
The solution is A ⫽ 8, B ⫽ ⫺3. This gives the answer (the solution of the initial
value problem)
y1 ⫽ 16 cos 2t ⫺ 6 sin 2t ⫹ et ⫺ 16eⴚt
y2 ⫽ ⫺3 cos 2t ⫺ 8 sin 2t ⫺ et ⫹ 4eⴚt.

15. y1 ⫽ e2t ⫹ 4t ⫹ 4 ⫺ 6et, y2 ⫽ 23e2t ⫹ 9t ⫹ 1 ⫺ 6et ⫹ 73eⴚt.

18. The equations are
(a) I1r ⫽ ⫺2I1 ⫹ 2I2 ⫹ 440 sin t
and

冮
8I2 ⫹ 2 I2 dt ⫹ 2(I2 ⫺ I1) ⫽ 0.
Thus

冮
I2 ⫽ ⫺0.25 I2 dt ⫹ 0.25(I1 ⫺ I2),

which, upon differentiation and insertion of I1r from (a) and simplification, gives
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86 Instructor’s Manual

(b) I r2 ⫽ ⫺0.4I1 ⫹ 0.2I2 ⫹ 88 sin t.

The general solution of the homogeneous system is as in Prob. 17, and the method
of undetermined coefficients gives, as a particular solution,

c d cos t ⫹ c d sin t.
1 352 1 616
⫺
3 44 3 132

c d,
⫺3 1.25
20. A ⫽
1 ⫺1

c d c d c d.
125 1 ⴚ0.5t 125 5 500 1
J⫽⫺ e ⫺ eⴚ3.5t ⫹
3 2 21 ⫺2 7 1

SOLUTIONS TO CHAPTER 4 REVIEW QUESTIONS AND PROBLEMS,

page 164

11. y1 ⫽ c1eⴚ3t ⫹ c2e3t, y2 ⫽ ⫺3c1eⴚ3t ⫹ 3c2e3t. Saddle point.

12. The matrix

c d
2 0

0 1

has the eigenvalues 2 and 1 with eigenvalues [1, 0]T and [0, 1]T, respectively. Hence,
a general solution is

y ⫽ c1 c d e2t ⫹ c2 c d et.
1 0

0 1
Since p ⫽ 3, q ⫽ 2, ¢ ⫽ p ⫺ 4q ⫽ 9 ⫺ 8 ⫽ 1 ⬎ 0, the critical point is an unstable
2

node.
13. y1 ⫽ eⴚ2t(c1 sin t ⫹ c2 cos t), y2 ⫽ 15eⴚ2t(3c1 sin t ⫹ c1 cos t ⫹ 3c2 cos t ⫺ c2 sin t).
Asymptotically stable spiral point.
14. Eigenvalues ⫺1, 6. Eigenvectors [1 ⫺1]T, [4 3]T. The corresponding general
solution is
y1 ⫽ c1e ⴚt ⫹ 4c2e6t, y2 ⫽ ⫺c1e ⴚt ⫹ 3c2e6t.
The critical point at (0, 0) is a saddle point, which is always unstable.
16. Eigenvalues ⫺2i and 2i. y1 ⫽ c1 sin 2t ⫹ c2 cos 2t, y2 ⫽ c1 cos 2t ⫺ c2 sin 2t.
Center, which is always stable.
18. Saddle point; see Table 4.1 (b) in Sec. 4.4.
20. y1 ⫽ 3 cosh t ⫹ sinh t ⫹ c1t ⫹ c2, y2 ⫽ ⫺3 cosh t ⫹ sinh t ⫹ c1 ⫺ c1t ⫺ c2
21. y1 ⫽ c1eⴚ2t ⫹ c2e2t ⫺ 4 ⫺ 8t 2, y2 ⫽ ⫺c1eⴚ2t ⫹ c2e2t ⫺ 8t.
22. The matrix

c d
1 1

4 1
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has the eigenvalues 3 and ⫺1 with eigenvectors [1 2]T and [1 ⫺2]T, respectively.
A particular solution of the nonhomogeneous system is obtained by the method of
undetermined coefficients. We set

y 1(p) ⫽ a1 cos t ⫹ b1 sin t, y 2(p) ⫽ a2 cos t ⫹ b2 sin t.

Differentiation and substitution gives the following two equations, where c ⫽ cos t
and s ⫽ sin t.
(1) ⫺a1s ⫹ b1c ⫽ a1c ⫹ b1s ⫹ a2c ⫹ b2s ⫹ s
(2) ⫺a2s ⫹ b2c ⫽ 4a1c ⫹ 4b1s ⫹ a2c ⫹ b2s.
From the cosine terms and the sine terms in (1) we obtain

b1 ⫽ a1 ⫹ a2, ⫺a1 ⫽ b1 ⫹ b2 ⫹ 1.

Similarly from (2),

b 2 ⫽ 4a1 ⫹ a2, ⫺a2 ⫽ 4b1 ⫹ b2.

Hence a1 ⫽ ⫺0.3, a2 ⫽ 0.4, b1 ⫽ 0.1, b2 ⫽ ⫺0.8. This gives the answer

y1 ⫽ c1e3t ⫹ c2e ⴚt ⫺ 0.3 cos t ⫹ 0.1 sin t

y2 ⫽ 2c1e3t ⫺ 2c2e ⴚt ⫹ 0.4 cos t ⫺ 0.8 sin t.

24. The balance equations are

16 6
y 1r ⫽ ⫺ y1 ⫹ y2
400 200
16 16
y r2 ⫽ y1 ⫺ y2
400 200
Note that the denominators differ. Note further that the outflow to the right must be
included in the balance equation for T2. The matrix is
It has the eigenvalues ⫺0.1 and ⫺0.02 with the eigenvectors [⫺1, 2]T and [3, 2]T
respectively. The initial condition is y1(0) ⫽ 120, y2(0) ⫽ 0. This gives the solution

y1 ⫽ 90eⴚ0.02t ⫹ 30eⴚ0.10t
y2 ⫽ 60eⴚ0.02t ⫺ 60eⴚ0.10t

25. I 1r ⫹ 2.5(I1 ⫺ I2) ⫽ 169 sin 2t, 2.5(I r2 ⫺ I 1r ) ⫹ 25I2 ⫽ 0,

I1(t) ⫽ (31.8 ⫹ 58.3t)e ⴚ5.0t ⫺ 31.8 cos (2.0t) ⫹ 50.2 sin (2.0t),
I2(t) ⫽ (⫺8.44 ⫺ 58.3t)e ⴚ5.0t ⫹ 8.04 sin (2.0t) ⫹ 8.44 cos (2.0t)
26. The ODEs of the system are

I1r ⫺ I2r ⫹ 5I1 ⫽ 0

I2 ⫺ I1 ⫹ 1.25I2r ⫽ 0,

written in the usual form

c dI
A⫺B ⫺A
Ir ⫽
A ⫺A
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where A ⫽ R>L ⫽ 0.8 and B ⫽ 1>(RC ) ⫽ 5. A general solution is

I ⫽ c1 c d eⴚt ⫹ c2 c d eⴚ4t.
1 4

⫺4 ⫺1
The initial conditions give c1 ⫽ ⫺13 and c2 ⫽ 13. Hence the particular solution
satisfying the initial conditions is
I1 ⫽ ⫺13eⴚt ⫹ 43eⴚ4t
I2 ⫽ 43eⴚt ⫺ 13eⴚ4t.

28. cos y2 ⫽ 0 when y2 ⫽ (2n ⫹ 1)p>2, where n is any integer. This gives the location
of the critical points, which lie on the y2-axis y1 ⫽ 0 in the phase plane.
For (0, 12p) the transformation is y2 ⫽ 12p ⫹ 苲 y 2. Now
cos y2 ⫽ cos (12p ⫹ 苲
y 2) ⫽ ⫺sin 苲 苲2.
y 2 ⬇ ⫺y
Hence the linearized system is
苲
y 1r ⫽ ⫺y苲2
苲
y 2r ⫽ 3y苲1.

For its matrix

c d
0 ⫺1

3 0
we obtain p苲 ⫽ 0, q苲 ⫽ 3 ⬎ 0, so that this is a center. Similarly, by periodicity, the
critical points at (4n ⫹ 1)p>2 are centers.
苲
For (0, ⫺12p) the transformation is y2 ⫽ ⫺12p ⫹ 苲y 2. From
cos y ⫽ cos (⫺1p ⫹ 苲
2 2 y ) ⫽ sin 苲y ⬇苲
2 y2 2

we obtain the linearized system

苲
y 1r ⫽ 苲
y2
苲 苲
y 2r ⫽ 3y 1

with matrix

c d
0 1

3 0
for which q苲 ⫽ ⫺3 ⬍ 0, so that this point and the points with y2 ⫽ (4n ⫺ 1)p>2 on
the y2-axis are saddle points.
29. (n2p, 0), Center when n is odd and saddle point when n is even.
30. Critical points at (0, 0) and (0, ⫺1). The linearized systems are:
y 1r ⫽ y2
y r2 ⫽ ⫺2y1
and
y 1r ⫽ ⫺y2
y r2 ⫽ ⫺2y1