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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 ϫ 2 matrices. Accordingly, Chap. 4 makes modest use of 2 ϫ 2 matrices. n ϫ 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 ϫ 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 ϫ 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 ϫ 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 Ϫ l a12 ϭ (a11 Ϫ l)(a22 Ϫ l) Ϫ a12a21 a21 a22 Ϫ l 2 ϭ l Ϫ (a11 ϩ a22)l ϩ a11a22 Ϫ a12a21 ϭ 0, 2 2 and then an eigenvector corresponding to l1 with components x 1, x 2 from

(a11 Ϫ l1)x 1 ϩ a12x 2 ϭ 0

and an eigenvector corresponding to l2 from

(a11 Ϫ l2)x 1 ϩ a12x 2 ϭ 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 ϫ 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 ϫ 2 matrices, which play the major role in Chap. 4; indeed, n ϫ 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 ϫ 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. c04.qxd 6/18/11 3:06 PM Page 66

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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 ϭ 0.02y2 Ϫ 0.01y1

yr2 ϭ 0.01y1 Ϫ 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

y1r ϭϪ0.01y1 ϩ 0.02y2

yr2 ϭ 0.01y1 Ϫ 0.02y2. As a single vector equation,

Ϫ0.01 0.02 yr ϭ Ay, where A ϭ . c 0.01 Ϫ0.02 d

A has the eigenvalues l1 ϭ 0 and l2 ϭϪ0.03 and corresponding eigenvectors 1 1 x(1) ϭ , x(2) ϭ , c 0.5 d c Ϫ1 d respectively. The corresponding general solution is ؊0.03t (2) (1) y ϭ c1x ϩ c2x e . From the initial values, 1 1 0 y(0) ϭ c1 ϩ c2 ϭ . c 0.5 d c Ϫ1 d c 150 d

In components this is c1 ϩ c2 ϭ 0, 0.5c1 Ϫ c2 ϭ 150 . Hence c1 ϭ 100, c2 ϭϪ100 . This gives the solution 1 1 .y ϭ 100 Ϫ 100 e؊0.03t c 0.5 d c Ϫ1 d In components, ؊0.03t y1 ϭ 100(1 Ϫ e ) ϭ 1 ϩ ؊0.03t y2 100(2 e ). c04.qxd 6/18/11 3:06 PM Page 67

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4. With Flow rate a ϭ Tank size we can write the system that models the process in the following form:

y1r ϭ ay2 Ϫ ay1

yr2 ϭ ay1 Ϫ ay2, ordered as needed for the proper vector form

y1r ϭϪay1 ϩ ay2

yr2 ϭ ay1 Ϫ ay2. In vector form, Ϫaa yr ϭ Ay, whereA ϭ . c a Ϫa d The characteristic equation is (l ϩ a)2 Ϫ a2 ϭ l2 ϩ 2al ϭ 0. Hence the eigenvalues are 0 and Ϫ2a . Corresponding eigenvectors are 1 1 and c 1 d c Ϫ1 d respectively. The corresponding “general solution” is 1 1 ؊2at y ϭ c1 ϩ c2 e . c 1 d c Ϫ1 d 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

Ϫ0.02 0.02 0

A ϭ 0.02 Ϫ0.04 0.02 . 0 0.02 Ϫ0.02

The characteristic polynomial isD T l3 ϩ 0.08l2 ϩ 0.0012l ϭ l(l ϩ 0.02)(l ϩ 0.06). This gives the eigenvalues and corresponding eigenvectors

1 1 1

(1) (2) (3) l1 ϭ 0, x ϭ 1 , l2 ϭϪ0.02, x ϭ 0 , l3 ϭϪ0.06, x ϭ Ϫ2 .

1 Ϫ1 1

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Hence a “general solution” is 1 1 1

؊0.02t ؊0.06t ϭ ϩ ϩ Ϫ y c1 1 c2 0 e c3 2 e .

1 Ϫ1 1

We use quotation marksD sinceT theD conceptT of a generalD solutionT 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

Ir2 ϭ 0.4I1r Ϫ 0.54I2. Substitution of the first ODE into the new second one, as in the text, gives

I2r ϭϪ1.6I1 ϩ 1.06I2 ϩ 4.8. Hence the matrix of the new system is

Ϫ44 A ϭ . c Ϫ1.6 1.06 d Its eigenvalues are Ϫ1.5 and Ϫ1.44 . The corresponding eigenvectors are x(1) ϭ [1 0.625]T and x(2) ϭ [1 0.64]T, respectively. The corresponding general solution is

؊1.5t (2) ؊1.44t (1) y ϭ c1x e ϩ c2x e .

10. The system is

y1r ϭ y2

yr2 ϭϪ3y1 Ϫ 4y2 The matrix has eigenvalues Ϫ1 and Ϫ3 and corresponding eigenvectors [1, Ϫ1]T and T ؊t ؊3t [1, Ϫ3] , respectively. From this y ϭ c1e ϩ c2e . r ϭ r ϭ ϩ 3 ϭ T 2t ϩ Ϫ T ؊t/2 11. y1 y2, y2 y1 2y2, y c1[1/2,1] e [ 2,1] e 12. The system is

yr1 ϭ y2

yr2 ϭ y3

y3 ϭϪ2y1 ϩ y2 ϩ 2y3 The eigenvalues of its matrix are Ϫ1, 1, 2. Eigenvalues are [1, Ϫ1, 1]T, [1, 1, 1]T, [1, 2, 4]T, respectively. The corresponding general solution is

T ؊t T t T 2t y ϭ c1[1, Ϫ1, 1] e ϩ c2[1, 1, 1] e ϩ c3[1, 2, 4] e .

؊4t 3t 13. yr1 ϭ y2, yr2 ϭ 12y1 Ϫ y2, y1 ϭ c1e ϩ c2e , y2 ϭ yr1 14. TEAM PROJECT. (a) From Sec. 2.5 we know that the undamped motions of a mass on an elastic spring are governed by mys ϩ ky ϭ 0 or mys ϭϪky c04.qxd 6/18/11 3:06 PM Page 69

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where y ϭ 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 ys ϭϪk y ϩ k ( y Ϫ y ) (11) 1 1 1 1 2 2 1 m 2ys2 ϭϪk 2( y2 Ϫ y1)

for the unknown displacements y1 ϭ y1(t) of the first mass m 1 and y2 ϭ 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 ϭ m 2 ϭ 1 , k 1 ϭ 3 , and k 2 ϭ 2 in Fig. 81 so that by ordering (11) we obtain

y1s ϭϪ5y1 ϩ 2y2

ys2 ϭ 2y1 Ϫ 2y2 or, written as a single vector equation,

ys1 Ϫ52y1 ys ϭ ϭ . c ys2 d c 2 Ϫ2 dcy2 d (b) As for a single equation, we try an exponential function of t, y ϭ xevt. Thenys ϭ v2xevt ϭ Axevt . Then, writing v2 ϭ l and dividing by evt , we get Ax ϭ lx. Eigenvalues and eigenvectors are 1 2 (1) (2) l1 ϭϪ1, x ϭ and l2 ϭϪ6 , x ϭ . c 2 d c Ϫ1 d Since v ϭϮ2l and 2Ϫ1 ϭ i and 2Ϫ6 ϭ i26 , we get it ؊it (2) i26t ؊i26t (1) y ϭ x (c1e ϩ c2e ) ϩ x (c3e ϩ c4e )

or, by (7) in Sec. 2.3,

(1) (1) (2) (2) y ϭ a1x cos t ϩ b1x sin t ϩ a2x cos 26t ϩ b2x sin 26t

where a1 ϭ c1 ϩ c2, b1 ϭ i(c1 Ϫ c2), a2 ϭ c3 ϩ c4, b2 ϭ i(c3 Ϫ c4) . These four arbitrary constants can be specified by four initial conditions. In components, this solution is

y1 ϭ a1 cos t ϩ b1 sin t ϩ 2a2 cos 26t ϩ 2b2 sin 26t

y2 ϭ 2a1 cos t ϩ 2b1 sin t Ϫ a2 cos 26t Ϫ 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 ϭ 1 and the three other con- stants are zero, we get y1 ϭ cos t, y2 ϭ 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 ϭ 1 and the other constants are zero, we c04.qxd 6/18/11 3:06 PM Page 70

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have y1 ϭ 2 cos 26t, y2 ϭϪcos 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 ϭ 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 c04.qxd 6/18/11 3:06 PM Page 71

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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 ϭ yr1 ϩ y1 . By differentiation and from the second ODE as well as from (a),

ys1 ϭϪyr1 ϩ yr2 ϭϪyr1 Ϫ y1 Ϫ (yr1 ϩ y1) ϭϪ2yr1 Ϫ 2y1. Complex solutions e(؊1Ϯi)t give the real solution ؊t y1 ϭ e (A cos t ϩ 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 ؊t t ؊t t 1. y1 ϭ c1e ϩ c2e , y2 ϭ 3c1e ϩ c2e 2. The eigenvalues are 3 and 9. Eigenvectors are [3 Ϫ1]T and [3 1]T , respectively. The corresponding general solution is 3t 9t y1 ϭ 3c1e ϩ 3c2e 3t 9t y2 ϭϪc1e ϩ c2e . ϭ ϩ t ϭ t ϩ 4 3. y1 c1 c2e , y2 2c2e 3c1 4. The matrix has the double eigenvalue Ϫ6 . An eigenvector is [1 Ϫ1]T . Hence the vector u needed is obtained from Ϫ2 Ϫ2 1 (A ϩ 6I) u ϭ u ϭ . c 22d c Ϫ1d We can take u ϭ 0, Ϫ1 . With this we obtain as a general solution 3 24 ؊6t ؊6t y1 ϭ c1e ϩ c2te ϭϪ ؊6t Ϫ ϩ 1 ؊6t y2 c1e c2(t 2)e . c04.qxd 6/18/11 3:06 PM Page 72

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6. The eigenvalues are complex, 2 ϩ 2i and 2 Ϫ 2i . Corresponding complex eigenvectors are [1 Ϫi]T and [1 i]T , respectively. Hence a complex general solution is (2ϩ2i)t (2–2i)t y1 ϭ c1e ϩ c2e (2ϩ2i)t (2–2i)t y2 ϭϪic1e ϩ ic2e . From this and the Euler formula we obtain a real general solution y ϭ e2t (c ϩ c ) cos 2t ϩ i(c Ϫ c ) sin 2t 1 3 1 2 1 2 4 ϭ e2t (A cos 2t ϩ B sin 2t), y ϭ e2t (Ϫic ϩ ic ) cos 2t ϩ i(Ϫic Ϫ ic ) sin 2t 2 3 1 2 1 2 4 ϭ e2t(ϪB cos 2t ϩ A sin 2t)

where A ϭ c1 ϩ c2 and B ϭ i(c1 Ϫ c2) .

7. y1 ϭϪ(1/2)c2 22 cos(22ax) ϩ (1/2)c3 sin(22ax)22 ϩ c1,

y2 ϭ c2 sin(22ax) ϩ c3 cos(22ax),

y3 ϭ (1/2)c2 22 cos(22ax) Ϫ (1/2)c3 sin(22ax)22 ϩ c1. 8. The eigenvalue of algebraic multiplicity 2 is 9. An eigenvector is [1 Ϫ1]T. There is no basis of eigenvectors. A first solution is

1 y(1) ϭ e9t. c Ϫ1 d

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

1 u1 y(2) ϭ te9t ϩ e9t c Ϫ1 d c u2 d

with u u T determined from 3 1 24

Ϫ1 Ϫ1 u1 1 (A Ϫ 9I) u ϭ ϭ . c 11dcu2 d c Ϫ1 d

Thus u1 ϩ u2 ϭϪ1 . We can take u1 ϭ 0, u2 ϭϪ1 . This gives the general solution

1 0 (1) (2) 9t 9t y ϭ c1y ϩ c2y ϭ (c1 ϩ c2t) e ϩ c2 e . c Ϫ1 d c Ϫ1 d

10. The eigenvalues are 1 and Ϫ3. The corresponding general solution is t ؊3t t ؊3t y1 ϭ c1e ϩ 5c2e and y2 ϭ c1e ϩ c2e Using the initial conditions we obtain

y1(0) ϭ c1 ϩ 5c2 ϭ 0 and y2(0) ϭ c1 ϩ c2 ϭ 4 Hence the solution of the IVP is t ؊3t y1 ϭ 5e Ϫ 5e t ؊3t y2 ϭ 5e Ϫ e c04.qxd 6/18/11 3:06 PM Page 73

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ϭ 8 t Ϫ 18 ؊t/4 ϭ 8 t Ϫ 8 ؊t/4 11. y1 5e 5 e and y2 5e 5e 12. The eigenvalues are 0 and 2. The corresponding general solution is ϭ ϩ 2t y1 c1 c2e ϭϪ1 ϩ 1 2t y2 3 c1 3 c2e . From this and the initial values we obtain

y1(0) ϭ c1 ϩ c2 ϭ 12 ϭϪ1 ϩ 1 ϭ y2(0) 3 c1 3 c2 2. Hence the solution of the IVP is

2t y1 ϭ 3 ϩ 9e 2t y2 ϭϪ1 ϩ 3e .

13. y1 ϭ sinh 2x, y2 ϭ cosh 2x 14. The eigenvalues are Ϫ1 ϩ i and Ϫ1 Ϫ i . The corresponding real general solution is

؊t y1 ϭ e (c1 cos t ϩ c2 sin t) ؊t y2 ϭ e (Ϫc2 cos t ϩ c1 sin t). From this and the initial conditions we obtain the solutions of the IVP

؊t y1 ϭ e cos t ؊t y2 ϭ e sin t.

؊t ؊t 15. y1 ϭ 0.25e , y2 ϭϪ0.25e 16. The system is

(a) y1r ϭ 8y1 Ϫ y2

(b) yr2 ϭ y1 ϩ 10y2.

10(a) ϩ (b) gives 10yr1 ϩ yr2 ϭ 81y1; hence

(c) yr2 ϭϪ10yr1 ϩ 81y1. Differentiating (a) and using (c) gives

0 ϭ ys1 Ϫ 8yr1 ϩ yr2

ϭ ys1 Ϫ 8yr1 Ϫ 10yr1 ϩ 81y1

ϭ ys1 Ϫ 18yr1 ϩ 81y1. A general solution is

9t y1 ϭ (c1 ϩ c2t)e . From this and (a) we obtain y ϭϪyr ϩ 8y ϭ Ϫc Ϫ 9(c ϩ c t) ϩ 8(c ϩ c t) e9t 2 1 1 3 2 1 2 1 2 4 9t ϭ (Ϫc2 Ϫ c1 Ϫ c2t)e , in agreement with the answer to Prob. 8. c04.qxd 6/18/11 3:06 PM Page 74

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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 ϭ Inflow Ϫ Outflow.

For Tank T1 this is (see Fig. 88) 4 16 . yr ϭ 12 ؒ 0 ϩ y Ϫ y 1 a 200 2b 200 1

For Tank T2 it is 16 4 ϩ 12 yr ϭ y Ϫ y . 2 200 1 200 2 Performing the divisions and ordering terms, we have

yr1 ϭϪ0.08y1 ϩ 0.02y2

yr2 ϭ 0.08y1 Ϫ 0.08y2. The eigenvalues of the matrix of this system are Ϫ0.04 and Ϫ0.12. Eigenvectors are [1 2]T and [1 Ϫ2]T, respectively. The corresponding general solution is 1 1 ؊0.04t ؊0.12t y ϭ c1 e ϩ c2 e . c 2 d c Ϫ2 d

The initial conditions are y1(0) ϭ 100, y2(0) ϭ 200. This gives c1 ϭ 100, c2 ϭ 0. In components the answer is

؊0.04t y1 ϭ 100e ؊0.04t y2 ϭ 200e . Both functions approach zero as t : ϱ, 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. c04.qxd 6/18/11 3:06 PM Page 75

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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 Ͻ 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 t 2t 1. Unstable improper node, y1 ϭ c1e , y2 ϭ e . 2. p ϭϪ7, q ϭ 12 Ͼ 0, ¢ ϭ 49 Ϫ 48 Ͼ 0, stable and attractive improper node, Ϫ4t ؊3t y1 ϭ c1e , y2 ϭ c2e

3. Center, always stable, y1 ϭ c1 sin 2t ϩ c2 cos 2t, y2 ϭ 2c1 cos 2t Ϫ 2c2 sin 2t 4. p ϭ 0, q ϭϪ9, saddle point, always unstable. A general solution is

؊3t 3t y1 ϭ c1e ϩ c2e ؊3t 3t y2 ϭϪ5c1e ϩ c2e .

؊t ؊t 5. Stable spiral, y1 ϭ e (c1 sin t ϩ c2 cos t), y2 ϭϪe (Ϫc1 cos t ϩ c2 sin t) 6. p ϭϪ12, q ϭ 27, ¢ ϭ 144 Ϫ108 Ͼ 0, stable and arrtactive node. A general solution is

؊3t ؊9t y1 ϭ c1e ϩ c2e ؊3t ؊9t y2 ϭϪ3c1e ϩ 3c2e .

؊2t 3t ؊2t 3t 7. Saddle point, always unstable, y1 ϭ c1e ϩ c2e , y2 ϭ c1e Ϫ 4c2e . 8. p ϭϪ3, q ϭϪ10, saddle point, always unstable. A general solution is

؊5t 2t y1 ϭ c1e ϩ 4c2e ؊5t 2t y2 ϭϪc1e ϩ 3c2e .

ϭ 2t ϩ 3t ϭϪ4 2t Ϫ 3t 9. Unstable node, y1 c1e c2e , y2 3c1e c2e . c04.qxd 6/18/11 3:06 PM Page 76

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10. p ϭϪ2, q ϭ 5, ¢ ϭ 4 Ϫ 20 Ͻ 0, stable and attractive spiral. The components of a general solution are

؊1ϩ2i)t (؊1؊2i)t) y1 ϭ c1e ϩ c2e ؊t ϭ e ((c1 ϩ c2) cos 2t ϩ i(c1 Ϫ c2) sin 2t) ,(ϭ e؊t(A cos 2t ϩ B sin 2t ؊1ϩ2i)t (؊1؊2i)t) y2 ϭ (Ϫ1 ϩ 2i)c1e ϩ (Ϫ1 Ϫ 2i)c2e ؊t ϭ e ((Ϫc1 Ϫ c2 ϩ 2ic1 Ϫ 2ic2) cos 2t ϩ i(2ic1 ϩ 2ic2 Ϫ c1 ϩ c2) sin 2t) (ϭ e؊t((ϪA ϩ 2B) cos 2t Ϫ (2A ϩ B) sin 2t

where A ϭ c1 ϩ c2 and B ϭ i(c1 Ϫ c2). ؊t 11. y ϭ e (c1 sin 2t ϩ c2 cos 2t). Stable and attractive spirals. ϭ 1 ϩ 1 12. y A cos 3t B sin 3t. The trajectories are the ellipses 1 2 ϩ 2 ϭ 9 y1 y2 const. This is obtained as in Example 4 in Sec. 4.3. 14. yr ϭϪdy dt, yr ϭϪdy dt, reversal of the direction of motion; to get the usual 1 1> 2 2> form, we have to multiply the transformed system by Ϫ1, which amounts to multiplying the matrix by Ϫ1, changing p into Ϫp, but leaving q and ¢ unchanged. In the example, we get an unstable node.

16. At a center, p ϭ a11 ϩ a22 ϭ 0, q ϭ det A Ͼ 0, hence ¢ Ͻ 0. Under the change, p changes into a11 ϩ k ϩ a22 ϩ k ϭ 2k 0 ; q remains positive because

2 (a11 ϩ k)(a22 ϩ k) Ϫ a12a21 ϭ q ϩ k Ͼ 0. Finally, ¢ remains unchanged because

(p ϩ 2k)2 Ϫ 4(q ϩ k 2) ϭ (2k)2 Ϫ 4(q ϩ k 2) ϭϪ4q Ͻ 0.

Hence we obtain a spiral point, which is unstable if k Ͼ 0 and stable and attractive if k Ͻ 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 ϩ k of A, causing a damped oscillation (when k Ͻ 0 ) or an increasing one (when k Ͼ 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 c04.qxd 6/18/11 3:06 PM Page 77

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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

yr1 ϭ y1(4 Ϫ y1)

yr2 ϭ y2 we see that the critical points are (0, 0) and (4, 0). For (0, 0) the linearized system is

yr1 ϭ 4y1

yr2 ϭ y2. The matrix is 40 . c 01d c04.qxd 6/18/11 3:06 PM Page 78

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Hence p ϭ 5, q ϭ 4, ¢ ϭ 25 Ϫ 16 ϭ 9. This shows that the critical point at (0, 0) is an unstable node. For (4, 0) the translation to the origin is ෂ ෂ y1 ϭ 4 ϩ y1, y2 ϭ y2. This gives the transformed system ෂ ෂ ෂ y1r ϭ (4 ϩ y1)(Ϫy1) ෂ ෂ y2r ϭ y2 and the corresponding linearized system ෂ ෂ y1r ϭϪ4y1 ෂ ෂ yr2 ϭ y2. Its matrix is Ϫ40 . c 01d ෂ ෂ Hence p ϭϪ3, q ϭϪ4. 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 ϭ 4 ϩ y1. Then yr2 ϭ y1.

6. At (0, 0), y1r ϭ y2, yr2 ϭϪy1, p ϭ 0, q ϭ 1, ¢ ϭϪ4, center. The other critical point ෂ ෂ 2 ෂ ෂ ෂ is at (Ϫ1, 0). We set y ϭϪ1 ϩ y , y ϭ y . Then Ϫy – y Ϸ y . Hence y1r ϭ y2, ෂr ෂ 1 1 2 2 1 1 1 y2 ϭ y1. This gives a saddle point. 1 7. The point (0, 0), y1r ϭϪy1 ϩ y2, y2r ϭϪy1 Ϫ y2, is stable and an attractive spiral 2 ~ ~ ~ point. The point (Ϫ1, 2) is a saddle point, y1 ϭϪ1 ϩ y1, y2 ϭ 2 ϩ y2, y1r ϭ Ϫ ~ Ϫ ~ ~r ϭϪ~ Ϫ 1~ 2y1 3y2, y 2 y1 2y2. 8. The system may be written

y1r ϭ y2(1 Ϫ y2)

yr2 ϭ y1(1 Ϫ 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

y1r ϭ y2

yr2 ϭ y1. The matrix is 01 . c 10d Hence p ϭ 0, q ϭϪ1, so that we have a saddle point. ෂ ෂ At (0, 1) the transformation is y1 ϭ y1, y2 ϭ 1 ϩ y2. This gives the transformed system ෂr ෂ ෂ y 1 ϭ (1 ϩ y2)(Ϫy2) ෂr ෂ ෂ y 2 ϭ y1(1 Ϫ y1). c04.qxd 6/18/11 3:06 PM Page 79

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Linearization gives ෂr ෂ y1 ϭϪy2 0 Ϫ1 with matrix ෂr ෂ y2 ϭ y1 c 10d ෂ ෂ ෂ for which p ϭ 0, q ϭ 1, ¢ ϭϪ4, and we have a center. ෂ ෂ At (1, 0) the transformation is y1 ϭ 1 ϩ y1, y2 ϭ y2. The transformed system is ෂr ෂ ෂ y1 ϭ y2(1 Ϫ y2) ෂr ෂ ෂ y 2 ϭ (1 ϩ y1)(Ϫy1). Its linearization is ෂr ෂ y1 ϭ y2 01 with matrix ෂr ෂ y2 ϭϪy1 c Ϫ10d ෂ ෂ ෂ for which p ϭ 0, q ϭ 1, ¢ ϭϪ4, so that we get another center. At (1, 1) the transformation is ෂ ෂ y1 ϭ 1 ϩ y1, y2 ϭ 1 ϩ y2. The transformed system is ෂ ෂ ෂ y1r ϭ (1 ϩ y2)(Ϫy2) ෂr ෂ ෂ y2 ϭ (1 ϩ y1)(Ϫy1). Linearization gives ෂr ෂ y1 ϭϪy2 0 Ϫ1 with matrix ෂr ෂ y2 ϭϪy1 c Ϫ10d ෂ ෂ for which p ϭ 0, q ϭϪ1, and we have another saddle point. 9. (0, 0) saddle point, (Ϫ2, 0) and (2, 0) centres. 10. ys ϩ y Ϫ y3 ϭ 0 written as a system is

y1r ϭ y2 3 y2r ϭϪy1 ϩ y1. 3 2 Now Ϫy1 ϩ y1 ϭ y1(Ϫ1 ϩ y1) ϭ 0 shows that there are three critical points, at (y1, y2) ϭ (0, 0), (Ϫ1, 0), and (1, 0). The linearized system at (0, 0) is

y1r ϭ y2 01 Matrix: . y2r ϭϪy1. c Ϫ10d

From the matrix we see that p ϭ a11 ϩ a22 ϭ 0, q ϭ 1. Hence (0, 0) is a center (see Sec. 3.4). For the next critical point we have to linearize at (Ϫ1, 0) by setting ෂ ෂ y1 ϭϪ1 ϩ y1, y2 ϭ y2. c04.qxd 6/18/11 3:06 PM Page 80

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Then

2 ෂ ෂ 2 y (Ϫ1 ϩ y ϭ (Ϫ1 ϩ y ) Ϫ1 ϩ (Ϫ1 ϩ y ) 1 1 1 3 1 4 ෂ ෂ ෂ2 ෂ ϭ (Ϫ1 ϩ y ) Ϫ2y ϩ y Ϸ 2y . 1 3 1 14 1 Hence the linearized system is ෂr ϭ ෂ y1 y2 01 Matrix: . ෂr ϭ ෂ y2 2y1. c 20d Hence q ϭ det A ϭϪ2 Ͻ 0, that is, the critical point at (Ϫ1, 0) is a saddle point. Similarly, to linearize at (1, 0), set ෂ ෂ y1 ϭ 1 ϩ y1, y2 ϭ y2. Then

3 ෂ Ϫy1 ϩ y1 Ϸ 2y1 and we obtain another saddle point, as just before. p Ϯ Ϫp Ϯ p 11. (4 np, 0) saddle points; ( 4 np, 0) centres. For the critical point (4 , 0) , after r ϭ p ϩ ~ r ϭ ~ linearizing and using the substitutions y1 4 y1, and y2 y2, we get ~ yr1 ϭ yr1 r ϭϪ p ϩ ~r y 2 cos 2(4 y 1) ~ ϭ sin 2y1 ~ ϭ 2y1 ϭ Ϫ 4 3 ϩ Á where we have used the fact 2y 2y 3y .

12. y1r ϭ y2, y2r ϭ y1(Ϫ9 Ϫ y1). (0, 0) is a critical point. The linearized system at (0, 0) is

y1r ϭ y2 01 with matrix yr2 ϭϪ9y1 c Ϫ90d for which p ϭ 0 and q ϭ 9 Ͼ 0, so that we have a center. A second critical point is at (Ϫ9, 0). The transformation is ෂ ෂ y1 ϭϪ9 ϩ y 1, y2 ϭ y 2. This gives the transformed system ෂr ϭ ෂ y1 y2 ෂr ϭ Ϫ ϩ ෂ Ϫෂ y2 ( 9 y1)( y1). Its linearization is ෂr ϭ ෂ y1 y2 01 with matrix ෂr ϭ ෂ y 2 9y1 c 90d ෂ for which q ϭϪ9 Ͻ 0, so that we have a saddle point. ~ 13. (Ϯ2np, 0) centres; y1 ϭ (2n ϩ 1)p ϩ y1, (p Ϯ 2np, 0) saddle points. c04.qxd 6/18/11 3:06 PM Page 81

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14. TEAM PROJECT. (a) Unstable node if ␮ м 2, unstable spiral point if 2 Ͼ ␮ Ͼ 0, center if ␮ ϭ 0, stable and attractive spiral point if 0 Ͼ ␮ ϾϪ2, stable and attractive node if ␮ ϹϪ2. (c) As a system we obtain

(A) yr1 ϭ y2 2 3 (B) yr2 ϭϪ(v0y1 ϩ by1). 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):

2 3 yr2y2 ϭϪ(v0y1 ϩ by1)yr1.

Integration on both sides and multiplication by 2 gives

2 ϩ 2 2 ϩ 1 4 ϭ y2 v0y1 2 by1 const.

For positive b these curves are closed because then by3 is a proper restoring term, adding to the restoring due to the y-term. If b is negative, the term by3 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 ,

؊2t y1s ϭϪ3y1r ϩ y2r ϩ 12e ؊2t ؊2t ϭϪ3y1r ϩ ( y1 Ϫ 3y2 ϩ 2e ) ϩ 12e ؊2t ؊2t ϭϪ3y1r ϩ y1 Ϫ 3( y1r ϩ 3y1 ϩ 6e ) ϩ 14e ؊2t ϭϪ6yr1 Ϫ 8y1 Ϫ 4e ,

solving this for y1 and then getting y2 from the solution. c04.qxd 6/18/11 3:06 PM Page 82

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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 Ϫ2. Eigenvectors are [1 1]T and [1 Ϫ3]T, respectively. Hence a general solution of the homogeneous system is

1 1 h) 2t ؊2t) y ϭ c1 e ϩ c2 e . c 1 d c Ϫ3 d We determine y(p) by the method of undetermined coefficients, starting from

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

Ϫa1 sin t ϩ b1 cos t ϭ (a1 ϩ a2) cos t ϩ (b1 ϩ b2) sin t ϩ 10 cos t

Ϫa2 sin t ϩ b2 cos t ϭ (3a1 Ϫ a2) cos t ϩ (3b1 Ϫ b2) sin t Ϫ 10 sin t. By equating the coefficients of the cosine and of the sine in the first of these two equations we obtain

b1 ϭ a1 ϩ a2 ϩ 10, Ϫa1 ϭ b1 ϩ b2. Similarly from the second equation,

b2 ϭ 3a1 Ϫ a2, Ϫa2 ϭ 3b1 Ϫ b2 Ϫ 10.

The solution is a1 ϭϪ2, b1 ϭ 4, a2 ϭϪ4, b2 ϭϪ2. This gives the answer

2t ؊2t y1 ϭ c1e ϩ c2e Ϫ 2 cos t ϩ 4 sin t 2t ؊2t y2 ϭ c1e Ϫ 3c2e Ϫ 4 cos t Ϫ 2 sin t. ϭ t ϩ ؊t ϭ t Ϫ ؊t Ϫ 5 2t 3. y1 c1e c2e , y2 c1e c2e 3e 4. From the characteristic equation,

4 1 (1) (2) l1 ϭ 2, x ϭ , l2 ϭϪ4, x ϭ . c 1 d c 1 d y(p) can be obtained by the method of undetermined coefficients, starting from

y(p) ϭ a cosh t ϩ b sinh t. c04.qxd 6/18/11 3:06 PM Page 83

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Substitution gives y(p)r ϭ a sinh t ϩ b cosh t 4 Ϫ8 2 0 ϭ (a cosh t ϩ b sinh t) ϩ cosh t ϩ sinh t. c 2 Ϫ6 d c 1 d c 2 d Comparing sinh terms and cosh terms (componentwise), we get from this ϭ a1 4b Ϫ 8b 1 2 sinh terms a2 ϭ 2b1 Ϫ 6b2 ϩ 2 f b ϭ 4a Ϫ 8a ϩ 2 1 1 2 cosh terms. b2 ϭ 2a1 Ϫ 6a2 ϩ 1 f To solve this, one can substitute the first two equations into the last two, solve for b1 ϭ 2, b2 ϭ 1, and then get from the first two equations a1 ϭ a2 ϭ 0. This gives the general solution

2t ؊4t 2t ؊4t y1 ϭ 4c1e ϩ c2e ϩ 2 sinh t, y2 ϭ c1e ϩ c2e ϩ sinh t. ϭϪ13 Ϫ 5 ϩ 2t ϩ t ϭ 7 Ϫ 2 2t Ϫ t ϩ 5. y1 4 2t c1e c2e , y2 2 3c1e c2e 3t 6. From the characteristic equation we obtain 1 1 (1) (2) l1 ϭ 4, x ϭ , l2 ϭϪ4, x ϭ . c 1 d c Ϫ1 d By the method of undetermined coefficients, we set y(p) ϭ u ϩ vt ϩ wt 2. By substitution, 04 0 0 y(p)r ϭ v ϩ 2wt ϭ (u ϩ vt ϩ wt 2) ϩ ϩ t 2. c 40d c 2 d c Ϫ16 d We now compare componentwise the constant terms, linear terms, and quadratic terms: v ϭ 4u 1 2 constant terms v2 ϭ 4u1 ϩ 2 f

2w1 ϭ 4v2 linear terms 2w2 ϭ 4v1 f

0 ϭ 4w2 quadratic terms. 0 ϭ 4w1 Ϫ 16 f We then obtain, in this order, ϭ ϭ ϭ ϭ ϭ 1 Ϫ ϭ ϭ w1 4, w2 0, v1 0, v2 2, u1 4 (v2 2) 0, u2 0. The corresponding general solution is

4t (2) ؊4t 2 (1) y ϭ c1x e ϩ c2x e ϩ vt ϩ wt ; in components,

4t ؊4t 2 4t ؊4t y1 ϭ c1e ϩ c2e ϩ 4t , y2 ϭ c1e Ϫ c2e ϩ 2t. c04.qxd 6/18/11 3:06 PM Page 84

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ϭ ؊2t ϩ 3t ϩ 3 ؊t ϩ 1 Ϫ 1 7. y1 c1e c2e 4e 6t 36 ϭ ؊2t Ϫ 3t ϩ 47 ϩ 175 y2 c1e 4c2e 6 t 36 1 4 (1) t (2) 2t (1) (2) t 10. y ϭ c1x e ϩ c2x e , x ϭ , x ϭ . Now e on the right is a c Ϫ1 d c Ϫ5 d solution of the homogeneous system. Hence, to find y(p), we have to proceed as in Example 1, setting y(p) ϭ utet ϩ vet. Substitution gives

Ϫ3 Ϫ4 5 y(p)r ϭ u(t ϩ 1)et ϩ vet ϭ (utet ϩ vet) ϩ et. c 56d c Ϫ6 d Equating the terms in et (componentwise) gives

u1 ϩ v1 ϭϪ3v1 Ϫ 4v2 ϩ 5

u2 ϩ v2 ϭ 5v1 ϩ 6v2 Ϫ 6 and the terms in tet give

u1 ϭϪ3u1 Ϫ 4u2

u2 ϭ 5u1 ϩ 6u2.

Hence u1 ϭ 1, u2 ϭϪ1, v1 ϭ 1, v2 ϭ 0. This gives the general solution

t 2t t t t 2t t y1 ϭ c1e ϩ 4c2e ϩ te ϩ e , y2 ϭϪc1e Ϫ 5c2e Ϫ te .

From the initial conditions we obtain c1 ϭϪ2, c2 ϭ 5. Answer:

t 2t t t t 2t t y1 ϭϪ2e ϩ 20e ϩ te ϩ e , y2 ϭ 2e Ϫ 25e Ϫ te . t ؊t 11. y1 Ϫ cosh t ϩ 2 sinh t ϩ e , y2 ϭ 2 cosh t Ϫ sinh t Ϫ e 12. A general solution of the homogeneous system is 2 2 (h) (1) 3t (2) ؊t (1) (2) y ϭ c1x e ϩ c2x e , x ϭ , x ϭ . c 1 d c Ϫ1 d Answer:

؊t 2 ؊t y1 ϭ 2e ϩ t , y2 ϭϪe Ϫ t. 13. y1: ϭϪsin 2t Ϫ cos 2t ϩ 2 sin t, y2: ϭ 2 cos 2t Ϫ 2 sin 2t 14. The matrix of the homogeneous system 04 c Ϫ10d has the eigenvalues 2i and Ϫ2i and eigenvectors [2 i]T and [2 Ϫi]T, respectively. Hence a complex general solution is 2 2 h) 2it ؊2it) y ϭ c1 e ϩ c2 e . c i d c Ϫi d c04.qxd 6/18/11 3:06 PM Page 85

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By Euler’s formula this becomes, in components,

(h) y1 ϭ (2c1 ϩ 2c2) cos 2t ϩ i(2c1 Ϫ 2c2) sin 2t (h) y2 ϭ (ic1 Ϫ ic2) cos 2t ϩ ( Ϫc1 Ϫ c2) sin 2t.

Setting A ϭ c1 ϩ c2 and B ϭ i(c1 Ϫ c2), we can write (h) y1 ϭ 2A cos 2t ϩ 2B sin 2t (h) y2 ϭ B cos 2t Ϫ 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 p) t ؊t) y1 ϭ a1e ϩ b1e p) t ؊t) y2 ϭ a2e ϩ b2e . Differentiation and substitution gives t ؊t t ؊t t a1e Ϫ b1e ϭ 4a2e ϩ 4b2e ϩ 5e t ؊t t ؊t ؊t a2e Ϫ b2e ϭϪa1e Ϫ b1e Ϫ 20e . Equating the coefficients of et on both sides, we get

a1 ϭ 4a2 ϩ 5, a2 ϭϪa1, hence a1 ϭ 1, a2 ϭϪ1. Equating the coefficients of eϪt, we similarly obtain

Ϫb1 ϭ 4b2, Ϫb2 ϭϪb1 Ϫ 20, hence b1 ϭϪ16, b2 ϭ 4. Hence a general solution of the given nonhomogeneous system is t ؊t y1 ϭ 2A cos 2t ϩ 2B sin 2t ϩ e Ϫ 16e t ؊t y2 ϭ B cos 2t Ϫ A sin 2t Ϫ e ϩ 4e . From this and the initial conditions we obtain

y1(0) ϭ 2A ϩ 1 Ϫ16 ϭ 1, y2(0) ϭ B Ϫ 1 ϩ 4 ϭ 0. The solution is A ϭ 8, B ϭϪ3. This gives the answer (the solution of the initial value problem) t ؊t y1 ϭ 16 cos 2t Ϫ 6 sin 2t ϩ e Ϫ 16e t ؊t y2 ϭϪ3 cos 2t Ϫ 8 sin 2t Ϫ e ϩ 4e . ϭ 2t ϩ ϩ Ϫ t ϭ 2 2t ϩ ϩ Ϫ t ϩ 7 ؊t 15. y1 e 4t 4 6e , y2 3e 9t 1 6e 3e . 18. The equations are

(a) I1r ϭϪ2I1 ϩ 2I2 ϩ 440 sin t and

8I2 ϩ 2 ΎI2 dt ϩ 2(I2 Ϫ I1) ϭ 0. Thus

I2 ϭϪ0.25 ΎI2 dt ϩ 0.25(I1 Ϫ I2),

which, upon differentiation and insertion of I1r from (a) and simplification, gives c04.qxd 6/18/11 3:06 PM Page 86

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(b) Ir2 ϭϪ0.4I1 ϩ 0.2I2 ϩ 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,

1 352 1 616 Ϫ cos t ϩ sin t. 3 c 44 d 3 c 132 d Ϫ3 1.25 20. A ϭ , c 1 Ϫ1 d 1 5 1 ؊ 125 ؊ 500 125 J ϭϪ e 0.5t Ϫ e 3.5t ϩ . 3 c 2 d 21 c Ϫ2 d 7 c 1 d

SOLUTIONS TO CHAPTER 4 REVIEW QUESTIONS AND PROBLEMS, page 164

؊3t 3t ؊3t 3t 11. y1 ϭ c1e ϩ c2e , y2 ϭϪ3c1e ϩ 3c2e . Saddle point. 12. The matrix 20 c 01d has the eigenvalues 2 and 1 with eigenvalues [1, 0]T and [0, 1]T, respectively. Hence, a general solution is 1 0 2t t y ϭ c1 e ϩ c2 e . c 0 d c 1 d Since p ϭ 3, q ϭ 2, ¢ ϭ p2 Ϫ 4q ϭ 9 Ϫ 8 ϭ 1 Ͼ 0, the critical point is an unstable node. ϭ ؊2t ϩ ϭ 1 ؊2t ϩ ϩ Ϫ 13. y1 e (c1 sin t c2 cos t), y2 5e (3c1 sin t c1 cos t 3c2 cos t c2 sin t). Asymptotically stable spiral point. 14. Eigenvalues Ϫ1, 6. Eigenvectors [1 Ϫ1]T, [4 3]T. The corresponding general solution is

؊t 6t ؊t 6t y1 ϭ c1e ϩ 4c2e , y2 ϭϪc1e ϩ 3c2e . The critical point at (0, 0) is a saddle point, which is always unstable.

16. Eigenvalues Ϫ2i and 2i. y1 ϭ c1 sin 2t ϩ c2 cos 2t, y2 ϭ c1 cos 2t Ϫ c2 sin 2t. Center, which is always stable. 18. Saddle point; see Table 4.1 (b) in Sec. 4.4.

20. y1 ϭ 3 cosh t ϩ sinh t ϩ c1t ϩ c2, y2 ϭϪ3 cosh t ϩ sinh t ϩ c1 Ϫ c1t Ϫ c2 ؊2t 2t 2 ؊2t 2t 21. y1 ϭ c1e ϩ c2e Ϫ 4 Ϫ 8t , y2 ϭϪc1e ϩ c2e Ϫ 8t. 22. The matrix 11 c 41d c04.qxd 6/18/11 3:06 PM Page 87

Instructor’s Manual 87

has the eigenvalues 3 and Ϫ1 with eigenvectors [1 2]T and [1 Ϫ2]T, respectively. A particular solution of the nonhomogeneous system is obtained by the method of undetermined coefficients. We set (p) ϭ ϩ (p) ϭ ϩ y1 a1 cos t b1 sin t, y2 a2 cos t b2 sin t. Differentiation and substitution gives the following two equations, where c ϭ cos t and s ϭ sin t.

(1) Ϫa1s ϩ b1c ϭ a1c ϩ b1s ϩ a2c ϩ b2s ϩ s

(2) Ϫa2s ϩ b2c ϭ 4a1c ϩ 4b1s ϩ a2c ϩ b2s. From the cosine terms and the sine terms in (1) we obtain

b1 ϭ a1 ϩ a2, Ϫa1 ϭ b1 ϩ b2 ϩ 1.

Similarly from (2),

b2 ϭ 4a1 ϩ a2, Ϫa2 ϭ 4b1 ϩ b2.

Hence a1 ϭϪ0.3, a2 ϭ 0.4, b1 ϭ 0.1, b2 ϭϪ0.8. This gives the answer

3t ؊t y1 ϭ c1e ϩ c2e Ϫ 0.3 cos t ϩ 0.1 sin t 3t ؊t y2 ϭ 2c1e Ϫ 2c2e ϩ 0.4 cos t Ϫ 0.8 sin t.

24. The balance equations are 16 6 yr ϭϪ y ϩ y 1 400 1 200 2 16 16 yr ϭ y Ϫ y 2 400 1 200 2 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 Ϫ0.1 and Ϫ0.02 with the eigenvectors [Ϫ1, 2]T and [3, 2]T respectively. The initial condition is y1(0) ϭ 120, y2(0) ϭ 0. This gives the solution

؊0.02t ؊0.10t y1 ϭ 90e ϩ 30e ؊0.02t ؊0.10t y2 ϭ 60e Ϫ 60e

25. I1r ϩ 2.5(I1 Ϫ I2) ϭ 169 sin 2t, 2.5(Ir2 Ϫ I1r) ϩ 25I2 ϭ 0, ؊5.0t I1(t) ϭ (31.8 ϩ 58.3t)e Ϫ 31.8 cos (2.0t) ϩ 50.2 sin (2.0t), ؊5.0t I2(t) ϭ (Ϫ8.44 Ϫ 58.3t)e ϩ 8.04 sin (2.0t) ϩ 8.44 cos (2.0t) 26. The ODEs of the system are

I1r Ϫ I2r ϩ 5I1 ϭ 0

I2 Ϫ I1 ϩ 1.25I2r ϭ 0,

written in the usual form

A Ϫ B ϪA Ir ϭ I c A ϪA d c04.qxd 6/18/11 3:06 PM Page 88

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where A ϭ R L ϭ 0.8 and B ϭ 1 (RC ) ϭ 5. A general solution is > > 1 4 ؊t ؊4t I ϭ c1 e ϩ c2 e . c Ϫ4 d c Ϫ1 d ϭϪ1 ϭ 1 The initial conditions give c1 3 and c2 3. Hence the particular solution satisfying the initial conditions is ϭϪ1 ؊t ϩ 4 ؊4t I1 3e 3e ϭ 4 ؊t Ϫ 1 ؊4t I2 3e 3e .

28. cos y ϭ 0 when y ϭ (2n ϩ 1)p 2, where n is any integer. This gives the location 2 2 > of the critical points, which lie on the y2 -axis y1 ϭ 0 in the phase plane. 1 ϭ 1 ϩ ෂ For (0, 2p) the transformation is y2 2p y2. Now ϭ 1 ϩ ෂ ϭϪ ෂ Ϸ Ϫෂ cos y2 cos (2p y2) sin y2 y2. Hence the linearized system is ෂr ෂ y1 ϭϪy2 ෂr ෂ y2 ϭ 3y1. For its matrix 0 Ϫ1 c 30d ෂ ෂ we obtain p ϭ 0, q ϭ 3 Ͼ 0, so that this is a center. Similarly, by periodicity, the ϩ critical points at (4n 1)p 2 are centers. ෂ Ϫ1 > ϭϪ1 ϩ ෂ For (0, 2p) the transformation is y2 2p y2. From ෂ ෂ ෂ ϭ Ϫ1 ϩ ෂ ϭ ෂ Ϸ ෂ cos y2 cos ( 2p y2 ) sin y2 y2 we obtain the linearized system ෂr ෂ y 1 ϭ y2 ෂr ෂ y2 ϭ 3y1 with matrix 01 c 30d ෂ for which q ϭϪ3 Ͻ 0, so that this point and the points with y ϭ (4n Ϫ 1)p 2 on 2 > the y2 -axis are saddle points. np 29. ( 2 , 0), Center when n is odd and saddle point when n is even. 30. Critical points at (0, 0) and (0, Ϫ1). The linearized systems are:

y1r ϭ y2

yr2 ϭϪ2y1 and

y1r ϭϪy2

yr2 ϭϪ2y1