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Section 9.2 Higher Order Constant Coefﬁcient Homogeneous Equations 487 (e) Now suppose n is a positive integer. Infer from (A) that if zk = cos 2kπ n + i sin 2kπ n , k = 0, 1, . . ., n −1, and ζk = cos (2k + 1)π n + i sin (2k + 1)π n , k = 0, 1, . . ., n −1, then zn k = 1 and ζn k = −1, k = 0, 1, . . ., n −1. (Why don’t we also consider other integer values for k?) (f) Let ρ be a positive number. Use (e) to show that zn −ρ = (z −ρ1/nz0)(z −ρ1/nz1) · · · (z −ρ1/nzn−1) and zn + ρ = (z −ρ1/nζ0)(z −ρ1/nζ1) · · ·(z −ρ1/nζn−1). 43. Use (e) of Exercise 42 to ﬁnd a fundamental set of solutions of the given equation. (a) y′′′ −y = 0 (b) y′′′ + y = 0 (c) y(4) + 64y = 0 (d) y(6) −y = 0 (e) y(6) + 64y = 0 (f) (D −1)6 −1 y = 0 (g) y(5) + y(4) + y′′′ + y′′ + y′ + y = 0 44. An equation of the form a0xny(n) + a1xn−1y(n−1) + · · · + an−1xy′ + any = 0, x > 0, (A) where a0, a1, ..., an are constants, is an Euler or equidimensional equation. Show that if x = et and Y (t) = y(x(t)), (B) then x dy dx = dY dt x2 d2y dx2 = d2Y dt2 −dY dt x3 d3y dx3 = d3Y dt3 −3d2Y dt2 + 2dY dt . In general, it can be shown that if r is any integer ≥2 then xr dry dxr = drY dtr + A1r dr−1Y dtr−1 + · · · + Ar−1,r dY dt where A1r, ..., Ar−1,r are integers. Use these results to show that the substitution (B) transforms (A) into a constant coefﬁcient equation for Y as a function of t.	Elementary Differential Equations with Boundary Value Problems_Page_497_Chunk2901
488 Chapter 9 Linear Higher Order Equations 45. Use Exercise 44 to show that a function y = y(x) satisﬁes the equation a0x3y′′′ + a1x2y′′ + a2xy′ + a3y = 0, (A) on (0, ∞) if and only if the function Y (t) = y(et) satisﬁes a0 d3Y dt3 + (a1 −3a0)d2Y dt2 + (a2 −a1 + 2a0)dY dt + a3Y = 0. Assuming that a0, a1, a2, a3 are real and a0 ̸= 0, ﬁnd the possible forms for the general solution of (A). 9.3 UNDETERMINED COEFFICIENTS FOR HIGHER ORDER EQUATIONS In this section we consider the constant coefﬁcient equation a0y(n) + a1y(n−1) + · · · + any = F (x), (9.3.1) where n ≥3 and F is a linear combination of functions of the form eαx	Elementary Differential Equations with Boundary Value Problems_Page_498_Chunk2902
Section 9.3 Undetermined Coefﬁcients for Higher Order Equations 489 into (9.3.2) and canceling ex yields (u′′′ + 3u′′ + 3u′ + u) + 3(u′′ + 2u′ + u) + 2(u′ + u) −u = 21 + 24x + 28x2 + 5x3, or u′′′ + 6u′′ + 11u′ + 5u = 21 + 24x + 28x2 + 5x3. (9.3.3) Since the unknown u appears on the left, we can see that (9.3.3) has a particular solution of the form up = A + Bx + Cx2 + Dx3. Then u′ p = B + 2Cx + 3Dx2 u′′ p = 2C + 6Dx u′′′ p = 6D. Substituting from the last four equations into the left side of (9.3.3) yields u′′′ p + 6u′′ p + 11u′ p + 5up = 6D + 6(2C + 6Dx) + 11(B + 2Cx + 3Dx2) +5(A + Bx + Cx2 + Dx3) = (5A + 11B + 12C + 6D) + (5B + 22C + 36D)x +(5C + 33D)x2 + 5Dx3. Comparing coefﬁcients of like powers of x on the right sides of this equation and (9.3.3) shows that up satisﬁes (9.3.3) if 5D = 5 5C + 33D = 28 5B + 22C + 36D = 24 5A + 11B + 12C + 6D = 21. Solving these equations successively yields D = 1, C = −1, B = 2, A = 1. Therefore up = 1 + 2x −x2 + x3 is a particular solution of (9.3.3), so yp = exup = ex(1 + 2x −x2 + x3) is a particular solution of (9.3.2) (Figure 9.3.1). Example 9.3.2 Find a particular solution of y(4) −y′′′ −6y′′ + 4y′ + 8y = e2x(4 + 19x + 6x2). (9.3.4) Solution Substituting y = ue2x, y′ = e2x(u′ + 2u), y′′ = e2x(u′′ + 4u′ + 4u), y′′′ = e2x(u′′′ + 6u′′ + 12u′ + 8u), y(4) = e2x(u(4) + 8u′′′ + 24u′′ + 32u′ + 16u)	Elementary Differential Equations with Boundary Value Problems_Page_499_Chunk2903
490 Chapter 9 Linear Higher Order Equations 1 2 10 20 30 40 50 x y Figure 9.3.1 yp = ex(1 + 2x −x2 + x3) into (9.3.4) and canceling e2x yields (u(4) + 8u′′′ + 24u′′ + 32u′ + 16u) −(u′′′ + 6u′′ + 12u′ + 8u) −6(u′′ + 4u′ + 4u) + 4(u′ + 2u) + 8u = 4 + 19x + 6x2, or u(4) + 7u′′′ + 12u′′ = 4 + 19x + 6x2. (9.3.5) Since neither u nor u′ appear on the left, we can see that (9.3.5) has a particular solution of the form up = Ax2 + Bx3 + Cx4. (9.3.6) Then u′ p = 2Ax + 3Bx2 + 4Cx3 u′′ p = 2A + 6Bx + 12Cx2 u′′′ p = 6B + 24Cx u(4) p = 24C. Substituting u′′ p, u′′′ p , and u(4) p into the left side of (9.3.5) yields u(4) p + 7u′′′ p + 12u′′ p = 24C + 7(6B + 24Cx) + 12(2A + 6Bx + 12Cx2) = (24A + 42B + 24C) + (72B + 168C)x + 144Cx2.	Elementary Differential Equations with Boundary Value Problems_Page_500_Chunk2904
Section 9.3 Undetermined Coefﬁcients for Higher Order Equations 491 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 .05 .10 .15 .20 .25 .30 x y Figure 9.3.2 yp = x2e2x 24 (−4 + 4x + x2) Comparing coefﬁcients of like powers of x on the right sides of this equation and (9.3.5) shows that up satisﬁes (9.3.5) if 144C = 6 72B + 168C = 19 24A + 42B + 24C = 4. Solving these equations successively yields C = 1/24, B = 1/6, A = −1/6. Substituting these into (9.3.6) shows that up = x2 24(−4 + 4x + x2) is a particular solution of (9.3.5), so yp = e2xup = x2e2x 24 (−4 + 4x + x2) is a particular solution of (9.3.4). (Figure 9.3.2). Forcing Functions of the Form eλx (P (x) cos ωx + Q(x) sinωx) We now consider equations of the form a0y(n) + a1y(n−1) + · · · + any = eλx (P (x) cos ωx + Q(x) sin ωx) , where P and Q are polynomials. Example 9.3.3 Find a particular solution of y′′′ + y′′ −4y′ −4y = ex[(5 −5x) cos x + (2 + 5x) sin x]. (9.3.7)	Elementary Differential Equations with Boundary Value Problems_Page_501_Chunk2905
492 Chapter 9 Linear Higher Order Equations Solution Substituting y = uex, y′ = ex(u′ + u), y′′ = ex(u′′ + 2u′ + u), y′′′ = ex(u′′′ + 3u′′ + 3u′ + u) into (9.3.7) and canceling ex yields (u′′′ + 3u′′ + 3u′ + u) + (u′′ + 2u′ + u) −4(u′ + u) −4u = (5 −5x) cos x + (2 + 5x) sin x, or u′′′ + 4u′′ + u′ −6u = (5 −5x) cos x + (2 + 5x) sinx. (9.3.8) Since cos x and sin x are not solutions of the complementary equation u′′′ + 4u′′ + u′ −6u = 0, a theorem analogous to Theorem 5.5.1 implies that (9.3.8) has a particular solution of the form up = (A0 + A1x) cos x + (B0 + B1x) sin x. (9.3.9) Then u′ p = (A1 + B0 + B1x) cos x + (B1 −A0 −A1x) sinx, u′′ p = (2B1 −A0 −A1x) cos x −(2A1 + B0 + B1x) sinx, u′′′ p = −(3A1 + B0 + B1x) cos x −(3B1 −A0 −A1x) sinx, so u′′′ p + 4u′′ p + u′ p −6up = −[10A0 + 2A1 −8B1 + 10A1x] cos x −[10B0 + 2B1 + 8A1 + 10B1x] sin x. Comparing the coefﬁcients of x cos x, x sin x, cos x, and sin x here with the corresponding coefﬁcients in (9.3.8) shows that up is a solution of (9.3.8) if −10A1 = −5 −10B1 = 5 −10A0 −2A1 + 8B1 = 5 −10B0 −2B1 −8A1 = 2. Solving the ﬁrst two equations yields A1 = 1/2, B1 = −1/2. Substituting these into the last two equations yields −10A0 = 5 + 2A1 −8B1 = 10 −10B0 = 2 + 2B1 + 8A1 = 5, so A0 = −1, B0 = −1/2. Substituting A0 = −1, A1 = 1/2, B0 = −1/2, B1 = −1/2 into (9.3.9) shows that up = −1 2 [(2 −x) cos x + (1 + x) sin x] is a particular solution of (9.3.8), so yp = exup = −ex 2 [(2 −x) cos x + (1 + x) sin x] is a particular solution of (9.3.7) (Figure 9.3.3).	Elementary Differential Equations with Boundary Value Problems_Page_502_Chunk2906
Section 9.3 Undetermined Coefﬁcients for Higher Order Equations 493 1 2 3 4 20 40 60 80 100 −20 x y Figure 9.3.3 yp = exup = −ex 2 [(2 −x) cos x + (1 + x) sin x] Example 9.3.4 Find a particular solution of y′′′ + 4y′′ + 6y′ + 4y = e−x [(1 −6x) cos x −(3 + 2x) sin x] . (9.3.10) Solution Substituting y = ue−x, y′ = e−x(u′ −u), y′′ = e−x(u′′ −2u′ + u), y′′′ = e−x(u′′′ −3u′′ + 3u′ −u) into (9.3.10) and canceling e−x yields (u′′′ −3u′′ + 3u′ −u) + 4(u′′ −2u′ + u) + 6(u′ −u) + 4u = (1 −6x) cos x −(3 + 2x) sinx, or u′′′ + u′′ + u′ + u = (1 −6x) cos x −(3 + 2x) sin x. (9.3.11) Since cos x and sin x are solutions of the complementary equation u′′′ + u′′ + u′ + u = 0, a theorem analogous to Theorem 5.5.1 implies that (9.3.11) has a particular solution of the form up = (A0x + A1x2) cos x + (B0x + B1x2) sin x. (9.3.12)	Elementary Differential Equations with Boundary Value Problems_Page_503_Chunk2907
494 Chapter 9 Linear Higher Order Equations Then u′ p = [A0 + (2A1 + B0)x + B1x2] cos x + [B0 + (2B1 −A0)x −A1x2] sin x, u′′ p = [2A1 + 2B0 −(A0 −4B1)x −A1x2] cos x +[2B1 −2A0 −(B0 + 4A1)x −B1x2] sinx, u′′′ p = −[3A0 −6B1 + (6A1 + B0)x + B1x2] cos x −[3B0 + 6A1 + (6B1 −A0)x −A1x2] sinx, so u′′′ p + u′′ p + u′ p + up = −[2A0 −2B0 −2A1 −6B1 + (4A1 −4B1)x] cos x −[2B0 + 2A0 −2B1 + 6A1 + (4B1 + 4A1)x] sinx. Comparing the coefﬁcients of x cos x, x sin x, cos x, and sin x here with the corresponding coefﬁcients in (9.3.11) shows that up is a solution of (9.3.11) if −4A1 + 4B1 = −6 −4A1 −4B1 = −2 −2A0 + 2B0 + 2A1 + 6B1 = 1 −2A0 −2B0 −6A1 + 2B1 = −3. Solving the ﬁrst two equations yields A1 = 1, B1 = −1/2. Substituting these into the last two equations yields −2A0 + 2B0 = 1 −2A1 −6B1 = 2 −2A0 −2B0 = −3 + 6A1 −2B1 = 4, 1 2 3 4 5 6 7 8 9 0.1 0.2 −0.1 −0.2 −0.3 −0.4 −0.5 x y Figure 9.3.4 yp = −xe−x 2 [(3 −2x) cos x + (1 + x) sin x]	Elementary Differential Equations with Boundary Value Problems_Page_504_Chunk2908
Section 9.3 Undetermined Coefﬁcients for Higher Order Equations 495 so A0 = −3/2 and B0 = −1/2. Substituting A0 = −3/2, A1 = 1, B0 = −1/2, B1 = −1/2 into (9.3.12) shows that up = −x 2 [(3 −2x) cos x + (1 + x) sinx] is a particular solution of (9.3.11), so yp = e−xup = −xe−x 2 [(3 −2x) cos x + (1 + x) sin x] (Figure 9.3.4) is a particular solution of (9.3.10). 9.3 Exercises In Exercises 1–59 ﬁnd a particular solution. 1. y′′′ −6y′′ + 11y′ −6y = −e−x(4 + 76x −24x2) 2. y′′′ −2y′′ −5y′ + 6y = e−3x(32 −23x + 6x2) 3. 4y′′′ + 8y′′ −y′ −2y = −ex(4 + 45x + 9x2) 4. y′′′ + 3y′′ −y′ −3y = e−2x(2 −17x + 3x2) 5. y′′′ + 3y′′ −y′ −3y = ex(−1 + 2x + 24x2 + 16x3) 6. y′′′ + y′′ −2y = ex(14 + 34x + 15x2) 7. 4y′′′ + 8y′′ −y′ −2y = −e−2x(1 −15x) 8. y′′′ −y′′ −y′ + y = ex(7 + 6x) 9. 2y′′′ −7y′′ + 4y′ + 4y = e2x(17 + 30x) 10. y′′′ −5y′′ + 3y′ + 9y = 2e3x(11 −24x2) 11. y′′′ −7y′′ + 8y′ + 16y = 2e4x(13 + 15x) 12. 8y′′′ −12y′′ + 6y′ −y = ex/2(1 + 4x) 13. y(4) + 3y′′′ −3y′′ −7y′ + 6y = −e−x(12 + 8x −8x2) 14. y(4) + 3y′′′ + y′′ −3y′ −2y = −3e2x(11 + 12x) 15. y(4) + 8y′′′ + 24y′′ + 32y′ = −16e−2x(1 + x + x2 −x3) 16. 4y(4) −11y′′ −9y′ −2y = −ex(1 −6x) 17. y(4) −2y′′′ + 3y′ −y = ex(3 + 4x + x2) 18. y(4) −4y′′′ + 6y′′ −4y′ + 2y = e2x(24 + x + x4) 19. 2y(4) + 5y′′′ −5y′ −2y = 18ex(5 + 2x) 20. y(4) + y′′′ −2y′′ −6y′ −4y = −e2x(4 + 28x + 15x2) 21. 2y(4) + y′′′ −2y′ −y = 3e−x/2(1 −6x) 22. y(4) −5y′′ + 4y = ex(3 + x −3x2) 23. y(4) −2y′′′ −3y′′ + 4y′ + 4y = e2x(13 + 33x + 18x2) 24. y(4) −3y′′′ + 4y′ = e2x(15 + 26x + 12x2) 25. y(4) −2y′′′ + 2y′ −y = ex(1 + x) 26. 2y(4) −5y′′′ + 3y′′ + y′ −y = ex(11 + 12x)	Elementary Differential Equations with Boundary Value Problems_Page_505_Chunk2909
496 Chapter 9 Linear Higher Order Equations 27. y(4) + 3y′′′ + 3y′′ + y′ = e−x(5 −24x + 10x2) 28. y(4) −7y′′′ + 18y′′ −20y′ + 8y = e2x(3 −8x −5x2) 29. y′′′ −y′′ −4y′ + 4y = e−x [(16 + 10x) cosx + (30 −10x) sinx] 30. y′′′ + y′′ −4y′ −4y = e−x [(1 −22x) cos 2x −(1 + 6x) sin2x] 31. y′′′ −y′′ + 2y′ −2y = e2x[(27 + 5x −x2) cos x + (2 + 13x + 9x2) sin x] 32. y′′′ −2y′′ + y′ −2y = −ex[(9 −5x + 4x2) cos 2x −(6 −5x −3x2) sin 2x] 33. y′′′ + 3y′′ + 4y′ + 12y = 8 cos 2x −16 sin2x 34. y′′′ −y′′ + 2y = ex[(20 + 4x) cos x −(12 + 12x) sinx] 35. y′′′ −7y′′ + 20y′ −24y = −e2x[(13 −8x) cos 2x −(8 −4x) sin2x] 36. y′′′ −6y′′ + 18y′ = −e3x[(2 −3x) cos 3x −(3 + 3x) sin3x] 37. y(4) + 2y′′′ −2y′′ −8y′ −8y = ex(8 cos x + 16 sin x) 38. y(4) −3y′′′ + 2y′′ + 2y′ −4y = ex(2 cos 2x −sin 2x) 39. y(4) −8y′′′ + 24y′′ −32y′ + 15y = e2x(15x cos 2x + 32 sin 2x) 40. y(4) + 6y′′′ + 13y′′ + 12y′ + 4y = e−x[(4 −x) cos x −(5 + x) sinx] 41. y(4) + 3y′′′ + 2y′′ −2y′ −4y = −e−x(cos x −sin x) 42. y(4) −5y′′′ + 13y′′ −19y′ + 10y = ex(cos 2x + sin 2x) 43. y(4) + 8y′′′ + 32y′′ + 64y′ + 39y = e−2x[(4 −15x) cos 3x −(4 + 15x) sin3x] 44. y(4) −5y′′′ + 13y′′ −19y′ + 10y = ex[(7 + 8x) cos 2x + (8 −4x) sin2x] 45. y(4) + 4y′′′ + 8y′′ + 8y′ + 4y = −2e−x(cos x −2 sinx) 46. y(4) −8y′′′ + 32y′′ −64y′ + 64y = e2x(cos 2x −sin 2x) 47. y(4) −8y′′′ + 26y′′ −40y′ + 25y = e2x[3 cos x −(1 + 3x) sin x] 48. y′′′ −4y′′ + 5y′ −2y = e2x −4ex −2 cos x + 4 sin x 49. y′′′ −y′′ + y′ −y = 5e2x + 2ex −4 cos x + 4 sinx 50. y′′′ −y′ = −2(1 + x) + 4ex −6e−x + 96e3x 51. y′′′ −4y′′ + 9y′ −10y = 10e2x + 20ex sin 2x −10 52. y′′′ + 3y′′ + 3y′ + y = 12e−x + 9 cos 2x −13 sin 2x 53. y′′′ + y′′ −y′ −y = 4e−x(1 −6x) −2x cos x + 2(1 + x) sinx 54. y(4) −5y′′ + 4y = −12ex + 6e−x + 10 cos x 55. y(4) −4y′′′ + 11y′′ −14y′ + 10y = −ex(sin x + 2 cos 2x) 56. y(4) + 2y′′′ −3y′′ −4y′ + 4y = 2ex(1 + x) + e−2x 57. y(4) + 4y = sinh x cos x −cosh x sinx 58. y(4) + 5y′′′ + 9y′′ + 7y′ + 2y = e−x(30 + 24x) −e−2x 59. y(4) −4y′′′ + 7y′′ −6y′ + 2y = ex(12x −2 cos x + 2 sin x) In Exercises 60–68 ﬁnd the general solution. 60. y′′′ −y′′ −y′ + y = e2x(10 + 3x) 61. y′′′ + y′′ −2y = −e3x(9 + 67x + 17x2)	Elementary Differential Equations with Boundary Value Problems_Page_506_Chunk2910
Section 9.3 Undetermined Coefﬁcients for Higher Order Equations 497 62. y′′′ −6y′′ + 11y′ −6y = e2x(5 −4x −3x2) 63. y′′′ + 2y′′ + y′ = −2e−x(7 −18x + 6x2) 64. y′′′ −3y′′ + 3y′ −y = ex(1 + x) 65. y(4) −2y′′ + y = −e−x(4 −9x + 3x2) 66. y′′′ + 2y′′ −y′ −2y = e−2x [(23 −2x) cos x + (8 −9x) sin x] 67. y(4) −3y′′′ + 4y′′ −2y′ = ex [(28 + 6x) cos 2x + (11 −12x) sin2x] 68. y(4) −4y′′′ + 14y′′ −20y′ + 25y = ex [(2 + 6x) cos 2x + 3 sin 2x] In Exercises 69–74 solve the initial value problem and graph the solution. 69. C/G y′′′ −2y′′ −5y′ + 6y = 2ex(1 −6x), y(0) = 2, y′(0) = 7, y′′(0) = 9 70. C/G y′′′ −y′′ −y′ + y = −e−x(4 −8x), y(0) = 2, y′(0) = 0, y′′(0) = 0 71. C/G 4y′′′ −3y′ −y = e−x/2(2 −3x), y(0) = −1, y′(0) = 15, y′′(0) = −17 72. C/G y(4) + 2y′′′ + 2y′′ + 2y′ + y = e−x(20 −12x), y(0) = 3, y′(0) = −4, y′′(0) = 7, y′′′(0) = −22 73. C/G y′′′ + 2y′′ + y′ + 2y = 30 cos x −10 sinx, y(0) = 3, y′(0) = −4, y′′(0) = 16 74. C/G y(4) −3y′′′ + 5y′′ −2y′ = −2ex(cos x −sin x), y(0) = 2, y′(0) = 0, y′′(0) = −1, y′′′(0) = −5 75. Prove: A function y is a solution of the constant coefﬁcient nonhomogeneous equation a0y(n) + a1y(n−1) + · · · + any = eαxG(x) (A) if and only if y = ueαx, where u satisﬁes the differential equation a0u(n) + p(n−1)(α) (n −1)! u(n−1) + p(n−2)(α) (n −2)! u(n−2) + · · · + p(α)u = G(x) (B) and p(r) = a0rn + a1rn−1 + · · · + an is the characteristic polynomial of the complementary equation a0y(n) + a1y(n−1) + · · · + any = 0. 76. Prove: (a) The equation a0u(n) + p(n−1)(α) (n −1)! u(n−1) + p(n−2)(α) (n −2)! u(n−2) + · · · + p(α)u =	Elementary Differential Equations with Boundary Value Problems_Page_507_Chunk2911
498 Chapter 9 Linear Higher Order Equations (b) If λ + iω is a zero of p with multiplicity m ≥1, then (A) can be written as a(u′′ + ω2u) =	Elementary Differential Equations with Boundary Value Problems_Page_508_Chunk2912
Section 9.4 Variation of Parameters for Higher Order Equations 499 These formulas are easy to remember, since they look as though we obtained them by differentiating (9.4.2) n −1 times while treating u1, u2, ..., un as constants. To see that (9.4.3) implies (9.4.4), we ﬁrst differentiate (9.4.2) to obtain y′ p = u1y′ 1 + u2y′ 2 + · · · + uny′ n + u′ 1y1 + u′ 2y2 + · · · + u′ nyn, which reduces to y′ p = u1y′ 1 + u2y′ 2 + · · · + uny′ n because of the ﬁrst equation in (9.4.3). Differentiating this yields y′′ p = u1y′′ 1 + u2y′′ 2 + · · · + uny′′ n + u′ 1y′ 1 + u′ 2y′ 2 + · · · + u′ ny′ n, which reduces to y′′ p = u1y′′ 1 + u2y′′ 2 + · · · + uny′′ n because of the second equation in (9.4.3). Continuing in this way yields (9.4.4). The last equation in (9.4.4) is y(n−1) p = u1y(n−1) 1 + u2y(n−1) 2 + · · · + uny(n−1) n . Differentiating this yields y(n) p = u1y(n) 1 + u2y(n) 2 + · · · + uny(n) n + u′ 1y(n−1) 1 + u′ 2y(n−1) 2 + · · · + u′ ny(n−1) n . Substituting this and (9.4.4) into (9.4.1) yields u1Ly1 + u2Ly2 + · · · + unLyn + P0(x) u′ 1y(n−1) 1 + u′ 2y(n−1) 2 + · · · + u′ ny(n−1) n = F (x). Since Lyi = 0 (1 ≤i ≤n), this reduces to u′ 1y(n−1) 1 + u′ 2y(n−1) 2 + · · · + u′ ny(n−1) n = F (x) P0(x). Combining this equation with (9.4.3) shows that yp = u1y1 + u2y2 + · · · + unyn is a solution of (9.4.1) if u′ 1y1 + u′ 2y2+ · · · +u′ nyn = 0 u′ 1y′ 1 + u′ 2y′ 2+ · · · +u′ ny′ n = 0 ... u′ 1y(n−2) 1 + u′ 2y(n−2) 2 + · · · +u′ ny(n−2) n = 0 u′ 1y(n−1) 1 + u′ 2y(n−1) 2 + · · · +u′ ny(n−1) n = F/P0, which can be written in matrix form as   y1 y2 · · · yn y′ 1 y′ 2 · · · y′ n ... ... ... ... y(n−2) 1 y(n−2) 2 · · · y(n−2) n y(n−1) 1 y(n−1) 2 · · · y(n−1) n     u′ 1 u′ 2 ... u′ n−1 u′ n   =   0 0 ... 0 F/P0   . (9.4.5)	Elementary Differential Equations with Boundary Value Problems_Page_509_Chunk2913
500 Chapter 9 Linear Higher Order Equations The determinant of this system is the Wronskian W of the fundamental set of solutions {y1, y2, . . ., yn}, which has no zeros on (a, b), by Theorem 9.1.4. Solving (9.4.5) by Cramer’s rule yields u′ j = (−1)n−j F Wj P0W , 1 ≤j ≤n, (9.4.6) where Wj is the Wronskian of the set of functions obtained by deleting yj from {y1, y2, . . ., yn} and keeping the remaining functions in the same order. Equivalently, Wj is the determinant obtained by deleting the last row and j-th column of W. Having obtained u′ 1, u′ 2, . . .,u′ n, we can integrate to obtain u1, u2, . . ., un. As in Section 5.7, we take the constants of integration to be zero, and we drop any linear combination of {y1, y2, . . ., yn} that may appear in yp. REMARK: For efﬁciency, it’s best to compute W1, W2, ..., Wn ﬁrst, and then compute W by expanding in cofactors of the last row; thus, W = n X j=1 (−1)n−jy(n−1) j Wj. Third Order Equations If n = 3, then W = y1 y2 y3 y′ 1 y′ 2 y′ 3 y′′ 1 y′′ 2 y′′ 3 . Therefore W1 = y2 y3 y′ 2 y′ 3 , W2 = y1 y3 y′ 1 y′ 3 , W3 = y1 y2 y′ 1 y′ 2 , and (9.4.6) becomes u′ 1 = F W1 P0W , u′ 2 = −F W2 P0W , u′ 3 = F W3 P0W . (9.4.7) Example 9.4.1 Find a particular solution of xy′′′ −y′′ −xy′ + y = 8x2ex, (9.4.8) given that y1 = x, y2 = ex, and y3 = e−x form a fundamental set of solutions of the complementary equation. Then ﬁnd the general solution of (9.4.8). Solution We seek a particular solution of (9.4.8) of the form yp = u1x + u2ex + u3e−x. The Wronskian of {y1, y2, y3} is W(x) = x ex e−x 1 ex −e−x 0 ex e−x ,	Elementary Differential Equations with Boundary Value Problems_Page_510_Chunk2914
Section 9.4 Variation of Parameters for Higher Order Equations 501 so W1 = ex e−x ex −e−x = −2, W2 = x e−x 1 −e−x = −e−x(x + 1), W3 = x ex 1 ex = ex(x −1). Expanding W by cofactors of the last row yields W = 0W1 −exW2 + e−xW3 = 0(−2) −ex	Elementary Differential Equations with Boundary Value Problems_Page_511_Chunk2915
502 Chapter 9 Linear Higher Order Equations Therefore W1 = y2 y3 y4 y′ 2 y′ 3 y′ 4 y′′ 2 y′′ 3 y′′ 4 , W2 = y1 y3 y4 y′ 1 y′ 3 y′ 4 y′′ 1 y′′ 3 y′′ 4 , W3 = y1 y2 y4 y′ 1 y′ 2 y′ 4 y′′ 1 y′′ 2 y′′ 4 , W4 = y1 y2 y3 y′ 1 y′ 2 y′ 3 y′′ 1 y′′ 2 y′′ 3 , and (9.4.6) becomes u′ 1 = −F W1 P0W , u′ 2 = F W2 P0W , u′ 3 = −F W3 P0W , u′ 4 = F W4 P0W . (9.4.9) Example 9.4.2 Find a particular solution of x4y(4) + 6x3y′′′ + 2x2y′′ −4xy′ + 4y = 12x2, (9.4.10) given that y1 = x, y2 = x2, y3 = 1/x and y4 = 1/x2 form a fundamental set of solutions of the complementary equation. Then ﬁnd the general solution of (9.4.10) on (−∞, 0) and (0, ∞). Solution We seek a particular solution of (9.4.10) of the form yp = u1x + u2x2 + u3 x + u4 x2 . The Wronskian of {y1, y2, y3, y4} is W(x) = x x2 1/x −1/x2 1 2x −1/x2 −2/x3 0 2 2/x3 6/x4 0 0 −6/x4 −24/x5 , so W1 = x2 1/x 1/x2 2x −1/x2 −2/x3 2 2/x3 6/x4 = −12 x4 , W2 = x 1/x 1/x2 1 −1/x2 −2/x3 0 2/x3 6/x4 = −6 x5 , W3 = x x2 1/x2 1 2x −2/x3 0 2 6/x4 = 12 x2 , W4 = x x2 1/x 1 2x −1/x2 0 2 2/x3 = 6 x.	Elementary Differential Equations with Boundary Value Problems_Page_512_Chunk2916
Section 9.4 Variation of Parameters for Higher Order Equations 503 Expanding W by cofactors of the last row yields W = −0W1 + 0W2 − −6 x4 W3 + −24 x5 W4 = 6 x4 12 x2 −24 x5 6 x = −72 x6 . Since F (x) = 12x2 and P0(x) = x4, F P0W = 12x2 x4 −x6 72 = −x4 6 . Therefore, from (9.4.9), u′ 1 = − −x4 6 W1 = x4 6 −12 x4 = −2, u′ 2 = −x4 6 W2 = −x4 6 −6 x5 = 1 x, u′ 3 = − −x4 6 W3 = x4 6 12 x2 = 2x2, u′ 4 = −x4 6 W4 = −x4 6 6 x = −x3. Integrating these and taking the constants of integration to be zero yields u1 = −2x, u2 = ln\|x\|, u3 = 2x3 3 , u4 = −x4 4 . Hence, yp = u1y1 + u2y2 + u3y3 + u4y4 = (−2x)x + (ln\|x\|)x2 + 2x3 3 1 x + −x4 4 1 x2 = x2 ln \|x\| −19x2 12 . Since −19x2/12 is a solution of the complementary equation, we redeﬁne yp = x2 ln\|x\|. Therefore y = x2 ln \|x\| + c1x + c2x2 + c3 x + c4 x2 is the general solution of (9.4.10) on (−∞, 0) and (0, ∞). 9.4 Exercises In Exercises 1–21 ﬁnd a particular solution, given the fundamental set of solutions of the complementary equation. 1. x3y′′′ −x2(x + 3)y′′ + 2x(x + 3)y′ −2(x + 3)y = −4x4; {x, x2, xex}	Elementary Differential Equations with Boundary Value Problems_Page_513_Chunk2917
504 Chapter 9 Linear Higher Order Equations 2. y′′′ + 6xy′′ + (6 + 12x2)y′ + (12x + 8x3)y = x1/2e−x2; {e−x2, xe−x2, x2e−x2} 3. x3y′′′ −3x2y′′ + 6xy′ −6y = 2x; {x, x2, x3} 4. x2y′′′ + 2xy′′ −(x2 + 2)y′ = 2x2; {1, ex/x, e−x/x} 5. x3y′′′ −3x2(x + 1)y′′ + 3x(x2 + 2x + 2)y′ −(x3 + 3x2 + 6x + 6)y = x4e−3x; {xex, x2ex, x3ex} 6. x(x2 −2)y′′′ + (x2 −6)y′′ + x(2 −x2)y′ + (6 −x2)y = 2(x2 −2)2; {ex, e−x, 1/x} 7. xy′′′ −(x −3)y′′ −(x + 2)y′ + (x −1)y = −4e−x; {ex, ex/x, e−x/x} 8. 4x3y′′′ + 4x2y′′ −5xy′ + 2y = 30x2; {√x, 1/√x, x2} 9. x(x2 −1)y′′′ + (5x2 + 1)y′′ + 2xy′ −2y = 12x2; {x, 1/(x −1), 1/(x + 1)} 10. x(1 −x)y′′′ + (x2 −3x + 3)y′′ + xy′ −y = 2(x −1)2; {x, 1/x, ex/x} 11. x3y′′′ + x2y′′ −2xy′ + 2y = x2; {x, x2, 1/x} 12. xy′′′ −y′′ −xy′ + y = x2; {x, ex, e−x} 13. xy(4) + 4y′′′ = 6 ln\|x\|; {1, x, x2, 1/x} 14. 16x4y(4) + 96x3y′′′ + 72x2y′′ −24xy′ + 9y = 96x5/2; {√x, 1/√x, x3/2, x−3/2} 15. x(x2 −6)y(4) + 2(x2 −12)y′′′ + x(6 −x2)y′′ + 2(12 −x2)y′ = 2(x2 −6)2; {1, 1/x, ex, e−x} 16. x4y(4) −4x3y′′′ + 12x2y′′ −24xy′ + 24y = x4; {x, x2, x3, x4} 17. x4y(4) −4x3y′′′ + 2x2(6 −x2)y′′ + 4x(x2 −6)y′ + (x4 −4x2 + 24)y = 4x5ex; {xex, x2ex, xe−x, x2e−x} 18. x4y(4) + 6x3y′′′ + 2x2y′′ −4xy′ + 4y = 12x2; {x, x2, 1/x, 1/x2} 19. xy(4) + 4y′′′ −2xy′′ −4y′ + xy = 4ex; {ex, e−x, ex/x, e−x/x} 20. xy(4)+(4−6x)y′′′+(13x−18)y′′+(26−12x)y′+(4x−12)y = 3ex; {ex, e2x, ex/x, e2x/x} 21. x4y(4) −4x3y′′′ + x2(12 −x2)y′′ + 2x(x2 −12)y′ + 2(12 −x2)y = 2x5; {x, x2, xex, xe−x} In Exercises 22–33 solve the initial value problem, given the fundamental set of solutions of the comple- mentary equation. Where indicated by C/G , graph the solution. 22. C/G x3y′′′−2x2y′′+3xy′−3y = 4x, y(1) = 4, y′(1) = 4, y′′(1) = 2; {x, x3, x lnx} 23. x3y′′′ −5x2y′′ + 14xy′ −18y = x3, y(1) = 0, y′(1) = 1, y′′(1) = 7; {x2, x3, x3 lnx} 24. (5 −6x)y′′′ + (12x −4)y′′ + (6x −23)y′ + (22 −12x)y = −(6x −5)2ex y(0) = −4, y′(0) = −3 2, y′′(0) = −19; {ex, e2x, xe−x} 25. x3y′′′ −6x2y′′ + 16xy′ −16y = 9x4, y(1) = 2, y′(1) = 1, y′′(1) = 5; {x, x4, x4 ln \|x\|} 26. C/G (x2 −2x + 2)y′′′ −x2y′′ + 2xy′ −2y = (x2 −2x + 2)2, y(0) = 0, y′(0) = 5, y′′(0) = 0; {x, x2, ex} 27. x3y′′′ + x2y′′ −2xy′ + 2y = x(x + 1), y(−1) = −6, y′(−1) = 43 6 , y′′(−1) = −5 2; {x, x2, 1/x}	Elementary Differential Equations with Boundary Value Problems_Page_514_Chunk2918
Section 9.4 Variation of Parameters for Higher Order Equations 505 28. (3x −1)y′′′ −(12x −1)y′′ + 9(x + 1)y′ −9y = 2ex(3x −1)2, y(0) = 3 4, y′(0) = 5 4, y′′(0) = 1 4; {x + 1, ex, e3x} 29. C/G (x2 −2)y′′′ −2xy′′ + (2 −x2)y′ + 2xy = 2(x2 −2)2, y(0) = 1, y′(0) = −5, y′′(0) = 5; {x2, ex, e−x} 30. C/G x4y(4) +3x3y′′′ −x2y′′ +2xy′ −2y = 9x2, y(1) = −7, y′(1) = −11, y′′(1) = −5, y′′′(1) = 6; {x, x2, 1/x, x ln x} 31. (2x −1)y(4) −4xy′′′ + (5 −2x)y′′ + 4xy′ −4y = 6(2x −1)2, y(0) = 55 4 , y′(0) = 0, y′′(0) = 13, y′′′(0) = 1; {x, ex, e−x, e2x} 32. 4x4y(4) +24x3y′′′ +23x2y′′ −xy′ +y = 6x, y(1) = 2, y′(1) = 0, y′′(1) = 4, y′′′(1) = −37 4 ; {x, √x, 1/x, 1/√x} 33. x4y(4) + 5x3y′′′ −3x2y′′ −6xy′ + 6y = 40x3, y(−1) = −1, y′(−1) = −7, y′′(−1) = −1, y′′′(−1) = −31; {x, x3, 1/x, 1/x2} 34. Suppose the equation P0(x)y(n) + P1(x)y(n−1) + · · · + Pn(x)y = F (x) (A) is normal on an interval (a, b). Let {y1, y2, . . ., yn} be a fundamental set of solutions of its com- plementary equation on (a, b), let W be the Wronskian of {y1, y2, . . ., yn}, and let Wj be the determinant obtained by deleting the last row and the j-th column of W. Suppose x0 is in (a, b), let uj(x) = (−1)(n−j) Z x x0 F (t)Wj(t) P0(t)W(t) dt, 1 ≤j ≤n, and deﬁne yp = u1y1 + u2y2 + · · · + unyn. (a) Show that yp is a solution of (A) and that y(r) p = u1y(r) 1 + u2y(r) 2 · · · + uny(r) n , 1 ≤r ≤n −1, and y(n) p = u1y(n) 1 + u2y(n) 2 + · · · + uny(n) n + F P0 . HINT: See the derivation of the method of variation of parameters at the beginning of the section. (b) Show that yp is the solution of the initial value problem P0(x)y(n) + P1(x)y(n−1) + · · · + Pn(x)y = F (x), y(x0) = 0, y′(x0) = 0, . . ., y(n−1)(x0) = 0. (c) Show that yp can be written as yp(x) = Z x x0 G(x, t)F (t) dt,	Elementary Differential Equations with Boundary Value Problems_Page_515_Chunk2919
506 Chapter 9 Linear Higher Order Equations where G(x, t) = 1 P0(t)W(t) y1(t) y2(t) · · · yn(t) y′ 1(t) y′ 2(t) · · · y′ n(t) ... ... ... ... y(n−2) 1 (t) y(n−2) 2 (t) · · · y(n−2) n (t) y1(x) y2(x) · · · yn(x) , which is called the Green’s function for (A). (d) Show that ∂jG(x, t) ∂xj = 1 P0(t)W(t) y1(t) y2(t) · · · yn(t) y′ 1(t) y′ 2(t) · · · y′ n(t) ... ... ... ... y(n−2) 1 (t) y(n−2) 2 (t) · · · y(n−2) n (t) y(j) 1 (x) y(j) 2 (x) · · · y(j) n (x) , 0 ≤j ≤n. (e) Show that if a < t < b then ∂jG(x, t) ∂xj x=t =    0, 1 ≤j ≤n −2, 1 P0(t), j = n −1. (f) Show that y(j) p (x) =          Z x x0 ∂jG(x, t) ∂xj F (t) dt, 1 ≤j ≤n −1, F (x) P0(x) + Z x x0 ∂(n)G(x, t) ∂xn F (t) dt, j = n. In Exercises 35–42 use the method suggested by Exercise 34 to ﬁnd a particular solution in the form yp = R x x0 G(x, t)F (t) dt, given the indicated fundamental set of solutions. Assume that x and x0 are in an interval on which the equation is normal. 35. y′′′ + 2y′ −y′ −2y = F (x); {ex, e−x, e−2x} 36. x3y′′′ + x2y′′ −2xy′ + 2y = F (x); {x, x2, 1/x} 37. x3y′′′ −x2(x + 3)y′′ + 2x(x + 3)y′ −2(x + 3)y = F (x); {x, x2, xex} 38. x(1 −x)y′′′ + (x2 −3x + 3)y′′ + xy′ −y = F (x); {x, 1/x, ex/x} 39. y(4) −5y′′ + 4y = F (x); {ex, e−x, e2x, e−2x} 40. xy(4) + 4y′′′ = F (x); {1, x, x2, 1/x} 41. x4y(4) + 6x3y′′′ + 2x2y′′ −4xy′ + 4y = F (x); {x, x2, 1/x, 1/x2} 42. xy(4) −y′′′ −4xy′ + 4y′ = F (x); {1, x2, e2x, e−2x}	Elementary Differential Equations with Boundary Value Problems_Page_516_Chunk2920
CHAPTER 10 Linear Systems of Differential Equations IN THIS CHAPTER we consider systems of differential equations involving more than one unknown function. Such systems arise in many physical applications. SECTION 10.1 presents examples of physical situations that lead to systems of differential equations. SECTION 10.2 discusses linear systems of differential equations. SECTION 10.3 deals with the basic theory of homogeneous linear systems. SECTIONS 10.4, 10.5, AND 10.6 present the theory of constant coefﬁcient homogeneous systems. SECTION 10.7 presents the method of variation of parameters for nonhomogeneous linear systems. 507	Elementary Differential Equations with Boundary Value Problems_Page_517_Chunk2921
508 Chapter 10 Linear Systems of Differential Equations 10.1 INTRODUCTION TO SYSTEMS OF DIFFERENTIAL EQUATIONS Many physical situations are modelled by systems of n differential equations in n unknown functions, where n ≥2. The next three examples illustrate physical problems that lead to systems of differential equations. In these examples and throughout this chapter we’ll denote the independent variable by t. Example 10.1.1 Tanks T1 and T2 contain 100 gallons and 300 gallons of salt solutions, respectively. Salt solutions are simultaneously added to both tanks from external sources, pumped from each tank to the other, and drained from both tanks (Figure 10.1.1). A solution with 1 pound of salt per gallon is pumped into T1 from an external source at 5 gal/min, and a solution with 2 pounds of salt per gallon is pumped into T2 from an external source at 4 gal/min. The solution from T1 is pumped into T2 at 2 gal/min, and the solution from T2 is pumped into T1 at 3 gal/min. T1 is drained at 6 gal/min and T2 is drained at 3 gal/min. Let Q1(t) and Q2(t) be the number of pounds of salt in T1 and T2, respectively, at time t > 0. Derive a system of differential equations for Q1 and Q2. Assume that both mixtures are well stirred. 6 gal/min 3 gal/min 2 gal/min 3 gal/min T1 T2 5 gal/min; 1 lb/gal 4 gal/min; 2 lb/gal 300 gal 100 gal Figure 10.1.1 Solution As in Section 4.2, let rate in and rate out denote the rates (lb/min) at which salt enters and leaves a tank; thus, Q′ 1 = (rate in)1 −(rate out)1, Q′ 2 = (rate in)2 −(rate out)2. Note that the volumes of the solutions in T1 and T2 remain constant at 100 gallons and 300 gallons, respectively.	Elementary Differential Equations with Boundary Value Problems_Page_518_Chunk2922
Section 10.1 Introduction to Systems of Differential Equations 509 T1 receives salt from the external source at the rate of (1 lb/gal) × (5 gal/min) = 5 lb/min, and from T2 at the rate of (lb/gal in T2) × (3 gal/min) = 1 300Q2 × 3 = 1 100Q2 lb/min. Therefore (rate in)1 = 5 + 1 100Q2. (10.1.1) Solution leaves T1 at the rate of 8 gal/min, since 6 gal/min are drained and 2 gal/min are pumped to T2; hence, (rate out)1 = ( lb/gal in T1) × (8 gal/min) = 1 100Q1 × 8 = 2 25Q1. (10.1.2) Eqns. (10.1.1) and (10.1.2) imply that Q′ 1 = 5 + 1 100Q2 −2 25Q1. (10.1.3) T2 receives salt from the external source at the rate of (2 lb/gal) × (4 gal/min) = 8 lb/min, and from T1 at the rate of (lb/gal in T1) × (2 gal/min) = 1 100Q1 × 2 = 1 50Q1 lb/min. Therefore (rate in)2 = 8 + 1 50Q1. (10.1.4) Solution leaves T2 at the rate of 6 gal/min, since 3 gal/min are drained and 3 gal/min are pumped to T1; hence, (rate out)2 = ( lb/gal in T2) × (6 gal/min) = 1 300Q2 × 6 = 1 50Q2. (10.1.5) Eqns. (10.1.4) and (10.1.5) imply that Q′ 2 = 8 + 1 50Q1 −1 50Q2. (10.1.6) We say that (10.1.3) and (10.1.6) form a system of two ﬁrst order equations in two unknowns, and write them together as Q′ 1 = 5 −2 25Q1 + 1 100Q2 Q′ 2 = 8 + 1 50Q1 −1 50Q2. Example 10.1.2 A mass m1 is suspended from a rigid support on a spring S1 and a second mass m2 is suspended from the ﬁrst on a spring S2 (Figure 10.1.2). The springs obey Hooke’s law, with spring constants k1 and k2. Internal friction causes the springs to exert damping forces proportional to the rates of change of their lengths, with damping constants c1 and c2. Let y1 = y1(t) and y2 = y2(t) be the displacements of the two masses from their equilibrium positions at time t, measured positive upward. Derive a system of differential equations for y1 and y2, assuming that the masses of the springs are negligible and that vertical external forces F1 and F2 also act on the objects.	Elementary Differential Equations with Boundary Value Problems_Page_519_Chunk2923
510 Chapter 10 Linear Systems of Differential Equations Mass m1 Mass m2 y1 y2 Spring S1 Spring S2 Figure 10.1.2 Solution In equilibrium, S1 supports both m1 and m2 and S2 supports only m2. Therefore, if ∆ℓ1 and ∆ℓ2 are the elongations of the springs in equilibrium then (m1 + m2)g = k1∆ℓ1 and m2g = k2∆ℓ2. (10.1.7) Let H1 be the Hooke’s law force acting on m1, and let D1 be the damping force on m1. Similarly, let H2 and D2 be the Hooke’s law and damping forces acting on m2. According to Newton’s second law of motion, m1y′′ 1 = −m1g + H1 + D1 + F1, m2y′′ 2 = −m2g + H2 + D2 + F2. (10.1.8) When the displacements are y1 and y2, the change in length of S1 is −y1 + ∆ℓ1 and the change in length of S2 is −y2 + y1 + ∆ℓ2. Both springs exert Hooke’s law forces on m1, while only S2 exerts a Hooke’s law force on m2. These forces are in directions that tend to restore the springs to their natural lengths. Therefore H1 = k1(−y1 + ∆ℓ1) −k2(−y2 + y1 + ∆ℓ2) and H2 = k2(−y2 + y1 + ∆ℓ2). (10.1.9) When the velocities are y′ 1 and y′ 2, S1 and S2 are changing length at the rates −y′ 1 and −y′ 2 + y′ 1, respec- tively. Both springs exert damping forces on m1, while only S2 exerts a damping force on m2. Since the force due to damping exerted by a spring is proportional to the rate of change of length of the spring and in a direction that opposes the change, it follows that D1 = −c1y′ 1 + c2(y′ 2 −y′ 1) and D2 = −c2(y′ 2 −y′ 1). (10.1.10)	Elementary Differential Equations with Boundary Value Problems_Page_520_Chunk2924
Section 10.1 Introduction to Systems of Differential Equations 511 From (10.1.8), (10.1.9), and (10.1.10), m1y′′ 1 = −m1g + k1(−y1 + ∆ℓ1) −k2(−y2 + y1 + ∆ℓ2) −c1y′ 1 + c2(y′ 2 −y′ 1) + F1 = −(m1g −k1∆ℓ1 + k2∆ℓ2) −k1y1 + k2(y2 −y1) −c1y′ 1 + c2(y′ 2 −y′ 1) + F1 (10.1.11) and m2y′′ 2 = −m2g + k2(−y2 + y1 + ∆ℓ2) −c2(y′ 2 −y′ 1) + F2 = −(m2g −k2∆ℓ2) −k2(y2 −y1) −c2(y′ 2 −y′ 1) + F2. (10.1.12) From (10.1.7), m1g −k1∆ℓ1 + k2∆ℓ2 = −m2g + k2∆ℓ2 = 0. Therefore we can rewrite (10.1.11) and (10.1.12) as m1y′′ 1 = −(c1 + c2)y′ 1 + c2y′ 2 −(k1 + k2)y1 + k2y2 + F1 m2y′′ 2 = c2y′ 1 −c2y′ 2 + k2y1 −k2y2 + F2. Example 10.1.3 Let X = X(t) = x(t) i + y(t) j + z(t) k be the position vector at time t of an object with mass m, relative to a rectangular coordinate system with origin at Earth’s center (Figure 10.1.3). According to Newton’s law of gravitation, Earth’s gravitational force F = F(x, y, z) on the object is inversely proportional to the square of the distance of the object from Earth’s center, and directed toward the center; thus, F = K ∥X∥2 −X ∥X∥ = −K x i + y j + z k (x2 + y2 + z2)3/2 , (10.1.13) where K is a constant. To determine K, we observe that the magnitude of F is ∥F∥= K ∥X∥ ∥X∥3 = K ∥X∥2 = K (x2 + y2 + z2). Let R be Earth’s radius. Since ∥F∥= mg when the object is at Earth’s surface, mg = K R2 , so K = mgR2. Therefore we can rewrite (10.1.13) as F = −mgR2 x i + y j + z k (x2 + y2 + z2)3/2 . Now suppose F is the only force acting on the object. According to Newton’s second law of motion, F = mX′′; that is, m(x′′ i + y′′ j + z′′ k) = −mgR2 x i + y j + z k (x2 + y2 + z2)3/2 . Cancelling the common factor m and equating components on the two sides of this equation yields the system x′′ = − gR2x (x2 + y2 + z2)3/2 y′′ = − gR2y (x2 + y2 + z2)3/2 z′′ = − gR2z (x2 + y2 + z2)3/2 . (10.1.14)	Elementary Differential Equations with Boundary Value Problems_Page_521_Chunk2925
512 Chapter 10 Linear Systems of Differential Equations x y z X(t) Figure 10.1.3 Rewriting Higher Order Systems as First Order Systems A system of the form y′ 1 = g1(t, y1, y2, . . ., yn) y′ 2 = g2(t, y1, y2, . . ., yn) ... y′ n = gn(t, y1, y2, . . ., yn) (10.1.15) is called a ﬁrst order system, since the only derivatives occurring in it are ﬁrst derivatives. The derivative of each of the unknowns may depend upon the independent variable and all the unknowns, but not on the derivatives of other unknowns. When we wish to emphasize the number of unknown functions in (10.1.15) we will say that (10.1.15) is an n × n system. Systems involving higher order derivatives can often be reformulated as ﬁrst order systems by intro- ducing additional unknowns. The next two examples illustrate this. Example 10.1.4 Rewrite the system m1y′′ 1 = −(c1 + c2)y′ 1 + c2y′ 2 −(k1 + k2)y1 + k2y2 + F1 m2y′′ 2 = c2y′ 1 −c2y′ 2 + k2y1 −k2y2 + F2. (10.1.16) derived in Example 10.1.2 as a system of ﬁrst order equations. Solution If we deﬁne v1 = y′ 1 and v2 = y′ 2, then v′ 1 = y′′ 1 and v′ 2 = y′′ 2 , so (10.1.16) becomes m1v′ 1 = −(c1 + c2)v1 + c2v2 −(k1 + k2)y1 + k2y2 + F1 m2v′ 2 = c2v1 −c2v2 + k2y1 −k2y2 + F2.	Elementary Differential Equations with Boundary Value Problems_Page_522_Chunk2926
Section 10.1 Introduction to Systems of Differential Equations 513 Therefore {y1, y2, v1, v2} satisﬁes the 4 × 4 ﬁrst order system y′ 1 = v1 y′ 2 = v2 v′ 1 = 1 m1 [−(c1 + c2)v1 + c2v2 −(k1 + k2)y1 + k2y2 + F1] v′ 2 = 1 m2 [c2v1 −c2v2 + k2y1 −k2y2 + F2] . (10.1.17) REMARK: The difference in form between (10.1.15) and (10.1.17), due to the way in which the unknowns are denoted in the two systems, isn’t important; (10.1.17) is a ﬁrst order system, in that each equation in (10.1.17) expresses the ﬁrst derivative of one of the unknown functions in a way that does not involve derivatives of any of the other unknowns. Example 10.1.5 Rewrite the system x′′ = f(t, x, x′, y, y′, y′′) y′′′ = g(t, x, x′, y, y′y′′) as a ﬁrst order system. Solution We regard x, x′, y, y′, and y′′ as unknown functions, and rename them x = x1, x′ = x2, y = y1, y′ = y2, y′′ = y3. These unknowns satisfy the system x′ 1 = x2 x′ 2 = f(t, x1, x2, y1, y2, y3) y′ 1 = y2 y′ 2 = y3 y′ 3 = g(t, x1, x2, y1, y2, y3). Rewriting Scalar Differential Equations as Systems In this chapter we’ll refer to differential equations involving only one unknown function as scalar differ- ential equations. Scalar differential equations can be rewritten as systems of ﬁrst order equations by the method illustrated in the next two examples. Example 10.1.6 Rewrite the equation y(4) + 4y′′′ + 6y′′ + 4y′ + y = 0 (10.1.18) as a 4 × 4 ﬁrst order system. Solution We regard y, y′, y′′, and y′′′ as unknowns and rename them y = y1, y′ = y2, y′′ = y3, and y′′′ = y4. Then y(4) = y′ 4, so (10.1.18) can be written as y′ 4 + 4y4 + 6y3 + 4y2 + y1 = 0.	Elementary Differential Equations with Boundary Value Problems_Page_523_Chunk2927
514 Chapter 10 Linear Systems of Differential Equations Therefore {y1, y2, y3, y4} satisﬁes the system y′ 1 = y2 y′ 2 = y3 y′ 3 = y4 y′ 4 = −4y4 −6y3 −4y2 −y1. Example 10.1.7 Rewrite x′′′ = f(t, x, x′, x′′) as a system of ﬁrst order equations. Solution We regard x, x′, and x′′ as unknowns and rename them x = y1, x′ = y2, and x′′ = y3. Then y′ 1 = x′ = y2, y′ 2 = x′′ = y3, and y′ 3 = x′′′. Therefore {y1, y2, y3} satisﬁes the ﬁrst order system y′ 1 = y2 y′ 2 = y3 y′ 3 = f(t, y1, y2, y3). Since systems of differential equations involving higher derivatives can be rewritten as ﬁrst order sys- tems by the method used in Examples 10.1.5 –10.1.7 , we’ll consider only ﬁrst order systems. Numerical Solution of Systems The numerical methods that we studied in Chapter 3 can be extended to systems, and most differential equation software packages include programs to solve systems of equations. We won’t go into detail on numerical methods for systems; however, for illustrativepurposes we’ll describe the Runge-Kutta method for the numerical solution of the initial value problem y′ 1 = g1(t, y1, y2), y1(t0) = y10, y′ 2 = g2(t, y1, y2), y2(t0) = y20 at equally spaced points t0, t1, ..., tn = b in an interval [t0, b]. Thus, ti = t0 + ih, i = 0, 1, . . ., n, where h = b −t0 n . We’ll denote the approximate values of y1 and y2 at these pointsby y10, y11, . . ., y1n and y20, y21, . . ., y2n.	Elementary Differential Equations with Boundary Value Problems_Page_524_Chunk2928
Section 10.1 Introduction to Systems of Differential Equations 515 The Runge-Kutta method computes these approximate values as follows: given y1i and y2i, compute I1i = g1(ti, y1i, y2i), J1i = g2(ti, y1i, y2i), I2i = g1 ti + h 2 , y1i + h 2I1i, y2i + h 2 J1i , J2i = g2 ti + h 2 , y1i + h 2I1i, y2i + h 2 J1i , I3i = g1 ti + h 2 , y1i + h 2I2i, y2i + h 2 J2i , J3i = g2 ti + h 2 , y1i + h 2I2i, y2i + h 2 J2i , I4i = g1(ti + h, y1i + hI3i, y2i + hJ3i), J4i = g2(ti + h, y1i + hI3i, y2i + hJ3i), and y1,i+1 = y1i + h 6 (I1i + 2I2i + 2I3i + I4i), y2,i+1 = y2i + h 6 (J1i + 2J2i + 2J3i + J4i) for i = 0, ..., n−1. Under appropriate conditions on g1 and g2, it can be shown that the global truncation error for the Runge-Kutta method is O(h4), as in the scalar case considered in Section 3.3. 10.1 Exercises 1. Tanks T1 and T2 contain 50 gallons and 100 gallons of salt solutions, respectively. A solution with 2 pounds of salt per gallon is pumped into T1 from an external source at 1 gal/min, and a solution with 3 pounds of salt per gallon is pumped into T2 from an external source at 2 gal/min. The solution from T1 is pumped into T2 at 3 gal/min, and the solution from T2 is pumped into T1 at 4 gal/min. T1 is drained at 2 gal/min and T2 is drained at 1 gal/min. Let Q1(t) and Q2(t) be the number of pounds of salt in T1 and T2, respectively, at time t > 0. Derive a system of differential equations for Q1 and Q2. Assume that both mixtures are well stirred. 2. Two 500 gallon tanks T1 and T2 initially contain 100 gallons each of salt solution. A solution with 2 pounds of salt per gallon is pumped into T1 from an external source at 6 gal/min, and a solution with 1 pound of salt per gallon is pumped into T2 from an external source at 5 gal/min. The solution from T1 is pumped into T2 at 2 gal/min, and the solution from T2 is pumped into T1 at 1 gal/min. Both tanks are drained at 3 gal/min. Let Q1(t) and Q2(t) be the number of pounds of salt in T1 and T2, respectively, at time t > 0. Derive a system of differential equations for Q1 and Q2 that’s valid until a tank is about to overﬂow. Assume that both mixtures are well stirred. 3. A mass m1 is suspended from a rigid support on a spring S1 with spring constant k1 and damping constant c1. A second mass m2 is suspended from the ﬁrst on a spring S2 with spring constant k2 and damping constant c2, and a third mass m3 is suspended from the second on a spring S3 with spring constant k3 and damping constant c3. Let y1 = y1(t), y2 = y2(t), and y3 = y3(t) be the displacements of the three masses from their equilibrium positions at time t, measured positive upward. Derive a system of differential equations for y1, y2 and y3, assuming that the masses of the springs are negligible and that vertical external forces F1, F2, and F3 also act on the masses.	Elementary Differential Equations with Boundary Value Problems_Page_525_Chunk2929
516 Chapter 10 Linear Systems of Differential Equations 4. Let X = x i + y j + z k be the position vector of an object with mass m, expressed in terms of a rectangular coordinate system with origin at Earth’s center (Figure 10.1.3). Derive a system of dif- ferential equations for x, y, and z, assuming that the object moves under Earth’s gravitational force (given by Newton’s law of gravitation, as in Example 10.1.3 ) and a resistive force proportional to the speed of the object. Let α be the constant of proportionality. 5. Rewrite the given system as a ﬁrst order system. (a) x′′′ = f(t, x, y, y′) y′′ = g(t, y, y′) (b) u′ = f(t, u, v, v′, w′) v′′ = g(t, u, v, v′, w) w′′ = h(t, u, v, v′, w, w′) (c) y′′′ = f(t, y, y′, y′′) (d) y(4) = f(t, y) (e) x′′ = f(t, x, y) y′′ = g(t, x, y) 6. Rewrite the system (10.1.14) of differential equations derived in Example 10.1.3 as a ﬁrst order system. 7. Formulate a version of Euler’s method (Section 3.1) for the numerical solution of the initial value problem y′ 1 = g1(t, y1, y2), y1(t0) = y10, y′ 2 = g2(t, y1, y2), y2(t0) = y20, on an interval [t0, b]. 8. Formulate a version of the improved Euler method (Section 3.2) for the numerical solution of the initial value problem y′ 1 = g1(t, y1, y2), y1(t0) = y10, y′ 2 = g2(t, y1, y2), y2(t0) = y20, on an interval [t0, b]. 10.2 LINEAR SYSTEMS OF DIFFERENTIAL EQUATIONS A ﬁrst order system of differential equations that can be written in the form y′ 1 = a11(t)y1 + a12(t)y2 + · · · + a1n(t)yn + f1(t) y′ 2 = a21(t)y1 + a22(t)y2 + · · · + a2n(t)yn + f2(t) ... y′ n = an1(t)y1 + an2(t)y2 + · · · + ann(t)yn + fn(t) (10.2.1) is called a linear system. The linear system (10.2.1) can be written in matrix form as   y′ 1 y′ 2... y′ n  =   a11(t) a12(t) · · · a1n(t) a21(t) a22(t) · · · a2n(t) ... ... ... ... an1(t) an2(t) · · · ann(t)     y1 y2 ... yn  +   f1(t) f2(t) ... fn(t)  ,	Elementary Differential Equations with Boundary Value Problems_Page_526_Chunk2930
Section 10.2 Linear Systems of Differential Equations 517 or more brieﬂy as y′ = A(t)y + f(t), (10.2.2) where y =   y1 y2 ... yn  , A(t) =   a11(t) a12(t) · · · a1n(t) a21(t) a22(t) · · · a2n(t) ... ... ... ... an1(t) an2(t) · · · ann(t)  , and f(t) =   f1(t) f2(t) ... fn(t)  . We call A the coefﬁcient matrix of (10.2.2) and f the forcing function. We’ll say that A and f are con- tinuous if their entries are continuous. If f = 0, then (10.2.2) is homogeneous; otherwise, (10.2.2) is nonhomogeneous. An initial value problem for (10.2.2) consists of ﬁnding a solution of (10.2.2) that equals a given constant vector k =   k1 k2 ... kn  . at some initial point t0. We write this initial value problem as y′ = A(t)y + f(t), y(t0) = k. The next theorem gives sufﬁcient conditions for the existence of solutions of initial value problems for (10.2.2). We omit the proof. Theorem 10.2.1 Suppose the coefﬁcient matrix A and the forcing function f are continuous on (a, b), let t0 be in (a, b), and let k be an arbitrary constant n-vector. Then the initial value problem y′ = A(t)y + f(t), y(t0) = k has a unique solution on (a, b). Example 10.2.1 (a) Write the system y′ 1 = y1 + 2y2 + 2e4t y′ 2 = 2y1 + y2 + e4t (10.2.3) in matrix form and conclude from Theorem 10.2.1 that every initial value problem for (10.2.3) has a unique solution on (−∞, ∞). (b) Verify that y = 1 5 8 7 e4t + c1 1 1 e3t + c2 1 −1 e−t (10.2.4) is a solution of (10.2.3) for all values of the constants c1 and c2. (c) Find the solution of the initial value problem y′ = 1 2 2 1 y + 2 1 e4t, y(0) = 1 5 3 22 . (10.2.5)	Elementary Differential Equations with Boundary Value Problems_Page_527_Chunk2931
518 Chapter 10 Linear Systems of Differential Equations SOLUTION(a) The system (10.2.3) can be written in matrix form as y′ = 1 2 2 1 y + 2 1 e4t. An initial value problem for (10.2.3) can be written as y′ = 1 2 2 1 y + 2 1 e4t, y(t0) = k1 k2 . Since the coefﬁcient matrix and the forcing function are both continuous on (−∞, ∞), Theorem 10.2.1 implies that this problem has a unique solution on (−∞, ∞). SOLUTION(b) If y is given by (10.2.4), then Ay + f = 1 5 1 2 2 1 8 7 e4t + c1 1 2 2 1 1 1 e3t +c2 1 2 2 1 1 −1 e−t + 2 1 e4t = 1 5 22 23 e4t + c1 3 3 e3t + c2 −1 1 e−t + 2 1 e4t = 1 5 32 28 e4t + 3c1 1 1 e3t −c2 1 −1 e−t = y′. SOLUTION(c) We must choose c1 and c2 in (10.2.4) so that 1 5 8 7 + c1 1 1 + c2 1 −1 = 1 5 3 22 , which is equivalent to 1 1 1 −1 c1 c2 = −1 3 . Solving this system yields c1 = 1, c2 = −2, so y = 1 5 8 7 e4t + 1 1 e3t −2 1 −1 e−t is the solution of (10.2.5). REMARK: The theory of n × n linear systems of differential equations is analogous to the theory of the scalar n-th order equation P0(t)y(n) + P1(t)y(n−1) + · · · + Pn(t)y = F (t), (10.2.6) as developed in Sections 9.1. For example, by rewriting (10.2.6) as an equivalent linear system it can be shown that Theorem 10.2.1 implies Theorem 9.1.1 (Exercise 12).	Elementary Differential Equations with Boundary Value Problems_Page_528_Chunk2932
Section 10.2 Linear Systems of Differential Equations 519 10.2 Exercises 1. Rewrite the system in matrix form and verify that the given vector function satisﬁes the system for any choice of the constants c1 and c2. (a) y′ 1 = 2y1 + 4y2 y′ 2 = 4y1 + 2y2; y = c1 1 1 e6t + c2 1 −1 e−2t (b) y′ 1 = −2y1 −2y2 y′ 2 = −5y1 + y2; y = c1 1 1 e−4t + c2 −2 5 e3t (c) y′ 1 = −4y1 −10y2 y′ 2 = 3y1 + 7y2; y = c1 −5 3 e2t + c2 2 −1 et (d) y′ 1 = 2y1 + y2 y′ 2 = y1 + 2y2; y = c1 1 1 e3t + c2 1 −1 et 2. Rewrite the system in matrix form and verify that the given vector function satisﬁes the system for any choice of the constants c1, c2, and c3. (a) y′ 1 = −y1 + 2y2 + 3y3 y′ 2 = y2 + 6y3 y′ 3 = −2y3; y = c1   1 1 0  et + c2   1 0 0  e−t + c3   1 −2 1  e−2t (b) y′ 1 = 2y2 + 2y3 y′ 2 = 2y1 + 2y3 y′ 3 = 2y1 + 2y2; y = c1   −1 0 1  e−2t + c2   0 −1 1  e−2t + c3   1 1 1  e4t (c) y′ 1 = −y1 + 2y2 + 2y3 y′ 2 = 2y1 −y2 + 2y3 y′ 3 = 2y1 + 2y2 −y3; y = c1   −1 0 1  e−3t + c2   0 −1 1  e−3t + c3   1 1 1  e3t (d) y′ 1 = 3y1 −y2 −y3 y′ 2 = −2y1 + 3y2 + 2y3 y′ 3 = 4y1 −y2 −2y3; y = c1   1 0 1  e2t + c2   1 −1 1  e3t + c3   1 −3 7  e−t 3. Rewrite the initial value problem in matrix form and verify that the given vector function is a solution. (a) y′ 1 = y1 + y2 y′ 2 = −2y1 + 4y2, y1(0) = 1 y2(0) = 0; y = 2 1 1 e2t − 1 2 e3t (b) y′ 1 = 5y1 + 3y2 y′ 2 = −y1 + y2, y1(0) = 12 y2(0) = −6; y = 3 1 −1 e2t + 3 3 −1 e4t	Elementary Differential Equations with Boundary Value Problems_Page_529_Chunk2933
520 Chapter 10 Linear Systems of Differential Equations 4. Rewrite the initial value problem in matrix form and verify that the given vector function is a solution. (a) y′ 1 = 6y1 + 4y2 + 4y3 y′ 2 = −7y1 −2y2 −y3, y′ 3 = 7y1 + 4y2 + 3y3 , y1(0) = 3 y2(0) = −6 y3(0) = 4 y =   1 −1 1  e6t + 2   1 −2 1  e2t +   0 −1 1  e−t (b) y′ 1 = 8y1 + 7y2 + 7y3 y′ 2 = −5y1 −6y2 −9y3, y′ 3 = 5y1 + 7y2 + 10y3, y1(0) = 2 y2(0) = −4 y3(0) = 3 y =   1 −1 1  e8t +   0 −1 1  e3t +   1 −2 1  et 5. Rewrite the system in matrix form and verify that the given vector function satisﬁes the system for any choice of the constants c1 and c2. (a) y′ 1 = −3y1 + 2y2 + 3 −2t y′ 2 = −5y1 + 3y2 + 6 −3t y = c1 2 cos t 3 cos t −sin t + c2 2 sint 3 sint + cos t + 1 t (b) y′ 1 = 3y1 + y2 −5et y′ 2 = −y1 + y2 + et y = c1 −1 1 e2t + c2 1 + t −t e2t + 1 3 et (c) y′ 1 = −y1 −4y2 + 4et + 8tet y′ 2 = −y1 −y2 + e3t + (4t + 2)et y = c1 2 1 e−3t + c2 −2 1 et + e3t 2tet (d) y′ 1 = −6y1 −3y2 + 14e2t + 12et y′ 2 = y1 −2y2 + 7e2t −12et y = c1 −3 1 e−5t + c2 −1 1 e−3t + e2t + 3et 2e2t −3et 6. Convert the linear scalar equation P0(t)y(n) + P1(t)y(n−1) + · · · + Pn(t)y(t) = F (t) (A) into an equivalent n × n system y′ = A(t)y + f(t), and show that A and f are continuous on an interval (a, b) if and only if (A) is normal on (a, b). 7. A matrix function Q(t) =   q11(t) q12(t) · · · q1s(t) q21(t) q22(t) · · · q2s(t) ... ... ... ... qr1(t) qr2(t) · · · qrs(t)  	Elementary Differential Equations with Boundary Value Problems_Page_530_Chunk2934
Section 10.2 Linear Systems of Differential Equations 521 is said to be differentiable if its entries {qij} are differentiable. Then the derivative Q′ is deﬁned by Q′(t) =   q′ 11(t) q′ 12(t) · · · q′ 1s(t) q′ 21(t) q′ 22(t) · · · q′ 2s(t) ... ... ... ... q′ r1(t) q′ r2(t) · · · q′ rs(t)  . (a) Prove: If P and Q are differentiable matrices such that P + Q is deﬁned and if c1 and c2 are constants, then (c1P + c2Q)′ = c1P ′ + c2Q′. (b) Prove: If P and Q are differentiable matrices such that P Q is deﬁned, then (P Q)′ = P ′Q + P Q′. 8. Verify that Y ′ = AY . (a) Y = e6t e−2t e6t −e−2t , A = 2 4 4 2 (b) Y = e−4t −2e3t e−4t 5e3t , A = −2 −2 −5 1 (c) Y = −5e2t 2et 3e2t −et , A = −4 −10 3 7 (d) Y = e3t et e3t −et , A = 2 1 1 2 (e) Y =   et e−t e−2t et 0 −2e−2t 0 0 e−2t  , A =   −1 2 3 0 1 6 0 0 −2   (f) Y =   −e−2t −e−2t e4t 0 e−2t e4t e−2t 0 e4t  , A =   0 2 2 2 0 2 2 2 0   (g) Y =   e3t e−3t 0 e3t 0 −e−3t e3t e−3t e−3t  , A =   −9 6 6 −6 3 6 −6 6 3   (h) Y =   e2t e3t e−t 0 −e3t −3e−t e2t e3t 7e−t  , A =   3 −1 −1 −2 3 2 4 −1 −2   9. Suppose y1 = y11 y21 and y2 = y12 y22 are solutions of the homogeneous system y′ = A(t)y, (A) and deﬁne Y = y11 y12 y21 y22 . (a) Show that Y ′ = AY . (b) Show that if c is a constant vector then y = Y c is a solution of (A). (c) State generalizations of (a) and (b) for n × n systems.	Elementary Differential Equations with Boundary Value Problems_Page_531_Chunk2935
522 Chapter 10 Linear Systems of Differential Equations 10. Suppose Y is a differentiable square matrix. (a) Find a formula for the derivative of Y 2. (b) Find a formula for the derivative of Y n, where n is any positive integer. (c) State how the results obtained in (a) and (b) are analogous to results from calculus concerning scalar functions. 11. It can be shown that if Y is a differentiable and invertible square matrix function, then Y −1 is differentiable. (a) Show that (Y −1)′ = −Y −1Y ′Y −1. (Hint: Differentiate the identity Y −1Y = I.) (b) Find the derivative of Y −n =	Elementary Differential Equations with Boundary Value Problems_Page_532_Chunk2936
Section 10.3 Basic Theory of Homogeneous Linear System 523 Theorem 10.3.1 Suppose the n×n matrix A = A(t) is continuouson (a, b). Then a set {y1, y2, . . ., yn} of n solutions of y′ = A(t)y on (a, b) is a fundamental set if and only if it’s linearly independent on (a, b). Example 10.3.1 Show that the vector functions y1 =   et 0 e−t  , y2 =   0 e3t 1  , and y3 =   e2t e3t 0   are linearly independent on every interval (a, b). Solution Suppose c1   et 0 e−t  + c2   0 e3t 1  + c3   e2t e3t 0  =   0 0 0  , a < t < b. We must show that c1 = c2 = c3 = 0. Rewriting this equation in matrix form yields   et 0 e2t 0 e3t e3t e−t 1 0     c1 c2 c3  =   0 0 0  , a < t < b. Expanding the determinant of this system in cofactors of the entries of the ﬁrst row yields et 0 e2t 0 e3t e3t e−t 1 0 = et e3t e3t 1 0 −0 0 e3t e−t 0 + e2t 0 e3t e−t 1 = et(−e3t) + e2t(−e2t) = −2e4t. Since this determinant is never zero, c1 = c2 = c3 = 0. We can use the method in Example 10.3.1 to test n solutions {y1, y2, . . ., yn} of any n × n system y′ = A(t)y for linear independence on an interval (a, b) on which A is continuous. To explain this (and for other purposes later), it’s useful to write a linear combination of y1, y2, ..., yn in a different way. We ﬁrst write the vector functions in terms of their components as y1 =   y11 y21 ... yn1  , y2 =   y12 y22 ... yn2  , . . ., yn =   y1n y2n ... ynn  . If y = c1y1 + c2y2 + · · · + cnyn then y = c1   y11 y21 ... yn1  + c2   y12 y22 ... yn2  + · · · + cn   y1n y2n ... ynn   =   y11 y12 · · · y1n y21 y22 · · · y2n ... ... ... ... yn1 yn2 · · · ynn     c1 c2 ... cn  .	Elementary Differential Equations with Boundary Value Problems_Page_533_Chunk2937
524 Chapter 10 Linear Systems of Differential Equations This shows that c1y1 + c2y2 + · · · + cnyn = Y c, (10.3.3) where c =   c1 c2 ... cn   and Y = [y1 y2 · · · yn] =   y11 y12 · · · y1n y21 y22 · · · y2n ... ... ... ... yn1 yn2 · · · ynn  ; (10.3.4) that is, the columns of Y are the vector functions y1, y2, . . ., yn. For reference below, note that Y ′ = [y′ 1 y′ 2 · · · y′ n] = [Ay1 Ay2 · · · Ayn] = A[y1 y2 · · · yn] = AY ; that is, Y satisﬁes the matrix differential equation Y ′ = AY. The determinant of Y , W = y11 y12 · · · y1n y21 y22 · · · y2n ... ... ... ... yn1 yn2 · · · ynn (10.3.5) is called the Wronskian of {y1, y2, . . ., yn}. It can be shown (Exercises 2 and 3) that this deﬁnition is analogous to deﬁnitions of the Wronskian of scalar functions given in Sections 5.1 and 9.1. The next theorem is analogous to Theorems 5.1.4 and 9.1.3. The proof is sketched in Exercise 4 for n = 2 and in Exercise 5 for general n. Theorem 10.3.2 [Abel’s Formula] Suppose the n × n matrix A = A(t) is continuous on (a, b), let y1, y2, ..., yn be solutions of y′ = A(t)y on (a, b), and let t0 be in (a, b). Then the Wronskian of {y1, y2, . . ., yn} is given by W(t) = W(t0) exp Z t t0 a11(s) + a22(s) + · · · + ann(s)] ds , a < t < b. (10.3.6) Therefore, either W has no zeros in (a, b) or W ≡0 on (a, b). REMARK: The sum of the diagonal entries of a square matrix A is called the trace of A, denoted by tr(A). Thus, for an n × n matrix A, tr(A) = a11 + a22 + · · · + ann,	Elementary Differential Equations with Boundary Value Problems_Page_534_Chunk2938
Section 10.3 Basic Theory of Homogeneous Linear System 525 and (10.3.6) can be written as W(t) = W(t0) exp Z t t0 tr(A(s)) ds , a < t < b. The next theorem is analogous to Theorems 5.1.6 and 9.1.4. Theorem 10.3.3 Suppose the n × n matrix A = A(t) is continuous on (a, b) and let y1, y2, ...,yn be solutions of y′ = A(t)y on (a, b). Then the following statements are equivalent; that is, they are either all true or all false: (a) The general solution of y′ = A(t)y on (a, b) is y = c1y1 + c2y2 + · · · + cnyn, where c1, c2, ..., cn are arbitrary constants. (b) {y1, y2, . . ., yn} is a fundamental set of solutions of y′ = A(t)y on (a, b). (c) {y1, y2, . . ., yn} is linearly independent on (a, b). (d) The Wronskian of {y1, y2, . . ., yn} is nonzero at some point in (a, b). (e) The Wronskian of {y1, y2, . . ., yn} is nonzero at all points in (a, b). We say that Y in (10.3.4) is a fundamental matrix for y′ = A(t)y if any (and therefore all) of the statements (a)-(e) of Theorem 10.3.2 are true for the columns of Y . In this case, (10.3.3) implies that the general solution of y′ = A(t)y can be written as y = Y c, where c is an arbitrary constant n-vector. Example 10.3.2 The vector functions y1 = −e2t 2e2t and y2 = −e−t e−t are solutions of the constant coefﬁcient system y′ = −4 −3 6 5 y (10.3.7) on (−∞, ∞). (Verify.) (a) Compute the Wronskian of {y1, y2} directly from the deﬁnition (10.3.5) (b) Verify Abel’s formula (10.3.6) for the Wronskian of {y1, y2}. (c) Find the general solution of (10.3.7). (d) Solve the initial value problem y′ = −4 −3 6 5 y, y(0) = 4 −5 . (10.3.8) SOLUTION(a) From (10.3.5) W(t) = −e2t −e−t 2e2t e−t = e2te−t −1 −1 2 1 = et. (10.3.9) SOLUTION(b) Here A = −4 −3 6 5 ,	Elementary Differential Equations with Boundary Value Problems_Page_535_Chunk2939
526 Chapter 10 Linear Systems of Differential Equations so tr(A) = −4 + 5 = 1. If t0 is an arbitrary real number then (10.3.6) implies that W(t) = W(t0) exp Z t t0 1 ds = −e2t0 −e−t0 2e2t0 e−t0 e(t−t0) = et0et−t0 = et, which is consistent with (10.3.9). SOLUTION(c) Since W(t) ̸= 0, Theorem 10.3.3 implies that {y1, y2} is a fundamental set of solutions of (10.3.7) and Y = −e2t −e−t 2e2t e−t is a fundamental matrix for (10.3.7). Therefore the general solution of (10.3.7) is y = c1y1 + c2y2 = c1 −e2t 2e2t + c2 −e−t e−t = −e2t −e−t 2e2t e−t c1 c2 . (10.3.10) SOLUTION(d) Setting t = 0 in (10.3.10) and imposing the initial condition in (10.3.8) yields c1 −1 2 + c2 −1 1 = 4 −5 . Thus, −c1 −c2 = 4 2c1 + c2 = −5. The solution of this system is c1 = −1, c2 = −3. Substituting these values into (10.3.10) yields y = − −e2t 2e2t −3 −e−t e−t = e2t + 3e−t −2e2t −3e−t as the solution of (10.3.8). 10.3 Exercises 1. Prove: If y1, y2, ..., yn are solutions of y′ = A(t)y on (a, b), then any linear combination of y1, y2, ..., yn is also a solution of y′ = A(t)y on (a, b). 2. In Section 5.1 the Wronskian of two solutions y1 and y2 of the scalar second order equation P0(x)y′′ + P1(x)y′ + P2(x)y = 0 (A) was deﬁned to be W = y1 y2 y′ 1 y′ 2 . (a) Rewrite (A) as a system of ﬁrst order equations and show that W is the Wronskian (as deﬁned in this section) of two solutions of this system. (b) Apply Eqn. (10.3.6) to the system derived in (a), and show that W(x) = W(x0) exp − Z x x0 P1(s) P0(s) ds , which is the form of Abel’s formula given in Theorem 9.1.3.	Elementary Differential Equations with Boundary Value Problems_Page_536_Chunk2940
Section 10.3 Basic Theory of Homogeneous Linear System 527 3. In Section 9.1 the Wronskian of n solutions y1, y2, ..., yn of the n−th order equation P0(x)y(n) + P1(x)y(n−1) + · · · + Pn(x)y = 0 (A) was deﬁned to be W = y1 y2 · · · yn y′ 1 y′ 2 · · · y′ n ... ... ... ... y(n−1) 1 y(n−1) 2 · · · y(n−1) n . (a) Rewrite (A) as a system of ﬁrst order equations and show that W is the Wronskian (as deﬁned in this section) of n solutions of this system. (b) Apply Eqn. (10.3.6) to the system derived in (a), and show that W(x) = W(x0) exp − Z x x0 P1(s) P0(s) ds , which is the form of Abel’s formula given in Theorem 9.1.3. 4. Suppose y1 = y11 y21 and y2 = y12 y22 are solutions of the 2 × 2 system y′ = Ay on (a, b), and let Y = y11 y12 y21 y22 and W = y11 y12 y21 y22 ; thus, W is the Wronskian of {y1, y2}. (a) Deduce from the deﬁnition of determinant that W ′ = y′ 11 y′ 12 y21 y22 + y11 y12 y′ 21 y′ 22 . (b) Use the equation Y ′ = A(t)Y and the deﬁnition of matrix multiplication to show that [y′ 11 y′ 12] = a11[y11 y12] + a12[y21 y22] and [y′ 21 y′ 22] = a21[y11 y12] + a22[y21 y22]. (c) Use properties of determinants to deduce from (a) and (a) that y′ 11 y′ 12 y21 y22 = a11W and y11 y12 y′ 21 y′ 22 = a22W. (d) Conclude from (c) that W ′ = (a11 + a22)W, and use this to show that if a < t0 < b then W(t) = W(t0) exp Z t t0 [a11(s) + a22(s)] ds a < t < b.	Elementary Differential Equations with Boundary Value Problems_Page_537_Chunk2941
528 Chapter 10 Linear Systems of Differential Equations 5. Suppose the n × n matrix A = A(t) is continuous on (a, b). Let Y =   y11 y12 · · · y1n y21 y22 · · · y2n ... ... ... ... yn1 yn2 · · · ynn  , where the columns of Y are solutions of y′ = A(t)y. Let ri = [yi1 yi2 . . . yin] be the ith row of Y , and let W be the determinant of Y . (a) Deduce from the deﬁnition of determinant that W ′ = W1 + W2 + · · · + Wn, where, for 1 ≤m ≤n, the ith row of Wm is ri if i ̸= m, and r′ m if i = m. (b) Use the equation Y ′ = AY and the deﬁnition of matrix multiplication to show that r′ m = am1r1 + am2r2 + · · · + amnrn. (c) Use properties of determinants to deduce from (b) that det(Wm) = ammW. (d) Conclude from (a) and (c) that W ′ = (a11 + a22 + · · · + ann)W, and use this to show that if a < t0 < b then W(t) = W(t0) exp Z t t0 a11(s) + a22(s) + · · · + ann(s)] ds , a < t < b. 6. Suppose the n × n matrix A is continuous on (a, b) and t0 is a point in (a, b). Let Y be a funda- mental matrix for y′ = A(t)y on (a, b). (a) Show that Y (t0) is invertible. (b) Show that if k is an arbitrary n-vector then the solution of the initial value problem y′ = A(t)y, y(t0) = k is y = Y (t)Y −1(t0)k. 7. Let A = 2 4 4 2 , y1 = e6t e6t , y2 = e−2t −e−2t , k = −3 9 . (a) Verify that {y1, y2} is a fundamental set of solutions for y′ = Ay. (b) Solve the initial value problem y′ = Ay, y(0) = k. (A)	Elementary Differential Equations with Boundary Value Problems_Page_538_Chunk2942
Section 10.3 Basic Theory of Homogeneous Linear System 529 (c) Use the result of Exercise 6(b) to ﬁnd a formula for the solution of (A) for an arbitrary initial vector k. 8. Repeat Exercise 7 with A = −2 −2 −5 1 , y1 = e−4t e−4t , y2 = −2e3t 5e3t , k = 10 −4 . 9. Repeat Exercise 7 with A = −4 −10 3 7 , y1 = −5e2t 3e2t , y2 = 2et −et , k = −19 11 . 10. Repeat Exercise 7 with A = 2 1 1 2 , y1 = e3t e3t , y2 = et −et , k = 2 8 . 11. Let A =   3 −1 −1 −2 3 2 4 −1 −2  , y1 =   e2t 0 e2t  , y2 =   e3t −e3t e3t  , y3 =   e−t −3e−t 7e−t  , k =   2 −7 20  . (a) Verify that {y1, y2, y3} is a fundamental set of solutions for y′ = Ay. (b) Solve the initial value problem y′ = Ay, y(0) = k. (A) (c) Use the result of Exercise 6(b) to ﬁnd a formula for the solution of (A) for an arbitrary initial vector k. 12. Repeat Exercise 11 with A =   0 2 2 2 0 2 2 2 0  , y1 =   −e−2t 0 e−2t  , y2 =   −e−2t e−2t 0  , y3 =   e4t e4t e4t  , k =   0 −9 12  . 13. Repeat Exercise 11 with A =   −1 2 3 0 1 6 0 0 −2  , y1 =   et et 0  , y2 =   e−t 0 0  , y3 =   e−2t −2e−2t e−2t  , k =   5 5 −1  .	Elementary Differential Equations with Boundary Value Problems_Page_539_Chunk2943
530 Chapter 10 Linear Systems of Differential Equations 14. Suppose Y and Z are fundamental matrices for the n × n system y′ = A(t)y. Then some of the four matrices Y Z−1, Y −1Z, Z−1Y , ZY −1 are necessarily constant. Identify them and prove that they are constant. 15. Suppose the columns of an n × n matrix Y are solutions of the n × n system y′ = Ay and C is an n × n constant matrix. (a) Show that the matrix Z = Y C satisﬁes the differential equation Z′ = AZ. (b) Show that Z is a fundamental matrix for y′ = A(t)y if and only if C is invertible and Y is a fundamental matrix for y′ = A(t)y. 16. Suppose the n × n matrix A = A(t) is continuous on (a, b) and t0 is in (a, b). For i = 1, 2, ..., n, let yi be the solution of the initial value problem y′ i = A(t)yi, yi(t0) = ei, where e1 =   1 0 ... 0  , e2 =   0 1 ... 0  , · · · en =   0 0 ... 1  ; that is, the jth component of ei is 1 if j = i, or 0 if j ̸= i. (a) Show that{y1, y2, . . ., yn} is a fundamental set of solutions of y′ = A(t)y on (a, b). (b) Conclude from (a) and Exercise 15 that y′ = A(t)y has inﬁnitely many fundamental sets of solutions on (a, b). 17. Show that Y is a fundamental matrix for the system y′ = A(t)y if and only if Y −1 is a funda- mental matrix for y′ = −AT (t)y, where AT denotes the transpose of A. HINT: See Exercise 11. 18. Let Z be the fundamental matrix for the constant coefﬁcient system y′ = Ay such that Z(0) = I. (a) Show that Z(t)Z(s) = Z(t + s) for all s and t. HINT: For ﬁxed s let Γ1(t) = Z(t)Z(s) and Γ2(t) = Z(t + s). Show that Γ1 and Γ2 are both solutions of the matrix initial value problem Γ′ = AΓ, Γ(0) = Z(s). Then conclude from Theorem 10.2.1 that Γ1 = Γ2. (b) Show that (Z(t))−1 = Z(−t). (c) The matrix Z deﬁned above is sometimes denoted by etA. Discuss the motivation for this notation. 10.4 CONSTANT COEFFICIENT HOMOGENEOUS SYSTEMS I We’ll now begin our study of the homogeneous system y′ = Ay, (10.4.1) where A is an n × n constant matrix. Since A is continuous on (−∞, ∞), Theorem 10.2.1 implies that all solutions of (10.4.1) are deﬁned on (−∞, ∞). Therefore, when we speak of solutions of y′ = Ay, we’ll mean solutions on (−∞, ∞). In this section we assume that all the eigenvalues of A are real and that A has a set of n linearly independent eigenvectors. In the next two sections we consider the cases where some of the eigenvalues of A are complex, or where A does not have n linearly independent eigenvectors. In Example 10.3.2 we showed that the vector functions y1 = −e2t 2e2t and y2 = −e−t e−t form a fundamental set of solutions of the system y′ = −4 −3 6 5 y, (10.4.2)	Elementary Differential Equations with Boundary Value Problems_Page_540_Chunk2944
Section 10.4 Constant Coefﬁcient Homogeneous Systems I 531 but we did not show how we obtained y1 and y2 in the ﬁrst place. To see how these solutions can be obtained we write (10.4.2) as y′ 1 = −4y1 −3y2 y′ 2 = 6y1 + 5y2 (10.4.3) and look for solutions of the form y1 = x1eλt and y2 = x2eλt, (10.4.4) where x1, x2, and λ are constants to be determined. Differentiating (10.4.4) yields y′ 1 = λx1eλt and y′ 2 = λx2eλt. Substituting this and (10.4.4) into (10.4.3) and canceling the common factor eλt yields −4x1 −3x2 = λx1 6x1 + 5x2 = λx2. For a given λ, this is a homogeneous algebraic system, since it can be rewritten as (−4 −λ)x1 −3x2 = 0 6x1 + (5 −λ)x2 = 0. (10.4.5) The trivial solution x1 = x2 = 0 of this system isn’t useful, since it corresponds to the trivial solution y1 ≡y2 ≡0 of (10.4.3), which can’t be part of a fundamental set of solutions of (10.4.2). Therefore we consider only those values of λ for which (10.4.5) has nontrivial solutions. These are the values of λ for which the determinant of (10.4.5) is zero; that is, −4 −λ −3 6 5 −λ = (−4 −λ)(5 −λ) + 18 = λ2 −λ −2 = (λ −2)(λ + 1) = 0, which has the solutions λ1 = 2 and λ2 = −1. Taking λ = 2 in (10.4.5) yields −6x1 −3x2 = 0 6x1 + 3x2 = 0, which implies that x1 = −x2/2, where x2 can be chosen arbitrarily. Choosing x2 = 2 yields the solution y1 = −e2t, y2 = 2e2t of (10.4.3). We can write this solution in vector form as y1 = −1 2 e2t. (10.4.6) Taking λ = −1 in (10.4.5) yields the system −3x1 −3x2 = 0 6x1 + 6x2 = 0, so x1 = −x2. Taking x2 = 1 here yields the solution y1 = −e−t, y2 = e−t of (10.4.3). We can write this solution in vector form as y2 = −1 1 e−t. (10.4.7) In (10.4.6) and (10.4.7) the constant coefﬁcients in the arguments of the exponential functions are the eigenvalues of the coefﬁcient matrix in (10.4.2), and the vector coefﬁcients of the exponential functions are associated eigenvectors. This illustrates the next theorem.	Elementary Differential Equations with Boundary Value Problems_Page_541_Chunk2945
532 Chapter 10 Linear Systems of Differential Equations Theorem 10.4.1 Suppose the n×n constant matrix A has n real eigenvalues λ1, λ2, . . ., λn (which need not be distinct) with associated linearly independent eigenvectors x1, x2, . . ., xn. Then the functions y1 = x1eλ1t, y2 = x2eλ2t, . . ., yn = xneλnt form a fundamental set of solutions of y′ = Ay; that is, the general solution of this system is y = c1x1eλ1t + c2x2eλ2t + · · · + cnxneλnt. Proof Differentiating yi = xieλit and recalling that Axi = λixi yields y′ i = λixieλit = Axieλit = Ayi. This shows that yi is a solution of y′ = Ay. The Wronskian of {y1, y2, . . ., yn} is x11eλ1t x12eλ2t · · · x1neλnt x21eλ1t x22eλ2t · · · x2neλnt ... ... ... ... xn1eλ1t xn2eλ2t · · · xnneλxnt = eλ1teλ2t · · · eλnt x11 x12 · · · x1n x21 x22 · · · x2n ... ... ... ... xn1 xn2 · · · xnn . Since the columns of the determinant on the right are x1, x2, ..., xn, which are assumed to be linearly independent, the determinant is nonzero. Therefore Theorem 10.3.3 implies that {y1, y2, . . ., yn} is a fundamental set of solutions of y′ = Ay. Example 10.4.1 (a) Find the general solution of y′ = 2 4 4 2 y. (10.4.8) (b) Solve the initial value problem y′ = 2 4 4 2 y, y(0) = 5 −1 . (10.4.9) SOLUTION(a) The characteristic polynomial of the coefﬁcient matrix A in (10.4.8) is 2 −λ 4 4 2 −λ = (λ −2)2 −16 = (λ −2 −4)(λ −2 + 4) = (λ −6)(λ + 2). Hence, λ1 = 6 and λ2 = −2 are eigenvalues of A. To obtain the eigenvectors, we must solve the system 2 −λ 4 4 2 −λ x1 x2 = 0 0 (10.4.10) with λ = 6 and λ = −2. Setting λ = 6 in (10.4.10) yields −4 4 4 −4 x1 x2 = 0 0 ,	Elementary Differential Equations with Boundary Value Problems_Page_542_Chunk2946
Section 10.4 Constant Coefﬁcient Homogeneous Systems I 533 which implies that x1 = x2. Taking x2 = 1 yields the eigenvector x1 = 1 1 , so y1 = 1 1 e6t is a solution of (10.4.8). Setting λ = −2 in (10.4.10) yields 4 4 4 4 x1 x2 = 0 0 , which implies that x1 = −x2. Taking x2 = 1 yields the eigenvector x2 = −1 1 , so y2 = −1 1 e−2t is a solution of (10.4.8). From Theorem 10.4.1, the general solution of (10.4.8) is y = c1y1 + c2y2 = c1 1 1 e6t + c2 −1 1 e−2t. (10.4.11) SOLUTION(b) To satisfy the initial condition in (10.4.9), we must choose c1 and c2 in (10.4.11) so that c1 1 1 + c2 −1 1 = 5 −1 . This is equivalent to the system c1 −c2 = 5 c1 + c2 = −1, so c1 = 2, c2 = −3. Therefore the solution of (10.4.9) is y = 2 1 1 e6t −3 −1 1 e−2t, or, in terms of components, y1 = 2e6t + 3e−2t, y2 = 2e6t −3e−2t. Example 10.4.2 (a) Find the general solution of y′ =   3 −1 −1 −2 3 2 4 −1 −2  y. (10.4.12)	Elementary Differential Equations with Boundary Value Problems_Page_543_Chunk2947
534 Chapter 10 Linear Systems of Differential Equations (b) Solve the initial value problem y′ =   3 −1 −1 −2 3 2 4 −1 −2  y, y(0) =   2 −1 8  . (10.4.13) SOLUTION(a) The characteristic polynomial of the coefﬁcient matrix A in (10.4.12) is 3 −λ −1 −1 −2 3 −λ 2 4 −1 −2 −λ = −(λ −2)(λ −3)(λ + 1). Hence, the eigenvalues of A are λ1 = 2, λ2 = 3, and λ3 = −1. To ﬁnd the eigenvectors, we must solve the system   3 −λ −1 −1 −2 3 −λ 2 4 −1 −2 −λ     x1 x2 x3  =   0 0 0   (10.4.14) with λ = 2, 3, −1. With λ = 2, the augmented matrix of (10.4.14) is   1 −1 −1 ... 0 −2 1 2 ... 0 4 −1 −4 ... 0  , which is row equivalent to   1 0 −1 ... 0 0 1 0 ... 0 0 0 0 ... 0  . Hence, x1 = x3 and x2 = 0. Taking x3 = 1 yields y1 =   1 0 1  e2t as a solution of (10.4.12). With λ = 3, the augmented matrix of (10.4.14) is   0 −1 −1 ... 0 −2 0 2 ... 0 4 −1 −5 ... 0  , which is row equivalent to   1 0 −1 ... 0 0 1 1 ... 0 0 0 0 ... 0  .	Elementary Differential Equations with Boundary Value Problems_Page_544_Chunk2948
Section 10.4 Constant Coefﬁcient Homogeneous Systems I 535 Hence, x1 = x3 and x2 = −x3. Taking x3 = 1 yields y2 =   1 −1 1  e3t as a solution of (10.4.12). With λ = −1, the augmented matrix of (10.4.14) is   4 −1 −1 ... 0 −2 4 2 ... 0 4 −1 −1 ... 0  , which is row equivalent to   1 0 −1 7 ... 0 0 1 3 7 ... 0 0 0 0 ... 0  . Hence, x1 = x3/7 and x2 = −3x3/7. Taking x3 = 7 yields y3 =   1 −3 7  e−t as a solution of (10.4.12). By Theorem 10.4.1, the general solution of (10.4.12) is y = c1   1 0 1  e2t + c2   1 −1 1  e3t + c3   1 −3 7  e−t, which can also be written as y =   e2t e3t e−t 0 −e3t −3e−t e2t e3t 7e−t     c1 c2 c3  . (10.4.15) SOLUTION(b) To satisfy the initial condition in (10.4.13) we must choose c1, c2, c3 in (10.4.15) so that   1 1 1 0 −1 −3 1 1 7     c1 c2 c3  =   2 −1 8  . Solving this system yields c1 = 3, c2 = −2, c3 = 1. Hence, the solution of (10.4.13) is y =   e2t e3t e−t 0 −e3t −3e−t e2t e3t 7e−t     3 −2 1   = 3   1 0 1  e2t −2   1 −1 1  e3t +   1 −3 7  e−t.	Elementary Differential Equations with Boundary Value Problems_Page_545_Chunk2949
536 Chapter 10 Linear Systems of Differential Equations Example 10.4.3 Find the general solution of y′ =   −3 2 2 2 −3 2 2 2 −3  y. (10.4.16) Solution The characteristic polynomial of the coefﬁcient matrix A in (10.4.16) is −3 −λ 2 2 2 −3 −λ 2 2 2 −3 −λ = −(λ −1)(λ + 5)2. Hence, λ1 = 1 is an eigenvalue of multiplicity 1, while λ2 = −5 is an eigenvalue of multiplicity 2. Eigenvectors associated with λ1 = 1 are solutions of the system with augmented matrix   −4 2 2 ... 0 2 −4 2 ... 0 2 2 −4 ... 0  , which is row equivalent to   1 0 −1 ... 0 0 1 −1 ... 0 0 0 0 ... 0  . Hence, x1 = x2 = x3, and we choose x3 = 1 to obtain the solution y1 =   1 1 1  et (10.4.17) of (10.4.16). Eigenvectors associated with λ2 = −5 are solutions of the system with augmented matrix   2 2 2 ... 0 2 2 2 ... 0 2 2 2 ... 0  . Hence, the components of these eigenvectors need only satisfy the single condition x1 + x2 + x3 = 0. Since there’s only one equation here, we can choose x2 and x3 arbitrarily. We obtain one eigenvector by choosing x2 = 0 and x3 = 1, and another by choosing x2 = 1 and x3 = 0. In both cases x1 = −1. Therefore   −1 0 1   and   −1 1 0  	Elementary Differential Equations with Boundary Value Problems_Page_546_Chunk2950
Section 10.4 Constant Coefﬁcient Homogeneous Systems I 537 are linearly independent eigenvectors associated with λ2 = −5, and the corresponding solutions of (10.4.16) are y2 =   −1 0 1  e−5t and y3 =   −1 1 0  e−5t. Because of this and (10.4.17), Theorem 10.4.1 implies that the general solution of (10.4.16) is y = c1   1 1 1  et + c2   −1 0 1  e−5t + c3   −1 1 0  e−5t. Geometric Properties of Solutions when n = 2 We’ll now consider the geometric properties of solutions of a 2 × 2 constant coefﬁcient system y′ 1 y′ 2 = a11 a12 a21 a22 y1 y2 . (10.4.18) It is convenient to think of a “y1-y2 plane," where a point is identiﬁed by rectangular coordinates (y1, y2). If y = y1 y2 is a non-constant solution of (10.4.18), then the point (y1(t), y2(t)) moves along a curve C in the y1-y2 plane as t varies from −∞to ∞. We call C the trajectory of y. (We also say that C is a trajectory of the system (10.4.18).) I’s important to note that C is the trajectory of inﬁnitely many solutions of (10.4.18), since if τ is any real number, then y(t −τ) is a solution of (10.4.18) (Exercise 28(b)), and (y1(t −τ), y2(t −τ)) also moves along C as t varies from −∞to ∞. Moreover, Exercise 28(c) implies that distinct trajectories of (10.4.18) can’t intersect, and that two solutions y1 and y2 of (10.4.18) have the same trajectory if and only if y2(t) = y1(t −τ) for some τ. From Exercise 28(a), a trajectory of a nontrivial solution of (10.4.18) can’t contain (0, 0), which we deﬁne to be the trajectory of the trivial solution y ≡0. More generally, if y = k1 k2 ̸= 0 is a constant solution of (10.4.18) (which could occur if zero is an eigenvalue of the matrix of (10.4.18)), we deﬁne the trajectory of y to be the single point (k1, k2). To be speciﬁc, this is the question: What do the trajectories look like, and how are they traversed? In this section we’ll answer this question, assuming that the matrix A = a11 a12 a21 a22 of (10.4.18) has real eigenvalues λ1 and λ2 with associated linearly independent eigenvectors x1 and x2. Then the general solution of (10.4.18) is y = c1x1eλ1t + c2x2eλ2t. (10.4.19) We’ll consider other situations in the next two sections. We leave it to you (Exercise 35) to classify the trajectories of (10.4.18) if zero is an eigenvalue of A. We’ll conﬁne our attention here to the case where both eigenvalues are nonzero. In this case the simplest situation is where λ1 = λ2 ̸= 0, so (10.4.19) becomes y = (c1x1 + c2x2)eλ1t. Since x1 and x2 are linearly independent, an arbitrary vector x can be written as x = c1x1 + c2x2. Therefore the general solution of (10.4.18) can be written as y = xeλ1t where x is an arbitrary 2- vector, and the trajectories of nontrivial solutions of (10.4.18) are half-lines through (but not including)	Elementary Differential Equations with Boundary Value Problems_Page_547_Chunk2951
538 Chapter 10 Linear Systems of Differential Equations the origin. The direction of motion is away from the origin if λ1 > 0 (Figure 10.4.1), toward it if λ1 < 0 (Figure 10.4.2). (In these and the next ﬁgures an arrow through a point indicates the direction of motion along the trajectory through the point.) y1 y2 Figure 10.4.1 Trajectories of a 2 × 2 system with a repeated positive eigenvalue y1 y2 Figure 10.4.2 Trajectories of a 2 × 2 system with a repeated negative eigenvalue Now suppose λ2 > λ1, and let L1 and L2 denote lines through the origin parallel to x1 and x2, respectively. By a half-line of L1 (or L2), we mean either of the rays obtained by removing the origin from L1 (or L2). Letting c2 = 0 in (10.4.19) yields y = c1x1eλ1t. If c1 ̸= 0, the trajectory deﬁned by this solution is a half-line of L1. The direction of motion is away from the origin if λ1 > 0, toward the origin if λ1 < 0. Similarly, the trajectory of y = c2x2eλ2t with c2 ̸= 0 is a half-line of L2. Henceforth, we assume that c1 and c2 in (10.4.19) are both nonzero. In this case, the trajectory of (10.4.19) can’t intersect L1 or L2, since every point on these lines is on the trajectory of a solution for which either c1 = 0 or c2 = 0. (Remember: distinct trajectories can’t intersect!). Therefore the trajectory of (10.4.19) must lie entirely in one of the four open sectors bounded by L1 and L2, but do not any point on L1 or L2. Since the initial point (y1(0), y2(0)) deﬁned by y(0) = c1x1 + c2x2 is on the trajectory, we can determine which sector contains the trajectory from the signs of c1 and c2, as shown in Figure 10.4.3. The direction of y(t) in (10.4.19) is the same as that of e−λ2ty(t) = c1x1e−(λ2−λ1)t + c2x2 (10.4.20) and of e−λ1ty(t) = c1x1 + c2x2e(λ2−λ1)t. (10.4.21) Since the right side of (10.4.20) approaches c2x2 as t →∞, the trajectory is asymptotically parallel to L2 as t →∞. Since the right side of (10.4.21) approaches c1x1 as t →−∞, the trajectory is asymptotically parallel to L1 as t →−∞. The shape and direction of traversal of the trajectory of (10.4.19) depend upon whether λ1 and λ2 are both positive, both negative, or of opposite signs. We’ll now analyze these three cases. Henceforth ∥u∥denote the length of the vector u.	Elementary Differential Equations with Boundary Value Problems_Page_548_Chunk2952
Section 10.4 Constant Coefﬁcient Homogeneous Systems I 539 x2 x1 c1 > 0, c2 < 0 c1 > 0, c2 > 0 c1 < 0, c2 > 0 c1 < 0, c2 < 0 L1 L2 Figure 10.4.3 Four open sectors bounded by L1 and L2 y1 y2 L1 L2 Figure 10.4.4 Two positive eigenvalues; motion away from origin Case 1: λ2 > λ1 > 0 Figure 10.4.4 shows some typical trajectories. In this case, limt→−∞∥y(t)∥= 0, so the trajectory is not only asymptotically parallel to L1 as t →−∞, but is actually asymptotically tangent to L1 at the origin. On the other hand, limt→∞∥y(t)∥= ∞and lim t→∞ y(t) −c2x2eλ2t = lim t→∞∥c1x1eλ1t∥= ∞, so, although the trajectory is asymptotically parallel to L2 as t →∞, it’s not asymptotically tangent to L2. The direction of motion along each trajectory is away from the origin. Case 2: 0 > λ2 > λ1 Figure 10.4.5 shows some typical trajectories. In this case, limt→∞∥y(t)∥= 0, so the trajectory is asymptotically tangent to L2 at the origin as t →∞. On the other hand, limt→−∞∥y(t)∥= ∞and lim t→−∞ y(t) −c1x1eλ1t = lim t→−∞∥c2x2eλ2t∥= ∞, so, although the trajectory is asymptotically parallel to L1 as t →−∞, it’s not asymptotically tangent to it. The direction of motion along each trajectory is toward the origin. Case 3: λ2 > 0 > λ1 Figure 10.4.6 shows some typical trajectories. In this case, lim t→∞∥y(t)∥= ∞ and lim t→∞ y(t) −c2x2eλ2t = lim t→∞∥c1x1eλ1t∥= 0, so the trajectory is asymptotically tangent to L2 as t →∞. Similarly, lim t→−∞∥y(t)∥= ∞ and lim t→−∞ y(t) −c1x1eλ1t = lim t→−∞∥c2x2eλ2t∥= 0, so the trajectory is asymptotically tangent to L1 as t →−∞. The direction of motion is toward the origin on L1 and away from the origin on L2. The direction of motion along any other trajectory is away from L1, toward L2.	Elementary Differential Equations with Boundary Value Problems_Page_549_Chunk2953
540 Chapter 10 Linear Systems of Differential Equations y1 y2 L1 L2 Figure 10.4.5 Two negative eigenvalues; motion toward the origin y1 y2 L1 L2 Figure 10.4.6 Eigenvalues of different signs 10.4 Exercises In Exercises 1–15 ﬁnd the general solution. 1. y′ = 1 2 2 1 y 2. y′ = 1 4 −5 3 3 −5 y 3. y′ = 1 5 −4 3 −2 −11 y 4. y′ = −1 −4 −1 −1 y 5. y′ = 2 −4 −1 −1 y 6. y′ = 4 −3 2 −1 y 7. y′ = −6 −3 1 −2 y 8. y′ =   1 −1 −2 1 −2 −3 −4 1 −1  y 9. y′ =   −6 −4 −8 −4 0 −4 −8 −4 −6  y 10. y′ =   3 5 8 1 −1 −2 −1 −1 −1  y 11. y′ =   1 −1 2 12 −4 10 −6 1 −7  y 12. y′ =   4 −1 −4 4 −3 −2 1 −1 −1  y 13. y′ =   −2 2 −6 2 6 2 −2 −2 2  y 14. y′ =   3 2 −2 −2 7 −2 −10 10 −5  y	Elementary Differential Equations with Boundary Value Problems_Page_550_Chunk2954
Section 10.4 Constant Coefﬁcient Homogeneous Systems I 541 15. y′ =   3 1 −1 3 5 1 −6 2 4  y In Exercises 16–27 solve the initial value problem. 16. y′ = −7 4 −6 7 y, y(0) = 2 −4 17. y′ = 1 6 7 2 −2 2 y, y(0) = 0 −3 18. y′ = 21 −12 24 −15 y, y(0) = 5 3 19. y′ = −7 4 −6 7 y, y(0) = −1 7 20. y′ = 1 6   1 2 0 4 −1 0 0 0 3  y, y(0) =   4 7 1   21. y′ = 1 3   2 −2 3 −4 4 3 2 1 0  y, y(0) =   1 1 5   22. y′ =   6 −3 −8 2 1 −2 3 −3 −5  y, y(0) =   0 −1 −1   23. y′ = 1 3   2 4 −7 1 5 −5 −4 4 −1  y, y(0) =   4 1 3   24. y′ =   3 0 1 11 −2 7 1 0 3  y, y(0) =   2 7 6   25. y′ =   −2 −5 −1 −4 −1 1 4 5 3  y, y(0) =   8 −10 −4   26. y′ =   3 −1 0 4 −2 0 4 −4 2  y, y(0) =   7 10 2   27. y′ =   −2 2 6 2 6 2 −2 −2 2  y, y(0) =   6 −10 7   28. Let A be an n × n constant matrix. Then Theorem 10.2.1 implies that the solutions of y′ = Ay (A) are all deﬁned on (−∞, ∞). (a) Use Theorem 10.2.1 to show that the only solution of (A) that can ever equal the zero vector is y ≡0.	Elementary Differential Equations with Boundary Value Problems_Page_551_Chunk2955
542 Chapter 10 Linear Systems of Differential Equations (b) Suppose y1 is a solution of (A) and y2 is deﬁned by y2(t) = y1(t −τ), where τ is an arbitrary real number. Show that y2 is also a solution of (A). (c) Suppose y1 and y2 are solutions of (A) and there are real numbers t1 and t2 such that y1(t1) = y2(t2). Show that y2(t) = y1(t −τ) for all t, where τ = t2 −t1. HINT: Show that y1(t −τ) and y2(t) are solutions of the same initial value problem for (A), and apply the uniqueness assertion of Theorem 10.2.1. In Exercises 29- 34 describe and graph trajectories of the given system. 29. C/G y′ = 1 1 1 −1 y 30. C/G y′ = −4 3 −2 −11 y 31. C/G y′ = 9 −3 −1 11 y 32. C/G y′ = −1 −10 −5 4 y 33. C/G y′ = 5 −4 1 10 y 34. C/G y′ = −7 1 3 −5 y 35. Suppose the eigenvalues of the 2 × 2 matrix A are λ = 0 and µ ̸= 0, with corresponding eigen- vectors x1 and x2. Let L1 be the line through the origin parallel to x1. (a) Show that every point on L1 is the trajectory of a constant solution of y′ = Ay. (b) Show that the trajectories of nonconstant solutions of y′ = Ay are half-lines parallel to x2 and on either side of L1, and that the direction of motion along these trajectories is away from L1 if µ > 0, or toward L1 if µ < 0. The matrices of the systems in Exercises 36-41 are singular. Describe and graph the trajectories of nonconstant solutions of the given systems. 36. C/G y′ = −1 1 1 −1 y 37. C/G y′ = −1 −3 2 6 y 38. C/G y′ = 1 −3 −1 3 y 39. C/G y′ = 1 −2 −1 2 y 40. C/G y′ = −4 −4 1 1 y 41. C/G y′ = 3 −1 −3 1 y 42. L Let P = P (t) and Q = Q(t) be the populations of two species at time t, and assume that each populationwould grow exponentially if the other didn’t exist; that is, in the absence of competition, P ′ = aP and Q′ = bQ, (A) where a and b are positive constants. One way to model the effect of competition is to assume that the growth rate per individual of each population is reduced by an amount proportional to the other population, so (A) is replaced by P ′ = aP −αQ Q′ = −βP + bQ,	Elementary Differential Equations with Boundary Value Problems_Page_552_Chunk2956
Section 10.5 Constant Coefﬁcient Homogeneous Systems II 543 where α and β are positive constants. (Since negative population doesn’t make sense, this system holds only while P and Q are both positive.) Now suppose P (0) = P0 > 0 and Q(0) = Q0 > 0. (a) For several choices of a, b, α, and β, verify experimentally (by graphing trajectories of (A) in the P -Q plane) that there’s a constant ρ > 0 (depending upon a, b, α, and β) with the following properties: (i) If Q0 > ρP0, then P decreases monotonically to zero in ﬁnite time, during which Q remains positive. (ii) If Q0 < ρP0, then Q decreases monotonically to zero in ﬁnite time, during which P remains positive. (b) Conclude from (a) that exactly one of the species becomes extinct in ﬁnite time if Q0 ̸= ρP0. Determine experimentally what happens if Q0 = ρP0. (c) Conﬁrm your experimental results and determine γ by expressing the eigenvalues and asso- ciated eigenvectors of A = a −α −β b in terms of a, b, α, and β, and applying the geometric arguments developed at the end of this section. 10.5 CONSTANT COEFFICIENT HOMOGENEOUS SYSTEMS II We saw in Section 10.4 that if an n × n constant matrix A has n real eigenvalues λ1, λ2, ..., λn (which need not be distinct) with associated linearly independent eigenvectors x1, x2, ..., xn, then the general solution of y′ = Ay is y = c1x1eλ1t + c2x2eλ2t + · · · + cnxneλnt. In this section we consider the case where A has n real eigenvalues, but does not have n linearly indepen- dent eigenvectors. It is shown in linear algebra that this occurs if and only if A has at least one eigenvalue of multiplicity r > 1 such that the associated eigenspace has dimension less than r. In this case A is said to be defective. Since it’s beyond the scope of this book to give a complete analysis of systems with defective coefﬁcient matrices, we will restrict our attention to some commonly occurring special cases. Example 10.5.1 Show that the system y′ = 11 −25 4 −9 y (10.5.1) does not have a fundamental set of solutions of the form {x1eλ1t, x2eλ2t}, where λ1 and λ2 are eigenval- ues of the coefﬁcient matrix A of (10.5.1) and x1, and x2 are associated linearly independent eigenvectors. Solution The characteristic polynomial of A is 11 −λ −25 4 −9 −λ = (λ −11)(λ + 9) + 100 = λ2 −2λ + 1 = (λ −1)2. Hence, λ = 1 is the only eigenvalue of A. The augmented matrix of the system (A −I)x = 0 is  10 −25 ... 0 4 −10 ... 0  ,	Elementary Differential Equations with Boundary Value Problems_Page_553_Chunk2957
544 Chapter 10 Linear Systems of Differential Equations which is row equivalent to   1 −5 2 ... 0 0 0 ... 0  . Hence, x1 = 5x2/2 where x2 is arbitrary. Therefore all eigenvectors of A are scalar multiples of x1 = 5 2 , so A does not have a set of two linearly independent eigenvectors. From Example 10.5.1, we know that all scalar multiples of y1 = 5 2 et are solutions of (10.5.1); however, to ﬁnd the general solution we must ﬁnd a second solution y2 such that {y1, y2} is linearly independent. Based on your recollection of the procedure for solving a constant coefﬁcient scalar equation ay′′ + by′ + cy = 0 in the case where the characteristic polynomial has a repeated root, you might expect to obtain a second solution of (10.5.1) by multiplying the ﬁrst solution by t. However, this yields y2 = 5 2 tet, which doesn’t work, since y′ 2 = 5 2 (tet + et), while 11 −25 4 −9 y2 = 5 2 tet. The next theorem shows what to do in this situation. Theorem 10.5.1 Suppose the n×n matrix A has an eigenvalue λ1 of multiplicity≥2 and the associated eigenspace has dimension 1; that is, all λ1-eigenvectors of A are scalar multiples of an eigenvector x. Then there are inﬁnitely many vectors u such that (A −λ1I)u = x. (10.5.2) Moreover, if u is any such vector then y1 = xeλ1t and y2 = ueλ1t + xteλ1t (10.5.3) are linearly independent solutions of y′ = Ay. A complete proof of this theorem is beyond the scope of this book. The difﬁculty is in proving that there’s a vector u satisfying (10.5.2), since det(A −λ1I) = 0. We’ll take this without proof and verify the other assertions of the theorem. We already know that y1 in (10.5.3) is a solution of y′ = Ay. To see that y2 is also a solution, we compute y′ 2 −Ay2 = λ1ueλ1t + xeλ1t + λ1xteλ1t −Aueλ1t −Axteλ1t = (λ1u + x −Au)eλ1t + (λ1x −Ax)teλ1t. Since Ax = λ1x, this can be written as y′ 2 −Ay2 = −((A −λ1I)u −x) eλ1t, and now (10.5.2) implies that y′ 2 = Ay2.	Elementary Differential Equations with Boundary Value Problems_Page_554_Chunk2958
Section 10.5 Constant Coefﬁcient Homogeneous Systems II 545 To see that y1 and y2 are linearly independent, suppose c1 and c2 are constants such that c1y1 + c2y2 = c1xeλ1t + c2(ueλ1t + xteλ1t) = 0. (10.5.4) We must show that c1 = c2 = 0. Multiplying (10.5.4) by e−λ1t shows that c1x + c2(u + xt) = 0. (10.5.5) By differentiating this with respect to t, we see that c2x = 0, which implies c2 = 0, because x ̸= 0. Substituting c2 = 0 into (10.5.5) yields c1x = 0, which implies that c1 = 0, again because x ̸= 0 Example 10.5.2 Use Theorem 10.5.1 to ﬁnd the general solution of the system y′ = 11 −25 4 −9 y (10.5.6) considered in Example 10.5.1. Solution In Example 10.5.1 we saw that λ1 = 1 is an eigenvalue of multiplicity 2 of the coefﬁcient matrix A in (10.5.6), and that all of the eigenvectors of A are multiples of x = 5 2 . Therefore y1 = 5 2 et is a solution of (10.5.6). From Theorem 10.5.1, a second solution is given by y2 = uet + xtet, where (A −I)u = x. The augmented matrix of this system is  10 −25 ... 5 4 −10 ... 2  , which is row equivalent to  1 −5 2 ... 1 2 0 0 ... 0  . Therefore the components of u must satisfy u1 −5 2u2 = 1 2, where u2 is arbitrary. We choose u2 = 0, so that u1 = 1/2 and u = 1 2 0 . Thus, y2 = 1 0 et 2 + 5 2 tet.	Elementary Differential Equations with Boundary Value Problems_Page_555_Chunk2959
546 Chapter 10 Linear Systems of Differential Equations Since y1 and y2 are linearly independent by Theorem 10.5.1, they form a fundamental set of solutions of (10.5.6). Therefore the general solution of (10.5.6) is y = c1 5 2 et + c2 1 0 et 2 + 5 2 tet . Note that choosing the arbitrary constant u2 to be nonzero is equivalent to adding a scalar multiple of y1 to the second solution y2 (Exercise 33). Example 10.5.3 Find the general solution of y′ =   3 4 −10 2 1 −2 2 2 −5  y. (10.5.7) Solution The characteristic polynomial of the coefﬁcient matrix A in (10.5.7) is 3 −λ 4 −10 2 1 −λ −2 2 2 −5 −λ = −(λ −1)(λ + 1)2. Hence, the eigenvalues are λ1 = 1 with multiplicity 1 and λ2 = −1 with multiplicity 2. Eigenvectors associated with λ1 = 1 must satisfy (A−I)x = 0. The augmented matrix of this system is   2 4 −10 ... 0 2 0 −2 ... 0 2 2 −6 ... 0  , which is row equivalent to   1 0 −1 ... 0 0 1 −2 ... 0 0 0 0 ... 0  . Hence, x1 = x3 and x2 = 2x3, where x3 is arbitrary. Choosing x3 = 1 yields the eigenvector x1 =   1 2 1  . Therefore y1 =   1 2 1  et is a solution of (10.5.7). Eigenvectors associated with λ2 = −1 satisfy (A + I)x = 0. The augmented matrix of this system is   4 4 −10 ... 0 2 2 −2 ... 0 2 2 −4 ... 0  ,	Elementary Differential Equations with Boundary Value Problems_Page_556_Chunk2960
Section 10.5 Constant Coefﬁcient Homogeneous Systems II 547 which is row equivalent to   1 1 0 ... 0 0 0 1 ... 0 0 0 0 ... 0  . Hence, x3 = 0 and x1 = −x2, where x2 is arbitrary. Choosing x2 = 1 yields the eigenvector x2 =   −1 1 0  , so y2 =   −1 1 0  e−t is a solution of (10.5.7). Since all the eigenvectors of A associated with λ2 = −1 are multiples of x2, we must now use Theo- rem 10.5.1 to ﬁnd a third solution of (10.5.7) in the form y3 = ue−t +   −1 1 0  te−t, (10.5.8) where u is a solution of (A + I)u = x2. The augmented matrix of this system is   4 4 −10 ... −1 2 2 −2 ... 1 2 2 −4 ... 0  , which is row equivalent to   1 1 0 ... 1 0 0 1 ... 1 2 0 0 0 ... 0  . Hence, u3 = 1/2 and u1 = 1 −u2, where u2 is arbitrary. Choosing u2 = 0 yields u =   1 0 1 2  , and substituting this into (10.5.8) yields the solution y3 =   2 0 1  e−t 2 +   −1 1 0  te−t of (10.5.7).	Elementary Differential Equations with Boundary Value Problems_Page_557_Chunk2961
548 Chapter 10 Linear Systems of Differential Equations Since the Wronskian of {y1, y2, y3} at t = 0 is 1 −1 1 2 1 0 1 0 1 2 = 1 2, {y1, y2, y3} is a fundamental set of solutions of (10.5.7). Therefore the general solution of (10.5.7) is y = c1   1 2 1  et + c2   −1 1 0  e−t + c3     2 0 1  e−t 2 +   −1 1 0  te−t  . Theorem 10.5.2 Suppose the n×n matrix A has an eigenvalue λ1 of multiplicity≥3 and the associated eigenspace is one–dimensional; that is, all eigenvectors associated with λ1 are scalar multiples of the eigenvector x. Then there are inﬁnitely many vectors u such that (A −λ1I)u = x, (10.5.9) and, if u is any such vector, there are inﬁnitely many vectors v such that (A −λ1I)v = u. (10.5.10) If u satisﬁes (10.5.9) and v satisﬁes (10.5.10), then y1 = xeλ1t, y2 = ueλ1t + xteλ1t, and y3 = veλ1t + uteλ1t + xt2eλ1t 2 are linearly independent solutions of y′ = Ay. Again, it’s beyond the scope of this book to prove that there are vectors u and v that satisfy (10.5.9) and (10.5.10). Theorem 10.5.1 implies that y1 and y2 are solutions of y′ = Ay. We leave the rest of the proof to you (Exercise 34). Example 10.5.4 Use Theorem 10.5.2 to ﬁnd the general solution of y′ =   1 1 1 1 3 −1 0 2 2  y. (10.5.11) Solution The characteristic polynomial of the coefﬁcient matrix A in (10.5.11) is 1 −λ 1 1 1 3 −λ −1 0 2 2 −λ = −(λ −2)3. Hence, λ1 = 2 is an eigenvalue of multiplicity 3. The associated eigenvectors satisfy (A −2I)x = 0. The augmented matrix of this system is   −1 1 1 ... 0 1 1 −1 ... 0 0 2 0 ... 0  ,	Elementary Differential Equations with Boundary Value Problems_Page_558_Chunk2962
Section 10.5 Constant Coefﬁcient Homogeneous Systems II 549 which is row equivalent to   1 0 −1 ... 0 0 1 0 ... 0 0 0 0 ... 0  . Hence, x1 = x3 and x2 = 0, so the eigenvectors are all scalar multiples of x1 =   1 0 1  . Therefore y1 =   1 0 1  e2t is a solution of (10.5.11). We now ﬁnd a second solution of (10.5.11) in the form y2 = ue2t +   1 0 1  te2t, where u satisﬁes (A −2I)u = x1. The augmented matrix of this system is   −1 1 1 ... 1 1 1 −1 ... 0 0 2 0 ... 1  , which is row equivalent to   1 0 −1 ... −1 2 0 1 0 ... 1 2 0 0 0 ... 0  . Letting u3 = 0 yields u1 = −1/2 and u2 = 1/2; hence, u = 1 2   −1 1 0   and y2 =   −1 1 0  e2t 2 +   1 0 1  te2t is a solution of (10.5.11). We now ﬁnd a third solution of (10.5.11) in the form y3 = ve2t +   −1 1 0  te2t 2 +   1 0 1  t2e2t 2	Elementary Differential Equations with Boundary Value Problems_Page_559_Chunk2963
550 Chapter 10 Linear Systems of Differential Equations where v satisﬁes (A −2I)v = u. The augmented matrix of this system is   −1 1 1 ... −1 2 1 1 −1 ... 1 2 0 2 0 ... 0  , which is row equivalent to   1 0 −1 ... 1 2 0 1 0 ... 0 0 0 0 ... 0  . Letting v3 = 0 yields v1 = 1/2 and v2 = 0; hence, v = 1 2   1 0 0  . Therefore y3 =   1 0 0  e2t 2 +   −1 1 0  te2t 2 +   1 0 1  t2e2t 2 is a solution of (10.5.11). Since y1, y2, and y3 are linearly independent by Theorem 10.5.2, they form a fundamental set of solutions of (10.5.11). Therefore the general solution of (10.5.11) is y = c1   1 0 1  e2t + c2     −1 1 0  e2t 2 +   1 0 1  te2t   +c3     1 0 0  e2t 2 +   −1 1 0  te2t 2 +   1 0 1  t2e2t 2  . Theorem 10.5.3 Suppose the n×n matrix A has an eigenvalue λ1 of multiplicity≥3 and the associated eigenspace is two–dimensional; that is, all eigenvectors of A associated with λ1 are linear combinations of two linearly independent eigenvectors x1 and x2. Then there are constants α and β (not both zero) such that if x3 = αx1 + βx2, (10.5.12) then there are inﬁnitely many vectors u such that (A −λ1I)u = x3. (10.5.13) If u satisﬁes (10.5.13), then y1 = x1eλ1t, y2 = x2eλ1t, and y3 = ueλ1t + x3teλ1t, (10.5.14) are linearly independent solutions of y′ = Ay. We omit the proof of this theorem.	Elementary Differential Equations with Boundary Value Problems_Page_560_Chunk2964
Section 10.5 Constant Coefﬁcient Homogeneous Systems II 551 Example 10.5.5 Use Theorem 10.5.3 to ﬁnd the general solution of y′ =   0 0 1 −1 1 1 −1 0 2  y. (10.5.15) Solution The characteristic polynomial of the coefﬁcient matrix A in (10.5.15) is −λ 0 1 −1 1 −λ 1 −1 0 2 −λ = −(λ −1)3. Hence, λ1 = 1 is an eigenvalue of multiplicity 3. The associated eigenvectors satisfy (A −I)x = 0. The augmented matrix of this system is   −1 0 1 ... 0 −1 0 1 ... 0 −1 0 1 ... 0  , which is row equivalent to   1 0 −1 ... 0 0 0 0 ... 0 0 0 0 ... 0  . Hence, x1 = x3 and x2 is arbitrary, so the eigenvectors are of the form x1 =   x3 x2 x3  = x3   1 0 1  + x2   0 1 0  . Therefore the vectors x1 =   1 0 1   and x2 =   0 1 0   (10.5.16) form a basis for the eigenspace, and y1 =   1 0 1  et and y2 =   0 1 0  et are linearly independent solutions of (10.5.15). To ﬁnd a third linearly independent solution of (10.5.15), we must ﬁnd constants α and β (not both zero) such that the system (A −I)u = αx1 + βx2 (10.5.17) has a solution u. The augmented matrix of this system is   −1 0 1 ... α −1 0 1 ... β −1 0 1 ... α  ,	Elementary Differential Equations with Boundary Value Problems_Page_561_Chunk2965
552 Chapter 10 Linear Systems of Differential Equations which is row equivalent to   1 0 −1 ... −α 0 0 0 ... β −α 0 0 0 ... 0  . (10.5.18) Therefore (10.5.17) has a solution if and only if β = α, where α is arbitrary. If α = β = 1 then (10.5.12) and (10.5.16) yield x3 = x1 + x2 =   1 0 1  +   0 1 0  =   1 1 1  , and the augmented matrix (10.5.18) becomes   1 0 −1 ... −1 0 0 0 ... 0 0 0 0 ... 0  . This implies that u1 = −1 + u3, while u2 and u3 are arbitrary. Choosing u2 = u3 = 0 yields u =   −1 0 0  . Therefore (10.5.14) implies that y3 = uet + x3tet =   −1 0 0  et +   1 1 1  tet is a solution of (10.5.15). Since y1, y2, and y3 are linearly independent by Theorem 10.5.3, they form a fundamental set of solutions for (10.5.15). Therefore the general solution of (10.5.15) is y = c1   1 0 1  et + c2   0 1 0  et + c3     −1 0 0  et +   1 1 1  tet  . Geometric Properties of Solutions when n = 2 We’ll now consider the geometric properties of solutions of a 2 × 2 constant coefﬁcient system y′ 1 y′ 2 = a11 a12 a21 a22 y1 y2 (10.5.19) under the assumptions of this section; that is, when the matrix A = a11 a12 a21 a22 has a repeated eigenvalue λ1 and the associated eigenspace is one-dimensional. In this case we know from Theorem 10.5.1 that the general solution of (10.5.19) is y = c1xeλ1t + c2(ueλ1t + xteλ1t), (10.5.20)	Elementary Differential Equations with Boundary Value Problems_Page_562_Chunk2966
Section 10.5 Constant Coefﬁcient Homogeneous Systems II 553 where x is an eigenvector of A and u is any one of the inﬁnitely many solutions of (A −λ1I)u = x. (10.5.21) We assume that λ1 ̸= 0. x u c2 > 0 c2 < 0 L Positive Half−Plane Negative Half−Plane Figure 10.5.1 Positive and negative half-planes Let L denote the line through the origin parallel to x. By a half-line of L we mean either of the rays obtained by removing the origin from L. Eqn. (10.5.20) is a parametric equation of the half-line of L in the direction of x if c1 > 0, or of the half-line of L in the direction of −x if c1 < 0. The origin is the trajectory of the trivial solution y ≡0. Henceforth, we assume that c2 ̸= 0. In this case, the trajectory of (10.5.20) can’t intersect L, since every point of L is on a trajectory obtained by setting c2 = 0. Therefore the trajectory of (10.5.20) must lie entirely in one of the open half-planes bounded by L, but does not contain any point on L. Since the initial point (y1(0), y2(0)) deﬁned by y(0) = c1x1 + c2u is on the trajectory, we can determine which half-plane contains the trajectory from the sign of c2, as shown in Figure 552. For convenience we’ll call the half-plane where c2 > 0 the positive half-plane. Similarly, the-half plane where c2 < 0 is the negative half-plane. You should convince yourself (Exercise 35) that even though there are inﬁnitely many vectors u that satisfy (10.5.21), they all deﬁne the same positive and negative half-planes. In the ﬁgures simply regard u as an arrow pointing to the positive half-plane, since wen’t attempted to give u its proper length or direction in comparison with x. For our purposes here, only the relative orientation of x and u is important; that is, whether the positive half-plane is to the right of an observer facing the direction of x (as in Figures 10.5.2 and 10.5.5), or to the left of the observer (as in Figures 10.5.3 and 10.5.4). Multiplying (10.5.20) by e−λ1t yields e−λ1ty(t) = c1x + c2u + c2tx. Since the last term on the right is dominant when \|t\| is large, this provides the following information on the direction of y(t): (a) Along trajectories in the positive half-plane (c2 > 0), the direction of y(t) approaches the direction of x as t →∞and the direction of −x as t →−∞. (b) Along trajectories in the negative half-plane (c2 < 0), the direction of y(t) approaches the direction of −x as t →∞and the direction of x as t →−∞.	Elementary Differential Equations with Boundary Value Problems_Page_563_Chunk2967
554 Chapter 10 Linear Systems of Differential Equations Since lim t→∞∥y(t)∥= ∞ and lim t→−∞y(t) = 0 if λ1 > 0, or lim t−→∞∥y(t)∥= ∞ and lim t→∞y(t) = 0 if λ1 < 0, there are four possible patterns for the trajectories of (10.5.19), depending upon the signs of c2 and λ1. Figures 10.5.2-10.5.5 illustrate these patterns, and reveal the following principle: If λ1 and c2 have the same sign then the direction of the traectory approaches the direction of −x as ∥y∥→0 and the direction of x as ∥y∥→∞. If λ1 and c2 have opposite signs then the direction of the trajectory approaches the direction of x as ∥y∥→0 and the direction of −x as ∥y∥→∞. y1 y2 u x L Figure 10.5.2 Positive eigenvalue; motion away from the origin y1 y2 u x L Figure 10.5.3 Positive eigenvalue; motion away from the origin y1 y2 u x L Figure 10.5.4 Negative eigenvalue; motion toward the origin y1 y2 x L u Figure 10.5.5 Negative eigenvalue; motion toward the origin	Elementary Differential Equations with Boundary Value Problems_Page_564_Chunk2968
Section 10.5 Constant Coefﬁcient Homogeneous Systems II 555 10.5 Exercises In Exercises 1–12 ﬁnd the general solution. 1. y′ = 3 4 −1 7 y 2. y′ = 0 −1 1 −2 y 3. y′ = −7 4 −1 −11 y 4. y′ = 3 1 −1 1 y 5. y′ = 4 12 −3 −8 y 6. y′ = −10 9 −4 2 y 7. y′ = −13 16 −9 11 y 8. y′ =   0 2 1 −4 6 1 0 4 2  y 9. y′ = 1 3   1 1 −3 −4 −4 3 −2 1 0  y 10. y′ =   −1 1 −1 −2 0 2 −1 3 −1  y 11. y′ =   4 −2 −2 −2 3 −1 2 −1 3  y 12. y′ =   6 −5 3 2 −1 3 2 1 1  y In Exercises 13–23 solve the initial value problem. 13. y′ = −11 8 −2 −3 y, y(0) = 6 2 14. y′ = 15 −9 16 −9 y, y(0) = 5 8 15. y′ = −3 −4 1 −7 y, y(0) = 2 3 16. y′ = −7 24 −6 17 y, y(0) = 3 1 17. y′ = −7 3 −3 −1 y, y(0) = 0 2 18. y′ =   −1 1 0 1 −1 −2 −1 −1 −1  y, y(0) =   6 5 −7   19. y′ =   −2 2 1 −2 2 1 −3 3 2  y, y(0) =   −6 −2 0   20. y′ =   −7 −4 4 −1 0 1 −9 −5 6  y, y(0) =   −6 9 −1  	Elementary Differential Equations with Boundary Value Problems_Page_565_Chunk2969
556 Chapter 10 Linear Systems of Differential Equations 21. y′ =   −1 −4 −1 3 6 1 −3 −2 3  y, y(0) =   −2 1 3   22. y′ =   4 −8 −4 −3 −1 −3 1 −1 9  y, y(0) =   −4 1 −3   23. y′ =   −5 −1 11 −7 1 13 −4 0 8  y, y(0) =   0 2 2   The coefﬁcient matrices in Exercises 24–32 have eigenvalues of multiplicity 3. Find the general solution. 24. y′ =   5 −1 1 −1 9 −3 −2 2 4  y 25. y′ =   1 10 −12 2 2 3 2 −1 6  y 26. y′ =   −6 −4 −4 2 −1 1 2 3 1  y 27. y′ =   0 2 −2 −1 5 −3 1 1 1  y 28. y′ =   −2 −12 10 2 −24 11 2 −24 8  y 29. y′ =   −1 −12 8 1 −9 4 1 −6 1  y 30. y′ =   −4 0 −1 −1 −3 −1 1 0 −2  y 31. y′ =   −3 −3 4 4 5 −8 2 3 −5  y 32. y′ =   −3 −1 0 1 −1 0 −1 −1 −2  y 33. Under the assumptions of Theorem 10.5.1, suppose u and ˆu are vectors such that (A −λ1I)u = x and (A −λ1I)ˆu = x, and let y2 = ueλ1t + xteλ1t and ˆy2 = ˆueλ1t + xteλ1t. Show that y2 −ˆy2 is a scalar multiple of y1 = xeλ1t. 34. Under the assumptions of Theorem 10.5.2, let y1 = xeλ1t, y2 = ueλ1t + xteλ1t, and y3 = veλ1t + uteλ1t + xt2eλ1t 2 . Complete the proof of Theorem 10.5.2 by showing that y3 is a solution of y′ = Ay and that {y1, y2, y3} is linearly independent.	Elementary Differential Equations with Boundary Value Problems_Page_566_Chunk2970
Section 10.6 Constant Coefﬁcient Homogeneous Systems III 557 35. Suppose the matrix A = a11 a12 a21 a22 has a repeated eigenvalue λ1 and the associated eigenspace is one-dimensional. Let x be a λ1- eigenvector of A. Show that if (A −λ1I)u1 = x and (A −λ1I)u2 = x, then u2 −u1 is parallel to x. Conclude from this that all vectors u such that (A−λ1I)u = x deﬁne the same positive and negative half-planes with respect to the line L through the origin parallel to x. In Exercises 36- 45 plot trajectories of the given system. 36. C/G y′ = −3 −1 4 1 y 37. C/G y′ = 2 −1 1 0 y 38. C/G y′ = −1 −3 3 5 y 39. C/G y′ = −5 3 −3 1 y 40. C/G y′ = −2 −3 3 4 y 41. C/G y′ = −4 −3 3 2 y 42. C/G y′ = 0 −1 1 −2 y 43. C/G y′ = 0 1 −1 2 y 44. C/G y′ = −2 1 −1 0 y 45. C/G y′ = 0 −4 1 −4 y 10.6 CONSTANT COEFFICIENT HOMOGENEOUS SYSTEMS III We now consider the system y′ = Ay, where A has a complex eigenvalue λ = α + iβ with β ̸= 0. We continue to assume that A has real entries, so the characteristic polynomial of A has real coefﬁcients. This implies that λ = α −iβ is also an eigenvalue of A. An eigenvector x of A associated with λ = α + iβ will have complex entries, so we’ll write x = u + iv where u and v have real entries; that is, u and v are the real and imaginary parts of x. Since Ax = λx, A(u + iv) = (α + iβ)(u + iv). (10.6.1) Taking complex conjugates here and recalling that A has real entries yields A(u −iv) = (α −iβ)(u −iv), which shows that x = u −iv is an eigenvector associated with λ = α −iβ. The complex conjugate eigenvalues λ and λ can be separately associated with linearly independent solutions y′ = Ay; however, we won’t pursue this approach, since solutions obtained in this way turn out to be complex–valued. Instead, we’ll obtain solutions of y′ = Ay in the form y = f1u + f2v (10.6.2) where f1 and f2 are real–valued scalar functions. The next theorem shows how to do this.	Elementary Differential Equations with Boundary Value Problems_Page_567_Chunk2971
558 Chapter 10 Linear Systems of Differential Equations Theorem 10.6.1 Let A be an n × n matrix with real entries. Let λ = α + iβ (β ̸= 0) be a complex eigenvalue of A and let x = u + iv be an associated eigenvector, where u and v have real components. Then u and v are both nonzero and y1 = eαt(u cos βt −v sin βt) and y2 = eαt(u sinβt + v cos βt), which are the real and imaginary parts of eαt(cos βt + i sin βt)(u + iv), (10.6.3) are linearly independent solutions of y′ = Ay. Proof A function of the form (10.6.2) is a solution of y′ = Ay if and only if f′ 1u + f′ 2v = f1Au + f2Av. (10.6.4) Carrying out the multiplicationindicated on the right side of (10.6.1) and collecting the real and imaginary parts of the result yields A(u + iv) = (αu −βv) + i(αv + βu). Equating real and imaginary parts on the two sides of this equation yields Au = αu −βv Av = αv + βu. We leave it to you (Exercise 25) to show from this that u and v are both nonzero. Substituting from these equations into (10.6.4) yields f′ 1u + f′ 2v = f1(αu −βv) + f2(αv + βu) = (αf1 + βf2)u + (−βf1 + αf2)v. This is true if f′ 1 = αf1 + βf2 f′ 2 = −βf1 + αf2, or, equivalently, f′ 1 −αf1 = βf2 f′ 2 −αf2 = −βf1. If we let f1 = g1eαt and f2 = g2eαt, where g1 and g2 are to be determined, then the last two equations become g′ 1 = βg2 g′ 2 = −βg1, which implies that g′′ 1 = βg′ 2 = −β2g1, so g′′ 1 + β2g1 = 0. The general solution of this equation is g1 = c1 cos βt + c2 sin βt. Moreover, since g2 = g′ 1/β, g2 = −c1 sin βt + c2 cos βt.	Elementary Differential Equations with Boundary Value Problems_Page_568_Chunk2972
Section 10.6 Constant Coefﬁcient Homogeneous Systems III 559 Multiplying g1 and g2 by eαt shows that f1 = eαt( c1 cos βt + c2 sin βt), f2 = eαt(−c1 sin βt + c2 cos βt). Substituting these into (10.6.2) shows that y = eαt [(c1 cos βt + c2 sin βt)u + (−c1 sin βt + c2 cos βt)v] = c1eαt(u cos βt −v sin βt) + c2eαt(u sinβt + v cos βt) (10.6.5) is a solution of y′ = Ay for any choice of the constants c1 and c2. In particular, by ﬁrst taking c1 = 1 and c2 = 0 and then taking c1 = 0 and c2 = 1, we see that y1 and y2 are solutions of y′ = Ay. We leave it to you to verify that they are, respectively, the real and imaginary parts of (10.6.3) (Exercise 26), and that they are linearly independent (Exercise 27). Example 10.6.1 Find the general solution of y′ = 4 −5 5 −2 y. (10.6.6) Solution The characteristic polynomial of the coefﬁcient matrix A in (10.6.6) is 4 −λ −5 5 −2 −λ = (λ −1)2 + 16. Hence, λ = 1 + 4i is an eigenvalue of A. The associated eigenvectors satisfy (A −(1 + 4i) I) x = 0. The augmented matrix of this system is  3 −4i −5 ... 0 5 −3 −4i ... 0  , which is row equivalent to  1 −3+4i 5 ... 0 0 0 ... 0  . Therefore x1 = (3 + 4i)x2/5. Taking x2 = 5 yields x1 = 3 + 4i, so x = 3 + 4i 5 is an eigenvector. The real and imaginary parts of et(cos 4t + i sin 4t) 3 + 4i 5 are y1 = et 3 cos 4t −4 sin4t 5 cos 4t and y2 = et 3 sin 4t + 4 cos 4t 5 sin4t , which are linearly independent solutions of (10.6.6). The general solution of (10.6.6) is y = c1et 3 cos 4t −4 sin 4t 5 cos 4t + c2et 3 sin 4t + 4 cos 4t 5 sin4t .	Elementary Differential Equations with Boundary Value Problems_Page_569_Chunk2973
560 Chapter 10 Linear Systems of Differential Equations Example 10.6.2 Find the general solution of y′ = −14 39 −6 16 y. (10.6.7) Solution The characteristic polynomial of the coefﬁcient matrix A in (10.6.7) is −14 −λ 39 −6 16 −λ = (λ −1)2 + 9. Hence, λ = 1 + 3i is an eigenvalue of A. The associated eigenvectors satisfy (A −(1 + 3i)I) x = 0. The augmented augmented matrix of this system is  −15 −3i 39 ... 0 −6 15 −3i ... 0  , which is row equivalent to  1 −5+i 2 ... 0 0 0 ... 0  . Therefore x1 = (5 −i)/2. Taking x2 = 2 yields x1 = 5 −i, so x = 5 −i 2 is an eigenvector. The real and imaginary parts of et(cos 3t + i sin 3t) 5 −i 2 are y1 = et sin 3t + 5 cos 3t 2 cos 3t and y2 = et −cos 3t + 5 sin3t 2 sin3t , which are linearly independent solutions of (10.6.7). The general solution of (10.6.7) is y = c1et sin 3t + 5 cos 3t 2 cos 3t + c2et −cos 3t + 5 sin3t 2 sin3t . Example 10.6.3 Find the general solution of y′ =   −5 5 4 −8 7 6 1 0 0  y. (10.6.8) Solution The characteristic polynomial of the coefﬁcient matrix A in (10.6.8) is −5 −λ 5 4 −8 7 −λ 6 1 0 −λ = −(λ −2)(λ2 + 1).	Elementary Differential Equations with Boundary Value Problems_Page_570_Chunk2974
Section 10.6 Constant Coefﬁcient Homogeneous Systems III 561 Hence, the eigenvalues of A are λ1 = 2, λ2 = i, and λ3 = −i. The augmented matrix of (A −2I)x = 0 is   −7 5 4 ... 0 −8 5 6 ... 0 1 0 −2 ... 0  , which is row equivalent to   1 0 −2 ... 0 0 1 −2 ... 0 0 0 0 ... 0  . Therefore x1 = x2 = 2x3. Taking x3 = 1 yields x1 =   2 2 1  , so y1 =   2 2 1  e2t is a solution of (10.6.8). The augmented matrix of (A −iI)x = 0 is   −5 −i 5 4 ... 0 −8 7 −i 6 ... 0 1 0 −i ... 0  , which is row equivalent to   1 0 −i ... 0 0 1 1 −i ... 0 0 0 0 ... 0  . Therefore x1 = ix3 and x2 = −(1 −i)x3. Taking x3 = 1 yields the eigenvector x2 =   i −1 + i 1  . The real and imaginary parts of (cos t + i sin t)   i −1 + i 1   are y2 =   −sin t −cos t −sin t cos t   and y3 =   cos t cos t −sin t sin t  ,	Elementary Differential Equations with Boundary Value Problems_Page_571_Chunk2975
562 Chapter 10 Linear Systems of Differential Equations which are solutions of (10.6.8). Since the Wronskian of {y1, y2, y3} at t = 0 is 2 0 1 2 −1 1 1 1 0 = 1, {y1, y2, y3} is a fundamental set of solutions of (10.6.8). The general solution of (10.6.8) is y = c1   2 2 1  e2t + c2   −sin t −cos t −sin t cos t  + c3   cos t cos t −sint sin t  . Example 10.6.4 Find the general solution of y′ =   1 −1 −2 1 3 2 1 −1 2  y. (10.6.9) Solution The characteristic polynomial of the coefﬁcient matrix A in (10.6.9) is 1 −λ −1 −2 1 3 −λ 2 1 −1 2 −λ = −(λ −2)	Elementary Differential Equations with Boundary Value Problems_Page_572_Chunk2976
Section 10.6 Constant Coefﬁcient Homogeneous Systems III 563 which is row equivalent to   1 0 −i ... 0 0 1 i ... 0 0 0 0 ... 0  . Therefore x1 = ix3 and x2 = −ix3. Taking x3 = 1 yields the eigenvector x2 =   i −i 1   The real and imaginary parts of e2t(cos 2t + i sin 2t)   i −i 1   are y2 = e2t   −sin 2t sin2t cos 2t   and y2 = e2t   cos 2t −cos 2t sin 2t  , which are solutions of (10.6.9). Since the Wronskian of {y1, y2, y3} at t = 0 is −1 0 1 −1 0 −1 1 1 0 = −2, {y1, y2, y3} is a fundamental set of solutions of (10.6.9). The general solution of (10.6.9) is y = c1   −1 −1 1  e2t + c2e2t   −sin 2t sin 2t cos 2t  + c3e2t   cos 2t −cos 2t sin 2t  . Geometric Properties of Solutions when n = 2 We’ll now consider the geometric properties of solutions of a 2 × 2 constant coefﬁcient system y′ 1 y′ 2 = a11 a12 a21 a22 y1 y2 (10.6.10) under the assumptions of this section; that is, when the matrix A = a11 a12 a21 a22 has a complex eigenvalue λ = α + iβ (β ̸= 0) and x = u + iv is an associated eigenvector, where u and v have real components. To describe the trajectories accurately it’s necessary to introduce a new rectangular coordinate system in the y1-y2 plane. This raises a point that hasn’t come up before: It is always possible to choose x so that (u, v) = 0. A special effort is required to do this, since not every eigenvector has this property. However, if we know an eigenvector that doesn’t, we can multiply it by a suitable complex constant to obtain one that does. To see this, note that if x is a λ-eigenvector of A and k is an arbitrary real number, then x1 = (1 + ik)x = (1 + ik)(u + iv) = (u −kv) + i(v + ku)	Elementary Differential Equations with Boundary Value Problems_Page_573_Chunk2977
564 Chapter 10 Linear Systems of Differential Equations is also a λ-eigenvector of A, since Ax1 = A((1 + ik)x) = (1 + ik)Ax = (1 + ik)λx = λ((1 + ik)x) = λx1. The real and imaginary parts of x1 are u1 = u −kv and v1 = v + ku, (10.6.11) so (u1, v1) = (u −kv, v + ku) = − (u, v)k2 + (∥v∥2 −∥u∥2)k −(u, v) . Therefore (u1, v1) = 0 if (u, v)k2 + (∥v∥2 −∥u∥2)k −(u, v) = 0. (10.6.12) If (u, v) ̸= 0 we can use the quadratic formula to ﬁnd two real values of k such that (u1, v1) = 0 (Exercise 28). Example 10.6.5 In Example 10.6.1 we found the eigenvector x = 3 + 4i 5 = 3 5 + i 4 0 for the matrix of the system (10.6.6). Here u = 3 5 and v = 4 0 are not orthogonal, since (u, v) = 12. Since ∥v∥2 −∥u∥2 = −18, (10.6.12) is equivalent to 2k2 −3k −2 = 0. The zeros of this equation are k1 = 2 and k2 = −1/2. Letting k = 2 in (10.6.11) yields u1 = u −2v = −5 5 and v1 = v + 2u = 10 10 , and (u1, v1) = 0. Letting k = −1/2 in (10.6.11) yields u1 = u + v 2 = 5 5 and v1 = v −u 2 = 1 2 −5 5 , and again (u1, v1) = 0. (The numbers don’t always work out as nicely as in this example. You’ll need a calculator or computer to do Exercises 29-40.) Henceforth, we’ll assume that (u, v) = 0. Let U and V be unit vectors in the directions of u and v, respectively; that is, U = u/∥u∥and V = v/∥v∥. The new rectangular coordinate system will have the same origin as the y1-y2 system. The coordinates of a point in this system will be denoted by (z1, z2), where z1 and z2 are the displacements in the directions of U and V, respectively. From (10.6.5), the solutions of (10.6.10) are given by y = eαt [(c1 cos βt + c2 sin βt)u + (−c1 sinβt + c2 cos βt)v] . (10.6.13) For convenience, let’s call the curve traversed by e−αty(t) a shadow trajectory of (10.6.10). Multiplying (10.6.13) by e−αt yields e−αty(t) = z1(t)U + z2(t)V,	Elementary Differential Equations with Boundary Value Problems_Page_574_Chunk2978
Section 10.6 Constant Coefﬁcient Homogeneous Systems III 565 where z1(t) = ∥u∥(c1 cos βt + c2 sin βt) z2(t) = ∥v∥(−c1 sinβt + c2 cos βt). Therefore (z1(t))2 ∥u∥2 + (z2(t))2 ∥v∥2 = c2 1 + c2 2 (verify!), which means that the shadow trajectories of (10.6.10) are ellipses centered at the origin, with axes of symmetry parallel to U and V. Since z′ 1 = β∥u∥ ∥v∥z2 and z′ 2 = −β∥v∥ ∥u∥z1, the vector from the origin to a point on the shadow ellipse rotates in the same direction that V would have to be rotated by π/2 radians to bring it into coincidence with U (Figures 10.6.1 and 10.6.2). y1 y2 V U Figure 10.6.1 Shadow trajectories traversed clockwise y1 y2 U V Figure 10.6.2 Shadow trajectories traversed counterclockwise If α = 0, then any trajectory of (10.6.10) is a shadow trajectory of (10.6.10); therefore, if λ is purely imaginary, then the trajectories of (10.6.10) are ellipses traversed periodically as indicated in Fig- ures 10.6.1 and 10.6.2. If α > 0, then lim t→∞∥y(t)∥= ∞ and lim t→−∞y(t) = 0, so the trajectory spirals away from the origin as t varies from −∞to ∞. The direction of the spiral depends upon the relative orientation of U and V, as shown in Figures 10.6.3 and 10.6.4. If α < 0, then lim t→−∞∥y(t)∥= ∞ and lim t→∞y(t) = 0, so the trajectory spirals toward the origin as t varies from −∞to ∞. Again, the direction of the spiral depends upon the relative orientation of U and V, as shown in Figures 10.6.5 and 10.6.6.	Elementary Differential Equations with Boundary Value Problems_Page_575_Chunk2979
566 Chapter 10 Linear Systems of Differential Equations y1 y2 V U Figure 10.6.3 α > 0; shadow trajectory spiraling outward y1 y2 U V Figure 10.6.4 α > 0; shadow trajectory spiraling outward y1 y2 V U Figure 10.6.5 α < 0; shadow trajectory spiraling inward y1 y2 U V Figure 10.6.6 α < 0; shadow trajectory spiraling inward 10.6 Exercises In Exercises 1–16 ﬁnd the general solution. 1. y′ = −1 2 −5 5 y 2. y′ = −11 4 −26 9 y 3. y′ = 1 2 −4 5 y 4. y′ = 5 −6 3 −1 y	Elementary Differential Equations with Boundary Value Problems_Page_576_Chunk2980
Section 10.6 Constant Coefﬁcient Homogeneous Systems III 567 5. y′ =   3 −3 1 0 2 2 5 1 1  y 6. y′ =   −3 3 1 1 −5 −3 −3 7 3  y 7. y′ =   2 1 −1 0 1 1 1 0 1  y 8. y′ =   −3 1 −3 4 −1 2 4 −2 3  y 9. y′ = 5 −4 10 1 y 10. y′ = 1 3 7 −5 2 5 y 11. y′ = 3 2 −5 1 y 12. y′ = 34 52 −20 −30 y 13. y′ =   1 1 2 1 0 −1 −1 −2 −1  y 14. y′ =   3 −4 −2 −5 7 −8 −10 13 −8  y 15. y′ =   6 0 −3 −3 3 3 1 −2 6  y′ 16. y′ =   1 2 −2 0 2 −1 1 0 0  y′ In Exercises 17–24 solve the initial value problem. 17. y′ = 4 −6 3 −2 y, y(0) = 5 2 18. y′ = 7 15 −3 1 y, y(0) = 5 1 19. y′ = 7 −15 3 −5 y, y(0) = 17 7 20. y′ = 1 6 4 −2 5 2 y, y(0) = 1 −1 21. y′ =   5 2 −1 −3 2 2 1 3 2  y, y(0) =   4 0 6   22. y′ =   4 4 0 8 10 −20 2 3 −2  y, y(0) =   8 6 5   23. y′ =   1 15 −15 −6 18 −22 −3 11 −15  y, y(0) =   15 17 10   24. y′ =   4 −4 4 −10 3 15 2 −3 1  y, y(0) =   16 14 6  	Elementary Differential Equations with Boundary Value Problems_Page_577_Chunk2981
568 Chapter 10 Linear Systems of Differential Equations 25. Suppose an n × n matrix A with real entries has a complex eigenvalue λ = α + iβ (β ̸= 0) with associated eigenvector x = u + iv, where u and v have real components. Show that u and v are both nonzero. 26. Verify that y1 = eαt(u cos βt −v sin βt) and y2 = eαt(u sinβt + v cos βt), are the real and imaginary parts of eαt(cos βt + i sin βt)(u + iv). 27. Show that if the vectors u and v are not both 0 and β ̸= 0 then the vector functions y1 = eαt(u cos βt −v sin βt) and y2 = eαt(u sin βt + v cos βt) are linearly independent on every interval. HINT: There are two cases to consider: (i) {u, v} linearly independent, and (ii) {u, v} linearly dependent. In either case, exploit the the linear independence of {cos βt, sin βt} on every interval. 28. Suppose u = u1 u2 and v = v1 v2 are not orthogonal; that is, (u, v) ̸= 0. (a) Show that the quadratic equation (u, v)k2 + (∥v∥2 −∥u∥2)k −(u, v) = 0 has a positive root k1 and a negative root k2 = −1/k1. (b) Let u(1) 1 = u −k1v, v(1) 1 = v + k1u, u(2) 1 = u −k2v, and v(2) 1 = v + k2u, so that (u(1) 1 , v(1) 1 ) = (u(2) 1 , v(2) 1 ) = 0, from the discussion given above. Show that u(2) 1 = v(1) 1 k1 and v(2) 1 = −u(1) 1 k1 . (c) Let U1, V1, U2, and V2 be unit vectors in the directions of u(1) 1 , v(1) 1 , u(2) 1 , and v(2) 1 , respectively. Conclude from (a) that U2 = V1 and V2 = −U1, and that therefore the counterclockwise angles from U1 to V1 and from U2 to V2 are both π/2 or both −π/2. In Exercises 29-32 ﬁnd vectors U and V parallel to the axes of symmetry of the trajectories, and plot some typical trajectories. 29. C/G y′ = 3 −5 5 −3 y 30. C/G y′ = −15 10 −25 15 y 31. C/G y′ = −4 8 −4 4 y 32. C/G y′ = −3 −15 3 3 y In Exercises 33-40 ﬁnd vectors U and V parallel to the axes of symmetry of the shadow trajectories, and plot a typical trajectory. 33. C/G y′ = −5 6 −12 7 y 34. C/G y′ = 5 −12 6 −7 y	Elementary Differential Equations with Boundary Value Problems_Page_578_Chunk2982
Section 10.7 Variation of Parameters for Nonhomogeneous Linear Systems 569 35. C/G y′ = 4 −5 9 −2 y 36. C/G y′ = −4 9 −5 2 y 37. C/G y′ = −1 10 −10 −1 y 38. C/G y′ = −1 −5 20 −1 y 39. C/G y′ = −7 10 −10 9 y 40. C/G y′ = −7 6 −12 5 y 10.7 VARIATION OF PARAMETERS FOR NONHOMOGENEOUS LINEAR SYSTEMS We now consider the nonhomogeneous linear system y′ = A(t)y + f(t), where A is an n × n matrix function and f is an n-vector forcing function. Associated with this system is the complementary system y′ = A(t)y. The next theorem is analogous to Theorems 5.3.2 and 9.1.5. It shows how to ﬁnd the general solution of y′ = A(t)y + f(t) if we know a particular solution of y′ = A(t)y + f(t) and a fundamental set of solutions of the complementary system. We leave the proof as an exercise (Exercise 21). Theorem 10.7.1 Suppose the n × n matrix function A and the n-vector function f are continuous on (a, b). Let yp be a particular solution of y′ = A(t)y + f(t) on (a, b), and let {y1, y2, . . ., yn} be a fundamental set of solutions of the complementary equation y′ = A(t)y on (a, b). Then y is a solution of y′ = A(t)y + f(t) on (a, b) if and only if y = yp + c1y1 + c2y2 + · · · + cnyn, where c1, c2, ..., cn are constants. Finding a Particular Solution of a Nonhomogeneous System We now discuss an extension of the method of variation of parameters to linear nonhomogeneous systems. This method will produce a particular solution of a nonhomogenous system y′ = A(t)y + f(t) provided that we know a fundamental matrix for the complementary system. To derive the method, suppose Y is a fundamental matrix for the complementary system; that is, Y =   y11 y12 · · · y1n y21 y22 · · · y2n ... ... ... ... yn1 yn2 · · · ynn  , where y1 =   y11 y21 ... yn1  , y2 =   y12 y22 ... yn2  , · · · , yn =   y1n y2n ... ynn   is a fundamental set of solutions of the complementary system. In Section 10.3 we saw that Y ′ = A(t)Y . We seek a particular solution of y′ = A(t)y + f(t) (10.7.1) of the form yp = Y u, (10.7.2)	Elementary Differential Equations with Boundary Value Problems_Page_579_Chunk2983
570 Chapter 10 Linear Systems of Differential Equations where u is to be determined. Differentiating (10.7.2) yields y′ p = Y ′u + Y u′ = AY u + Y u′ (since Y ′ = AY ) = Ayp + Y u′ (since Y u = yp). Comparing this with (10.7.1) shows that yp = Y u is a solution of (10.7.1) if and only if Y u′ = f. Thus, we can ﬁnd a particular solution yp by solving this equation for u′, integrating to obtain u, and computing Y u. We can take all constants of integration to be zero, since any particular solution will sufﬁce. Exercise 22 sketches a proof that this method is analogous to the method of variation of parameters discussed in Sections 5.7 and 9.4 for scalar linear equations. Example 10.7.1 (a) Find a particular solution of the system y′ = 1 2 2 1 y + 2e4t e4t , (10.7.3) which we considered in Example 10.2.1. (b) Find the general solution of (10.7.3). SOLUTION(a) The complementary system is y′ = 1 2 2 1 y. (10.7.4) The characteristic polynomial of the coefﬁcient matrix is 1 −λ 2 2 1 −λ = (λ + 1)(λ −3). Using the method of Section 10.4, we ﬁnd that y1 = e3t e3t and y2 = e−t −e−t are linearly independent solutions of (10.7.4). Therefore Y = e3t e−t e3t −e−t is a fundamental matrix for (10.7.4). We seek a particular solution yp = Y u of (10.7.3), where Y u′ = f; that is, e3t e−t e3t −e−t u′ 1 u′ 2 = 2e4t e4t .	Elementary Differential Equations with Boundary Value Problems_Page_580_Chunk2984
Section 10.7 Variation of Parameters for Nonhomogeneous Linear Systems 571 The determinant of Y is the Wronskian e3t e−t e3t −e−t = −2e2t. By Cramer’s rule, u′ 1 = −1 2e2t 2e4t e−t e4t −e−t = 3e3t 2e2t = 3 2et, u′ 2 = −1 2e2t e3t 2e4t e3t e4t = e7t 2e2t = 1 2e5t. Therefore u′ = 1 2 3et e5t . Integrating and taking the constants of integration to be zero yields u = 1 10 15et e5t , so yp = Y u = 1 10 " e3t e−t e3t −e−t # 15et e5t = 1 5 8e4t 7e4t is a particular solution of (10.7.3). SOLUTION(b) From Theorem 10.7.1, the general solution of (10.7.3) is y = yp + c1y1 + c2y2 = 1 5 8e4t 7e4t + c1 e3t e3t + c2 e−t −e−t , (10.7.5) which can also be written as y = 1 5 8e4t 7e4t + e3t e−t e3t −e−t c, where c is an arbitrary constant vector. Writing (10.7.5) in terms of coordinates yields y1 = 8 5e4t + c1e3t + c2e−t y2 = 7 5e4t + c1e3t −c2e−t, so our result is consistent with Example 10.2.1. . If A isn’t a constant matrix, it’s usually difﬁcult to ﬁnd a fundamental set of solutions for the system y′ = A(t)y. It is beyond the scope of this text to discuss methods for doing this. Therefore, in the following examples and in the exercises involving systems with variable coefﬁcient matrices we’ll provide fundamental matrices for the complementary systems without explaining how they were obtained. Example 10.7.2 Find a particular solution of y′ = 2 2e−2t 2e2t 4 y + 1 1 , (10.7.6)	Elementary Differential Equations with Boundary Value Problems_Page_581_Chunk2985
572 Chapter 10 Linear Systems of Differential Equations given that Y = e4t −1 e6t e2t is a fundamental matrix for the complementary system. Solution We seek a particular solution yp = Y u of (10.7.6) where Y u′ = f; that is, e4t −1 e6t e2t u′ 1 u′ 2 = 1 1 . The determinant of Y is the Wronskian e4t −1 e6t e2t = 2e6t. By Cramer’s rule, u′ 1 = 1 2e6t 1 −1 1 e2t = e2t + 1 2e6t = e−4t + e−6t 2 u′ 2 = 1 2e6t e4t 1 e6t 1 = e4t −e6t 2e6t = e−2t −1 2 . Therefore u′ = 1 2 e−4t + e−6t e−2t −1 . Integrating and taking the constants of integration to be zero yields u = −1 24 3e−4t + 2e−6t 6e−2t + 12t , so yp = Y u = −1 24 e4t −1 e6t e2t 3e−4t + 2e−6t 6e−2t + 12t = 1 24 4e−2t + 12t −3 −3e2t(4t + 1) −8 is a particular solution of (10.7.6). Example 10.7.3 Find a particular solution of y′ = −2 t2 t −3t2 1 −2t y + t2 1 1 , (10.7.7) given that Y = 2t 3t2 1 2t is a fundamental matrix for the complementary system on (−∞, 0) and (0, ∞). Solution We seek a particular solution yp = Y u of (10.7.7) where Y u′ = f; that is, 2t 3t2 1 2t u′ 1 u′ 2 = t2 t2 .	Elementary Differential Equations with Boundary Value Problems_Page_582_Chunk2986
Section 10.7 Variation of Parameters for Nonhomogeneous Linear Systems 573 The determinant of Y is the Wronskian 2t 3t2 1 2t = t2. By Cramer’s rule, u′ 1 = 1 t2 t2 3t2 t2 2t = 2t3 −3t4 t2 = 2t −3t2, u′ 2 = 1 t2 2t t2 1 t2 = 2t3 −t2 t2 = 2t −1. Therefore u′ = 2t −3t2 2t −1 . Integrating and taking the constants of integration to be zero yields u = t2 −t3 t2 −t , so yp = Y u = 2t 3t2 1 2t t2 −t3 t2 −t = t3(t −1) t2(t −1) is a particular solution of (10.7.7). Example 10.7.4 (a) Find a particular solution of y′ =   2 −1 −1 1 0 −1 1 −1 0  y +   et 0 e−t  . (10.7.8) (b) Find the general solution of (10.7.8). SOLUTION(a) The complementary system for (10.7.8) is y′ =   2 −1 −1 1 0 −1 1 −1 0  y. (10.7.9) The characteristic polynomial of the coefﬁcient matrix is 2 −λ −1 −1 1 −λ −1 1 −1 −λ = −λ(λ −1)2. Using the method of Section 10.4, we ﬁnd that y1 =   1 1 1  , y2 =   et et 0  , and y3 =   et 0 et  	Elementary Differential Equations with Boundary Value Problems_Page_583_Chunk2987
574 Chapter 10 Linear Systems of Differential Equations are linearly independent solutions of (10.7.9). Therefore Y =   1 et et 1 et 0 1 0 et   is a fundamental matrix for (10.7.9). We seek a particular solution yp = Y u of (10.7.8), where Y u′ = f; that is,   1 et et 1 et 0 1 0 et     u′ 1 u′ 2 u′ 3  =   et 0 e−t  . The determinant of Y is the Wronskian 1 et et 1 et 0 1 0 et = −e2t. Thus, by Cramer’s rule, u′ 1 = −1 e2t et et et 0 et 0 e−t 0 et = −e3t −et e2t = e−t −et u′ 2 = −1 e2t 1 et et 1 0 0 1 e−t et = −1 −e2t e2t = 1 −e−2t u′ 3 = −1 e2t 1 et et 1 et 0 1 0 e−t = e2t e2t = 1. Therefore u′ =   e−t −et 1 −e−2t 1  . Integrating and taking the constants of integration to be zero yields u =   −et −e−t e−2t 2 + t t  , so yp = Y u =   1 et et 1 et 0 1 0 et     −et −e−t e−2t 2 + t t   =   et(2t −1) −e−t 2 et(t −1) −e−t 2 et(t −1) −e−t   is a particular solution of (10.7.8).	Elementary Differential Equations with Boundary Value Problems_Page_584_Chunk2988
Section 10.7 Variation of Parameters for Nonhomogeneous Linear Systems 575 SOLUTION(a) From Theorem 10.7.1 the general solution of (10.7.8) is y = yp + c1y1 + c2y2 + c3y3 =   et(2t −1) −e−t 2 et(t −1) −e−t 2 et(t −1) −e−t  + c1   1 1 1  + c2   et et 0  + c3   et 0 et  , which can be written as y = yp + Y c =   et(2t −1) −e−t 2 et(t −1) −e−t 2 et(t −1) −e−t  +   1 et et 1 et 0 1 0 et  c where c is an arbitrary constant vector. Example 10.7.5 Find a particular solution of y′ = 1 2   3 e−t −e2t 0 6 0 −e−2t e−3t −1  y +   1 et e−t  , (10.7.10) given that Y =   et 0 e2t 0 e3t e3t e−t 1 0   is a fundamental matrix for the complementary system. Solution We seek a particular solution of (10.7.10) in the form yp = Y u, where Y u′ = f; that is,   et 0 e2t 0 e3t e3t e−t 1 0     u′ 1 u′ 2 u′ 3  =   1 et e−t  . The determinant of Y is the Wronskian et 0 e2t 0 e3t e3t e−t 1 0 = −2e4t. By Cramer’s rule, u′ 1 = −1 2e4t 1 0 e2t et e3t e3t e−t 1 0 = e4t 2e4t = 1 2 u′ 2 = −1 2e4t et 1 e2t 0 et e3t e−t e−t 0 = e3t 2e4t = 1 2e−t u′ 3 = −1 2e4t et 0 1 0 e3t et e−t 1 e−t = −e3t −2e2t 2e4t = 2e−2t −e−t 2 .	Elementary Differential Equations with Boundary Value Problems_Page_585_Chunk2989
576 Chapter 10 Linear Systems of Differential Equations Therefore u′ = 1 2   1 e−t 2e−2t −e−t  . Integrating and taking the constants of integration to be zero yields u = 1 2   t −e−t e−t −e−2t  , so yp = Y u = 1 2   et 0 e2t 0 e3t e3t e−t 1 0     t −e−t e−t −e−2t  = 1 2   et(t + 1) −1 −et e−t(t −1)   is a particular solution of (10.7.10). 10.7 Exercises In Exercises 1–10 ﬁnd a particular solution. 1. y′ = −1 −4 −1 −1 y + 21e4t 8e−3t 2. y′ = 1 5 −4 3 −2 −11 y + 50e3t 10e−3t 3. y′ = 1 2 2 1 y + 1 t 4. y′ = −4 −3 6 5 y + 2 −2et 5. y′ = −6 −3 1 −2 y + 4e−3t 4e−5t 6. y′ = 0 1 −1 0 y + 1 t 7. y′ =   3 1 −1 3 5 1 −6 2 4  y +   3 6 3  8. y′ =   3 −1 −1 −2 3 2 4 −1 −2  y +   1 et et   9. y′ =   −3 2 2 2 −3 2 2 2 −3  y +   et e−5t et   10. y′ = 1 3   1 1 −3 −4 −4 3 −2 1 0  y +   et et et   In Exercises 11–20 ﬁnd a particular solution, given that Y is a fundamental matrix for the complementary system. 11. y′ = 1 t 1 t −t 1 y + t cos t sin t ; Y = t cos t sin t −sin t cos t 12. y′ = 1 t 1 t t 1 y + t t2 ; Y = t et e−t et −e−t	Elementary Differential Equations with Boundary Value Problems_Page_586_Chunk2990
Section 10.7 Variation of Parameters for Nonhomogeneous Linear Systems 577 13. y′ = 1 t2 −1 t −1 −1 t y + t 1 −1 ; Y = t 1 1 t 14. y′ = 1 3 1 −2e−t 2et −1 y + e2t e−2t ; Y = 2 e−t et 2 15. y′ = 1 2t4 3t3 t6 1 −3t3 y + 1 t t2 1 ; Y = 1 t2 t3 t4 −1 t 16. y′ =   1 t −1 −e−t t −1 et t + 1 1 t + 1  y + t2 −1 t2 −1 ; Y = t e−t et t 17. y′ = 1 t   1 1 0 0 2 1 −2 2 2  y +   1 2 1   Y =   t2 t3 1 t2 2t3 −1 0 2t3 2   18. y′ =   3 et e2t e−t 2 et e−2t e−t 1  y +   e3t 0 0  ; Y =   e5t e2t 0 e4t 0 et e3t −1 −1   19. y′ = 1 t   1 t 0 0 1 t 0 −t 1  y +   t t t  ; Y = t   1 cos t sin t 0 −sin t cos t 0 −cos t −sin t   20. y′ = −1 t   e−t −t 1 −e−t e−t 1 −t −e−t e−t −t 1 −e−t  y + 1 t   et 0 et  ; Y = 1 t   et e−t t et −e−t e−t et e−t 0   21. Prove Theorem 10.7.1. 22. (a) Convert the scalar equation P0(t)y(n) + P1(t)y(n−1) + · · · + Pn(t)y = F (t) (A) into an equivalent n × n system y′ = A(t)y + f(t). (B) (b) Suppose (A) is normal on an interval (a, b) and {y1, y2, . . ., yn} is a fundamental set of solutions of P0(t)y(n) + P1(t)y(n−1) + · · · + Pn(t)y = 0 (C) on (a, b). Find a corresponding fundamental matrix Y for y′ = A(t)y (D) on (a, b) such that y = c1y1 + c2y2 + · · · + cnyn is a solution of (C) if and only if y = Y c with c =   c1 c2 ... cn   is a solution of (D).	Elementary Differential Equations with Boundary Value Problems_Page_587_Chunk2991
578 Chapter 10 Linear Systems of Differential Equations (c) Let yp = u1y1 + u1y2 + · · · + unyn be a particular solution of (A), obtained by the method of variation of parameters for scalar equations as given in Section 9.4, and deﬁne u =   u1 u2 ... un  . Show that yp = Y u is a solution of (B). (d) Let yp = Y u be a particular solution of (B), obtained by the method of variation of param- eters for systems as given in this section. Show that yp = u1y1 + u1y2 + · · · + unyn is a solution of (A). 23. Suppose the n × n matrix function A and the n–vector function f are continuous on (a, b). Let t0 be in (a, b), let k be an arbitrary constant vector, and let Y be a fundamental matrix for the homogeneous system y′ = A(t)y. Use variation of parameters to show that the solution of the initial value problem y′ = A(t)y + f(t), y(t0) = k is y(t) = Y (t) Y −1(t0)k + Z t t0 Y −1(s)f(s) ds .	Elementary Differential Equations with Boundary Value Problems_Page_588_Chunk2992
CHAPTER 11 Boundary Value Problems and Fourier Expansions IN THIS CHAPTER we develop series representations of functions that will be used to solve partial differential equations in Chapter 12. SECTION 11.1 deals with ﬁve boundary value problems for the differential equation y′′ + λy = 0. They are related to problems in partial differential equations that will be discussed in Chapter 12. We deﬁne what is meant by eigenvalues and eigenfunctions of the boundary value problems, and show that the eigenfunctions have a property called orthogonality. SECTION 11.2 introduces Fourier series, which are expansions of given functions in term of sines and cosines. SECTION 11.3 deals with expansions of functions in terms of the eigenfunctionsof four of the eigenvalue problems discussed in Section 11.1. They are all related to the Fourier series discussed in Section 11.2. 581	Elementary Differential Equations with Boundary Value Problems_Page_590_Chunk2993
Section 11.1 Eigenvalue Problems for y′′ + λy = 0 581 11.1 EIGENVALUE PROBLEMS FOR y′′ + λy = 0 In Chapter 12 we’ll study partial differential equations that arise in problems of heat conduction, wave propagation, and potential theory. The purpose of this chapter is to develop tools required to solve these equations. In this section we consider the following problems, where λ is a real number and L > 0: Problem 1: y′′ + λy = 0, y(0) = 0, y(L) = 0 Problem 2: y′′ + λy = 0, y′(0) = 0, y′(L) = 0 Problem 3: y′′ + λy = 0, y(0) = 0, y′(L) = 0 Problem 4: y′′ + λy = 0, y′(0) = 0, y(L) = 0 Problem 5: y′′ + λy = 0, y(−L) = y(L), y′(−L) = y′(L) In each problem the conditionsfollowing the differential equation are called boundary conditions. Note that the boundary conditions in Problem 5, unlike those in Problems 1-4, don’t require that y or y′ be zero at the boundary points, but only that y have the same value at x = ±L , and that y′ have the same value at x = ±L. We say that the boundary conditions in Problem 5 are periodic. Obviously, y ≡0 (the trivial solution) is a solution of Problems 1-5 for any value of λ. For most values of λ, there are no other solutions. The interesting question is this: For what values of λ does the problem have nontrivial solutions, and what are they? A value of λ for which the problem has a nontrivial solution is an eigenvalue of the problem, and the nontrivial solutions are λ-eigenfunctions, or eigenfunctions associated with λ. Note that a nonzero constant multiple of a λ-eigenfunction is again a λ-eigenfunction. Problems 1-5 are called eigenvalue problems. Solving an eigenvalue problem means ﬁnding all its eigenvalues and associated eigenfunctions. We’ll take it as given here that all the eigenvalues of Prob- lems 1-5 are real numbers. This is proved in a more general setting in Section 13.2. Theorem 11.1.1 Problems 1–5 have no negative eigenvalues. Moreover, λ = 0 is an eigenvalue of Problems 2 and 5, with associated eigenfunction y0 = 1, but λ = 0 isn’t an eigenvalue of Problems 1, 3, or 4. Proof We consider Problems 1-4, and leave Problem 5 to you (Exercise 1). If y′′ + λy = 0, then y(y′′ + λy) = 0, so Z L 0 y(x)(y′′(x) + λy(x)) dx = 0; therefore, λ Z L 0 y2(x) dx = − Z L 0 y(x)y′′(x) dx. (11.1.1) Integration by parts yields Z L 0 y(x)y′′(x) dx = y(x)y′(x) L 0 − Z L 0 (y′(x))2 dx = y(L)y′(L) −y(0)y′(0) − Z L 0 (y′(x))2 dx. (11.1.2) However, if y satisﬁes any of the boundary conditions of Problems 1-4, then y(L)y′(L) −y(0)y′(0) = 0;	Elementary Differential Equations with Boundary Value Problems_Page_591_Chunk2994
582 Chapter 11 Boundary Value Problems and Fourier Expansions hence, (11.1.1) and (11.1.2) imply that λ Z L 0 y2(x) dx = Z L 0 (y′(x))2 dx. If y ̸≡0, then R L 0 y2(x) dx > 0. Therefore λ ≥0 and, if λ = 0, then y′(x) = 0 for all x in (0, L) (why?), and y is constant on (0, L). Any constant function satisﬁes the boundary conditions of Problem 2, so λ = 0 is an eigenvalue of Problem 2 and any nonzero constant function is an associated eigenfunction. However, the only constant function that satisﬁes the boundary conditions of Problems 1, 3, or 4 is y ≡0. Therefore λ = 0 isn’t an eigenvalue of any of these problems. Example 11.1.1 (Problem 1) Solve the eigenvalue problem y′′ + λy = 0, y(0) = 0, y(L) = 0. (11.1.3) Solution From Theorem 11.1.1, any eigenvalues of (11.1.3) must be positive. If y satisﬁes (11.1.3) with λ > 0, then y = c1 cos √ λ x + c2 sin √ λ x, where c1 and c2 are constants. The boundary condition y(0) = 0 implies that c1 = 0. Therefore y = c2 sin √ λ x. Now the boundary condition y(L) = 0 implies that c2 sin √ λ L = 0. To make c2 sin √ λ L = 0 with c2 ̸= 0, we must choose √ λ = nπ/L, where n is a positive integer. Therefore λn = n2π2/L2 is an eigenvalue and yn = sin nπx L is an associated eigenfunction. For future reference, we state the result of Example 11.1.1 as a theorem. Theorem 11.1.2 The eigenvalue problem y′′ + λy = 0, y(0) = 0, y(L) = 0 has inﬁnitely many positive eigenvalues λn = n2π2/L2, with associated eigenfunctions yn = sin nπx L , n = 1, 2, 3, . . .. There are no other eigenvalues. We leave it to you to prove the next theorem about Problem 2 by an argument like that of Exam- ple 11.1.1 (Exercise 17). Theorem 11.1.3 The eigenvalue problem y′′ + λy = 0, y′(0) = 0, y′(L) = 0 has the eigenvalue λ0 = 0, with associated eigenfunction y0 = 1, and inﬁnitely many positive eigenvalues λn = n2π2/L2, with associated eigenfunctions yn = cos nπx L , n = 1, 2, 3 . . .. There are no other eigenvalues.	Elementary Differential Equations with Boundary Value Problems_Page_592_Chunk2995
Section 11.1 Eigenvalue Problems for y′′ + λy = 0 583 Example 11.1.2 (Problem 3) Solve the eigenvalue problem y′′ + λy = 0, y(0) = 0, y′(L) = 0. (11.1.4) Solution From Theorem 11.1.1, any eigenvalues of (11.1.4) must be positive. If y satisﬁes (11.1.4) with λ > 0, then y = c1 cos √ λ x + c2 sin √ λ x, where c1 and c2 are constants. The boundary condition y(0) = 0 implies that c1 = 0. Therefore y = c2 sin √ λ x. Hence, y′ = c2 √ λ cos √ λ x and the boundary condition y′(L) = 0 implies that c2 cos √ λ L = 0. To make c2 cos √ λ L = 0 with c2 ̸= 0 we must choose √ λ = (2n −1)π 2L , where n is a positive integer. Then λn = (2n −1)2π2/4L2 is an eigenvalue and yn = sin (2n −1)πx 2L is an associated eigenfunction. For future reference, we state the result of Example 11.1.2 as a theorem. Theorem 11.1.4 The eigenvalue problem y′′ + λy = 0, y(0) = 0, y′(L) = 0 has inﬁnitely many positive eigenvalues λn = (2n −1)2π2/4L2, with associated eigenfunctions yn = sin (2n −1)πx 2L , n = 1, 2, 3, . . .. There are no other eigenvalues. We leave it to you to prove the next theorem about Problem 4 by an argument like that of Exam- ple 11.1.2 (Exercise 18). Theorem 11.1.5 The eigenvalue problem y′′ + λy = 0, y′(0) = 0, y(L) = 0 has inﬁnitely many positive eigenvalues λn = (2n −1)2π2/4L2, with associated eigenfunctions yn = cos (2n −1)πx 2L , n = 1, 2, 3, . . .. There are no other eigenvalues. Example 11.1.3 (Problem 5) Solve the eigenvalue problem y′′ + λy = 0, y(−L) = y(L), y′(−L) = y′(L). (11.1.5)	Elementary Differential Equations with Boundary Value Problems_Page_593_Chunk2996
584 Chapter 11 Boundary Value Problems and Fourier Expansions Solution From Theorem 11.1.1, λ = 0 is an eigenvalue of (11.1.5) with associated eigenfunction y0 = 1, and any other eigenvalues must be positive. If y satisﬁes (11.1.5) with λ > 0, then y = c1 cos √ λ x + c2 sin √ λ x, (11.1.6) where c1 and c2 are constants. The boundary condition y(−L) = y(L) implies that c1 cos(− √ λ L) + c2 sin(− √ λ L) = c1 cos √ λ L + c2 sin √ λ L. (11.1.7) Since cos(− √ λ L) = cos √ λ L and sin(− √ λ L) = −sin √ λ L, (11.1.8) (11.1.7) implies that c2 sin √ λ L = 0. (11.1.9) Differentiating (11.1.6) yields y′ = √ λ −c1 sin √ λx + c2 cos √ λx . The boundary condition y′(−L) = y′(L) implies that −c1 sin(− √ λ L) + c2 cos(− √ λ L) = −c1 sin √ λ L + c2 cos √ λ L, and (11.1.8) implies that c1 sin √ λ L = 0. (11.1.10) Eqns. (11.1.9) and (11.1.10) imply that c1 = c2 = 0 unless √ λ = nπ/L, where n is a positive integer. In this case (11.1.9) and (11.1.10) both hold for arbitrary c1 and c2. The eigenvalue determined in this way is λn = n2π2/L2, and each such eigenvalue has the linearly independent associated eigenfunctions cos nπx L and sin nπx L . For future reference we state the result of Example 11.1.3 as a theorem. Theorem 11.1.6 The eigenvalue problem y′′ + λy = 0, y(−L) = y(L), y′(−L) = y′(L), has the eigenvalue λ0 = 0, with associated eigenfunction y0 = 1 and inﬁnitely many positive eigenvalues λn = n2π2/L2, with associated eigenfunctions y1n = cos nπx L and y2n = sin nπx L , n = 1, 2, 3, . . .. There are no other eigenvalues. Orthogonality We say that two integrable functions f and g are orthogonal on an interval [a, b] if Z b a f(x)g(x) dx = 0. More generally, we say that the functions φ1, φ2, ..., φn, ...(ﬁnitely or inﬁnitely many) are orthogonal on [a, b] if Z b a φi(x)φj(x) dx = 0 whenever i ̸= j. The importance of orthogonality will become clear when we study Fourier series in the next two sections.	Elementary Differential Equations with Boundary Value Problems_Page_594_Chunk2997
Section 11.1 Eigenvalue Problems for y′′ + λy = 0 585 Example 11.1.4 Show that the eigenfunctions 1, cos πx L , sin πx L , cos 2πx L , sin 2πx L , . . ., cos nπx L , sin nπx L , . . . (11.1.11) of Problem 5 are orthogonal on [−L, L]. Solution We must show that Z L −L f(x)g(x) dx = 0 (11.1.12) whenever f and g are distinct functions from (11.1.11). If r is any nonzero integer, then Z L −L cos rπx L dx = L rπ sin rπx L L −L = 0. (11.1.13) and Z L −L sin rπx L dx = −L rπ cos rπx L L −L = 0. Therefore (11.1.12) holds if f ≡1 and g is any other function in (11.1.11). If f(x) = cos mπx/L and g(x) = cos nπx/L where m and n are distinct positive integers, then Z L −L f(x)g(x) dx = Z L −L cos mπx L cos nπx L dx. (11.1.14) To evaluate this integral, we use the identity cos A cos B = 1 2[cos(A −B) + cos(A + B)] with A = mπx/L and B = nπx/L. Then (11.1.14) becomes Z L −L f(x)g(x) dx = 1 2 "Z L −L cos (m −n)πx L dx + Z L −L cos (m + n)πx L dx # . Since m −n and m + n are both nonzero integers, (11.1.13) implies that the integrals on the right are both zero. Therefore (11.1.12) is true in this case. If f(x) = sin mπx/L and g(x) = sinnπx/L where m and n are distinct positive integers, then Z L −L f(x)g(x) dx = Z L −L sin mπx L sin nπx L dx. (11.1.15) To evaluate this integral, we use the identity sinA sin B = 1 2[cos(A −B) −cos(A + B)] with A = mπx/L and B = nπx/L. Then (11.1.15) becomes Z L −L f(x)g(x) dx = 1 2 "Z L −L cos (m −n)πx L dx − Z L −L cos (m + n)πx L dx # = 0.	Elementary Differential Equations with Boundary Value Problems_Page_595_Chunk2998
586 Chapter 11 Boundary Value Problems and Fourier Expansions If f(x) = sin mπx/L and g(x) = cos nπx/L where m and n are positive integers (not necessarily distinct), then Z L −L f(x)g(x) dx = Z L −L sin mπx L cos nπx L dx = 0 because the integrand is an odd function and the limits are symmetric about x = 0. Exercises 19-22 ask you to verify that the eigenfunctions of Problems 1-4 are orthogonal on [0, L]. However, this also follows from a general theorem that we’ll prove in Chapter 13. 11.1 Exercises 1. Prove that λ = 0 is an eigenvalue of Problem 5 with associated eigenfunction y0 = 1, and that any other eigenvalues must be positive. HINT: See the proof of Theorem 11.1.1. In Exercises 2-16 solve the eigenvalue problem. 2. y′′ + λy = 0, y(0) = 0, y(π) = 0 3. y′′ + λy = 0, y′(0) = 0, y′(π) = 0 4. y′′ + λy = 0, y(0) = 0, y′(π) = 0 5. y′′ + λy = 0, y′(0) = 0, y(π) = 0 6. y′′ + λy = 0, y(−π) = y(π), y′(−π) = y′(π) 7. y′′ + λy = 0, y′(0) = 0, y′(1) = 0 8. y′′ + λy = 0, y′(0) = 0, y(1) = 0 9. y′′ + λy = 0, y(0) = 0, y(1) = 0 10. y′′ + λy = 0, y(−1) = y(1), y′(−1) = y′(1) 11. y′′ + λy = 0, y(0) = 0, y′(1) = 0 12. y′′ + λy = 0, y(−2) = y(2), y′(−2) = y′(2) 13. y′′ + λy = 0, y(0) = 0, y(2) = 0 14. y′′ + λy = 0, y′(0) = 0, y(3) = 0 15. y′′ + λy = 0, y(0) = 0, y′(1/2) = 0 16. y′′ + λy = 0, y′(0) = 0, y′(5) = 0 17. Prove Theorem 11.1.3. 18. Prove Theorem 11.1.5. 19. Verify that the eigenfunctions sin πx L , sin 2πx L , . . ., sin nπx L , . . . of Problem 1 are orthogonal on [0, L]. 20. Verify that the eigenfunctions 1, cos πx L , cos 2πx L , . . ., cos nπx L , . . . of Problem 2 are orthogonal on [0, L].	Elementary Differential Equations with Boundary Value Problems_Page_596_Chunk2999
Section 11.2 Fourier Expansions I 587 21. Verify that the eigenfunctions sin πx 2L, sin 3πx 2L , . . ., sin (2n −1)πx 2L , . . . of Problem 3 are orthogonal on [0, L]. 22. Verify that the eigenfunctions cos πx 2L, cos 3πx 2L , . . ., cos (2n −1)πx 2L , . . . of Problem 4 are orthogonal on [0, L]. In Exercises 23-26 solve the eigenvalue problem. 23. y′′ + λy = 0, y(0) = 0, Z L 0 y(x) dx = 0 24. y′′ + λy = 0, y′(0) = 0, Z L 0 y(x) dx = 0 25. y′′ + λy = 0, y(L) = 0, Z L 0 y(x) dx = 0 26. y′′ + λy = 0, y′(L) = 0, Z L 0 y(x) dx = 0 11.2 FOURIER EXPANSIONS I In Example 11.1.4 and Exercises 11.1.4–11.1.22 we saw that the eigenfunctions of Problem 5 are orthog- onal on [−L, L] and the eigenfunctions of Problems 1–4 are orthogonal on [0, L]. In this section and the next we introduce some series expansions in terms of these eigenfunctions. We’ll use these expansions to solve partial differential equations in Chapter 12. Theorem 11.2.1 Suppose the functions φ1, φ2, φ3, ..., are orthogonal on [a, b] and Z b a φ2 n(x) dx ̸= 0, n = 1, 2, 3, . . .. (11.2.1) Let c1, c2, c3,...be constants such that the partial sums fN(x) = PN m=1 cmφm(x) satisfy the inequalities \|fN(x)\| ≤M, a ≤x ≤b, N = 1, 2, 3, . . . for some constant M < ∞. Suppose also that the series f(x) = ∞ X m=1 cmφm(x) (11.2.2) converges and is integrable on [a, b]. Then cn = Z b a f(x)φn(x) dx Z b a φ2 n(x) dx , n = 1, 2, 3, . . .. (11.2.3)	Elementary Differential Equations with Boundary Value Problems_Page_597_Chunk3000

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