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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 −... | 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 re... | 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 p... | 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 neith... | 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 + 168... | 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 + ... | 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... | 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 = ... | 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... | 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) ... | 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... | 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... | 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 o... | 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 ... | 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 ... | 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} ... | 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′′... | 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) · · · y... | 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... | 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 equati... | 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 ... | 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 H... | 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 ... | 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 derivati... | 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 (1... | 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′ = y... | 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 , I... | 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 m... | 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 t... | 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 ar... | 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 = ... | 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 + ... | 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 an... | 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 c... | 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 y... | 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 functi... | 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(... | 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, y... | 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... | 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 le... | 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 ... | 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 solut... | 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 ... | 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 funda... | 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... | 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 eigen... | 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... | 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 multiplic... | 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 ... | 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 Figu... | 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 traject... | 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... | 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... | 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 t... | 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 gra... | 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, w... | 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 differe... | 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 b... | 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 eige... | 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... | 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 soluti... | 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... | 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 eigenvalu... | 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 =... | 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 lin... | 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, a... | 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... | 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 multipl... | 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 tha... | 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 ... | 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 ... | 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 sa... | 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.... | 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 ... | 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 −s... | 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 (... | 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 para... | 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 inwar... | 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. ... | 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... | 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 ... | 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 pa... | 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 ... | 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 e... | 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 i... | 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 = ... | 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 =... | 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) ... | 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 ... | 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 =... | 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 i... | 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. I... | 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... | 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 a... | 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... | 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 disti... | 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 ... | 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 sol... | Elementary Differential Equations with Boundary Value Problems_Page_597_Chunk3000 |
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