Section 3.4. Second Order Nonhomogeneous. The corresponding homogeneous equation

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4 Section 3.4. Second Order Nonhomogeneous Equations y + p(x)y + q(x)y = f(x) (N) The corresponding homogeneous equation y + p(x)y + q(x)y = 0 (H) is called the reduced equation of (N). 1

5 General Results THEOREM 1: If z = z 1 (x) and z = z 2 (x) are solutions of equation (N), then y(x) = z 1 (x) z 2 (x) is a solution of equation (H). 2

6 THEOREM 2: Let y = y 1 (x) and y = y 2 (x) be linearly independent solutions of the reduced equation (H) and let z = z(x) be a particular solution of (N). Then y(x) = C 1 y 1 (x) + C 2 y 2 (x) + z(x) is the general solution of (N). 3

7 The general solution of (N) consists of the general solution of the reduced equation (H) plus a particular solution of (N): y = C 1 y 1 (x) + C 2 y 2 (x) }{{} + z(x). }{{} gen soln (H) + part soln (N) 4

8 To find the general solution of (N) you need to find: (i) a fundamental set of solutions y 1, y 2 of the reduced equation (H), and (ii) a particular solution z of (N). 5

9 THEOREM: (Superposition Principle) If z = z f (x) is a particular solution of y + p(x)y + q(x)y = f(x) and z = z g (x) is a particular solution of y + p(x)y + q(x)y = g(x), then z(x) = z f (x) + z g (x) 6

10 is a particular solution of y + p(x)y + q(x)y = f(x) + g(x).

11 Variation of Parameters Let y = y 1 (x) and y = y 2 (x) be independent solutions of the reduced equation (H) and let W (x) = y 1 y 2 y 1 y 2 = y 1y 2 y 2y 1 be their Wronskian. Set z(x) = y 1 (x)u(x) + y 2 (x)v(x) where u and v are to be determined so that z is a solution of (N). 7

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16 u (x) = y 2(x)f(x), W (x) u(x) = y 2 (x)f(x) W (x) dx; v (x) = y 1(x)f(x) W (x) v(x) = y 1 (x)f(x) W (x) dx and z(x) = y 1 (x) y 2 (x)f(x) W (x) dx + y 2 (x) y 1 (x)f(x) W (x) dx 8

17 Examples: 1. {y 1 (x) = x 2, y 2 (x) = x 4 } is a fundamental set of solutions of y 5 x y + 8 x2y = 0. Find a particular solution z of the equation. y 5 x y + 8 x2y = 4x3 9

18 y 1 = x 2, y 2 = x 4 W [y 1, y 2 ] = 10

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20 2. Find the general solution of y 4y + 4y = e2x x 11

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23 3. Find the general solution of y + y 6y = 3e 2x 12

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26 4. Find the general solution of y + 4y = 2 tan 2x 13

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29 Section 3.5. Undetermined Coefficients aka Guessing NOTE: This method can be used only when: 1. the de has constant coefficients 2. f an exponential function y + ay + by = f(x) a, b constants, f an exponential function. 14

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31 Case 1: If y + ay + by = ce rx Set z(x) = Ae rx Example 1: Find a particular solution of y 5y + 6y = 7e 4x. Set z = Ae 4x where A is to be determined. 15

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33 Answer: z = 1 6 e 4x. The general solution of the differential equation is: y = C 1 e 2x + C 2 e 3x e 4x. Note: If L[y] = y + ay + by, then L[Ae rx ] = A ( r 2 + ar + b ) e rx = Ke rx That is, L[Ae rx ] is a constant multiple of e rx 16

34 Example 2: Find a particular solution of y + 2y 3y = 9e 2x. 17

35 Case 2: If y + ay + by = c cos βx, or y + ay + by = c sin βx, or y +ay +by = c cos βx+d sin βx, Example: Find a particular solution of y 2y + y = 5 cos 2x. 18

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38 Note: If L[y] = y + ay + by, then L[A cos βx] = K cos βx + M sin βx That is, L[A cos βx] involves BOTH cosine and sine. Similarly for L[B sin βx] and L[A cos βx + B sin βx] Therefore, if f(x) = c cos βx or f(x) = d sin βx or f(x) = c cos βx + d sin βx 19

39 set z(x) = A cos βx + B sin βx where A, B are to be determined 20

40 Example 3: Find a particular solution of y 2y + y = 5 cos 2x. Set z = A cos 2x + B sin 2x 21

41 Answer: z = 3 5 cos 2x 4 5 sin 2x. The general solution of the differential equation is: y = C 1 e x + C 2 xe x 3 5 cos 2x 4 5 sin 2x. 22

42 Example 4: Find a particular solution of y 2y + 5y = 2 cos 3x 4 sin 3x + e 2x Set z = A cos 3x + B sin 3x + Ce 2x where A, B, C are to be determined. 23

43 y 2y + 5y = 2 cos 3x 4 sin 3x + e 2x 24

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45 Answer: z = 8 13 cos 3x sin 3x e2x. 25

46 If f(x) = ce αx cos βx, de αx sin βx or ce αx cos βx + de αx sin βx set z(x) = Ae αx cos βx + Be αx sin βx where A, B are to be determined. 26

47 Example 5: Find a particular solution of y + 9y = 4e x sin 2x. Set z = Ae x cos 2x + Be x sin 2x Answer: z = 4 13 ex cos 2x ex sin 2x. 27

48 Example 6: Find a particular solution z of y + y 6y = 3e 2x. 28

49 A BIG Difficulty: The trial solution z is a solution of the reduced equation. In this case, y 1 = e 3x and y 2 = e 2x are solutions of the reduced equation y + y 6y = 0 From Example 3, Section 3.4: z = 3 5 x e2x 29

50 Example 7: Find a particular solution z of y + y 6y = 3e 2x. Since z = Ae 2x? NO this satisfies the reduced equation, Set z = Axe 2x 30

51 Example 8: Find a particular solution of y 2y 15y = 6e 3x 31

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53 Example 9: Find a particular solution z of y + 4y = 2 cos 2x. 32

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56 Example 10: Find a particular solution z of y + 6y + 9y = 4e 3x. 33

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58 Example 11: Find a particular solution z of y 2y 8y = 3e 2x

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60 Example 12: Find a particular solution z of y 3y = 4e 3x

61 Answers: z 8 = 3 4 x e 3x z 9 = 1 2 x sin 2x z 10 = 2x 2 e 3x z 11 = 1 2 xe 2x 3 4 z 12 = 4 3 x e3x 2 3 x 36

62 The Method of Undetermined Coefficients A. Applies only to equations of the form y + ay + by = f(x) where a, b are constants and f is an exponential function. c.f Variation of Parameters 37

63 B. Basic Case: If: f(x) = ae rx set z = Ae rx. f(x) = c cos βx, d sin βx, or c cos βx + d sin βx, set z = A cos βx + B sin βx. f(x) = ce αx cos βx, de αx sin βx or ce αx cos βx + de αx sin βx, set z = Ae αx cos βx + Be αx sin βx. 38

64 BUT: If z satisfies the reduced equation, use xz; if xz also satisfies the reduced equation, then x 2 z will give a particular solution. 39

65 C. General Case: If f(x) = p(x)e rx where p is a polynomial of degree n, then set z = P (x)e rx where P is a polynomial of degree n with undetermined coefficients. 40

66 Example 13: Find a particular solution of y 2y 8y = (4x + 5)e 2x. Set z = (Ax + B)e 2x. 41

67 Example 14: Find a particular solution of y 3y + 2y = (2x 2 1)e x. Set z = (Ax 2 + Bx + C)e x. Answer: z = ( 1 3 x x ) e x. 42

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70 If f(x) = p(x) cos βx + q(x) sin βx where p, q are polynomials of degree n, then set z = P (x) cos βx + Q(x) sin βx where P, Q are polynomials of degree n with undetermined coefficients. 43

71 Example 15: y 2y 3y = 3 cos x + (x 2) sin x. Set z = (Ax+B) cos x+(cx+d) sin x. 44

72 Answer: z = ( 1 10 x 47 ) ( 1 cos x 50 5 x 2 ) sin x

73 If f(x) = p(x)e αx cos βx + q(x)e αx sin βx where p, q are polynomials of degree n, then set z = P (x)e αx cos βx + Q(x)e αx sin βx where P, Q are polynomials of degree n with undetermined coefficients. 46

74 Example 16: y + 4y = 2x e x cos x. Set z = (Ax + B)e x cos x + (Cx + D)e x sin x Answer: z = 1 25 (10x 7)ex cos x (5x 1)ex sin x. 47

75 Example 17: Find a particular solution of y 2y 8y = (4x + 5)e 2x. Set z = (Ax + B)e 2x. 48

76 BUT: Warning!!! If any part of z satisfies the reduced equation, try xz; if any part of xz also satisfies the reduced equation, then x 2 z will give a particular solution. 49

77 Examples: 1. Give the form of a particular solution of y 4y 5y = 2 cos 3x 5e 5x + 4. Answer: z = A cos 3x + B sin 3x + Cxe 5x + D 50

78 Answer: z = Ax + B + Cx 2 e 4x Give the form of a particular solution of y + 8y + 16y = 2x 1 + 7e 4x.

79 C cos 2x + D sin 2x + Ee 2x Give the form of a particular solution of y + y = 4 sin x cos 2x + 2e 2x. Answer: z = Ax cos x + Bx sin x +

80 4. Give the form of the general solution of y + 9y = 4 cos 2x + 3 sin 2x Answer: y = C 1 cos 3x + C 2 sin 3x + A cos 2x + B sin 2x 53

81 5. Give the form of a particular solution of y + 9y = 4 cos 3x + 3 sin 2x Answer: z = Ax cos 3x + Bx sin 3x + C cos 2x + D sin 2x 54

82 6. Give the form of the general solution of y + 4y + 4y = 4xe 2x + 3 Answer: y = C 1 e 2x + C 2 xe 2x + (Ax 3 + Bx 2 )e 2x + C 55

83 7. Give the form of the general solution of y + 4y + 4y = 4e 2x sin 2x + 3x Answer: y = C 1 e 2x + C 2 xe 2x + Ae 2x cos 2x + Be 2x sin 2x + Cx + D 56

84 8. Give the form of the general solution of y + 4y = 4 sin 2x + 3 Answer: y = C 1 e 4x +C 2 +A cos 2x+B sin 2x+Cx 57

85 9. Give the form of a particular solution of y + 2y + 10y = 2e 3x sin x + 4e 3x Answer: z = Ae 3x cos x + Be 3x sin x + Ce 3x 58

86 10. Give the form of the general solution of y + 2y + 10y = 2e x sin 3x + 2e x Answer: y = C 1 e x cos 3x+C 2 e x sin 3x+ Axe x cos 3x + Bxe x sin 3x + Ce x 59

87 11. Give the form of a particular solution of y 2y 8y = 2 cos 3x (3x+1)e 2x 4 Answer: z = A cos 3x+B sin 3x+(Cx 2 +Dx)e 2x +E 60

88 12. Give the form of a particular solution of y 2y 8y = 2 cos 3x 3xe 2x 3x Answer: z = A cos 3x+B sin 3x+(Cx 2 +Dx)e 2x +Ex + F 61

89 13. Find the general solution of y 4y + 4y = 4e 2x + e2x x Answer: y = C 1 e 2x + C 2 xe 2x 2x 2 e 2x + xe 2x ln x 62

90 Summary: 1. Variation of parameters: Can be applied to any linear nonhomogeneous equations, but requires a fundamental set of solutions of the reduced equation. 63

91 2. Undetermined coefficients: Is limited to linear nonhomogeneous equations with constant coefficients, and f must be an exponential function, f(x) = ae rx, f(x) = c cos βx + d sin βx, f(x) = ce αx cos βx + de αx sin βx, or p(x)f(x) p a polynomial. 64

92 In cases where both methods are applicable, the method of undetermined coefficients is usually more efficient and, hence, the preferable method. 65

93 Section 3.6. Vibrating Mechanical Systems undamped (no friction) damped (with friction) free (e.g., only the spring) forcing (add external force) 66

94 I. Free Vibrations Hooke s Law: The restoring force of a spring is proportional to the displacement: F = ky, k > 0. Newton s Second Law: Force equals mass times acceleration: F = ma = m d2 y dt 2. Mathematical model: m d2 y dt 2 = ky 67

95 which can be written d 2 y dt 2 + ω2 y = 0 where ω = k/m. ω is called the natural frequency of the system.

96 The general solution of this equation is: y = C 1 sin ωt + C 2 cos ωt which can be written y = A sin (ωt + φ). A is called the amplitude, φ is called the phase shift. 68

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99 y = A sin(ωt + φ) 69

100 II. Forced Free Vibrations Apply an external force G to the freely vibrating system Force Equation: F = ky + G. Mathematical Model: my = ky+g or y + k m y = G m, a nonhomogeneous equation. 70

101 A periodic external force: G = a cos γt, a, γ > 0 const. Force Equation: F = ky + a cos γt Mathematical Model: y + k m y = a m cos γt y + ω 2 y = α cos γt where ω = k/m, α = a m. 71

102 ω is called the natural frequency of the system, γ is called the applied frequency. if omega = gamma, will have a RESONANCE 72

103 Case 1: γ ω. y + ω 2 y = α cos γt General solution, reduced equation: y = C 1 cos ωt+c 2 sin ωt = A sin (ωt+φ 0 ). Form of particular solution (undetermined coefficients): z = A cos γt + B sin γt. A particular solution: z = α ω 2 γ 2 cos γt. 73

104 General solution: y = A sin (ωt + φ 0 ) + α ω 2 cos γt. γ2 forcing by a periodic force, no damping NO RESONANCE superposition of two frequencies

105 ω/γ rational: periodic motion ω/γ irrational: not periodic 74

106 Case 2: γ = ω. RESONANCE!!!! y + ω 2 y = α cos ωt General solution, reduced equation: y = C 1 cos ωt + C 2 sin ωt = A sin (ωt + φ 0 ). Form of particular solution (undetermined coefficients): z = A t cos γt + B t sin γt. 75

107 A particular solution: α 2ω t sin ωt. General solution: y = A sin (ωt + φ 0 ) + α 2ω t sin ωt. Unbounded oscillation This is known as resonance 76

108 y = A sin (ωt + φ 0 ) + α 2ω t sin ωt. amplitude grows approx. linearly 77

109 resistance (e.g friction), but no forcing III. Damped Free Vibrations: A resistance force R (e.g., friction) proportional to the velocity v = y and acting in a direction opposite to the motion: R = cy with c > 0. Force Equation: F = ky(t) cy (t). Newton s Second Law: F = ma = my 78

110 Mathematical Model: my (t) = ky(t) cy (t) or y + c m y + k m y = 0 (c, k, m constant) or y + αy + βy = 0 α = c m, β = k m α, β positive constants. alpha not zero => roots have a nonzero real part

111 Characteristic equation: r 2 + α r + β = 0. Roots r = α ± α 2 4β. 2 There are three cases to consider: α 2 4β < 0, complex roots α 2 4β > 0, two real roots α 2 4β = 0. double (real) root 79

112 Case 1: α 2 4β < 0. Complex roots: (Underdamped) r 1 = α 2 + iω, r 2 = α 2 iω where ω = 4β α 2. 2 General solution: y = e ( α/2)t (C 1 cos ωt + C 2 sin ωt) 80

113 or decaying oscillation y(t) = A e ( α/2)t sin (ωt + φ 0 ) where A and φ 0 are constants, NOTE: The motion is oscillatory AND y(t) 0 as t. 81

114 Underdamped Case: 82

115 Case 2: α 2 4β > 0. Two distinct real roots: (Overdamped) r 1 = α + α 2 4β, r 2 = α α 2 4β. 2 2 General solution: y(t) = y = C 1 e r 1t + C 2 e r 2t. The motion is nonoscillatory. decays 83

116 NOTE: Since α 2 4β < α 2 = α, r 1 and r 2 are both negative and y(t) 0 as t. 84

117 Case 3: α 2 4β = 0. One real root: (Critically Damped) resonance!!! r 1 = r 2 = α 2, General solution: y(t) = y = C 1 e (α/2) t + C 2 t e (α/2) t. The motion is nonoscillatory and y(t) 0 as t. linear growth defeated by the exponential decay 85

118 Overdamped and Critically Damped Cases: 86

119 Summary of Case III: All solutions of y + ay + by = 0 have limit 0 as t. That is, in the presence of a resistant force (friction), all solutions ultimately return to the equilibrium position. 87

120 IV. Forced Damped Vibrations Apply an external force G to a damped, freely vibrating system Force Equation: F = ky cy + G. Mathematical Model: my = ky cy + G or y + c m y + k m y = G m, 88

121 which we write as y + αy + βy = g, where α = c/m, β = k/m, g = G/m A periodic external force: g = a cos γt, a, γ > 0 const. Mathematical Model: y + αy + βy = a cos γt 89

122 General solution: y(t) = C 1 y 1 (t) + C 2 y 2 (t) + Z(t) = Y c (t) + Z(t), Note: From Case III lim t Y c(t) = 0. as t so lim t y(t) = Z(t). 90

123 Particular solution of (N) y + α y + β y = a cos γt will have the form: Z(t) = A cos γt + B sin γt. General solution of (N): Note: y(t) = Y c (t) + Z(t) LARGE amplification if gamma is close to omega lim y(t) = Z(t) t 91

124 Y c (t), the general solution of the reduced equation, is called the transient solution. Z(t) the particular solution of (N), is called the steady state solution. 92

125 Example 1: y y y = cos t General solution y = C 1 e t/4 +C 2 e t/ cos t sin t Transient solution: y(t) = 2e t/4 +e t/2 (C 1 = 2, C 2 = 1) Steady-state solution: Z(t) = cos t sin t what "survives" for large t 93

126 Transient solution: Steady-state solution: 94

127 y = 2e t/4 + e t/ cos t sin t 95

128 Example 2: y + 2y + 5y = cos t General solution y = C 1 e t cos 2t + C 2 e t sin 2t + Transient solution: 1 5 cos t sin t y(t) = 2e t cos 2t (C 1 = 2, C 2 = 0) Steady-State solution: Z(t) = 1 5 cos t sin t 96 b/c of damping, the sols of the hom. eq's decay (exponentially fast)

129 Transient solution. Steady-State solution: 97

130 higher order linear ODE's with (non-)constant coefficients very similar to the case of 2nd order ODEs Ch 4. Laplace transforms

131 have one part that is undet coeff and one that is variation of constants at 13:

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