1. Higher-Order Differential Equations
CHAPTER 3
Higher-Order Differential Equations

2. Contents 3.1 Preliminary Theory: Linear Equations
3.2 Reduction of Order
3.3 Homogeneous Linear Equations with Constants Coefficients
3.4 Undetermined Coefficients
3.5 Variation of Parameters
3.6 Cauchy-Euler Equations
3.7 Nonlinear Equations
3.8 Linear Models: Initial-Value Problems
3.9 Linear Models: Boundary-Value Problems
3.10 Nonlinear Models
3.11 Solving Models of Linear Equations

3. 3.1 Preliminary Theory: Linear Equ.
Initial-value Problem An nth-order initial problem is Solve: Subject to: (1) with n initial conditions.

4. Let an(x), an-1(x), …, a0(x), and g(x) be continuous on I,
an(x)  0 for all x on I. If x = x0 is any point in this
interval, then a solution y(x) of (1) exists on the interval
and is unique.
THEOREM 3.1
Existence and Uniqueness

5. Example 1
The problem possesses the trivial solution y = 0. Since this DE with constant coefficients, from Theorem 3.1, hence y = 0 is the only one solution on any interval containing x = 1.

6. Example 2
Please verify y = 3e2x + e–2x – 3x, is a solution of This DE is linear and the coefficients and g(x) are all continuous, and a2(x)  0 on any I containing x = 0. This DE has an unique solution on I.

7. Boundary-Value Problem
Solve: Subject to: is called a boundary-value problem (BVP). See Fig 3.1.

8. Fig 3.1

9. Example 3
In example 4 of Sec 1.1, we see the solution of is x = c1 cos 4t + c2 sin 4t (2)
(a) Suppose x(0) = 0, then c1 = 0, x(t) = c2 sin 4t Furthermore, x(/2) = 0, we obtain 0 = 0, hence (3) has infinite many solutions. See Fig 3.2.
(b) If (4) we have c1 = 0, c2 = 0, x = 0 is the only solution.

10. Example 3 (2)
(c) If (5) we have c1 = 0, and 1 = 0 (contradiction). Hence (5) has no solutions.

11. Fig 3.2

12. The following DE. (6) is said to be homogeneous;
The following DE (6) is said to be homogeneous; (7) with g(x) not identically zero, is nonhomogeneous.

13. Differential Operators
Let dy/dx = Dy. This symbol D is called a differential operator. We define an nth-order differential operator as (8) In addition, we have (9) so the differential operator L is a linear operator.
Differential Equations We can simply write the DEs as L(y) = 0 and L(y) = g(x)

14. Let y1, y2, …, yk be a solutions of the homogeneous
Nth-order differential equation (6) on an interval I.
Then the linear combination y = c1y1(x) + c2y2(x) + …+ ckyk(x) where the ci, i = 1, 2, …, k are arbitrary constants, is
also a solution on the interval.
THEOREM 3.2
Superposition Principles – Homogeneous Equations

15. (A) y = cy1 is also a solution if y1 is a solution.
(B) A homogeneous linear DE always possesses the trivial solution y = 0.
COROLLARY
Corollaries to Theorem 3.2

16. Example 4
The function y1 = x2, y2 = x2 ln x are both solutions of Then y = x2 + x2 ln x is also a solution on (0, ).
A set of f1(x), f2(x), …, fn(x) is linearly dependent on
an interval I, if there exists constants c1, c2, …, cn,
not all zero, such that c1f1(x) + c2f2(x) + … + cn fn(x) = 0 If not linearly dependent, it is linearly independent.
DEFINITION 3.1
Linear Dependence and Linear Independence

17. In other words, if the set is linearly independent, when c1f1(x) + c2f2(x) + … + cn fn(x) = 0 then c1 = c2 = … = cn = 0
Referring to Fig 3.3, neither function is a constant multiple of the other, then these two functions are linearly independent.

18. Fig 3.3

19. Example 5
The functions f1 = cos2 x, f2 = sin2 x, f3 = sec2 x, f4 = tan2 x are linearly dependent on the interval (-/2, /2) since c1 cos2 x +c2 sin2 x +c3 sec2 x +c4 tan2 x = 0 when c1 = c2 = 1, c3 = -1, c4 = 1.

20. Example 6
The functions f1 = x½ + 5, f2 = x½ + 5x, f3 = x – 1, f4 = x2 are linearly dependent on the interval (0, ), since f2 = 1 f1 + 5 f3 + 0 f4

21. Suppose each of the functions f1(x), f2(x), …, fn(x)
possesses at least n – 1 derivatives. The determinant is called the Wronskian of the functions.
DEFINITION 3.2
Wronskian

22. Let y1(x), y2(x), …, yn(x) be solutions of the nth-order
homogeneous DE (6) on an interval I. This set of
solutions is linearly independent if and on if
W(y1, y2, …, yn)  0 for every x in the interval.
THEOREM 3.3
Criterion for Linear Independence
Any set y1(x), y2(x), …, yn(x) of n linearly independent
solutions is said to be a fundamental set of solutions.
DEFINITION 3.3
Fundamental Set of a Solution

23. There exists a fundamental set of solutions for (6) on an interval I.
THEOREM 3.4
Existence of a Fundamental Set
Let y1(x), y2(x), …, yn(x) be a fundamental set of
solutions of homogeneous DE (6) on an interval I. Then
the general solution is y = c1y1(x) + c2y2(x) + … + cnyn(x) where ci are arbitrary constants.
THEOREM 3.5
General Solution – Homogeneous Equations

24. Example 7
The functions y1 = e3x, y2 = e-3x are solutions of y” – 9y = 0 on (-, ) Now for every x. So y = c1y1 + c2y2 is the general solution.

25. Example 8
The functions y = 4 sinh 3x - 5e3x is a solution of example 7 (Verify it). Observer = 4 sinh 3x – 5e-3x

26. Example 9
The functions y1 = ex, y2 = e2x , y3 = e3x are solutions of y’’’ – 6y” + 11y’ – 6y = 0 on (-, ). Since for every real value of x. So y = c1ex + c2 e2x + c3e3x is the general solution on (-, ).

27. Any yp free of parameters satisfying (7) is called a
particular solution. If y1(x), y2(x), …, yk(x) be a
fundamental set of solutions of (6), then the general
solution of (7) is y= c1y1 + c2y2 +… + ckyk + yp (10)
THEOREM 3.6
General Solution – Nonhomogeneous Equations
Complementary Function y = c1y1 + c2y2 +… + ckyk + yp = yc + yp = complementary + particular

28. Example 10
The function yp = -(11/12) – ½ x is a particular solution of (11) From previous discussions, the general solution of (11) is

29. DE (12) with gi(x), then (13) is a particular solution of (14)
Given (12) where i = 1, 2, …, k. If ypi denotes a particular solution corresponding to the
DE (12) with gi(x), then (13) is a particular solution of (14)
THEOREM 3.7

30. Example 11
We find yp1 = -4x2 is a particular solution of yp2 = e2x is a particular solution of yp3 = xex is a particular solution of
From Theorem 3.7, is a solution of

31. Note:
If ypi is a particular solution of (12), then is also a particular solution of (12) when the right-hand member is

32. 3.2 Reduction of Order
Introduction: We know the general solution of (1) is y = c1y1 + c2y1. Suppose y1(x) denotes a known solution of (1). We assume the other solution y2 has the form y2 = uy1. Our goal is to find a u(x) and this method is called reduction of order.

33. Example 1
Given y1 = ex is a solution of y” – y = 0, find a second solution y2 by the method of reduction of order.
Solution If y = uex, then And Since ex  0, we let w = u’, then

34. Example 1 (2)
Thus (2) Choosing c1 = 0, c2 = -2, we have y2 = e-x. Because W(ex, e-x)  0 for every x, they are independent.

35. General Case
Rewrite (1) as the standard form (3) Let y1(x) denotes a known solution of (3) and y1(x)  0 for every x in the interval.
If we define y = uy1, then we have

36. This implies that or (4) where we let w = u’. Solving (4), we have or

37. then Let c1 = 1, c2 = 0, we find (5)

38. Example 2
The function y1= x2 is a solution of Find the general solution on (0, ).
Solution: The standard form is From (5) The general solution is

39. 3.3 Homogeneous Linear Equation with Constant Coefficients
Introduction: (1) where ai are constants, an  0.
Auxiliary Equation: For n = 2, (2) Try y = emx, then (3) is called an auxiliary equation.

40. From (3) the two roots are (1)
From (3) the two roots are (1) b2 – 4ac > 0: two distinct real numbers. (2) b2 – 4ac = 0: two equal real numbers. (3) b2 – 4ac < 0: two conjugate complex numbers.

41. Case 1: Distinct real roots The general solution is (4)
Case 2: Repeated real roots and from (5) of Sec 3.2, (5) The general solution is (6)

42. Case 3: Conjugate complex roots We write. , a general solution is
Case 3: Conjugate complex roots We write , a general solution is From Euler’s formula: and (7) and

43. Since is a solution then set C1 = C1 = 1 and C1 = 1, C2 = -1 , we have two solutions: So, ex cos x and ex sin x are a fundamental set of solutions, that is, the general solution is (8)

44. Example 1
Solve the following DEs:
(a)
(b)
(c)

45. Example 2
Solve
Solution: See Fig 3.4.

46. Fig 3.4

47. Two Equations worth Knowing
For the first equation: (9) For the second equation: (10) Let Then (11)

48. Higher-Order Equations
Given (12) we have (13) as an auxiliary equation.

49. Example 3
Solve
Solution:

50. Example 4
Solve
Solution:

51. Repeated complex roots
If m1 =  + i is a complex root of multiplicity k, then m2 =  − i is also a complex root of multiplicity k. The 2k linearly independent solutions:

52. 3.4 Undetermined Coefficients
Introduction If we want to solve (1) we have to find y = yc + yp. Thus we introduce the method of undetermined coefficients.

53. Example 1
Solve
Solution: We can get yc as described in Sec Now, we want to find yp.
Since the right side of the DE is a polynomial, we set After substitution, 2A + 8Ax + 4B – 2Ax2 – 2Bx – 2C = 2x2 – 3x + 6

54. Example 1 (2)
Then

55. Example 2 Find a particular solution of
Solution: Let yp = A cos 3x + B sin 3x After substitution,
Then

56. Example 3
Solve (3)
Solution: We can find Let After substitution,
Then

57. Example 4
Find yp of
Solution: First let yp = Ae2x After substitution, 0 = 8e2x, (wrong guess)
Let yp = Axe2x After substitution, -3Ae2x = 8e2x Then A = -8/3, yp = (−8/3)xe2x

58. Rule of Case 1:
No function in the assumed yp is part of yc Table 3.1 shows the trial particular solutions.

59. Example 5 Find the form of yp of (a)
Solution: We have and try There is no duplication between yp and yc .
(b) y” + 4y = x cos x
Solution: We try There is also no duplication between yp and yc .

60. Example 6 Find the form of yp of
Solution: For 3x2: For -5 sin 2x: For 7xe6x:
No term in duplicates a term in yc

61. Rule of Case 2:
If any term in yp duplicates a term in yc, it should be multiplied by xn, where n is the smallest positive integer that eliminates that duplication.

62. Example 8
Solve
Solution: First trial: yp = Ax + B + C cos x + E sin x (5) However, duplication occurs. Then we try yp = Ax + B + Cx cos x + Ex sin x After substitution and simplification, A = 4, B = 0, C = -5, E = 0 Then y = c1 cos x + c2 sin x + 4x – 5x cos x Using y() = 0, y’() = 2, we have y = 9 cos x + 7 sin x + 4x – 5x cos x

63. Example 9
Solve
Solution: yc = c1e3x + c2xe3x After substitution and simplification, A = 2/3, B = 8/9, C = 2/3, E = -6
Then

64. Example 10
Solve
Solution: m3 + m2 = 0, m = 0, 0, -1 yc = c1+ c2x + c3e-x yp = Aex cos x + Bex sin x After substitution and simplification, A = -1/10, B = 1/5
Then

65. Example 11 Find the form of yp of
Solution: yc = c1+ c2x + c3x2 + c4e-x Normal trial: Multiply A by x3 and (Bx2e-x + Cxe-x + Ee-x) by x Then yp = Ax3 + Bx3e-x + Cx2e-x + Exe-x

66. 3.5 Variation of Parameters
Some Assumptions For the DE (1) we put (1) in the form (2) where P, Q, f are continuous on I.

67. Method of Variation of Parameters
We try (3) After we obtain yp’, yp”, we put them into (2), then (4)

68. Making further assumptions:. y1u1’ + y2u2’ = 0, then from (4),
Making further assumptions: y1u1’ + y2u2’ = 0, then from (4), y1’u1’ + y2’u2’ = f(x) Express the above in terms of determinants
and (5) where (6)

69. Example 1
Solve
Solution: m2 – 4m + 4 = 0, m = 2, 2 y1 = e2x, y2 = xe2x,
Since f(x) = (x + 1)e2x, then

70. Example 1 (2)
From (5), Then u1 = (-1/3)x3 – ½ x2, u2 = ½ x2 + x And

71. Example 2
Solve
Solution: y” + 9y = (1/4) csc 3x m2 + 9 = 0, m = 3i, -3i y1 = cos 3x, y2 = sin 3x, f = (1/4) csc 3x
Since

72. Example 2 (2)
Then And

73. Example 3
Solve
Solution: m2 – 1 = 0, m = 1, -1 y1 = ex, y2 = e-x, f = 1/x, and W(ex, e-x) = -2 Then The low and up bounds of the integral are x0 and x, respectively.

74. Example 3 (2)

75. Higher-Order Equations
For the DEs of the form (8) then yp = u1y1 + u2y2 + … + unyn, where yi , i = 1, 2, …, n, are the elements of yc. Thus we have (9) and uk’ = Wk/W, k = 1, 2, …, n.

76. For the case n = 3, (10)

77. 3.6 Cauchy-Eulaer Equation
Form of Cauchy-Euler Equation
Method of Solution We try y = xm, since

78. An Auxiliary Equation
For n = 2, y = xm, then am(m – 1) + bm + c = 0, or am2 + (b – a)m + c = 0 (1)
Case 1: Distinct Real Roots (2)

79. Example 1
Solve
Solution: We have a = 1, b = -2 , c = -4 m2 – 3m – 4 = 0, m = -1, 4, y = c1x-1 + c2x4

80. Case 2: Repeated Real Roots
Using (5) of Sec 3.2, we have Then (3)

81. Example 2
Solve
Solution: We have a = 4, b = 8, c = 1 4m2 + 4m + 1 = 0, m = -½ , -½

82. Case 3: Conjugate Complex Roots
Higher-Order: multiplicity
Case 3: Conjugate Complex Roots m1 =  + i, m2 =  – i, y = C1x( + i) + C2x( - i) Since xi = (eln x)i = ei ln x = cos( ln x) + i sin( ln x) x-i = cos ( ln x) – i sin ( ln x) Then y = c1x cos( ln x) + c2x sin( ln x) = x [c1 cos( ln x) + c2 sin( ln x)] (4)

83. Example 3
Solve
Solution: We have a = 4, b = 0 , c = 17 4m2 − 4m + 17 = 0, m = ½ + 2i Apply y(1) = -1, y’(1) = 0, then c1 = -1, c2 = 0, See Fig 3.15.

84. Fig 3.15

85. Example 4
Solve
Solution: Let y = xm, Then we have xm(m + 2)(m2 + 4) = 0 m = -2, m = 2i, m = -2i y = c1x-2 + c2 cos(2 ln x) + c3 sin(2 ln x)

86. Example 5
Solve
Solution: We have (m – 1)(m – 3) = 0, m = 1, 3 yc = c1x + c2x3 , use variation of parameters, yp = u1y1 + u2y2, where y1 = x, y2 = x3 Rewrite the DE as Then P = -3/x, Q = 3/x2, f = 2x2ex

87. Example 5 (2)
Thus
We find

88. Example 5 (3)
Then

89. 3.7 Nonlinear Equations Example 1 Solve
Solution: This nonlinear equation misses y term. Let u(x) = y’, then du/dx = y”, or (This form is just for convenience) Since u-1 = 1/y’, So,

90. Example 2
Solve
Solution: This nonlinear equation misses x term. Let u(x) = y’, then y” = du/dx = (du/dy)(dy/dx) = u du/dy or ln|u| = ln|y| + c1, u = c2y (where ) Since u = dy/dx = c2y, dy/y = c2 dx ln|y| = c2x + c3,

91. Example 3
Assume (1) exists. If we further assume y(x) possesses a Taylor series centered at 0: (2) Remember that y(0) = -1, y’(0) = 1. From the original DE, y”(0) = 0 + y(0) – y(0)2 = −2. Then (3)

92. Example 3 (2)
(4) (5) and so on. So we can use the same method to obtain y(3)(0) = 4, y(4)(0) = −8, …… Then

93. Example 4
The DE in example 3 is equivalent to With the aid of a solver, Fig 3.16 shows the graph of this DE. For comparison, the curve of fifth-degree Taylor series is also shown.

94. Fig 3.16

95. 3.8 Linear Models: IVP
Newton’s Law See Fig 3.18, we have (1)

96. Fig 3.18

97. Fig3.19

98. Free Undamped Motion
From (1), we have (2) where  = k/m. (2) is called a simple harmonic motion, or free undamped motion.

99. Solution and Equation of Motion
From (2), the general solution is (3) Period T = 2/, frequency f = 1/T = /2.

100. Example 1
A mass weighing 2 pounds stretches a spring 6 inches. At t = 0, the mass is released from a 8 inches below the equilibrium position with an upward velocity 4/3 ft/s. Determine the equation of motion.
Solution: Unit convert: 6 in = 1/2 ft; 8 in = 2/3 ft, m = W/g = 1/16 slug From Hooke’s Law, 2 = k(1/2), k = 4 lb/ft Hence (1) gives

101. Example 1 (2)
together with x(0) = 2/3, x’(0) = -4/3. Since 2 = 64,  = 8, the solution is x(t) = c1 cos 8t + c2 sin 8t (4) Applying the initial condition, we have (5)

102. Alternate form of x(t)
(4) can be written as x(t) = A sin(t + ) (6) where and  is a phase angle, (7) (8) (9)

103. Fig 3.20

104. Example 2
Solution (5) is x(t) = (2/3) cos 8t − (1/6) sin 8t = A sin(t + ) Then However it is not the solution, since we know tan-1 (+/−) will locate in the second quadrant Then so (9) The period is T = 2/8 = /4.

105. Fig 3.21
Fig 3.21 shows the motion.

106. Free Damped Motion
If the DE is as (10) where  is a positive damping constant. Then x”(t) + (/m)x’ + (k/m)x = 0 can be written as (11) where 2 = /m, 2 = k/m (12) The auxiliary equation is m2 + 2m + 2 = 0, and the roots are

107. Case 1:
2 – 2 > 0. Let then (13) It is said to be overdamped. See Fig 3.23.

108. Fig3.23

109. Case 2:
2 – 2 = 0. then (14) It is said to be critically damped. See Fig 3.24.

110. Fig3.24

111. Case 3:
2 – 2 < 0. Let then (15) It is said to be underdamped. See Fig 3.25.

112. Fig 3.25

113. Example 3
The solution of is (16) See Fig 3.26.

114. Fig 3.26

115. Example 4
A mass weighing 8 pounds stretches a spring 2 feet. Assuming a damping force equal to 2 times the instantaneous velocity exists. At t = 0, the mass is released from the equilibrium position with an upward velocity 3 ft/s. Determine the equation of motion.
Solution: From Hooke’s Law, 8 = k (2), k = 4 lb/ft, and m = W/g = 8/32 = ¼ slug, hence (17)

116. Example 4 (2)
m2 + 8m + 16 = 0, m = −4, −4 x(t) = c1 e-4t + c2t e-4t (18) Initial conditions: x(0) = 0, x’(0) = −3, then x(t) = −3t e-4t (19) See Fig 3.27.

117. Fig 3.27

118. Example 5
A mass weighing 16 pounds stretches a spring from 5 feet to 8.2 feet. t. Assuming a damping force is equal to the instantaneous velocity exists. At t = 0, the mass is released from rest at a point 2 feet above the equilibrium position. Determine the equation of motion.
Solution: From Hooke’s Law, 16 = k (3.2), k = 5 lb/ft, and m = W/g = 16/32 = ½ slug, hence (20) m2 + 2m + 10 = 0, m = −3 + 3i, −3 − 3i

119. Example 5 (2)
(21) Initial conditions: x(0) = −2, x’(0) = 0, then (22)

120. Alternate form of x(t)
(22) can be written as (23) where and

121. DE of Driven Motion with Damping
As in Fig 3.28, (24) (25) where

122. Fig 3.28

123. Example 6 Interpret and solve (26)
Solution: Interpret: m = 1/5, k = 2,  = 1.2, f(t) = 5 cos 4t release from rest at a point ½ below Sol:

124. Example 6 (2)
Assuming xp(t) = A cos 4t + B sin 4t, we have A = −25/102, B = 50/51, then Using x(0) = 1/2, x’(0) = 0 c1 = 38/51, c2 = −86/51, (28)

125. Transient and Steady-State
Graph of (28) is shown in Fig 3.29.
xc(t) will vanish at t  : transient term xp(t) will still remain at t  : steady-state term

126. Fig 3.29

127. Example 7
The solution of is See Fig 3.30.

128. Fig 3.30

129. Example 8 Solve where F0 is a constant and   .
Solution: xc = c1 cos t + c2 sin t Let xp = A cos t + B sin t, after substitution, A = 0, B = F0/(2− 2),

130. Example 8 (2)
Since x(0) = 0, x’(0) = 0, then
Thus (30)

131. Pure Resonance
When  = , we consider the case    (31)

132. When t  , the displacements become large In fact, |x(tn)|   when tn = n/, n = 1, 2, ….. As shown in Fig 3.31, it is said to be pure resonance.

133. Fig 3.31

134. LRC-Series Circuits
The following equation is the DE of forced motion with damping: (32) If i(t) denotes the current shown in Fig 3.32, then (33) Since i = dq/dt, we have (34)

135. Fig 3.32

136. Example 9
Find q(t) in Fig 3.32, where L = henry, R = 10 ohms, C = farad, E(t) = 0, q(0) = q0 coulombs, and i(0) = 0 ampere.
Solution: Using the given data: As described before, Using q(0) = q0, i(0) = q’(0) = 0, c1 = q0, c2 = q0/3

137. Example 10
Find the steady-state qp(t) and the steady-state current, when E(t) = E0 sin t .
Solution: Let qp(t) = A sin t + B cos t,

138. Example 10 (2) If Using the similar method, we have So
Note: X and Z are called the reactance and impedance, respectively.

139. 3.9 Linear Models: BVP
Deflection of a Beam The bending moment M(x) at a point x along the beam is related to the load per unit length w(x) by (1) In addition, M(x) is proportional to the curvature  of the elastic curve M(x) = EI (2) where E, I are constants.

140. From calculus, we have   y”, when the deflection y(x) is small
From calculus, we have   y”, when the deflection y(x) is small. Finally we have (3) Then (4)

141. Terminology Ends of the beam Boundary Conditions embedded
y = 0, y’ = 0
free
y” = 0, y’’’ = 0
simply supported (hinged)
y = 0, y” = 0
See Fig 3.41

142. Fig 3.41

143. Example 1
A beam of length L is embedded at both ends. Find the deflection of the beam if a constant load w0 is uniformly distributed aling its length, that is, w(x)= w0 , 0 < x < L
Solution: From (4) we have Embedded ends means We have m4 = 0, yc(x) = c1 + c2x + c3x2 + c4x3, and

144. Example 1 (2)
So Using the boundary conditions, we have c1 = 0, c2 = 0, c3 = w0L2/24EI, c4 = −w0L/12EI Choosing w0 = 24EI and L = 1, we have Fig 3.42.

145. Fig 3.42

146. Example 2
Solve
Solution: Case 1 :  = 0 y = c1x + c2, y(0) = c2 = 0, y(L) = c1L = 0, c1 = 0 then y = 0, trivial solution.
Case 2 :  < 0,  = −2,  > 0 Choose y = c1 cosh x + c2 sinh x y(0) = 0, c1 = 0; y(L) = 0, c2 = 0 then y = 0, trivial solution.

147. Example 2 (2)
Case 3 :  > 0,  = 2,  > 0 Choose y = c1 cos x + c2 sin x y(0) = 0, c1 = 0; y(L) = 0, c2 sin L= 0 If c2 = 0, y = 0, trivial solution. So c2  0, sin L = 0, L = n,  = n/L Thus, y = c2 sin (nx/L) is a solution for each n.

148. Example 2 (3)
Simply take c2 = 1, for each: the corresponding function:
Note: n = (n/L)2, n = 1, 2, 3, … are known as characteristic values or eigenvalues. yn = sin (nx/L) are called characteristic functions or eigenfunctions.

149. Bulking of a Thin Vertical Column
Referring to Fig 3.43, the DE is (5) where P is a constant vertical compressive force applied to the column’s top.

150. Fig 3.43

151. Example 3
Referring to Fig 3.43, when the column is hinged at both ends, find the deflection.
Solution: The boundary-value problem is From the intuitive view, if the load P is not great enough, there is no deflection. The question is: For what values of P does the given BVP possess nontrivial solutions?

152. Example 3 (2)
By writing  = P/EI, we see is identical to example 2. From Case 3, the deflection curves are yn = c2 sin (nx/L), corresponding to eigenvalues n = Pn/EI = n22/L2, n = 1, 2, 3, …
Physically, only Pn = EIn22/L2, deflection occurs. We call these Pn the critical loads and the smallest P = P1 = EI2/L2 is called the Euler load, and y1 = c2 sin(x/L) is known as the first buckling mode. See Fig 3.44

153. Fig 3.44

154. Rotating String
The simple DE y” + y = 0 (6) occurs again as a model of a rotating string. See Fig 3.45.

155. Fig 3.45

156. We have. F = T sin 2 – T sin 1. (7) When 1 and 2 are small,
We have F = T sin 2 – T sin 1 (7) When 1 and 2 are small, sin 2  tan 2 , sin 1  tan 1 Since tan2, tan1 are slopes of the lines containing the vectors T1 and T2, then tan 2 = y’(x + x), tan 1 = y’(x) Thus (7) becomes (8) Because F = ma, m = x, a = r2. With x small, we take r = y.

157. Thus. (9) Letting (8) = (9), we have
Thus (9) Letting (8) = (9), we have (10) For x close to zero, we have (11) And the boundary conditions are y(0) = y(L) = 0.

158. 3.10 Nonlinear Models
Nonlinerar Springs The model (1) when F(x) = kx is said to be linear. However, (2) is a nonlinear spring. Another model (3)

159. Hard and Soft Springs
F(x) = kx + k1x3 is said to be hard if k1 > 0; and is soft, if k1 < 0. See Fig 3.50.
Fig 3.50

160. Example 1
The DEs (4) and (5) are special cases of (2). Fig3.51 shows the graph from a numerical solver.

161. Fig 3.51

162. Nonlinear Pendulum
The model of a simple pendulum is shown in Fig From the figure, We have the angle acceleration a = s” = l”, the force Then (6)

163. Fig 3.52

164. Linearization
Since If we use only the first two terms, If  is small, (7)

165. Example 2
Fig 3.53 shows some results with different initial conditions by a solver. We can see if the initial velocity is great enough, it will go out of bounds.

166. Fig 3.53

167. Telephone Wire
Recalling from (17) in Sec 1.3 and Fig dy/dx = W/T1, can be modified as (8) where  is the density and s is the arc length. Since the length s is (9)

168. then (10) Differentiating (8) w.s.t x and using (10), then (11)

169. Example 3
From Fig 1.26, we obtain From Fig 1.26, we obtain y(0) = a, y’(0) = 0. Let u = y’, equation (11) becomes Thus Now y’(0) = u(0) = 0, sinh-10 = 0 = c1 Since u = sinh(x/T1) = dy/dx, then Using y(0) = a, c2 = a − (T1/)

170. Rocket Motion
From Fig 3.54, we have (12) when y = R, kMm/R2 = Mg, k = gR2/M, then (13)

171. Fig 3.54

172. Variable Mass
Assuming the mass is variable, then F = ma should be modified as (14)

173. Example 4
A uniform 10-foot-long chain is coiled loosely on the ground. On end is pulled vertically by a force of 5 lb. The chain weigh 1 lb per foot. Determine the height of the end at time t.
Solution: Let x(t) = the height v(t) = dx/dt (velocity) W = x1 = x (weight) m = W/g = x/32 (mass) F = 5 – W (net force)

174. Example 4 (2)
Then (15) Since v = dx/dt (16) is of the form F(x, x’, x”) = 0 Since v = x’, and then (15) becomes (17)

175. Example 4 (3)
Rewriting (17) as (v2+32x – 160) dx + xv = 0 (18) (18) can be multiplied by an integrating factor to become exact, where we can find the integrating factor is (x) = x (please verify). Then Use the method in Sec (19) Since x(0) = 0, then c1 = 0. By solving (19) = 0, for v = dx/dt > 0, we get

176. Example 4 (4)
Thus please verify that (20) Using x(0) = 0 again, , we square both sides of (20) and solve for x (21)

177. 3.11 Solving Systems of Linear Equations
Coupled Spring/Mass System From Fig 3.58 and Newton’s Law (1)

178. Fig 3.58

179. Method of Solution
Consider dx/dt = 3y, dy/dt = 2x or Dx – 3y = 0, 2x – Dy = 0 (2) Then, multiplying the first by D, the second by −3, and then eliminating y, gives D2x – 6x = (3) Similar method can give (4)

180. Return to the original equations,
Return to the original equations, dx/dt = 3y then after simplification, we have (5)

181. Example 1 Solve Dx + (D + 2)y = 0 (D – 3)x – 2y = 0 (6)
Solution: Multiplying the first by D – 3, the second by D, then subtracting, [(D – 3)(D + 2) + 2D]y = 0 (D2 + D – 6)y = 0 then y(t) = c1e2t + c2e-3t (7)

182. Example 1 (2)
Using the similar method, x(t) = c3e2t + c4e-3t (8) Substituting (7) and (8) into the first equation of (6), (4c1 + 2c3)e2t + (−c2 – 3c4)e−3t = 0 Then 4c1 + 2c3 = 0 = −c2 – 3c4 c3 = –2c1, c4 = – ⅓c2

183. Example 2 Solve x’ – 4x + y” = t2 x’ + x + y’ = 0 (9)
Solution: (D – 4)x + D2y = t2 (D + 1)x + Dy = (10) By eliminating x, then and m = 0, 2i, −2i Let then we can get A = 1/12, B = ¼ , C = −1/8.

184. Example 2 (2)
Thus (11) Similar method to get x(t) Then m= 2i, −2i, Let xp(t) = At2 + Bt + C, then we can get A = −1/4, B = 0, C = 1/8

185. Example 2 (3)
Thus (12) By using the second equation of (9), we have

186. Example 3
In (3) of Sec. 2.9, we have Together with the given initial conditions, we can use the same method to solve x1 and x2, not mentioned here.

187. Example 4
Solve (13) with
Solution: Then

188. Example 4 (2)
Using the same method, we have (14)

189. Fig 3.59

190. Thank You !