CHAPTER 3 HigherOrder Differential Equations

1
CHAPTER 3Higher-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 Equation

• Initial-value Problem An nth-order initial
  problem isSolve Subject to (
  1)with n initial conditions.

4
(No Transcript)
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 e2x 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

• SolveSubject to is called a boundary-value
  problem (BVP).See Fig. 3.1.

8
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.

9
Example 3 (2)

• (c) If (5)we have c1 0, and 1 0
  (contradiction).Hence (5) has no solutions.

10
Fig. 3.2

11

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

12
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)

13
(No Transcript)
14
(No Transcript)
15
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, ?).

16

• In other words, if the set is linearly
  independent, when c1f1(x) c2f2(x) cn
  fn(x) 0then 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.

17
Example 5

• The functions f1 cos2 x, f2 sin2 x, f3
  sec2 x, f4 tan2 x are linearly dependent on
  the interval (-?/2, ?/2) sincec1 cos2 x c2
  sin2 x c3 sec2 x c4 tan2 x 0when c1 c2
  1, c3 -1, c4 1.

18
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

19
(No Transcript)
20
(No Transcript)
21
(No Transcript)
22
Example 7

• The functions y1 e3x, y2 e-3x are solutions
  of y? 9y 0 on (-?, ?)Now for every x.
  So y c1e3x c2e-3x is the general solution.

23
Example 8

• The functions y 4 sinh 3x - 5e3x is a solution
  of example 7 (Verify it). Observer 4
  sinh 3x 5e-3x

24
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 (-?, ?).

25
THEOREM 3.6

• Complementary Function y c1y1 c2y2 cnyn
  yp yc yp complementary particular

26
Example 10

• The function yp -(11/12) ½ x is a particular
  solution of (11)From previous
  discussions, the general solution of (11) is

27
(No Transcript)
28
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

29
Note

• If ypi is a particular solution of (12), then
  is also a particular solution of (14) when
  the right-hand member is

30
3.2 Reduction of Order

• IntroductionWe know the general solution of
  (1)is y c1y1 c2y2. 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.

31
Example 1

• Given y1 ex is a solution of y? y 0, find a
  second solution y2 by the method of reduction of
  order.
• SolutionIf y u(x) y1(x) u(x) ex, then And
  Since ex ? 0, we let w u?, then

32
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.

33
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

34

• This implies that or (4)where we
  let w u?. Solving (4), we have or

35

• then Let c1 1, c2 0, we find (5)
  

36
Example 2

• The function y1 x2 is a solution of Find
  the general solution on (0, ?).
• SolutionThe standard form is From (5) The
  general solution is

37
3.3 Homogeneous Linear Equation with Constant
Coefficients

• Introduction (1)where ai are
  constants, an ? 0.
• Auxiliary EquationFor n 2,
  (2)Try y emx, then (3)is
  called an auxiliary equation.

38

• From (3) the two roots are(1) b2 4ac gt 0
  two distinct real numbers.(2) b2 4ac 0 two
  equal real numbers.(3) b2 4ac lt 0 two
  conjugate complex numbers.

39

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

40

• Case 3 Conjugate complex rootsWe write
  , a general solution is From Eulers
  formula and
  (7)
  and

41

• Since is a solution then setC1
  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)

42
Example 1

• Solve the following DEs
• (a)
• (b)
• (c)

43
Example 2

• Solve
• SolutionHence the solution of the IVP is
• See Fig. 3.4.

44
Fig. 3.4
45
Two Equations worth Knowing

• For the first equation (9)
• For the second equation
  
  (10)
• If c1 c2½ and c1 ½, c2 -½ in (10)
• Then, an alternative form for the general
  solution is (11)

46
Higher-Order Equations

• Given (12)we have (13)as
  an auxiliary equation.

47
Example 3

• Solve
• Solution

48
Example 4

• Solve
• Solution

49
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

50
3.4 Undetermined Coefficients

• IntroductionIf we want to solve (1)we
  have to find y yc yp. Thus we introduce the
  method of undetermined coefficients.

51
Example 1

• Solve
• Solution We can get yc as described in Sec 3.3.
  
• Now, we want to find yp.

52
Example 1 (2)

• Since the right side of the DE is a polynomial,
  we set
• After substitution, 2A 8Ax 4B 2Ax2 2Bx
  2C 2x2 3x 6
• Then

53
Example 2

• Find a particular solution of
• Solution Let yp A cos 3x B sin 3xAfter
  substitution,
• Then

54
Example 3

• Solve (3)
• Solution We can find Let After
  substitution,
• Then

55
Example 4

• Find yp of
• Solution First let yp AexAfter substitution,
  0 8ex, (wrong guess)
• Let yp AxexAfter substitution, -3Aex 8ex
  Then A -8/3, yp (-8/3)xex

56
Rule of Case 1

• No function in the assumed yp is part of ycTable
  3.1 Trial particular solutions.

57
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 .

58
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

59
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.

60
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 0Then y c1 cos x c2 sin x 4x
  5x cos xUsing y(?) 0, y?(?) 2, we have y
  9? cos x 7 sin x 4x 5x cos x

61
Example 9

• Solve
• Solution yc c1e3x c2xe3x After
  substitution and simplification, A 2/3, B
  8/9, C 2/3, E -6
• Then

62
Example 10

• Solve
• Solution m3 m2 0, m 0, 0, -1 yc c1
  c2x c3e-x yp Aex cos x Bex sin xAfter
  substitution and simplification, A -1/10, B
  1/5
• Then

63
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 xThen yp Ax3 Bx3e-x Cx2e-x
  Exe-x

64
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.

65
Method of Variation of Parameters

• We try (3) After we obtain y?p,
  y?p we put them into (2), then
  

66

• Making further assumptions y1u?1 y2u? 2 0,
  then from (4), y? 1u? 1 y? 2u? 2
  f(x)Express the above in terms of determinants
• and (5)where (6)

67
Example 1

• Solve
• Solution m2 4m 4 0, m 2, 2 y1 e2x,
  y2 xe2x,
• Since f(x) (x 1)e2x, then

68
Example 1 (2)

• From (5), Then u1 (-1/3)x3 ½ x2, u2
  ½ x2 xAnd

69
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

70
Example 2 (2)

• Then And

71
Example 3

• Solve
• Solution m2 1 0, m 1, -1 y1 ex, y2
  e-x, f 1/x, and W(ex, e-x) -2 ThenThe
  low and up bounds of the integral are x0 and x,
  respectively.

72
Example 3 (2)
73
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 u?k Wk/W, k 1,
  2, , n.

74

• For the case n 3, (10)

75
3.6 Cauchy-Euler Equation

• Form of Cauchy-Euler Equation
• Method of SolutionWe try y xm, since

76
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)
  

77
Example 1

• Solve
• SolutionWe have a 1, b -2 , c -4 m2 3m
  4 0, m -1, 4
• y c1x-1 c2x4

78
Case 2 Repeated Real Roots

• Using (5) of Sec 3.2,
• Then (3)

79
Example 2

• Solve
• SolutionWe have a 4, b 8, c 1 4m2 4m
  1 0, m -½ , -½

80
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)

81

• For C1 C2 1, y1 x?(x i? x - i? ) 2
  x?cos(? ln x)
• For C1 1, C2 -1, y2 x?(x i? -x - i? )
  2ix?sin(? ln x)
• Since W(x?cos(? ln x), x?sin(? ln x) ) ? x2?-1
  ? 0
• Then y c1x? cos(? ln x) c2x? sin(? ln x)
  x? c1 cos(? ln x) c2 sin(? ln x) (4)

82
Example 3

• Solve
• SolutionWe have a 4, b 0 , c 17 4m2 - 4m
  17 0, m ½ 2i Apply y(1) -1, y?(1)
  -1/2, then c1 -1, c2 0See Fig. 3.15.

83
Fig. 3.15
84
Example 4

• Solve
• SolutionLet 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)

85
Example 5

• Solve
• SolutionWe 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 asThen P -3/x, Q 3/x2, f
  2x2ex

86
Example 5 (2)

• Thus
• We find

87
Example 5 (3)

• Then

88
3.7 Nonlinear Equations

• Example 1 Solve
• SolutionThis 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,

89
Example 2

• Solve
• SolutionThis nonlinear equation misses x term.
  Let u(x) y?, then y? du/dx
  (du/dy)(dy/dx) u du/dy or lnu
  lny c1, u c2y (where )Since
  u dy/dx c2y, dy/y c2 dx lny c2x
  c3,

90
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)

91
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

92
3.8 Linear Models IVP

• Newtons LawSee Fig. 3.18, we have
  (1)

Fig. 3.19
Fig. 3.18
93
Free Undamped Motion

• From (1), we have (2)where ? k/m. (2)
  is called a simple harmonic motion, or free
  undamped motion.
• From (2), the general solution is
  (3)Period T 2?/?, frequency f 1/T ?/2?.

94
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.
• SolutionUnit convert 6 in 1/2 ft 8 in
  2/3 ft, m W/g 1/16 slugFrom Hookes Law,
  2 k(1/2), k 4 lb/ft Hence (1) gives

95
Example 1 (2)

• together with x(0) 2/3, x?(0) -4/3.Since ?2
  k/m 64, ? 8, the solution is x(t) c1
  cos 8t c2 sin 8t (4)Applying the initial
  condition, we have (5)

96
Alternate form of x(t)

• (3) can be written as x(t) A sin(?t ?)
  (6)where and ? is a phase angle,
  (7) (8)
  (9)

97
Fig. 3.20
98
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 quadrantThen
  so (9)The period is T 2?/8 ?/4.

99
Fig. 3.21

• Fig. 3.21 shows the motion.

100
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

101
Case 1

• ?2 ?2 gt 0, then (13)It is said to
  be overdamped. See Fig. 3.23.

Fig. 3.23
102
Case 2

• ?2 ?2 0. then
  (14)
• It is said to be critically damped. See Fig.
  3.24.

Fig. 3.24
103
Case 3

• ?2 ?2 lt 0, then (15)It is said
  to be underdamped. See Fig. 3.25.

104
Fig. 3.25
105
Example 3

• The solution of is (16)See Fig.
  3.26.

106
Fig. 3.26
107
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.
• SolutionFrom Hookes Law, 8 k (2), k 4
  lb/ft, and m W/g 8/32 ¼ slug, hence
  (17)

108
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.

109
Fig. 3.27
110
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.
• SolutionFrom Hookes Law, 16 k (3.2), k 5
  lb/ft, and m W/g 16/32 ½ slug, hence
  (20) m2 2m 10 0, m -1 3i,
  -1 - 3i

111
Example 5 (2)

• (21)Initial conditions x(0) -2,
  x?(0) 0, then (22)

112
Alternate form of x(t)

• (22) can be written as (23)where
  and
• Thus

113
DE of Driven Motion with Damping

• As in Fig. 3.28, (24) (25)wh
  ere

114
Fig. 3.28
115
Example 6

• Interpret and solve (26)
• SolutionInterpret m 1/5, k 2, ? 1.2,
  f(t) 5 cos 4t release from rest at a point ½
  below Sol

116
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)

117
Transient and Steady-State

• Graph of (28) is shown in Fig. 3.29.
• xc(t) will vanish at t ? ? transient termxp(t)
  will still remain at t ? ? steady-state
  term

Fig. 3.29
118
Example 7

• The solution of isSee Fig. 3.30.

119
Fig. 3.30
120
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),

121
Example 8 (2)

• Since x(0) 0, x?(0) 0, then
• Thus (30)

122
Pure Resonance

• When ? ?, we consider the case ? ? ?.
  (31)

123
Fig. 3.31

• 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.

124
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)

125
Fig. 3.32

• Overdampered if R2-4L/C gt 0
• Critically dampered if R2-4L/C 0
• Underdampered if R2-4L/C gt 0

126
Example 9

• Find q(t) in Fig. 3.32, where L 0.25 henry, R
  10 ohms, C 0.001 farad, E(t) 0, q(0) q0
  coulombs, and i(0) 0 ampere.
• SolutionUsing the given dataAs described
  before, Using q(0) q0, i(0) q?(0) 0, c1
  q0, c2 q0/3

127
Example 10

• Find the steady-state qp(t) and the steady-state
  current in an LRC-series circuit when E(t) E0
  sin ?t .
• SolutionLet qp(t) A sin ?t B cos ?t,

128
Example 10 (2)

• If
• If
• Using the similar method, we haveSo
• Note X and Z are called the reactance and
  impedance, respectively.

129
3.9 Linear Models BVP

• Deflection of a BeamThe 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.

130

• From calculus, we have ? ? y?, when the
  deflection y(x) is small. Finally we
  have (3)Then (4)

131
Terminology
132
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 along its length, that
  is, w(x) w0 , 0 lt x lt L
• SolutionFrom (4) we have Embedded ends
  means We have m4 0, yc(x) c1 c2x c3x2
  c4x3, and

133
Example 1 (2)

• So Using the boundary conditions, we have c1
  0, c2 0, c3 w0L2/24EI, c4
  -w0L/12EIChoosing w0 24EI and L 1, we
  have Fig. 3.42.

134
Fig. 3.42
135
Example 2

• Solve
• SolutionCase 1 ? 0 y c1x c2, y(0)
  c2 0, y(L) c1L 0, c1 0then y 0,
  trivial solution.
• Case 2 ? lt 0, ? -?2, ? gt 0Choose y c1 cosh
  ?x c2 sinh ?x y(0) 0, c1 0 y(L) 0, c2
  0 then y 0, trivial solution.

136
Example 2 (2)

• Case 3 ? gt 0, ? ?2, ? gt 0Choose 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.

137
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.

138
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 columns top.

Fig. 3.43
139
Example 3

• Referring to Fig. 3.43, when the column is hinged
  at both ends, find the deflection.
• SolutionThe 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?

140
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, andy1 c2 sin(?x/L) is known as the first
  buckling mode.See Fig. 3.44

141
Fig. 3.44

142
Rotating String

• The simple DEy? ?y 0 (6)occurs again as a
  model of a rotating string. See Fig. 3.45.

Fig 3.45
143

• We have the net vertical force F T sin ?2 T
  sin ?1 (7)When ?1 and ?2 are small, sin ?2
  ? tan ?2 , sin ?1 ? tan ?1Since 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, centripetal acceleration a r?2.
  With ?x small, we take r y.

144

• 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.

145
3.10 Nonlinear Models

• Nonlinerar SpringsThe model (1)when
  F(x) kx is said to be a linear spring.
  However, (2)is a nonlinear spring.
  Another model of a free/mass system with
  damping
• 
  (3)

146
Hard and Soft Springs

• F(x) kx k1x3 is said to be hard if k1 gt
  0and is soft, if k1 lt 0. See Fig. 3.50.

Fig. 3.50
147
Example 1

• The DEs (4)and (5)are special
  cases of (2). Fig. 3.51 shows the graph from a
  numerical solver.

148
Fig. 3.51
149
Nonlinear Pendulum

• The model of a simple pendulum is shown in Fig.
  3.52. From the figure, the arc s of a circle of
  radius l is related to the center ? by s l?.
  Angle acceleration is a s? l??, the force
  Then (6)

Fig. 3.52
150
Linearization

• Since If we use only the first two terms,
  If ? is small, (7)

151
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.

Fig. 3.53
152
Telephone Wire

• Recalling from (17) in Sec. 1.3 and Fig. 1.26
  dy/dx W/T1, it can be modified
  as (8)where ? is the density and s is
  the arc length.Since the length s is
  (9)

153

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

154
Example 3

• From Fig. 1.26, we obtain y(0) a, y ?(0) 0.
  Let u y?, equation (11) becomesThus 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/?)

155
Rocket Motion

• From Fig 3.54, we have
  
• (12)
• when y R, kMm/R2 Mg, k gR2/M, then
  (13)

Fig. 3.54
156
Variable Mass

• Assuming the mass is variable, then F ma should
  be modified as (14)

157
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.
• SolutionLet 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)

158
Example 4 (2)

• Then (15)Since v dx/dt (16)
  is of the form F(x, x?, x?) 0Since v x?,
  andthen (15) becomes (17)

159
Example 4 (3)

• Rewriting (17) as (v232x 160) dx xv dv
  0 (18)(18) can be multiplied by an integrating
  factor to become exact, where we can find the
  integrating factor is ?(x) x. ThenUse the
  method in Sec. 2.4 (19)Since x(0) 0,
  then c1 0. By solving (19) 0, for v dx/dt
  gt 0, we get

160
Example 4 (4)

• Thus (20)Using x(0) 0 again,
  , we square both sides of (20)
  and solve for x (21)

161
3.11 Solving Systems of Linear Equations

• Coupled Spring/Mass SystemFrom Fig. 3.58 and
  Newtons Law
• (1)

Fig. 3.58
162
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 0 (3)Sim
  ilar method can give (4)

163

• Return to the original equations, dx/dt
  3yafter substituting and simplification, we
  have (5)

164
Example 1

• Solve Dx (D 2)y 0 (D 3)x
  2y 0 (6)
• SolutionMultiplying the first by D 3, the
  second by D, then subtracting, (D 3)(D 2)
  2Dy 0 (D2 D 6)y 0then y(t) c1e2t
  c2e-3t (7)

165
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 0Then 4c1 2c3 0 -c2 3c4 c3
  2c1, c4 ?c2

166
Example 2

• Solve x? 4x y? t2 x? x y? 0 (9)
• Solution (D 4)x D2y t2 (D 1)x Dy
  0 (10)By eliminating x, then
  and m 0, 2i, -2iLet
  then we can get A 1/12, B ¼ ,
  C -1/8.

167
Example 2 (2)

• Thus (11)Similar method to get
  x(t) Then m 2i, -2i, Let xp(t) At2 Bt
  C, thenwe can get A -1/4, B 0, C 1/8

168
Example 2 (3)

• Thus (12)By using the second equation
  of (9), we have

169
Example 4

• Solve (13) with
• Solution Then

170
Example 4 (2)

• Using the same method, we have (14)

171
Fig. 3.59
172
Thank You for Your Attention