Higher-Order Differential Equations

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Higher-Order Differential Equations

• CHAPTER 3

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Contents

• 3.1 Preliminary Theory Linear Equations
• 3.3 Homogeneous Linear Equations with Constant
  Coefficients
• 3.4 Undetermined Coefficients
• 3.5 Variation of Parameters
• 3.8 Linear Models Initial-Value Problems

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3.1 Preliminary Theory Linear Equ.

• Initial-value Problem An initial value problem
  for nth-order linear DE is with
  (1)as n initial conditions.

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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.

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• The following DE (6)is
  homogeneous (7)with g(x) 0, is
  nonhomogeneous.

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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
  n-th order linear DEs as L(y) 0 and L(y)
  g(x)

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Linear Dependence and Independence

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

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Fig 3.3
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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 0when c1 c2
  1, c3 -1, c4 1.

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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

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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.

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Example 8

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

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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 (-?, ?).

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

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Example 10

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

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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

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Note

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

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3.3 Homogeneous Linear Equation with Constant
Coefficients

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

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• 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.

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• Case 1 Distinct real rootsThe general solution
  is (why?) (4)
• Case 2 Repeated real roots ,
  (why?) (5)The general solution is
  (why?) (6)

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

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• 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)

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Example 1

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

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Example 2

• Solve
• Solution See Fig 3.4.

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Fig 3.4
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Higher-Order Equations

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

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Example 3

• Solve
• Solution

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Example 4

• Solve
• Solution

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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

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3.4 Undetermined Coefficients

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

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Example 1

• Solve
• Solution We can get yc as described in Sec 3.3.
  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

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Example 1 (2)

• Then

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Example 2

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

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Example 3

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

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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

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Rule of Case 1

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

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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 .

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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

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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.

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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

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Example 9

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

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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

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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

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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. Let and be
  two linearly independent solutions of the
  associated homogeneous equation of (2).

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Method of Variation of Parameters

• We try (3) After we obtain yp,
  yp, we put them into (2), then
  (4)

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• Making further assumptions y1u1 y2u2 0,
  then from (4), y1u1 y2u2 f(x)Express
  the above in terms of determinants
• and (5)where (6)

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Example 1

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

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Example 1 (2)

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

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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

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Example 2 (2)

• Then And

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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.

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Example 3 (2)
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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.

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• For the case n 3, (10)

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3.8 Linear Models IVP

• Newtons LawSee Fig 3.18, we have (1)

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Fig 3.18
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Fig3.19
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Free Undamped Motion

• From (1), if and f(t)0, we
  have (2)where ? k/m. (2) is called a
  simple harmonic motion, or free undamped motion.

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Solution and Equation of Motion

• From (2), the general solution is
  (3)Period T 2?/?, frequency f 1/T ?/2?.

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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

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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)

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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)

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Fig 3.20
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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.

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Fig 3.21

• Fig 3.21 shows the motion.

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Free Damped Motion

• If and f(t)0, 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

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Case 1

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

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Fig3.23
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Case 2

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

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Fig3.24
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Case 3

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

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Fig 3.25
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Example 3

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

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Fig 3.26
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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)

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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.

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Fig 3.27
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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 -3 3i,
  -3 - 3i

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Example 5 (2)

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

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Alternate form of x(t)

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

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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 , we
  have (34)

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Fig 3.32
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Forced Undamped and Damped

• If R0, (34) is forced undamped.
• If R 0, (34) is forced damped.
• Q Can you find the relationship between R, L, C
  to
• distinguish the cases of forced overdamped,
  forced
• critically damped and forced underdamped?

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Thank You !