CHAPTER 3
Higher-Order Differential
Equations
(3.7~3.12)
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1
Chapter Contents
3.7 Nonlinear Equations
3.8 Linear Models: Initial-Value Problems
3.9 Linear Models: Boundary-Value Problems
3.10 Green’s Functions
3.11 Nonlinear Models
3.12 Solving Models of Linear Equations
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Ch3_2
3.7 Nonlinear Equations
Example 1 Solve y  2 x( y) 2
Solution:
This nonlinear equation misses y term. Let u(x) = y’,
then du/dx = y”,
du
du
or
2
 2xu
 2 x dx
u2
dx
 u 1  x 2  c12
Since
So,
u-1
(This form is just for convenience)
dy
1
 2 2
= 1/y’,
dx
x  c1
dx
1
1 x
y    2 2   tan
 c2
x  c1
c1
c1
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Ch3_3
Example 2 Independent Variable x Is
Missing
Solve yy  ( y' ) 2
Solution:
This nonlinear equation misses x term. Let u(x) = y’,
then y” = du/dx = (du/dy)(dy/dx) = u du/dy
 du 
y u   u 2 or
 dy 
du dy

u
y
ln|u| = ln|y| + c1, u = c2y (where c2  ec )
Since u = dy/dx = c2y, dy/y = c2 dx
ln|y| = c2x + c3,
1
y  c4ec2 x
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Ch3_4
Example 3 Taylor Series Solution of an IVP
Assume
y  x  y  y 2 ,
y(0)  1 ,
y(0)  1
(1)
exists. If we further assume y(x) possesses a Taylor
series centered at 0:
y ( x)
(2)
y(0)
y(0) 2 y(0) 3 y ( 4 ) (0) 4 y ( 5) (0) 5
 y ( 0) 
x
x 
x 
x 
x 
1!
2!
3!
4!
5!
Remember that y(0) = -1, y’(0) = 1. From the original
DE, y”(0) = 0 + y(0) – y(0)2 = −2.
Then
d
(3)
y( x)  ( x  y  y 2 )  1  y  2 yy
dx
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Ch3_5
Example 3 (2)
d
y ( x)  (1  y  2 yy)  y  2 yy  2( y) 2
dx
( 4)
y ( 5) ( x) 
d
( y  2 yy  2( y) 2 )  y  2 yy  6 yy
dx
(4)
(5)
and so on. So we can use the same method to obtain
y(3)(0) = 4, y(4)(0) = −8, ……
Then
2
1
1
y ( x)  1  x  x 2  x 3  x 4  x 5  
3
3
5
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Ch3_6
Example 4 General Analysis of Ex 3
The DE in example 3 is equivalent to
dy
u
dx
du
 x  y  y 2 , y (0)  1, u (0)  1
dx
With the aid of a solver, Fig 3.7.1 shows the graph of
this DE. For comparison, the curve of fifth-degree
Taylor series
2
1
1
T5 ( x)  1  x  x 2  x 3  x 4  x 5
3
3
5
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Ch3_7
Fig 3.7.1 Comparison of two approximate
solution in Ex 4
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Ch3_8
Figure 3.7.2 Numerical solution curve of IVP
in (1) of Ex 3
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Ch3_9
3.8 Linear Models: IVP
Newton’s Law
See Fig 3.8.1, we have
d2 x
m 2  k ( s  x)  mg  kx  mg  ks  kx

dt
zero
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(1)
Ch3_10
Fig 3.8.1 Spring/mass system
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Ch3_11
Fig 3.8.2 Position direction is below
equilibrium position
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Ch3_12
DE of Free Undamped Motion
From (1), we have
d 2x
2


x0
2
dt
(2)
where  = k/m. (2) is called a simple harmonic
motion, or free undamped motion.
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Ch3_13
Solution and Equation of Motion
From (2), the general solution is
x(t )  c1 cos t  c2 sin t
(3)
Period T = 2/, frequency f = 1/T = /2.
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Ch3_14
Example 1 Free Undamped Motion
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
2
1 d 2x
d
x


4
x
,
 64 x  0
2
2
16 dt
dt
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Ch3_15
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
Applying the initial condition, we have
2
1
x(t )  cos 8t  sin 8t
3
6
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(4)
(5)
Ch3_16
Alternate Form of x(t)
(4) can be written as
x(t) = A sin(t + )
where A  c12  c22 , and  is a phase angle,
c1 
sin   
A tan   c1

c2 
c2
cos  
A
(6)
(7)
A sin t cos   A cos t sin   ( A sin  ) cos t  ( A cos  ) sin t (8)
c
c
A 1 cos t  A 2 sin t  c1 cos t  c2 sin t  x(t )
A
A
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(9)
Ch3_17
Fig 3.8.3 a relationship between c1>0, c2>0
and phase angle 
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Ch3_18
Example 2 Alternate Form of Solution (5)
Solution (5) is
x(t) = (2/3) cos 8t − (1/6) sin 8t = A sin(8t + )
Then
2
2
A  ( 2 3 )  ( 16 ) 
17
36
 0.69
 tan 1 (4)  1.326 rad
However it is not the solution, since we know
tan-1 (+/−) will locate in the second quadrant
Then     (1.326)  1.816 rad, so
17
x(t ) 
sin( 8t  1.816)
6
The period is T = 2/8 = /4.
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(9)
Ch3_19
Fig 3.8.4 Simple harmonic motion
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Ch3_20
Figure 3.8.5 Damping devices
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Ch3_21
DE of Free Damped Motion
If the DE is as
d 2x
dx
m 2  kx  
dt
dt
(10)
where  is a positive damping constant.
Then x”(t) + (/m)x’ + (k/m)x = 0 can be written as
d 2x
dx
2

2



x0
2
dt
dt
(11)
where
2 = /m, 2 = k/m
(12)
The auxiliary equation is m2 + 2m + 2 = 0, and the
roots are
m1    2   2 , m2    2   2
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Ch3_22
Case 1: 2 – 2 > 0
It is said to be overdamped.
Because the damped coefficient b is large when
compared to the spring constant k. The corresponding
solution of (11) is x(t )  c1em t c m t or
1
 t
x(t )  e (c1e
2  2 t
2
 c2 e
2
 2  2 t
)
(13)
See Fig 3.8.6.
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Ch3_23
Fig 3.8.6 Motion of an overdamped system
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Ch3_24
Case 2: 2 – 2 = 0
It is said to be critically damped.
Because any slight decrease in he damping force
would result in oscillatory motion. The general
solution of (11) is
x(t )  e t (c1  c2t )
(14)
See Fig 3.8.7.
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Ch3_25
Fig 3.8.7 Motion of an critically damped
system
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Ch3_26
Case 3: 2 – 2 < 0
It is said to be underdamped.
Because the damping coefficient is small compared to
the spring constants. The roots m1 and m2 are now
complex:
m1     2  2 i ,
m2     2  2 i
The general solution of equation (11) is
x(t )  e t (c1 cos  2  2 t  c2 sin  2  2 t )
(15)
See Fig 3.8.8.
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Ch3_27
Fig 3.8.8 Motion of an underdamped system
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Ch3_28
Example 3 Overdamped Motion
The solution of the IVP
d 2x
dx
 5  4x  0 ,
2
dt
dt
is
x(0)  1 ,
5 t 2 4t
x(t )  e  e
3
3
x(0)  1
(16)
See Fig 3.8.9.
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Ch3_29
Fig 3.8.9 Overdamped system in Ex 3
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Ch3_30
Example 4 Critically Damped Motion
A 8-pounds weight 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
1 d 2x
dx
 4 x  2 ,
2
4 dt
dt
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d 2x
dx
 8  16 x  0
2
dt
dt
(17)
Ch3_31
Example 4 (2)
m2 + 8m + 16 = 0, m = −4, −4
x(t) = c1 e-4t + c2t e-4t
Initial conditions: x(0) = 0, x’(0) = −3, then
x(t) = −3t e-4t
See Fig 3.8.10.
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(18)
(19)
Ch3_32
Example 5 Underdamped Motion
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
1 d 2x
dx
 5 x  ,
2
2 dt
dt
d 2x
dx
 2  10 x  0
2
dt
dt
(20)
m2 + 2m + 10 = 0, m = −3 + 3i, −3 − 3i
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Ch3_33
Example 5 (2)
x(t )  e t (c1 cos 3t  c2 sin 3t )
(21)
Initial conditions: x(0) = −2, x’(0) = 0, then
2

x(t )  e   2 cos 3t  sin 3t 
3


t
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(22)
Ch3_34
Alternate form of x(t)
(22) can be written as
x(t )  Aet sin(  2  2 t   )
where A  c12  c22 , and tan  
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(23)
c1
c2
Ch3_35
DE of Driven Motion with Damping
As in Fig 3.8.11,
d 2x
dx
m 2  kx    f (t )
dt
dt
(24)
d 2x
dx
2

2



x  F (t )
2
dt
dt
(25)
where F (t )  f (t )/m, 2   /m,  2  k /m
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Ch3_36
Fig 3.8.11 Oscillatory vertical motion of the
support
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Ch3_37
Example 6 Interpretation of an IVP
Interpret and solve
1 d 2x
dx
1
 1.2  2 x  5 cos 4t , x(0)  , x(0)  0
2
5 dt
dt
2
Solution:
Interpret:
Sol:
(26)
m = 1/5, k = 2,  = 1.2, f(t) = 5 cos 4t
release from rest at a point ½ below
dx 2
dx
 6  10 x  0
2
dt
dt
xc (t )  e 3t (c1 cos t  c2 sin t )
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Ch3_38
Example 6 (2)
Assuming xp(t) = A cos 4t + B sin 4t,
we have A = −25/102, B = 50/51, then
x(t )  e
3t
25
50
(c1 cos t  c2 sin t ) 
cos 4t  sin 4t
102
51
Using x(0) = 1/2, x’(0) = 0
c1 = 38/51, c2 = −86/51,
38
86
25
50


x(t )  e  cos t  sin t  
cos 4t  sin 4t
51
51
 51
 102
3t
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(28)
Ch3_39
Transient and Steady-State Terms
Graph of (28) is shown in Fig 3.8.12.
xc(t) will vanish at t  : transient term
xp(t) will still remain at t  : steady-state term
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Ch3_40
Fig 3.8.12 Graph of solution (28) in Ex 6
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Ch3_41
Example 7 Transient/Steady-State Solutions
The solution of
d 2x
dx

2
 2 x  4 cos t  2 sin t ,
2
dt
dt
x(0)  0 , x(0)  x1
is
x(t )  ( x1  2)e t sin t  2 sin t
 
transient
steady-state
See Fig 3.8.13.
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Ch3_42
Fig 3.8.13 Graph of solution in Ex 7 for
various values of x1
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Ch3_43
Example 8 Dmdamped Forced Motion
Solve
d 2x
2


x  F0 sin  t ,
2
dt
x(0)  0 ,
x(0)  0
(29)
where F0 is a constant and   .
Solution:
xc (t) = c1 cos t + c2 sin t
Let xp = A cos t + B sin t, after substitution,
A = 0, B = F0/(2− 2),
x p (t ) 
F0
sin  t
2
2
 
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Ch3_44
Example 8 (2)
F0
x(t )  xc  x p  c1 cos t  c2 sin t  2
sin  t
2
 
Since x(0) = 0, x’(0) = 0, then
c1  0, c2  F0 /  ( 2   2 )
Thus
F0
x(t ) 
( sin t   sin  t ) ,   
2
2
 (   )
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(30)
Ch3_45
Pure Resonance
When  = , we consider the case   .
d
( sin t   sin  t )
  sin t   sin  t
d
x(t )  lim F0

F
lim
0
 
 
d
 ( 2   2 )
( 3   2 )
d
 sin t  t cos  t
 F0 lim
 
 2
 sin t  t cos t
 F0
 2 2
F0
F0

sin t 
t cos t
2
2
2
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(31)
Ch3_46
When t  , the displacements become large
In fact, |x(tn)|   when tn = n/, n = 1, 2, …..
As shown in Fig 3.8.14, it is said to be pure
resonance.
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Ch3_47
Fig 3.8.14 Graph of solution in (31)
illustrating pure resonance
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Ch3_48
LRC-Series Circuits
The following equation is the DE of forced motion
with damping:
d 2x
dx
m 2   kx  f (t )
dt
dt
(32)
If i(t) denotes the current shown in Fig 3.8.15, then
L
di
1
 Ri  q  E (t )
dt
C
(33)
Since i = dq/dt, we have
d 2q
dq 1
L 2  R  q  E (t )
dt
dt C
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(34)
Ch3_49
Fig 3.8.15 LRC-series circuit
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Ch3_50
Example 9 Underdamped Series Circuit
Find q(t) in Fig 3.8.15, where L = 0.025 henry (h), R =
10 ohms (), C = 0.001 farad (f), E(t) = 0, q(0) = q0
coulombs (C), and i(0) = 0 ampere (A).
Solution:
Using the given data:
1
q  10q  1000q  0,
4
q  40q  4000q  0
As described before,
q(t )  e20t (c1 cos 60t  c2 sin 60t )
Using q(0) = q0, i(0) = q’(0) = 0, c1 = q0, c2 = q0/3
q0 10 20 t
q(t ) 
e sin( 60t  1.249)
3
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Ch3_51
Example 10 Steady-State Current
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,
1 

E0  L  

C 

A
,
2L 1
   L2 2   2 2  R2 


C C
E0 R
B


2
L
1
   L2 2  
 R2 


2
2
C
C 


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Ch3_52
Example 10 (2)
1
If X  L  ,
C
2L
1
X L 
 2 2
C C
2
2
2
2L
1
2
Z

L




R
If Z  X  R ,
C C 2 2
2
2
2
2
2
Using the similar method, we have
So
A  E0 X /( Z 2 ), B  E0 R /( Z 2 )
E X
ER
q p (t )   0 2 sin  t  0 2 cos  t
Z
Z
E0  R
X

i p (t )  q p (t )   sin  t  cos  t 
Z Z
Z

Note: X and Z are called the reactance and
impedance, respectively.
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Ch3_53
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
d 2M
 w( x)
2
dx
(1)
In addition, M(x) is proportional to the curvature  of
the elastic curve
M(x) = EI
(2)
where E, I are constants.
The product EI is called the flexural rigidity of the
beam.
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Ch3_54
Fig 3.9.1 Deflection of a homogeneous beam
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Ch3_55
From calculus, we have   y”, when the deflection
y(x) is small. Finally we have
Then
d 2M
d2
d4y
 EI 2 y  EI 4
2
dx
dx
dx
d4y
EI  4  w( x)
dx
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(3)
(4)
Ch3_56
Terminology
Ends of the beam
Boundary Conditions
embedded
y = 0, y’ = 0
free
y” = 0, y’’’ = 0
simply supported or hinged y = 0, y” = 0
See Fig 3.9.2.
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Ch3_57
Fig 3.9.2 Beams with various end conditions
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Ch3_58
Example 1 An Embedded Beam
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:
d4y
From (4) we have EI 4  w0
dx
Embedded ends means
y (0)  0 , y(0)  0 , y ( L)  0 , y( L)  0
We have m4 = 0, yc(x) = c1 + c2x + c3x2 + c4x3, and
w0 4
yp 
x
24 EI
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Ch3_59
Example 1 (2)
So
y ( x)  c1  c2 x  c3 x 2  c4 x 3 
w0 4
x
24 EI
Using the boundary conditions, we have
c1 = 0, c2 = 0, c3 = w0L2/24EI, c4 = −w0L/12EI
w0 L2 2 w0 L 3
w
w
y ( x) 
x 
x  0 x 4  0 x 2 ( x  L) 2
24 EI
12 EI
24 EI
24 EI
Choosing w0 = 24EI and L = 1, we have Fig 3.9.3.
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Ch3_60
Example 2 Nontrivial Solution of a BVP
Solve y   y  0, y (0)  0, y ( L)  0
Solution:
Case I:  = 0
y = c1x + c2,
y(0) = c2 = 0, y(L) = c1L = 0, c1 = 0
then y = 0, trivial solution.
Case II:  < 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.
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Ch3_61
Example 2 (2)
Case III:  > 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
2 2
n

2
n   n  2 ,
L
n  1, 2, 3, 
Thus, y = c2 sin (nx/L) is a solution for each n.
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Ch3_62
Example 2 (3)
Simply take c2 = 1, for each:
2
4 2
,
,
2
2
L
L
9 2
,
2
L
the corresponding function:
sin

L
x,
2
sin
x,
L
3
sin
x ,
L
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.
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Ch3_63
Bulking of a Thin Vertical Column
Referring to Fig 3.9.4, the DE is
d2y
EI 2   Py ,
dx
d2y
EI 2  Py  0
dx
(7)
where P is a constant vertical compressive force
applied to the column’s top.
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Ch3_64
Fig 3.9.4 Elastic column bucking under a
compressive force
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Ch3_65
Example 3 The Euler Load
Referring to Fig 3.9.4, when the column is hinged at
both ends, find the deflection.
Solution:
The boundary-value problem is
d2y
EI 2  Py  0 ,
dx
y (0)  0 ,
y ( L)  0
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?
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Ch3_66
Example 3 (2)
By writing  = P/EI, we see
y  y  0 ,
y (0)  0 ,
y ( L)  0
is identical to example 2. From Case III, 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 = n22EI/L2, deflection occurs.
We call these Pn the critical loads and the smallest P
= P1 = 2EI/L2 is called the Euler load, and
y1 = c2 sin(x/L) is known as the first buckling
mode. See Fig 3.9.5
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Ch3_67
Fig 3.9.5 Deflecting curves for compressive
forces P1, P2, P3
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Ch3_68
Rotating String
The simple DE
y” + y = 0
(8)
occurs again as a model of a rotating string. See Fig
3.9.6.
We have
F = T sin 2 – T sin 1
(9)
When 1 and 2 are small, sin 2  tan 2 , sin 1 
tan 1.
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Ch3_69
Fig 3.9.6 Rotating rope and forces acting on
it
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Ch3_70
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
F  T [ y( x  x)  y( x)]
(10)
Because F = ma, m = x, a = r2. With x small,
we take r = y.
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Ch3_71
Thus
F  ( x) y 2
(11)
Letting (8) = (9), we have
T [ y( x  x)  y( x)]  ( x) y 2
y( x  x)  y( x)
T
   2 y
x
(12)
For x close to zero, we have
2
d y
T 2    2 y,
dx
d2y
T 2   2 y  0
dx
(13)
And the boundary conditions are y(0) = y(L) = 0.
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Ch3_72
3.10 Green’s Functions
The IVP
The solution of the second-order IVP
y  P( x) y  Q( x) y  f ( x), y( x0 )  y0 , y( x0 )  y1 (3)
can be press as the superposition of two solution: the
solution yh of the associated homogeneous DE with
nonhomogeneous initial conditions
y  P( x) y  Q( x) y  0, y ( x0 )  y0 , y( x0 )  y1
(4)
and the solution yp of the nonhomogeneous DE with
homogeneous initial condition
y  P( x) y  Q( x) y  f ( x), y( x0 )  0, y( x0 )  0 (5)
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Ch3_73
Green’s Function
If y1(x) and y2(x) from a fundamental set of solution
of the interval I of the associated homogeneous from
of (2), then a particular solution of the
nonhomogeneous equation (2) on the interval I can be
found by variation of parameters.
The form of this solution is
y p ( x)  u1 ( x) y1 ( x)  u2 ( x) y2 ( x)
(6)
The variable coefficient u1(x) and u2(x) in (6) are
defined by (5) of Section 3.5:
y ( x) f ( x)
y ( x) f ( x)
(7)
u ( x)   2
, u ( x)  1
1
W
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2
W
Ch3_74
The linear independent of y1(x) and y2(x) on the
interval I guarantees that the Wronskian W = W(y1(x),
y2(x))  0 for all x in I. If x and x0 are numbers in I,
then integrating the derivatives in (7) on the interval
[x0, x] and substituting the results in (6) give
x y (t ) f (t )
 y2 (t ) f (t )
y p ( x)  y1 ( x) 
dt  y2 ( x)  1
dt
x0
x0
W (t )
W (t )
x  y (t ) y (t )
x y (t ) y (t )
1
2
2

f (t )dt   1
f (t )dt ,
x0
x0
W (t )
W (t )
x
where
y1 (t )
W (t )  W ( y1 (t ), y2 (t )) 
y1 (t )
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y2 (t )
y2 (t )
Ch3_75
From the properties of the definite integral, the two
integrals in the second line of (8) can be rewritten as a
single integral
x
y p ( x)   G ( x, t ) f (t )dt
x0
(9)
The function G(x, t) in (9),
y1 (t ) y2 ( x)  y1 ( x) y2 (t )
G ( x, t ) 
W (t )
(10)
is called the Green’s function for the differential
equation (2).
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Ch3_76
Observe (10), y1(x) and y2(x) of the associated
homogeneous differential equation for (2) and not on
the forcing function f(x). Therefore all linear secondorder differential equations (2) with the same lefthand sode but different forcing functions have the
same Green’s function. So an alternative title for (10)
of the Green’s function for the second-order
differential operator L = D2 + P(x)D + Q(x).
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Ch3_77
Example 1 Particular Solution
Use (9) and (10) to find a particular solution of y  y  f (x).
Solution:
The solution of the associates homogeneous equation
y  y  0 are y1 = ex, y2 = e-x, and W(y1(x), y2(x)) = -2.
It follows from (10) that the Green’s function is
x
y p ( x)   sinh( x  t ) f (t )dt
x0
(11)
Thus from (9), a particular solution of the DE is
et e  x  e x e t e xt  e ( xt )
G ( x, t ) 

 sinh( x  t )
2
2
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(12)
Ch3_78
Example 2 General Solutions
Find the general solution of the following
nonhomogeneous differential equations.
2x


y

y

1
/
x


y

y

e
(a)
(b)
Solution:
(a) With the identifications f(x) = 1/x amd f(t) = 1/t we
see from (12) that a particular solution of y  y  1 / x is
sinh( x  t )
y p ( x)  
dt
x0
t
x
Thus the general solution y = yc + yp of the DE on any
interval [x0, x] not containing the origin is
x sinh( x  t )
x
x
(13)
y  c1e  c2 e  
dt
x
0
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t
Ch3_79
Example 2 (2)
(b) With f(x) = e2x in (12), a particular solution of y  y  e2 x
x
is y p ( x)   sinh( x  t )e 2t dt.
x
Thus the general solution y = yc + yp is then
0
x
y  c1e  c2e   sinh( x  t )e 2t dt.
x
x
x0
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(14)
Ch3_80
Theorem 3.10.1 Solution of the IVP in (5)
The function yp(x) defined in (9) in the solution of the
initial-value problem (5).
Proof:
By construction we know that yp(x) in (9) satisfies the
nonhomogeneous DE. Because a definite integral has
a
the property   0 we have
a
x
y p ( x0 )   G( x0 , t ) f (t )dt  0
x0
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Ch3_81
Theorem 3.10.1 proof
To show that yp ( x0 )  0 we utilize the Leibniz formula
for the derivative of an integral:
0 from (10)
 x y (t ) y (t )  y (t ) y (t )
2
1
2
yp ( x0 )  G ( x, x) f ( x)   1
f (t )dt
x0
w(t )
Hence,
yp ( x0 )  
x0
x0
y1 (t ) y2 ( x0 )  y1 ( x0 ) y2 (t )
f (t )dt  0
W (t )
Leibniz formula:
v( x) 
d v( x)
F ( x, t )dt  F ( x, v( x))v( x)  F ( x, u ( x))u( x)  
F ( x, t )dt

u
(
x
)
u
(
x
)
dx
x
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Ch3_82
Example 3 Example 2 Revisited
Solve the IVP
(a) y  y  1 / x, y (1)  0, y(1)  0
(b) y  y  e2 x , y(0)  0, y(0)  0
Solution:
(a) With x0 = 0 and f(t) = 1/t, it follows from (13) of Ex
2 and Theorem 3.10.1 that the solution of the IVP is
y p ( x)  
x
1
sinh( x  t )
dt
t
where [1, x], x > 0.
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Ch3_83
Example 3 (2)
(b) Identifying x0 = 0 and f(t) = e2t, we see from (14)
that the solution of the IVP is
x
y p ( x)   sinh( x  t )e 2t dt
0
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(15)
Ch3_84
We can carry out the integration in (15), but bear in
mind that x is held constant throughout the integration
with respect to t:
x
x
0
0
y p ( x)   sinh( x  t )e 2t dt  
e x t  e  ( x t ) 2 t
e dt
2
1 x x t
1  x x 3t
 e  e dt  e  e dt
0
2 0
2
1 2 x 1 x 1 x
 e  e  e
3
2
6
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Ch3_85
Example 4 Another IVP
Solve the IVP
y  4 y  x, y (0)  0, y(0)  0.
Solution:
The two linearly independent solutions of y  4 y  0 are
y1 ( x)  cos 2 x and y2 ( x)  sin 2 x. From (10), with W(cos2t,
sin2t) = 2, we find
cos 2t sin 2 x  cos 2 x sin 2t 1
G ( x, t ) 
 sin 2( x  t ).
2
2
With the identification x0 = 0, a solution of the given
IVP is
1 x
y p ( x) 
t sin 2( x  t )dt

2
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0
Ch3_86
Example 4 (2)
If we wish to evaluate the integral, we first write
x
x
1
1
y p ( x)  sin 2 x  t cos 2tdt  cos 2 x  t sin 2tdt
0
0
2
2
and the use integration by parts:
x
x
1
1
1
1
1
1



y p ( x)  sin 2 x  t sin 2t  cos 2t   cos 2 x  t cos 2t  sin 2t 
2
4
4
2
0 2
 2
0
or
1
1
y p ( x)  x  sin 2 x
4
8
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Ch3_87
Theorem 3.10.2 Solution of the IVP in (3)
If yh is the solution of the initial-vaule problem (4) and
yp is the solution (9) of the initial-value problem (5) on
the interval I, then
y = yh + yp
(16)
Is the solution of the initial-value problem (3).
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Ch3_88
Theorem 3.10.2 proof
Proof:
From (10) of Section 3.3, y = yh + yp is a solution of the
nonhomogeneous DE. Since yh satisfied the initial
conditions in (4) and yp satisfies the initial conditions in
(5), we have
y ( x0 )  yh ( x0 )  y p ( x0 )  y0  0  y0
y( x0 )  yh ( x0 )  yp ( x0 )  y1  0  y1
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Ch3_89
Keeping in mind the absence of a forcing function in
(4) and the presence of such a term in (5), we see
from (16) that the response y(x) of a physical system
described by the initial-value problem (3) can be
separated into two different response:
y ( x) 
y ( x)

y p ( x)
h

response of system
response of system
duo to initial conditions duo to the forcing
y ( x0 ) y0 , y( x0 ) y1
function f
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(17)
Ch3_90
Example 5 Using Theorem 3.10.2
Solve the initial-value problem
y  4 y  sin 2 x, y (0)  1, y(0)  2
Solution:
First, we solve y  4 y  0, y (0)  1, y(0)  2. By the lefthand side of the differential equation y(x) = c1cos2x +
c2sin2x of the homogeneous DE, we find that c1 = 1 and
c2 = -1. therefore, yh(x) = cos2x – sin2x.
Next we solve y  4 y  sin 2 x, y (0)  0, y(0)  0. Since the
left-hand side of the differential equation is the same as
the DE in Ex 4, the Green’s function is the same second
problem is y p ( x)  1 x sin 2( x  t ) sin 2tdt
2 0
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Ch3_91
Example 5 (2)
Finally, in view of (16) in Theorem 3.10.2, the
solution of the original IVP is
y ( x)  yh ( x)  y p ( x)
1 x
 cos 2 x  sin 2 x   sin 2( x  t ) sin 2tdt
2 0
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(18)
Ch3_92
We can integrate the definite integral in (18) by using
the trigonometric identity
1
sin A sin B  cos( A  B)  cos( A  B)
2
with A = 2(x – t) and B = 2t:
1 x
y p ( x)   sin 2( x  t ) sin 2tdt
2 0
1 x
  [cos( 2 x  4t )  cos 2 x]dt
4 0
(19)
x
1 1
1
1

  sin( 2 x  4t )  t cos 2 x   sin 2 x  x cos 2 x
4 4
4
0 8
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Ch3_93
Hence, the solution (18) can be rewritten as
1
1

y ( x)  yh ( x)  y p ( x)  cos 2 x  sin 2 x   sin 2 x  x cos 2 x 
4
8

or
7
1
y ( x)  cos 2 x  sin 2 x  x cos 2 x
(20)
8
4
Note that the physical significance indicated in (17) is
lost in (20) after combining like terms in the two parts
of the solution y(x) = yh(x) + yp(x).
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Ch3_94
The beauty of the solution given in (18) is that we can
immediately write down the response of a system if
the initial conditions remain the same but the forcing
function is changed. For example, if the problem in
Ex 5 is changed to
y  4 y  x, y (0)  1, y(0)  2
we simply replace sin 2t in the integral in (18) by t
and the solution is then
y ( x)  yh ( x)  y p ( x )
1 x
1
9
 cos 2 x  sin 2 x   t sin 2( x  t )dt  x  cos 2 x  sin 2 x.
2 0
4
8
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Ch3_95
Example 6 An Initial-Value Problem
Solve the initial-value problem
y  4 y  f ( x), y (0)  1, y(0)  2
when the forcing function f is a piecewise defined:
x0
0,

f ( x)  sin 2 x, 0  x  2
0,
x  2

Solution:
From (18), with sin2t replaced by f(t), we can write
1 x
y ( x)  cos 2 x  sin 2 x   sin 2( x  t ) f (t )dt
2 0
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Ch3_96
Example 6 (2)
Because f is defined in three pieces, we consider three
cases in the evaluation of the definite integral.
For x < 0,
1 x
y p ( x) 
sin 2( x  t )0dt  0

2
0
For 0  x  2,
1 x
y p ( x)   sin 2( x  t ) sin 2tdt
2 0
1
1
 sin 2t  x cos 2 x
8
4
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Ch3_97
Example 6 (3)
For x > 2,
1 2
1 x
y p ( x)   sin 2( x  t ) sin 2tdt   sin 2( x  t )0dt
2 0
2 2
1 2
  sin 2( x  t ) sin 2tdt
2 0
2
1 1
  sin( 2 x  4t )  t cos 2 x 
4 4
0
1
1
1
  sin( 2 x  8 )   cos 2 x  sin 2 x
16
2
16
1
   cos 2 x
2
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Ch3_98
Example 6 (4)
Hence yp(x) is
and so
x0
0,

y p ( x)  18 sin 2 x  14 x cos 2 x, 0  x  2
 1  cos 2 x,
x  2
 2
y ( x)  yh ( x)  y p ( x)  cos 2 x  sin 2 x  y p ( x)
Putting all the pieces we get
x0
cos 2 x  sin 2 x,

y p ( x)  (1  14 x) cos 2 x  78 sin 2 x, 0  x  2
(1  1  ) cos 2 x  sin 2 x, x  2

2
The graph y(x) is given in Fig 3.10.1.
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Ch3_99
Fig 3.10.1 Graph of y(x) in Ex 6
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Ch3_100
Boundary-Value Problems
In contrast to a second-order IVP in which y(x) and
y’(x) are specified at the same point, a BVP for a
second-order DE involves conditions on y(x) and y’(x)
that are specified at two different x = a and x = b.
Conditions such as
y (a)  0, y (b)  0;
y (a)  0, y(b)  0;
y(a)  0, y(b)  0
are just special cases of the more general
homogeneous boundary conditions
A1 y(a)  B1 y(a)  0
A2 y(b)  B2 y(b)  0
and
where A1, A2, B1, and B2 are constants.
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(21)
(22)
Ch3_101
Specifically, our goal is to find an integral solution
yp(x) that is analogous to (9) for nonhomogeneous
boundary-value problems of the form
y  P( x) y  Q( x) y  f ( x),
A1 y (a)  B1 y(a)  0,
A2 y (b)  B2 y(b)  0.
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(23)
Ch3_102
In addition to the usual assumptions that P(x), Q(x),
and f(x) are continuous on [a, b], we assume that the
homogeneous problem
y  P( x) y  Q( x) y  0,
A1 y (a)  B1 y(a)  0,
A2 y (b)  B2 y(b)  0.
possesses only the trivial solution y = 0. This latter
assumption is sufficient to guarantee that a unique
solution bof (23) exists and it gives by an integral
y p ( x)   G( x, t ) f (t )dt , where G(x, t) is a Green’s
a
function.
The starting point in the construction of G(x, t) is
again the variation of parameters formula (6) and (7).
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Ch3_103
Another Green’s Function
Suppose y1(x) and y2(x) are linearly independent
solutions on [a, b] of the associated homogeneous
form of the DE in (23) and that x is a number in the
interval [a, b]. Unlike the construction of (8) where
we started by integrating the derivatives in (7) over
the same interval, we now integrate the first equation
in (7) on [b, x] and the second equation in (7) on
[a, x]:
u1 ( x)   
x
b
x y (t ) f (t )
y2 (t ) f (t )
dt and u2 ( x)   1
dt
a
W (t )
W (t )
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(24)
Ch3_104
The reason for integrating u1 ( x) and u2 ( x) over different
intervals will become clear shortly. From (24), a
particular solution y p ( x)  u1 ( x) y1 ( x)  u2 ( x) y2 ( x) of the
DE is
here we used the minus
sign in (24) to reverse
the limits of integratio n


b y (t ) f (t )
x y (t ) f (t )
2
y p ( x)  y1 ( x) 
dt  y2 ( x)  1
dt
x
a
W (t )
W (t )
or
y p ( x)  
x
a
b y ( x ) y (t )
y2 ( x) y1 (t )
2
f (t )dt   1
f (t )dt
x
W (t )
W (t )
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(25)
Ch3_105
The right-hand side of (25) can be written compactly
as a single integral b
y p ( x)   G ( x, t ) f (t )dt
(26)
a
where the function G(x, t) is
 y1 (t ) y2 ( x) , a  t  x
 W (t )
G ( x, t )  
y1 ( x) y2 (t )

, xt b
 W (t )
(27)
The Piecewise-defined function (27) is called a
Green’s function for the boundary-value problem
(23). It can be proved that G(x, t) is a continuous
function of x on the interval [a, b].
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Ch3_106
If the solutions y1(x) and y2(x) used in the
construction of (27) are chosen in such a manner that
at x = a, y1(x) satisfies A1 y1 (a)  B1 y1 (a)  0, and at x  b, y2 ( x)
satisfies A2 y2 (b)  B2 y2 (b)  0, then, wondrously, yp(x)
defined in (26) satisfied both homogeneous boundary
conditions in (23).
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Ch3_107
To see this we need
y p ( x)  u1 ( x) y1 ( x)  u2 ( x) y2 ( x)
(28)
and
yp ( x)  u1 ( x) y1 ( x)  y1 ( x)u1 ( x)  u2 ( x) y2 ( x)  y2 ( x)u2 ( x)
 u1 ( x) y1 ( x)  u2 ( x) y2 ( x)
(29)
Before proceeding, observe in (24) that u1(b) = 0 and
u2(a) = 0. In view of the second of these two
properties we can show that yp(x) satisfies (21)
whenever y1(x) satisfies the same boundary condition.
From (28) and (29) we have
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Ch3_108
A1 y p (a )  B1 yp (a )
0
0


 A1[u1 (a ) y1 (a )  u2 (a ) y2 (a )]  B1[u1 (a ) y1 (a )  u2 (a ) y2 (a )]
 u1 (a )[ A1 y1 (a )  B1 y1 (a )]  0

0 from (21)
Likewise, u1(b) = 0 implies that whenever y2(x)
satisfies (22) so does yp(x):
A2 y p (b)  B2 yp (b)
0
0


 A2 [u1 (b) y1 (b)  u2 (b) y2 (b)]  B2 [u1 (b) y1 (b)  u2 (b) y2 (b)]
 u2 (b)[ A2 y2 (b)  B2 y2 (b)]  0



0 from (22)
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Ch3_109
Theorem 3.10.3 Solution of a BVP
Let y1(x) and y2(x) be linearly independent solution of
y  P( x) y  Q( x) y  0
on [a, b], and suppose y1(x) and y2(x) satisfy (21) and
(22), respectively. Then the function yp(x) defined in
(26) is a solution of the boundary-value problem (23).
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Ch3_110
Example 7 Using Theorem 3.10.3
Solve the BVP
y  4 y  0, y(0)  0, y ( / 2)  0
Solution:
The solution of the associated homogeneous equation
y  4 y  0 are y1(x) = cos2x and y2(x) = sin2x and y1(x)
satisfies y’(0)=0, whereas y2(x) satisfies y(/2) = 0. The
Wronskian is W(y1, y2) = 2, and so from (27) we see that
Green’s function for the BVP is
 1 cos 2t sin 2 x, 0  t  x
2
G ( x, t )  
1
 cos 2 x sin 2t , x  t   / 2
2
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Ch3_111
Example 7 (2)
It follows from Theorem 3.10.3 that a solution of the
BVP is (26) with the identifications a = 0, b = /2, and
f(t) = 3:
 /2
y p ( x)  3
0
G ( x, t )dt
x
 /2
1
1
 3  sin 2 x  cos 2tdt  3  cos 2 x  sin 2tdt
0
x
2
2
or after evaluating the definite integrals,
3 3
y p ( x)   cos 2 x.
4 4
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Ch3_112
Example 8 A BVP
Solve the BVP
x 2 y  3xy  3 y  24 x5 , y(1)  0, y(2)  0.
Solution:
The differential equation is recognized as a CauchyEuler DE. From the auxiliary equation m(m-1)-3m+3 =
(m-1)(m-3) = 0 the general solution of the associated
homogeneous equation is y = c1x + c2x3.
Apply y(1) = 0 to this solution implies c1+c2=0 or c1=-c2.
By choosing c2 = -1 we get c1 = 1 and y1= x – x3. On the
other hand, y(2)=0 applied to the general solution 2c1 +
8c2 = 0 or c1 = -4c2.
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Ch3_113
Example 8 (2)
The choice c2 = -1 now gives c1 = 4 and so t2(x) = 4x –
x3. The Wronskian of these two function is
x  x3
W ( y1 ( x), y2 ( x)) 
1  3x 2
4 x  x3
3

6
x
4  3x 2
Hence the Green’s function for the BVP is
 (t  t 3 )(4 x  x 3 )
, 0t  x
3

6t
G ( x, t )  
3
3
(
x

x
)(
4
t

t
)

, xt 2
3

6t
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Ch3_114
Example 8 (3)
In order to identify the correct forcing function f we
write the DE in standard form:
3
3
y  y  2 y  24 x 3
x
x
From this equation we see that f(t) = 24t3 and so (26)
becomes
2
y p ( x)  24 G ( x, t )t 3 dt
1
x
2
 4(4 x  x )  (t  t )dt  4( x  x )  (4t  t 3 )dt
3
1
3
3
x
Straightforward definite integration and algebraic
simplification yield the solution yp(x) = 12x – 15x3 + 3x5.
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Ch3_115
3.11 Nonlinear Models
Nonlinerar Springs
The model
d 2x
m 2  F ( x)  0
dt
(1)
when F(x) = kx is said to be linear.
However,
2
d 2x
d
x
3
m 2  kx  0, m 2  kx  k1 x 3  0
dt
dt
(2)
is a nonlinear spring.
Another model
d 2x
dx dx
m 2 
 kx  0
dt
dt dt
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(3)
Ch3_116
Hard and Soft Springs
F(x) = kx + k1x3 is said to be hard if k1 > 0; and is
soft, if k1 < 0. See Fig 3.11.1.
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Ch3_117
Example 1 Comparison of Hard and Soft
Springs
The DEs
and
d 2x
3

x

x
0
2
dt
(4)
d 2x
3

x

x
0
2
dt
(5)
are special cases of (2). Fig 3.11.2 shows the graph
from a numerical solver.
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Ch3_118
Fig 3.11.2 Numerical solution curves
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Ch3_119
Nonlinear Pendulum
The model of a simple pendulum is shown in Fig
3.11.3.
From the figure, We have the angle acceleration a = s”
= l”, the force
d 2
F  ma  ml 2
dt
Then
d 2 g
 sin   0
2
dt
l
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(6)
Ch3_120
Fig 3.11.3 Simple pendulum
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Ch3_121
Linearization
Since
sin    
3 5
3!

5!

If we use only the first two terms,
d 2 /dt 2  ( g /l )  ( g / 6l ) 3  0
If  is small,
d 2 g
  0
2
dt
l
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(7)
Ch3_122
Example 2 Two Initial-Value Problems
Fig 3.11.4 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.
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Ch3_123
Fig 3.11.4 Numerical solution curves in (a);
oscillating pendulum in (b); whirling
pendulum in (c) in Ex 2
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Ch3_124
Telephone Wire
Recalling from (17) in Sec 1.3 and Fig 1.3.8
dy/dx = W/T1, can be modified as
dy s

dx T1
where  is the density and s is the arc length.
Since the length s is
s
x
0
2
dy
1    dx
 dx 
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(8)
(9)
Ch3_125
then
ds
dy 

 1  
dx
 dx 
2
(10)
Differentiating (8) w.s.t x and using (10), then
d y  ds

,
2
dx
T1 dx
2
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d y 
 dy 

1

 
2
dx
T1
 dx 
2
2
(11)
Ch3_126
Example 3 An Initial-Value Problem
From Fig 1.3.8, we obtain y(0) = a, y’(0) = 0. Let u =
y’, equation (11) becomes
du

du 
2
  dx

1 u , 
2
T1
1 u
dx T1

Thus
1
sinh u  x  c
1
T1
Now y’(0) = u(0) = 0, sinh-10 = 0 = c1
Since u = sinh(x/T1) = dy/dx, then
dy

 sinh x,
dx
T1
y
T1

cosh
Using y(0) = a, c2 = a − (T1/)
y
T1

cosh

T1
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xa

T1
x  c2
T1

Ch3_127
Rocket Motion
From Fig 3.11.5, we have
d2y
Mm
m 2  k 2 ,
dt
y
d2y
M


k
dt 2
y2
when y = R, kMm/R2 = mg, k = gR2/M,
then
2
2
d y
R
 g 2
2
dt
y
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(12)
(13)
Ch3_128
Fig 3.11.5 Distance to rocket is large
compared to R
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Ch3_129
Variable Mass
Assuming the mass is variable, then F = ma should
be modified as
d
F  (mv)
dt
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(14)
Ch3_130
Example 4 Chain Pulled Upward by a
Constant Firce
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 – x (net force)
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Ch3_131
Example 4 (2)
Then
d x 
 v   5  x,
dt  32 
x
dv
dx
 v  160  32 x
dt
dt
(15)
Since v = dx/dt
2
d x  dx 
    32 x  160
2
dt
 dt 
2
x
(16)
is of the form F ( x, x, x)  0
Since v = x’, and
dv dv dx
then (15) becomes
dv

v
dt dx dt
dx
dv 2
xv  v  160  32 x
dt
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(17)
Ch3_132
Example 4 (3)
Rewriting (17) as
(v2+32x – 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 (please verify). Then
f /x  xv 2  32 x 2  160 x, f /v  xv 2
Use the method in Sec. 2.4
(19)
1 2 2 32 3
x v  x  80 x 2  c1
2
3
Since x(0) = 0, then c1 = 0. By solving (19) = 0, for
v = dx/dt > 0, we get dx
64
dt
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 160 
3
x
Ch3_133
Example 4 (4)
Thus please verify that
1/ 2
3
64 
 160  x 
32 
3 
 t  c2
(20)
Using x(0) = 0 again, c2  3 10 / 8 , we square both
sides of (20) and solve for x
x(t ) 
15 15  4 10 
 1 
t
2 2
15 
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2
(21)
Ch3_134
3.12 Solving Systems of Linear Equations
Coupled Spring/Mass System
From Fig 3.12.1 and Newton’s Law
m1 x1  k1 x1  k 2 ( x2  x1 )
m2 x2  k 2 ( x2  x1 )
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(1)
Ch3_135
Fig 3.12.1 Coupled spring/mass system
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Ch3_136
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
x(t )  c1e 
6t
 c2e
6t
(3)
Similar method can give
y (t )  c3e 
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6t
 c4 e
6t
(4)
Ch3_137
Return to the original equations,
dx/dt = 3y
then after simplification,
( 6c1  3c3 )e 
we have
 ( 6c2  3c4 )e
6t
0
6
6
c3   c1 , c4 
c2
3
3
Hence
x(t )  c1e 
6t
6t
 c2e 6 t ,
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y (t )  
6 
c1e
3
(5)
6t

6
c2e
3
6t
Ch3_138
Example 1 Solution by Elimination
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)
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Ch3_139
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
1
x(t )  2c1e 2t  c2 e 3t ,
3
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y (t )  c1e 2t c2 e 3t
Ch3_140
Example 2 Solution by Elimination
x’ – 4x + y” = t2
x’ + x + y’ = 0
Solve
Solution:
(D – 4)x + D2y = t2
(D + 1)x + Dy = 0
By eliminating x,
(9)
(10)
[(D  1) D 2  ( D  4) D] y  ( D  1)t 2  ( D  4)0
then ( D3  4D) y  t 2  2t , and m = 0, 2i, −2i
yc  c1  c2 cos 2t  c3 sin 2t
Let y p  At 3  Bt 2  Ct , then we can get A = 1/12, B = ¼ ,
C = −1/8.
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Ch3_141
Example 2 (2)
Thus
y  yc  y p
1 3 1 2 1
 c1  c2 cos 2t  c3 sin 2t  t  t  t
12
4
8
(11)
Similar method to get x(t)
[( D  4)  D( D  1)]x  t 2 ,
( D 2  4) x  t 2
Then m= 2i, −2i,
xc  c4 cos 2t  c5 sin 2t
Let xp(t) = At2 + Bt + C, then we can get A = −1/4, B =
0, C = 1/8
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Ch3_142
Example 2 (3)
Thus
1
1
x  xc  x p  c4 cos 2t  c5 sin 2t  t 2 
4
8
(12)
By using the second equation of (9), we have
(c5  2c4  2c2 ) sin 2t  (2c5  c4  2c3 ) cos 2t  0
c4  1/5(4c2  2c3 ), c5  1/5(2c2  4c3 )
1
1
1
1
x(t )   (4c2  2c3 ) cos 2t  (2c2  4c3 ) sin 2t  t 2 
5
5
4
8
1 3 1 2 1
y (t )  c1  c2 cos 2t  c3 sin 2t  t  t  t
12
4
8
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Ch3_143
Example 3 A Mathematical Model Revisited
In (3) of Sec. 2.9, we have
1
D 2  x 
x2  0

 1
25 
50

2
2

 x1   D   x2  0
25
25 

Together with the given initial conditions, we can use
the same method to solve x1 and x2, not mentioned
here.
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Ch3_144
Example 4
Solve
x"1 10 x1
 4 x2  0
4 x1  x"2 4 x2  0
with x1 (0)  0, x1 (0)  1, x2 (0)  0, x2 (0)  1
Solution:
( D 2  10) x1
(13)
 4 x2  0
 4 x1  ( D 2  4) x2  0
Then
( D 2  2)( D 2  12) x1  0, ( D 2  2)( D 2  12) x2  0
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Ch3_145
Example 4 (2)
Using the same method, we have
2
3
x1 (t )  
sin 2t 
sin 2 3t
10
5
2
3
x2 (t )  
sin 2t  sin 2 3t
5
10
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(14)
Ch3_146
Fig 3.12.2 Displacement of the two masses in
Ex 4
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Ch3_147
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148