CHAPTER 3
Higher-Order Differential
Equations
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
CH3_2
3.1 Preliminary Theory: Linear Equ.
Initial-value Problem
An initial value problem for nth-order linear DE is
n
n1
d y
d
dy
an ( x) n  an1 ( x) n1    a1 ( x)  a0 ( x) y  g ( x)
dx
dx
dx
with
y ( x0 )  y0 , y( x0 )  y1 ,  , y ( n1) ( x0 )  yn1
(1)
as n initial conditions.
CH3_3
THEOREM 3.1
Existence and Uniqueness
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 a point in this
interval, then a solution y(x) of (1) exists on the interval
and is unique.
CH3_4
Example 1
The problem
3 y  5 y  y  7 y  0, y (1)  0 , y(1)  0, y(1)  0
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.
CH3_5
The following DE
n
n1
d y
d y
dy
an ( x) n  an1 ( x) n1    a1 ( x)  a0 ( x) y  0
dx
dx
dx
(6)
is homogeneous;
n
n1
d y
d y
dy
an ( x) n  an1 ( x) n1    a1 ( x)  a0 ( x) y  g ( x) (7)
dx
dx
dx
with g(x)  0, is nonhomogeneous.
CH3_6
Differential Operators
Let dy/dx = Dy. This symbol D is called a differential
operator.
We define an nth-order differential operator as
L  an ( x) D n  an1 ( x) D n1    a1 ( x) D  a0 ( x) (8)
In addition, we have
L{f ( x)  g ( x)}  L( f ( x))  L( g ( x))
(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)
CH3_7
THEOREM 3.2
Superposition Principles – Homogeneous Equations
Let y1, y2, …, yn be n solutions of the homogeneous
nth-order differential equation (6) on an interval I.
Then the linear combination
y = c1y1(x) + c2y2(x) + …+ cnyn(x)
where the ci, i = 1, 2, …, n are arbitrary constants, is
also a solution on the interval.
CH3_8
COROLLARY
Corollaries to Theorem 3.2
(A) y = cy1 is also a solution if y1 is a solution.
(B) A homogeneous linear DE always possesses the
trivial solution y = 0.
CH3_9
Linear Dependence and Independence
DEFINITION 3.1
Linear Dependence and Linear Independence
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.
CH3_10
In other words, if the set is linearly independent, then
c1f1(x) + c2f2(x) + … + cn fn(x) = 0
implies 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.
CH3_11
Fig 3.3
CH3_12
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.
CH3_13
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
CH3_14
DEFINITION 3.2
Wronskian
Suppose each of the functions f1(x), f2(x), …, fn(x)
possesses at least n – 1 derivatives. The determinant
W ( f1 ,..., f n ) 
f1
f2

fn
f1 '
f2 '

fn '


f 1( n1)
f 2( n1) 

f n( n1)
is called the Wronskian of the functions.
CH3_15
THEOREM 3.3
Criterion for Linear Independence
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.
Corollary 3.3
If W(y1, y2, …, yn)  0 for some x in the interval then y1, y2, …, yn are linearly
independent in the interval. If W(y1, y2, …, yn) = 0 for some x in the interval then
y1, y2, …, yn are linearly dependent in the interval.
.
CH3_16
DEFINITION 3.3
Fundamental Set of a Solution
Any set y1(x), y2(x), …, yn(x) of n linearly independent
solutions is said to be a fundamental set of solutions.
CH3_17
THEOREM 3.4
Existence of a Fundamental Set
There exists a fundamental set of solutions for (6) on an
interval I.
THEOREM 3.5
General Solution – Homogeneous Equations
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.
CH3_18
Example 7
The functions y1 = e3x, y2 = e-3x are solutions of
y” – 9y = 0 on (-, )
Now
3x
e
W ( e 3 x , e 3 x ) 
3e3 x
e 3 x
 6  0
3 x
 3e
for every x.
So y = c1y1 + c2y2 is the general solution.
CH3_19
Example 8
The functions y = 4 sinh 3x - 5e3x is a solution of
example 7 (Verify it). Observe
y  2e  2e
3x
3 x
 5e
3 x
 e3 x  e3 x 
3 x


 4
 5e

2


= 4 sinh 3x – 5e-3x
CH3_20
Example 9
The functions y1 = ex, y2 = e2x , y3 = e3x are solutions
of y’’’ – 6y” + 11y’ – 6y = 0 on (-, ).
Since
e x e2 x
e3 x
W (e x , e 2 x , e3 x )  e x
2e2 x
3e3 x  2e6 x  0
ex
4e2 x
9e3 x
for every real value of x.
So y = c1ex + c2 e2x + c3e3x is the general solution on
(-, ).
CH3_21
THEOREM 3.6
General Solution – Nonhomogeneous Equations
Any yp free of parameters satisfying (7) is called a
particular solution. If y1(x), y2(x), …, yn(x) be a
fundamental set of solutions of (6), then the general
solution of (7) is
y= c1y1 + c2y2 +… + cnyn + yp
(10)
Complementary Function: yc
y = c1y1 + c2y2 +… + cnyn + yp = yc + yp
= complementary + particular
CH3_22
Example 10
The function yp = -(11/12) – ½ x is a particular
solution of
y  6 y  11 y  6 y  3x
(11)
From previous discussions, the general solution of
(11) is
11 1
y  yc  y p  c1e  c2e  c3e   x
12 2
x
2x
3x
CH3_23
THEOREM 3.7
Given
an ( x) y ( n )  an1 ( x) y ( n1)    a1 ( x) y  a0 ( x) y (12)
 gi ( x)
where i = 1, 2, …, k.
If ypi denotes a particular solution corresponding to the
DE (12) with gi(x), then
y p  y p1 ( x)  y p2 ( x)    y pk ( x)
(13)
is a particular solution of
an ( x) y ( n )  an1 ( x) y ( n1)    a1 ( x) y  a0 ( x) y (14)
 g1 ( x)  g 2 ( x)    g k ( x)
CH3_24
Example 11
We find
yp1 = -4x2 is a particular solution of
y"3 y'4 y  16 x 2  24 x  8
yp2 = e2x is a particular solution of
y"3 y '4 y  2e2 x
yp3 = xex is a particular solution of
y"3 y'4 y  2 xe x  e x
From Theorem 3.7, y  y p1  y p2  y p3 is a solution of
2
2x
x
x
y  3 y  4 y  
16
x

24
x

8

2
e

2
xe

e
  

g1 ( x )
g2 ( x )
g3 ( x )
CH3_25
Note:
If ypi is a particular solution of (12), then
y p  c1 y p1  c2 y p2    ck y pk ,
is also a particular solution of (12) when the righthand member is
c1g1 ( x)  c2 g2 ( x)    ck gk ( x)
CH3_26
3.3 Homogeneous Linear Equation with Constant
Coefficients
Introduction:
(n)
( n 1)
an y  an1 y
   a2 y  a1 y  a0 y  0
where ai are constants, an  0.
Auxiliary Equation (Characteristic Equation):
For n = 2,
ay  by  cy  0
Try y = emx, then
emx (am2  bm  c)  0
am2  bm  c  0
is called an auxiliary equation or characteristic
equation.
(1)
(2)
(3)
CH3_27
From (3) the two roots are
m1  (b  b 2  4ac) / 2a
m2  (b  b 2  4ac ) / 2a
(1) b2 – 4ac > 0: two distinct real numbers.
(2) b2 – 4ac = 0: two equal real numbers.
(3) b2 – 4ac < 0: two conjugate complex numbers.
CH3_28
Case 1: Distinct real roots
The general solution is
y  c1e m1x  c2em2 x (why?)
Case 2: Repeated real roots
y1  e m1x,
(why?)
m1x
y2  xe
(4)
(5)
The general solution is
y  c1em1x  c2 xem1x
(why?)
(6)
CH3_29
Case 3: Conjugate complex roots
We write m1    i , m2    i , a general
solution is
y  C1e( i ) x  C2e( i ) x
From Euler’s formula:
ei  cos   i sin
eix  cos x  i sin x and eix  cos x  i sin x
eix  eix  2 cos x and eix  eix  2i sin x
(7)
CH3_30
( i ) x
( i ) x
y

C
e

C
e
Since
is a solution then set
1
2
C1 = C1 = 1 and C1 = 1, C2 = -1 , we have two solutions:
y1  ex (eix  eix )  2ex cos x
y2  ex (eix  eix )  2iex sin x
So, ex cos x and ex sin x are a fundamental set of
solutions, that is, the general solution is
y  c1ex cos x  c2ex sin x
(8)
x
 e (c1 cos x  c2 sin x)
CH3_31
Example 1
Solve the following DEs:
(a) 2 y"5 y '3 y  0
2
2m  5m  3  (2m  1)(m  3) , m1  1/2 , m2  3
y  c1e x2  c2e3 x
(b) y"10 y '25 y  0
m2  10m  25  (m  5)2 , m1  m2  5
y  c1e5 x  c2 xe5 x
(c) y"4 y '7 y  0
m2  4m  7  0 , m1  2  3i , m2  2  3i
  2 ,   3 , y  e2 x (c1 cos 3x  c2 sin 3x)
CH3_32
Example 2
Solve 4 y"4 y '17 y  0, y (0)  1, y ' (0)  2
Solution:
4m2  4m  17  0, m1  1/2  2i
y  e x / 2 (c1 cos 2 x  c2 sin 2 x)
y (0)  1, c1  1, and y ' (0)  2, c2  3/4
See Fig 3.4.
CH3_33
Fig 3.4
CH3_34
Higher-Order Equations
Given
an y ( n )  an1 y ( n1)    a2 y  a1 y  a0 y  0 (12)
we have
an mn  an1mn1    a2m2  a1m  a0  0
(13)
as an auxiliary equation (or characteristic equation).
CH3_35
Example 3
Solve y  3 y  4 y  0
Solution:
m  3m  4  (m  1)(m  4m  4)  (m  1)(m  2)
m2  m3  2
y  c1e x  c2e 2 x  c3 xe2 x
3
2
2
2
CH3_36
Example 4
Solve
d4y
d2y
2 2  y 0
4
dx
dx
Solution:
m4  2m2  1  (m2  1)2  0
m1  m3  i, m2  m4  i
y  C1e  C2e
ix
ix
 C3 xe  C4 xe
ix
ix
 c1 cos x  c2 sin x  c3 x cos x  c4 x sin x
CH3_37
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:
ex cos x , xex cos x , x 2ex cos x ,  , x k 1ex cos x
ex sin x , xex sin x , x 2ex sin x ,  , x k 1ex sin x
CH3_38
3.4 Undetermined Coefficients
Introduction
If we want to solve the nonhomogeneous linear DE
an y ( n )  an1 y ( n1)    a1 y  a0 y  g ( x)
(1)
we have to find y = yc + yp. Thus we introduce the
method of undetermined coefficients to solve for yp.
CH3_39
Example 1
2
y
"

4
y
'

2
y

2
x
 3x  6
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
y p  Ax 2  Bx  C , y p '  2 Ax  B, y p " 2 A
After substitution,
2A + 8Ax + 4B – 2Ax2 – 2Bx – 2C = 2x2 – 3x + 6
CH3_40
Example 1 (2)
Then
 2 A  2 , 8 A  2 B  3 , 2 A  4 B  2C  6
A  1, B  5/2, C  9
5
2
yp  x  x  9
2
CH3_41
Example 2
Find a particular solution of
y" y ' y  2 sin 3x
Solution:
Let yp = A cos 3x + B sin 3x
After substitution,
(8 A  3B) cos 3x  (3 A  8B) sin 3x  2 sin 3x
Then
A  6/73, B  16/73
6
16
y p  cos 3 x  sin 3 x
73
73
CH3_42
Example 3
y"2 y '3 y  4 x  5  6 xe2 x
Solve
Solution:
We can find
yc  c1e  x  c2e3 x
Let y p  Ax  B  Cxe2 x  Ee2 x
After substitution,
2x
2x
 3 Ax  2 A  3B  3Cxe  (2C  3E )e
 4 x  5  6 xe
Then
(3)
2x
A  4/3, B  23/9, C  2, E  4/3
4
23
4 2x
2x
y p   x   2 xe  e
3
9
3
4
23 
4  2x
x
3x
y  c1e  c2e  x    2 x   e
3
9 
3
CH3_43
Example 4
Find yp of y"5 y '4 y  8e x
Solution:
First let yp = Aex
After substitution, 0 = 8ex, (wrong guess)
Let yp = Axex
After substitution, -3Aex = 8ex
Then A = -8/3, yp = (−8/3)xex
CH3_44
Rule of Case 1:
No function in the assumed yp is part of yc
Table 3.1 shows the trial particular solutions.
g (x )
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
1 (any constant)
5x  7
3x 2  2
x3  x  1
sin 4 x
cos 4 x
e5 x
(9 x  2)e5 x
x 2 e5 x
e3 x sin 4 x
5 x 2 sin 4 x
xe3 x cos 4 x
Form of
yp
A
Ax  B
Ax 2  Bx  C
Ax3  Bx 2  Cx  E
A cos 4 x  B sin 4 x
A cos 4 x  B sin 4 x
Ae5 x
( Ax  B)e5 x
( Ax 2  Bx  C )e5 x
Ae3 x cos 4 x  Be3 x sin 4 x
( Ax 2  Bx  C ) cos 4 x  ( Ex 2  Fx  G) sin 4 x
CH3_45
( Ax  B)e3 x cos 4 x  (Cx  E )e3 x sin 4 x
Example 5
Find the form of yp of
(a) y"8 y '25 y  5 x3e x  7e x
Solution:
We have g ( x)  (5 x3  7)e x and try
y p  ( Ax3  Bx 2  Cx  E )e x
There is no duplication between yp and yc .
(b) y” + 4y = x cos x
Solution:
We try x p  ( Ax  B) cos x  (Cx  E ) sin x
There is also no duplication between yp and yc .
CH3_46
Example 6
Find the form of yp of
y  9 y  14 y  3x 2  5 sin 2 x  7 xe6 x
Solution:
2
2
y

Ax
 Bx  C
For 3x :
p1
For -5 sin 2x: y p2  E cos 2 x  F sin 2 x
For
7xe6x:
y p3  (Gx  H )e6 x
No term in y p  y p1  y p2  y p3 duplicates a term in yc
CH3_47
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.
CH3_48
Example 8
Solve y" y  4 x  10 sin x, y ( )  0, y ' ( )  2
Solution:
yc  c1 cos x  c2 sin x
First trial: yp = Ax + B + C cos x + E sin x
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
(5)
CH3_49
Example 9
2
3x
y
"

6
y
'

9
y

6
x

2

12
e
Solve
Solution:
yc = c1e3x + c2xe3x
2
2 3x
yp  
Ax

Bx

C

Ex
 
 e
y p1
y p2
After substitution and simplification,
A = 2/3, B = 8/9, C = 2/3, E = -6
Then
2 2 8
2
3x
3x
y  c1e  c2 xe  x  x   6 x 2e3 x
3
9
3
CH3_50
Example 10
x



y

y
"
e
cos x
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
1 x
1 x
x
y  yc  y p  c1  c2 x  c3e  e cos x  e sin x
10
5
CH3_51
Example 11
Find the form of yp of
y ( 4)  y  1  x 2e x
Solution:
yc = c1+ c2x + c3x2 + c4e-x
2 x
x
x
Normal trial: y p  
A 
Bx
e Cxe

Ee

y p1
y p2
Cxe-x +
Multiply A by x3 and (Bx2e-x +
Ee-x) by x
Then
yp = Ax3 + Bx3e-x + Cx2e-x + Exe-x
CH3_52
3.5 Variation of Parameters
Some Assumptions
For the DE
a2 ( x) y  a1 ( x) y  a0 ( x) y  g ( x)
we put (1) in the form
y  P( x) y  Q( x) y  f ( x)
(1)
(2)
where P, Q, f are continuous on I.
Let y1 ( x) and y2 ( x) be two linearly independent
solutions of the associated homogeneous equation of
(2).
CH3_53
Method of Variation of Parameters
We try
y p  u1 ( x) y1 ( x)  u2 ( x) y2 ( x)
(3)
After we obtain yp’, yp”, we put them into (2), then
yp  P( x) yp  Q( x) y p
 u1[ y1  Py1  Qy1 ]  u2[ y2  Py2  Qy2 ]
 y1u1  u1 y1  y2u2  u2 y2  P[ y1u1  y2u2 ]  y1u1  y2 u2
d
d
 [ y1u1 ]  [ y2u2 ]  P[ y1u1  y2u2 ]  y1u1  y2 u2
dx
dx
d
 [ y1u1  y2u2 ]  P[ y1u1  y2u2 ]  y1u1  y2 u2  f ( x) (4)
dx
CH3_54
Making further assumptions:
y1u1’ + y2u2’ = 0, then from (4),
y1’u1’ + y2’u2’ = f(x)
Express the above in terms of determinants
W1
y2 f ( x )
W
y
f
(
x
)
2
1

and
u1 

u 2 

W
W
W
W
where
y1 y2
0
y2
y1
0
W
, W1 
, W2 
y1 y2
f ( x)1 y2
y1 f ( x)
(5)
(6)
CH3_55
Example 1
Solve y"4 y'4 y  ( x  1)e2 x
Solution:
m2 – 4m + 4 = 0, m = 2, 2
y1 = e2x, y2 = xe2x,
W (e 2 x
2x
e
, xe2 x ) 
2e 2 x
xe2 x
4x

e
0
2x
2x
2 xe  e
Since f(x) = (x + 1)e2x, then
0
W1 
( x  1)e 2 x
xe2 x
e2 x
4x
 ( x  1) xe , W2  2 x
2x
2 xe
2e
0
4x

(
x

1
)
e
( x  1)e 2 x
CH3_56
Example 1 (2)
From (5),
4x
( x  1) xe4 x
(
x

1
)
e
2
u1  


x
 x , u2 
 x 1
4x
4x
e
e
Then
u1 = (-1/3)x3 – ½ x2, u2 = ½ x2 + x
And
1 3 1 2  2x  1 2
1 3 2x 1 2 2x


2x
x p    x  x e   x  x  xe  x e  x e
2 
6
2
 3
2

1 3 2x 1 2 2x
2x
2x
y  yc  y p  c1e  c2 xe  x e  x e
6
2
CH3_57
Example 2
Solve 4 y"36 y  csc 3 x
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
cos 3x
sin 3x
W (cos 3x , sin 3x) 
3
 3sin 3x 3 cos 3x
0
sin 3x
cos 3x
0
1
1 cos 3x
W1 
  , W2 

4
1/4 csc 3x 3 cos 3x
 3sin 3x 1/4 csc 3x 4 sin 3x
CH3_58
Example 2 (2)
W1
1
u1 

W
12
W2 1 cos 3 x
u2 

W 12 sin 3 x
Then
u1  1/12 x, u2  1/36 ln | sin 3x |
And
1
1
y p   x cos 3 x  (sin 3 x) ln | sin 3 x |
12
36
1
1
y  yc  y p  c1 cos 3 x  c2 sin 3 x  x cos 3 x  (sin 3 x) ln | sin 3 x |
12
36
CH3_59
Example 3
1
Solve y" y 
x
Solution:
m2 – 1 = 0, m = 1, -1
y1 = ex, y2 = e-x, f = 1/x, and W(ex, e-x) = -2
Then
e x (1/ x)
1 x et
u1  
, u1  
dt
x
0
2
2
t
e x (1/ x)
1 x et
u2  
, u2   
dt
x
0
2
2 t
The low and up bounds of the integral are x0 and x,
respectively.
CH3_60
Example 3 (2)
1 x x et
1  x x et
yp  e 
dt  e 
dt
x
x
0 t
0 t
2
2
1
y  yc  y p  c1e  e
2
x
t
e
1
x0 t dt  2 e
x x
t
e
x0 t dt
x x
CH3_61
Higher-Order Equations
For the DEs of the form
y ( n )  Pn1 ( x) y ( n1)    P1 ( x) y  P0 ( x) y  f ( x)
(8)
then yp = u1y1 + u2y2 + … + unyn, where yi , i = 1, 2, …,
n, are the elements of yc. Thus we have
y1u1  y2u2    ynun  0
y1u1  y2 u2    yn un  0


( n 1)
( n 1)
( n 1)


y1 u1  y2 u2    yn un  f ( x)
(9)
and uk’ = Wk/W, k = 1, 2, …, n.
CH3_62
For the case n = 3,
W1
W2
W3
u1  , u2  , u3 
W
W
W
(10)
CH3_63
3.8 Linear Models: IVP
Newton’s Law
See Fig 3.18, we have
d 2x
dx
m 2  k (s  x)  mg    f (t )
dt
dt
dx
dx
 kx  mg  ks    f (t )  kx    f (t )


 dt
dt
zero
(1)
CH3_64
Fig 3.18
CH3_65
Fig3.19
CH3_66
Free Undamped Motion
From (1), if   0 and f(t)=0, we have
d 2x
2
 x  0
2
dt
(2)
where  = k/m. (2) is called a simple harmonic
motion, or free undamped motion.
CH3_67
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.
CH3_68
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
2
1 d 2x
d
x
 4 x,
 64 x  0
2
2
16 dt
dt
CH3_69
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
(4)
(5)
CH3_70
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   
c1
A
 tan  
c2
c2 
cos 
A
A sin t cos   A cos t sin 
 ( A sin  ) cos t  ( A cos  ) sin t
c1
c2
A cos t  A sin t  c1 cos t  c2 sin t  x(t )
A
A
(6)
(7)
(8)
(9)
CH3_71
Fig 3.20
CH3_72
Example 2
Solution (5) is
x(t) = (2/3) cos 8t − (1/6) sin 8t = A sin(t + )
Then
A  ( 2 3 )2  ( 16 )2  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)
(9)
6
The period is T = 2/8 = /4.
CH3_73
Fig 3.21
Fig 3.21 shows the motion.
CH3_74
Free Damped Motion
If   0 and f(t)=0, the DE is as
d 2x
dx
m 2  kx  
(10)
dt
dt
where  is a positive damping constant.
Then x”(t) + (/m)x’ + (k/m)x = 0 can be written as
2
d x
dx
2

2



x0
2
(11)
dt
dt
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
CH3_75
Case 1:
 2 – 2 > 0. Let h  2   2 , then
x(t )  e
 t
(c1e
2  2t
 c2e
 2  2t
)
(13)
It is said to be overdamped. See Fig 3.23.
CH3_76
Fig3.23
CH3_77
Case 2:
 2 – 2 = 0. then
x(t )  et (c1  c2t )
(14)
It is said to be critically damped. See Fig 3.24.
CH3_78
Fig3.24
CH3_79
Case 3:
 2 – 2 < 0. Let h  2   2 , then
m1       i ,
2
x(t )  e
 t
2
m2       i
2
2
(c1 cos    t  c2 sin    t ) (15)
2
2
2
2
It is said to be underdamped. See Fig 3.25.
CH3_80
Fig 3.25
CH3_81
Example 3
The solution of
2
d x
dx
 5  4 x  0 , x(0)  1 ,
2
dt
dt
5 t 2 4t
x(t )  e  e
is
3
3
x(0)  1
(16)
See Fig 3.26.
CH3_82
Fig 3.26
CH3_83
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
1 d 2x
dx d 2 x
dx
 4 x  2 ,
 8  16 x  0
2
2
4 dt
dt dt
dt
(17)
CH3_84
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 3.27.
(18)
(19)
CH3_85
Fig 3.27
CH3_86
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
1 d 2x
dx d 2 x
dx
 5 x  ,
 2  10 x  0
2
2
(20)
2 dt
dt dt
dt
m2 + 2m + 10 = 0, m = −3 + 3i, −3 − 3i
CH3_87
Example 5 (2)
t
x(t )  e (c1 cos 3t  c2 sin 3t )
Initial conditions: x(0) = −2, x’(0) = 0, then
2
t 
x(t )  e   2 cos 3t  sin 3t 
3


(21)
(22)
CH3_88
Alternate form of x(t)
(22) can be written as
x(t )  Ae t sin(  2  2 t   )
where A 
c12
 c22 ,
(23)
c1
and tan  
c2
CH3_89
LRC-Series Circuits
The following equation is the DE of forced motion
with damping:
d 2x
dx
m 2   kx  f (t )
(32)
dt
dt
If i(t) denotes the current shown in Fig 3.32, then
di
L  Ri  vC  E (t )
dt
(33)
Since i = C dvdt , we have
C
d 2vC
dvC
LC 2  RC
 vC  E (t )
dt
dt
(34)
CH3_90
Fig 3.32
CH3_91
Forced Undamped and Damped
If R=0, (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?
CH3_92