CHAPTER 3
Higher-Order Differential
Equations
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
CH3_2
3.1 Preliminary Theory: Linear Equ.
Initial-value Problem
An nth-order initial problem is
Solve:
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
Subject to:
y ( x0 )  y0 , y( x0 )  y1 ,  , y ( n1) ( x0 )  yn1
(1)
with 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 any 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
Example 2
Please verify y = 3e2x + e–2x – 3x, is a solution of
y"4 y  12 x, y (0)  4, y ' (0)  1
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.
CH3_6
Boundary-Value Problem
d2y
dy
Solve: a2 ( x) 2  a1 ( x)  a0 ( x) y  g ( x)
dx
dx
Subject to: y (a)  y0 , y (b)  y1
is called a boundary-value problem (BVP).
See Fig 3.1.
CH3_7
Fig 3.1
CH3_8
Example 3
 In example 4 of Sec 1.1, we see the solution of
x"16 x  0 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


x  16 x  0 , x(0)  0 , x   0
(3)
2
has infinite many solutions. See Fig 3.2.


(b) If
x  16 x  0 , x(0)  0 , x   0
(4)
8
we have c1 = 0, c2 = 0, x = 0 is the only solution.
CH3_9
Example 3 (2)


x  16 x  0 , x(0)  0 , x   1
2
we have c1 = 0, and 1 = 0 (contradiction).
Hence (5) has no solutions.
(c) If
(5)
CH3_10
Fig 3.2
CH3_11
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 said to be 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) not identically zero, is nonhomogeneous.
CH3_12
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 DEs as
L(y) = 0 and L(y) = g(x)
CH3_13
THEOREM 3.2
Superposition Principles – Homogeneous Equations
Let y1, y2, …, yk be a solutions of the homogeneous
Nth-order differential equation (6) on an interval I.
Then the linear combination
y = c1y1(x) + c2y2(x) + …+ ckyk(x)
where the ci, i = 1, 2, …, k are arbitrary constants, is
also a solution on the interval.
CH3_14
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_15
Example 4
The function y1 = x2, y2 = x2 ln x are both solutions of
x3 y  2 xy  4 y  0
Then y = x2 + x2 ln x is also a solution on (0, ).
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_16
In other words, if the set is linearly independent,
when c1f1(x) + c2f2(x) + … + cn fn(x) = 0
then c1 = c2 = … = cn = 0
Referring to Fig 3.3, neither function is a constant
multiple of the other, then these two functions are
linearly independent.
CH3_17
Fig 3.3
CH3_18
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_19
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_20
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_21
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.
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_22
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_23
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_24
Example 8
The functions y = 4 sinh 3x - 5e3x is a solution of
example 7 (Verify it). Observer
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_25
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 2e 2 x 3e3 x  2e6 x  0
e x 4e 2 x 9e 3 x
for every real value of x.
So y = c1ex + c2 e2x + c3e3x is the general solution on
(-, ).
CH3_26
THEOREM 3.6
General Solution – Nonhomogeneous Equations
Any yp free of parameters satisfying (7) is called a
particular solution. If y1(x), y2(x), …, yk(x) be a
fundamental set of solutions of (6), then the general
solution of (7) is
y= c1y1 + c2y2 +… + ckyk + yp
(10)
Complementary Function
y = c1y1 + c2y2 +… + ckyk + yp = yc + yp
= complementary + particular
CH3_27
Example 10
The function yp = -(11/12) – ½ x is a particular
solution of
y  6 y  11y  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_28
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_29
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_30
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
c1 g1 ( x)  c2 g 2 ( x)    ck g k ( x)
CH3_31
3.2 Reduction of Order
Introduction:
We know the general solution of
(1)
a2 ( x) y  a1 ( x) y  a0 ( x) y  0
is y = c1y1 + c2y1.
Suppose y1(x) denotes a known solution of (1). We
assume the other solution y2 has the form y2 = uy1.
Our goal is to find a u(x) and this method is called
reduction of order.
CH3_32
Example 1
Given y1 = ex is a solution of y” – y = 0, find a second
solution y2 by the method of reduction of order.
Solution
If y = uex, then
x
x
x
x
x





y  ue  e u , y  ue  2e u  e e
And
y" y  e x (u"2u ' )  0
Since ex  0, we let w = u’, then
w  c1e2 x  u
u  1/2 c1e2 x  c2
CH3_33
Example 1 (2)
Thus
c1  x
x
y  u ( x)e   e  c2e
2
x
(2)
Choosing c1 = 0, c2 = -2, we have y2 = e-x.
Because W(ex, e-x)  0 for every x, they are independent.
CH3_34
General Case
Rewrite (1) as the standard form
(3)
y  P( x) y  Q( x) y  0
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
y  uy1  y1u , y  uy1  2 y1u  y1u
y  Py  Qy
 u[ y1  Py1  Qy1 ]  y1u  (2 y1  Py1 )u  0



zero
CH3_35
This implies that
y1u  (2 y1  Py1 )u  0
or
y1w  (2 y1  Py1 )2w  0
where we let w = u’.
Solving (4), we have
dw
y1
 2 dx  Pdx  0
w
y1
ln | wy12 |   Pdx  c or
(4)
wy12  c1e Pdx
CH3_36
then
e Pdx
u  c1  2 dx  c2
y1
Let c1 = 1, c2 = 0, we find
e P ( x ) dx
y2  y1 ( x)  2
dx
y1 ( x)
(5)
CH3_37
Example 2
The function y1= x2 is a solution of
x 2 y"3xy'4 y  0
Find the general solution on (0, ).
Solution:
The standard form is
3
4
y  y  2  0
x
x
3  dx / x
From (5)
2 e
y2  x  4 dx  x 2 ln x
x
The general solution is
y  c1x 2  c2 x 2 ln x
CH3_38
3.3 Homogeneous Linear Equation with Constant
Coefficients
Introduction:
an y ( n )  an1 y ( n1)    a2 y  a1 y  a0 y  0
where ai are constants, an  0.
Auxiliary 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.
(1)
(2)
(3)
CH3_39
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_40
Case 1: Distinct real roots
The general solution is
y  c1e m1x  c2em2 x
Case 2: Repeated real roots
y1  em1x and from (5) of Sec 3.2,
2 m1x
e
y2  em1x  2 m1x dx  em1x  dx  xe m1x
e
The general solution is
y  c1em1x  c2 xem1x
(4)
(5)
(6)
CH3_41
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_42
( 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_43
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_44
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_45
Fig 3.4
CH3_46
Two Equations worth Knowing
2
2




y

k
y

0
,
y

k
y  0, k  0

For the first equation:
y  c1 cos kx  c2 sin kx
For the second equation:
y  c1ekx  c2ekx
Let
y1  1/2(ekx  ekx )  cosh kx
Then
(9)
(10)
y2  1/2(ekx  ekx )  sinh kx
y  c1 cosh kx  c2 sinh kx
(11)
CH3_47
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.
CH3_48
Example 3
Solve y  3 y  4 y  0
Solution:
m3  3m2  4  (m  1)(m2  4m  4)  (m  1)(m  2)2
m2  m3  2
y  c1e x  c2e2 x  c3 xe 2 x
CH3_49
Example 4
4
Solve
2
d y
d y
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_50
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_51
3.4 Undetermined Coefficients
Introduction
If we want to solve
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.
CH3_52
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 , y p '  2 Ax  B, y p " 2 A
After substitution,
2A + 8Ax + 4B – 2Ax2 – 2Bx – 2C = 2x2 – 3x + 6
CH3_53
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_54
Example 2
Find a particular solution of
y" y ' y  2 sin 3 x
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 3x  sin 3x
73
73
CH3_55
Example 3
y"2 y'3 y  4 x  5  6 xe2 x
Solve
Solution:
x
3x
We can find
yc  c1e  c2e
Let y p  Ax  B  Cxe 2 x  Ee2 x
After substitution,
2x
2x
 3 Ax  2 A  3B  3Cxe  (2C  3E )e
(3)
 4 x  5  6 xe 2 x
Then
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_56
Example 4
Find yp of y"5 y'4 y  8e x
Solution:
First let yp = Ae2x
After substitution, 0 = 8e2x, (wrong guess)
Let yp = Axe2x
After substitution, -3Ae2x = 8e2x
Then A = -8/3, yp = (−8/3)xe2x
CH3_57
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 2e5 x
e3 x sin 4 x
5 x 2 sin 4 x
xe3 x cos 4 x
Form of
yp
A
Ax  B
Ax2  Bx  C
Ax3  Bx2  Cx  E
A cos 4 x  B sin 4 x
A cos 4 x  B sin 4 x
Ae5 x
( Ax  B)e5 x
( Ax2  Bx  C )e5 x
Ae3 x cos 4 x  Be3 x sin 4 x
( Ax2  Bx  C ) cos 4 x  ( Ex2  Fx  G) sin 4 x
CH3_58
( 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_59
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
p1
For 3x :
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_60
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_61
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_62
Example 9
2
3x
y
"

6
y
'

9
y

6
x

2

12
e
Solve
Solution:
yc = c1e3x + c2xe3x
2
3x
yp  
Ax

Bx

C

Ee

 
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
2 3x
y  c1e  c2 xe  x  x   6 x e
3
9
3
CH3_63
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_64
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_65
3.5 Variation of Parameters
Some Assumptions
For the DE
a2 ( x) y  a1 ( x) y  a0 ( x) y  g ( x)
(1)
we put (1) in the form
y  P( x) y  Q( x) y  f ( x)
(2)
where P, Q, f are continuous on I.
CH3_66
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_67
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 )
W2
y1 f ( x)

and
u1 

u2 

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_68
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
, xe 2 x ) 
2e 2 x
xe 2 x
4x

e
0
2x
2x
2 xe  e
Since f(x) = (x + 1)e2x, then
0
W1 
( x  1)e 2 x
xe 2 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_69
Example 1 (2)
From (5),
4x
( x  1) xe 4 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_70
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 3 x
sin 3 x
W (cos 3 x , sin 3 x) 
3
 3 sin 3 x 3 cos 3 x
0
sin 3 x
cos 3 x
0
1
1 cos 3 x
W1 
  , W2 

4
4 sin 3 x
1/4 csc 3 x 3 cos 3 x
 3 sin 3 x 1/4 csc 3 x
CH3_71
Example 2 (2)
W1
1
u1 

W
12
W2 1 cos 3x

u2 

W 12 sin 3x
Then
u1  1/12 x, u2  1/36 ln | sin 3x |
And
1
1
y p   x cos 3x  (sin 3x) ln | sin 3x |
12
36
1
1
y  yc  y p  c1 cos 3x  c2 sin 3x  x cos 3x  (sin 3x) ln | sin 3x |
12
36
CH3_72
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
2
2 x0 t
The low and up bounds of the integral are x0 and x,
respectively.
CH3_73
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_74
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_75
For the case n = 3,
W1
W2
W3
u1  , u2  , u3 
W
W
W
(10)
CH3_76
3.6 Cauchy-Eulaer Equation
Form of Cauchy-Euler Equation
n
n1
d
y
d
y
dy
n
n1
an x
 an1x
   a1x  a0 y  g ( x)
n
n1
dx
dx
dx
Method of Solution
We try y = xm, since
k
d y
k
mk
ak x

a
x
m
(
m

1
)(
m

2
)

(
m

k

1
)
x
k
k
dx
 ak m(m  1)(m  2)(m  k  1) x m
k
CH3_77
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
y  c1x m1  c2 x m2
(2)
CH3_78
Example 1
2
d
y
dy
2
 2x  4 y  0
Solve x
2
dx
dx
Solution:
We have a = 1, b = -2 , c = -4
m2 – 3m – 4 = 0, m = -1, 4,
y = c1x-1 + c2x4
CH3_79
Case 2: Repeated Real Roots
m1
y

x
ln x
Using (5) of Sec 3.2, we have 2
Then
y  c1x m1  c2 x m1 ln x
(3)
CH3_80
Example 2
2
d
y
dy
2
 8x  y  0
Solve 4 x
2
dx
dx
Solution:
We have a = 4, b = 8, c = 1
4m2 + 4m + 1 = 0, m = -½ , -½
y  c1x
1/ 2
 c2 x
1/ 2
ln x
CH3_81
Case 3: Conjugate Complex Roots
Higher-Order: multiplicity
x m1 , x m1 , x m1 (ln x)2 ,  , x m1 (ln x)k 1
Case 3: Conjugate Complex Roots
m1 =  + i, m2 =  – i,
y = C1x( + i) + C2x( - i)
Since
xi = (eln x)i = ei ln x = cos( ln x) + i sin( ln x)
x-i = cos ( ln x) – i sin ( ln x)
Then
y = c1x cos( ln x) + c2x sin( ln x)
= x [c1 cos( ln x) + c2 sin( ln x)]
(4)
CH3_82
Example 3
1
Solve 4 x y  17 y  0, y (1)  1, y ' (1)  
2
Solution:
We have a = 4, b = 0 , c = 17
4m2 − 4m + 17 = 0, m = ½ + 2i
1/ 2
y  x [c1 cos( 2 ln x)  c2 sin(2 ln x)]
Apply y(1) = -1, y’(1) = 0, then c1 = -1, c2 = 0,
y   x1/2 cos( 2 ln x)
See Fig 3.15.
2
CH3_83
Fig 3.15
CH3_84
Example 4
3
2
d
y
d
y
dy
3
2
 5x
 7x  8y  0
Solve x
3
2
dx
dx
dx
Solution:
Let y = xm,
2
dy
d
y
m 1
 mx , 2  m(m  1) x m2 ,
dx
dx
d3y
m 3

m
(
m

1
)(
m

2
)
x
dx3
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)
CH3_85
Example 5
Solve x 2 y"3xy'3 y  2 x 4e x
Solution:
We have (m – 1)(m – 3) = 0, m = 1, 3
yc = c1x + c2x3 , use variation of parameters,
yp = u1y1 + u2y2, where y1 = x, y2 = x3
Rewrite the DE as
3
3
2 x



y  y  2 y  2x e
x
x
Then P = -3/x, Q = 3/x2, f = 2x2ex
CH3_86
Example 5 (2)
Thus
x x3
3
W

2
x
,
2
1 3x
x
0
0
x3
5 x
3 x
W1 


2
x
e
,
W


2
x
e
2
2 x
2 x
2
1 2x e
2 x e 3x
We find
5 x
2 x 5e x
2
x
e
2 x
x

u1  


x
e
,
u


e
2
3
2x
2 x3
x
2 x
x
x
u   x e  2 xe  2e , u2  e
1
CH3_87
Example 5 (3)
Then
y p  u1 y1  u2 y2  ( x e  2 xe  2e ) x  e x
2 x
 2 x e  2 xe
2 x
x
x
x 3
x
y  yc  y p  c1x  c2 x3  2 x 2e x  2 xe x
CH3_88
3.7 Nonlinear Equations
2
y
"

2
x
(
y
'
)
Example 1 Solve
Solution:
This nonlinear equation misses y term. Let u(x) = y’,
then du/dx = y”,
du
2 or du
 2 x dx
 2xu
2
u
dx
1
2
2 (This form is just for convenience)
 u  x  c1
dy
1
 2 2
Since
= 1/y’,
dx
x  c1
dx
1
1 x
y    2 2   tan
 c2
So,
c1
c1
x  c1
u-1
CH3_89
Example 2
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 dy
 du 
2

y u   u or
u
y
 dy 
c1
ln|u| = ln|y| + c1, u = c2y (where c2  e )
Since u = dy/dx = c2y, dy/y = c2 dx
ln|y| = c2x + c3,
y  c4ec2 x
CH3_90
Example 3
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:
(2)
y ( x)
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
y( x)  ( x  y  y 2 )  1  y  2 yy (3)
dx
CH3_91
Example 3 (2)
d
y ( x)  (1  y  2 yy)  y  2 yy  2( y)2 (4)
dx
( 4)
d
y ( x)  ( y  2 yy  2( y)2 )  y  2 yy  6 yy (5)
dx
( 5)
and so on. So we can use the same method to obtain
y(3)(0) = 4, y(4)(0) = −8, ……
Then
2 3 1 4 1 5
2
y ( x)  1  x  x  x  x  x  
3
3
5
CH3_92
Example 4
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 3.16 shows the graph of
this DE. For comparison, the curve of fifth-degree
Taylor series is also shown.
CH3_93
Fig 3.16
CH3_94
3.8 Linear Models: IVP
Newton’s Law
See Fig 3.18, we have
d2 x
m 2  k ( s  x)  mg  kx  mg  ks  kx



dt
(1)
zero
CH3_95
Fig 3.18
CH3_96
Fig3.19
CH3_97
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.
CH3_98
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_99
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_100
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_101
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 
(6)
(7)
(8)
 ( A sin  ) cos t  ( A cos  ) sin t
c1
c2
A cos t  A sin t  c1 cos t  c2 sin t  x(t ) (9)
A
A
CH3_102
Fig 3.20
CH3_103
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_104
Fig 3.21
Fig 3.21 shows the motion.
CH3_105
Free Damped Motion
If 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_106
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_107
Fig3.23
CH3_108
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_109
Fig3.24
CH3_110
Case 3:
 2 – 2 < 0. Let h  2   2 , then
m1     2  2 i ,
x(t )  e
 t
m2     2  2 i
(c1 cos    t  c2 sin    t ) (15)
2
2
2
2
It is said to be underdamped. See Fig 3.25.
CH3_111
Fig 3.25
CH3_112
Example 3
The solution of
2
d x
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 3.26.
CH3_113
Fig 3.26
CH3_114
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_115
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_116
Fig 3.27
CH3_117
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_118
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_119
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_120
DE of Driven Motion with Damping
As in Fig 3.28,
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
CH3_121
Fig 3.28
CH3_122
Example 6
Interpret and solve
1 d 2x
dx
1
 1.2  2 x  5 cos 4t , x(0)  , x(0)  0 (26)
2
5 dt
dt
2
Solution:
Interpret:
Sol:
m = 1/5, k = 2,  = 1.2, f(t) = 5 cos 4t
release from rest at a point ½ below
dx2
dx
 6  10 x  0
2
dt
dt
xc (t )  e3t (c1 cos t  c2 sin t )
CH3_123
Example 6 (2)
Assuming xp(t) = A cos 4t + B sin 4t,
we have A = −25/102, B = 50/51, then
25
50
3t
x(t )  e (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,
x(t )
(28)
86
25
50

3t  38
 e  cos t  sin t  
cos 4t  sin 4t
51
51
 51
 102
CH3_124
Transient and Steady-State
Graph of (28) is shown in Fig 3.29.
xc(t) will vanish at t  : transient term
xp(t) will still remain at t  : steady-state term
CH3_125
Fig 3.29
CH3_126
Example 7
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 3.30.
CH3_127
Fig 3.30
CH3_128
Example 8
Solve
d 2x
2


x  F0 sin  t , x(0)  0 , x(0)  0
2
dt
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),
F0
x p (t )  2
sin  t
2
 
CH3_129
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
Thus
2
F0
x(t ) 
( sin t   sin  t ) ,
2
2
 (   )
 
(30)
CH3_130
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
(31)
2
2
2
CH3_131
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.
CH3_132
Fig 3.31
CH3_133
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
1
L  Ri  q  E (t )
(33)
dt
C
Since i = dq/dt, we have
2
d q
dq 1
L 2  R  q  E (t )
dt C
dt
(34)
CH3_134
Fig 3.32
CH3_135
Example 9
Find q(t) in Fig 3.32, where L = 0.025 henry, R = 10
ohms, C = 0.001 farad, E(t) = 0, q(0) = q0 coulombs,
and i(0) = 0 ampere.
Solution:
Using the given data:
1
q  10q  1000q  0, q  40q  4000q  0
4
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 20t
q(t ) 
e sin( 60t  1.249)
3
CH3_136
Example 10
Find the steady-state qp(t) and the steady-state current,
when E(t) = E0 sin t .
Solution:
Let qp(t) = A sin t + B cos t,
1 

E0  L 

C 

A
,
   L2 2  2CL  21 2  R 2 
C 


E0 R
B
   L2 2  2CL  21 2  R 2 
C 


CH3_137
Example 10 (2)
2L
1
X L 
 2 2
If
C C
2L
1
2
2 2
2
2
2
Z

L




R
If Z  X  R ,
C C 2 2
Using the similar method, we have
2
2
A  E0 X /(  Z ), B  E0 R /(  Z )
So
E0 X
E0 R
q p (t )   2 sin  t  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.
CH3_138
1
X  L 
,
C
2
2 2
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
(1)
dx
In addition, M(x) is proportional to the curvature  of
the elastic curve
M(x) = EI
(2)
where E, I are constants.
CH3_139
From calculus, we have   y”, when the deflection
y(x) is small. Finally we have
d 2M
d2
d4y
 EI 2 y  EI 4
2
(3)
dx
dx
dx
Then
d4y
EI  4  w( x)
dx
(4)
CH3_140
Terminology
Ends of the beam
embedded
free
simply supported (hinged)
Boundary Conditions
y = 0, y’ = 0
y” = 0, y’’’ = 0
y = 0, y” = 0
See Fig 3.41
CH3_141
Fig 3.41
CH3_142
Example 1
A beam of length L is embedded at both ends. Find the
deflection of the beam if a constant load w0 is uniformly
distributed aling its length, that is,
w(x)= w0 , 0 < x < L
Solution:
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
CH3_143
Example 1 (2)
So
w0 4
y ( x)  c1  c2 x  c3 x  c4 x 
x
24 EI
2
3
Using the boundary conditions, we have
c1 = 0, c2 = 0, c3 = w0L2/24EI, c4 = −w0L/12EI
w0 L2 2 w0 L 3
w0 4
w0 2
y ( x) 
x 
x 
x 
x ( x  L) 2
24 EI
12 EI
24 EI
24 EI
Choosing w0 = 24EI and L = 1, we have Fig 3.42.
CH3_144
Fig 3.42
CH3_145
Example 2
Solve y" y  0, y (0)  0, y ( L)  0
Solution:
Case 1 :  = 0
y = c1x + c2,
y(0) = c2 = 0, y(L) = c1L = 0, c1 = 0
then y = 0, trivial solution.
Case 2 :  < 0,  = −2,  > 0
Choose
y = c1 cosh x + c2 sinh x
y(0) = 0, c1 = 0; y(L) = 0, c2 = 0
then y = 0, trivial solution.
CH3_146
Example 2 (2)
Case 3 :  > 0,  = 2,  > 0
Choose
y = c1 cos x + c2 sin x
y(0) = 0, c1 = 0; y(L) = 0, c2 sin L= 0
If c2 = 0,
y = 0, trivial solution.
So c2  0, sin L = 0, L = n,  = n/L
2 2
n

2
   n  2 , n  1, 2, 3, 
L
Thus, y = c2 sin (nx/L) is a solution for each n.
CH3_147
Example 2 (3)
Simply take c2 = 1, for each:
2
2
2

4
9
,
,
,
2
2
2
L
L
L
the corresponding function:

2
3
sin x , sin
x , sin x , 
L
L
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.
CH3_148
Bulking of a Thin Vertical Column
Referring to Fig 3.43, the DE is
2
d y
EI 2   Py ,
dx
2
d y
EI 2  Py  0
dx
(5)
where P is a constant vertical compressive force
applied to the column’s top.
CH3_149
Fig 3.43
CH3_150
Example 3
Referring to Fig 3.43, when the column is hinged at
both ends, find the deflection.
Solution:
The boundary-value problem is
2
d y
EI 2  Py  0 , y (0)  0 , y ( L)  0
dx
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?
CH3_151
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 3, the deflection
curves are yn = c2 sin (nx/L), corresponding to
eigenvalues n = Pn/EI = n22/L2, n = 1, 2, 3, …
Physically, only Pn = EIn22/L2, deflection occurs.
We call these Pn the critical loads and the smallest P
= P1 = EI2/L2 is called the Euler load, and
y1 = c2 sin(x/L) is known as the first buckling mode.
See Fig 3.44
CH3_152
Fig 3.44
CH3_153
Rotating String
The simple DE
y” + y = 0
(6)
occurs again as a model of a rotating string. See Fig
3.45.
CH3_154
Fig 3.45
CH3_155
We have
F = T sin 2 – T sin 1
(7)
When 1 and 2 are small,
sin 2  tan 2 , sin 1  tan 1
Since tan2, tan1 are slopes of the lines containing
the vectors T1 and T2, then
tan 2 = y’(x + x), tan 1 = y’(x)
Thus (7) becomes
F  T [ y( x  x)  y( x)]
(8)
Because F = ma, m = x, a = r2. With x small,
we take r = y.
CH3_156
Thus
F  ( x) y 2
(9)
Letting (8) = (9), we have
T [ y( x  x)  y( x)]  ( x) y 2
(10)
y( x  x)  y( x)
T
   2 y
x
For x close to zero, we have
2
d2y
d
y
2
T 2    y, T 2   2 y  0
(11)
dx
dx
And the boundary conditions are y(0) = y(L) = 0.
CH3_157
3.10 Nonlinear Models
Nonlinerar Springs
The model
d 2x
m 2  F ( x)  0
dt
when F(x) = kx is said to be linear.
However,
2
d 2x
d
x
3
m 2  kx  0, m 2  kx  k1x3  0
dt
dt
is a nonlinear spring.
Another model
d 2x
dx dx
m 2 
 kx  0
dt dt
dt
(1)
(2)
(3)
CH3_158
Hard and Soft Springs
F(x) = kx + k1x3 is said to be hard if k1 > 0;
and is soft, if k1 < 0. See Fig 3.50.
Fig 3.50
CH3_159
Example 1
The DEs
d 2x
3
(4)

x

x
0
2
dt
2
d
x
and
(5)
3
xx 0
2
dt
are special cases of (2). Fig3.51 shows the graph from
a numerical solver.
CH3_160
Fig 3.51
CH3_161
Nonlinear Pendulum
The model of a simple pendulum is shown in Fig
3.52.
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
(6)
2
l
dt
CH3_162
Fig 3.52
CH3_163
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
l
dt
(7)
CH3_164
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.
CH3_165
Fig 3.53
CH3_166
Telephone Wire
Recalling from (17) in Sec 1.3 and Fig 1.26
dy/dx = W/T1, can be modified as
dy ws

dx T1
where  is the density and s is the arc length.
Since the length s is
s
x
0
(8)
2
dy 

1    dx
 dx 
(9)
CH3_167
then
ds
dy 

 1  
dx
 dx 
2
(10)
Differentiating (8) w.s.t x and using (10), then
2
d y w ds d y w
dy 


,

1  
2
2
T1 dx dx
dx
T1
 dx 
2
2
(11)
CH3_168
Example 3
From Fig 1.26, we obtain From Fig 1.26, we obtain
y(0) = a, y’(0) = 0. Let u = y’, equation (11) becomes
du

du 
2
  dx

1 u ,

2
T1
dx T1
1 u
w
Thus
1
sinh u  x  c1
T1
Now y’(0) = u(0) = 0, sinh-10 = 0 = c1
Since u = sinh(x/T1) = dy/dx, then
dy

T1

 sinh x, y  cosh x  c2
dx
T1
w
T1
Using y(0) = a, c2 = a − (T1/)
T1

T1
y  cosh x  a 
CH3_169

T1

Rocket Motion
From Fig 3.54, we have
2
2
d
s
M
d s
Mm
 k 2
m 2  k 2 ,
2
dt
y
dt
y
when y = R, kMm/R2 = Mg, k = gR2/M,
then
d 2s
R2
 g 2
2
dt
y
(12)
(13)
CH3_170
Fig 3.54
CH3_171
Variable Mass
Assuming the mass is variable, then F = ma should
be modified as
d
F  (mv)
dt
(14)
CH3_172
Example 4
A uniform 10-foot-long chain is coiled loosely on the
ground. On end is pulled vertically by a force of 5 lb.
The chain weigh 1 lb per foot. Determine the height of
the end at time t.
Solution:
Let
x(t) = the height
v(t) = dx/dt (velocity)
W = x1 = x (weight)
m = W/g = x/32 (mass)
F = 5 – W (net force)
CH3_173
Example 4 (2)
Then
d x 
dv
dx
 v   5  x, x  v  160  32 x
dt  32 
dt
dt
Since v = dx/dt
2
2
d x  dx 
x 2     32 x  160
dt
 dt 
is of the form F(x, x’, x”) = 0
Since v = x’, and
dv dv dx
dv

v
dt dx dt
dx
then (15) becomes
dv 2
xv  v  160  32 x
dt
(15)
(16)
(17)
CH3_174
Example 4 (3)
Rewriting (17) as
(v2+32x – 160) dx + xv = 0
(18)
(18) can be multiplied by an integrating factor to
become exact, where we can find the integrating
factor is (x) = x (please verify). Then
f /x  xv2  32 x 2  160 x, f /v  xv2
Use the method in Sec. 2.4
(19)
1 2 2 32 3
2
x v  x  80 x  c1
2
3
Since x(0) = 0, then c1 = 0. By solving (19) = 0, for
v = dx/dt > 0, we get dx
64
 160  x
dt
3
CH3_175
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
15 15  4 10 
x(t )   1 
t
2 2
15 
2
(21)
CH3_176
3.11 Solving Systems of Linear Equations
Coupled Spring/Mass System
From Fig 3.58 and Newton’s Law
m1 x1  k1 x1  k2 ( x2  x1 )
m2 x2  k2 ( x2  x1 )
(1)
CH3_177
Fig 3.58
CH3_178
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
 6t
 c4e
6t
(4)
CH3_179
Return to the original equations,
dx/dt = 3y
then after simplification,
( 6c1  3c3 )e 
6t
 ( 6c2  3c4 )e
6t
0
we have
6
6
c3   c1, c4 
c2
3
3
(5)
CH3_180
Example 1
Solve
Dx + (D + 2)y = 0
(D – 3)x –
2y = 0
(6)
Solution:
Multiplying the first by D – 3, the second by D, then
subtracting,
[(D – 3)(D + 2) + 2D]y = 0
(D2 + D – 6)y = 0
then
y(t) = c1e2t + c2e-3t
(7)
CH3_181
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 3t
2t
2t
3t
x(t )  2c1e  c2e , y (t )  c1e c2e
3
CH3_182
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,
[( D  1) D2  ( D  4) D] y  ( D  1)t 2  ( D  4)0
then ( D3  4 D) 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.
CH3_183
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
Similar method to get x(t)
[( D  4)  D( D  1)]x  t 2 , ( D2  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
(11)
CH3_184
Example 2 (3)
Thus
1 2 1
x  xc  x p  c4 cos 2t  c5 sin 2t  t 
4
8
By using the second equation of (9), we have
(12)
(c5  2c4  2c2 ) sin 2t  (2c5  c4  2c3 ) cos 2t  0
c4  1/ 5(4c2  2c3 ), c5  1/ 5(2c2  4c3 )
1
1
1 2 1
x(t )   (4c2  2c3 ) cos 2t  (2c2  4c3 ) sin 2t  t 
5
5
4
8
1 3 1 2 1
y (t )  c1  c2 cos 2t  c3 sin 2t  t  t  t
12
4
8
CH3_185
Example 3
In (3) of Sec. 2.9, we have
D  2  x  1 x  0

 1
2
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.
CH3_186
Example 4
Solve
x"1 10 x1  4 x2  0
4 x1  x"2 4 x2  0
with x1 (0)  0, x'1 (0)  1, x2 (0)  0, x'2 (0)  1
Solution:
(13)
( D 2  10) x1  4 x2  0
4 x1  ( D  4) x2  0
2
Then
( D  2)(D  12) x1  0, ( D  2)(D  12) x2  0
2
2
2
2
CH3_187
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
(14)
CH3_188
Fig 3.59
CH3_189