CHAPTER 3
Higher-Order Differential
Equations
(3.1~3.6)
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1
Chapter Contents
3.1 Theory of 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
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Ch3_2
3.1 Preliminary Theory: Linear Equ.
Initial-value Problem
An nth-order initial problem is
Solve:
dny
d n1 y
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.
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Ch3_3
Theorem 3.1.1 Existence of a Unique Solution
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.
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Ch3_4
Example 1 Unique Solution of an IVP
The IVP
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 3.1.1, hence
y = 0 is the only one solution on any interval
containing x = 1.
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Ch3_5
Example 2 Unique Solution of an VP
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.
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Ch3_6
Boundary-Value Problem
Solve:
d2y
dy
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 3.1.1.
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Ch3_7
Fig 3.1.1 Colored curves are solution of a
BVP
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Ch3_8
Example 3 A BVP Can Have Money, Ine, or
Not Solutions
 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
2
(3)
has infinite many solutions. See Fig 3.1.2.


(b) If
x  16 x  0 , x(0)  0 , x   0
8
(4)
we have c1 = 0, c2 = 0, x = 0 is the only solution.
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Ch3_9
Example 3 (2)
(c) If

x  16 x  0 , x(0)  0 , x   1
2
(5)
we have c1 = 0, and 1 = 0 (contradiction).
Hence (5) has no solutions.
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Ch3_10
Fig 3.1.2 The BVP in (3) of Ex 3 has many
solutions
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Ch3_11
Homogeneous Equations
The following DE
dny
d n1 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;
dny
d n1 y
dy
an ( x) n  an1 ( x) n1    a1 ( x)  a0 ( x) y  g ( x)
dx
dx
dx
(7)
with g(x) not identically zero, is nonhomogeneous.
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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)
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Ch3_13
Theorem 3.1.2 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.
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Ch3_14
Corollaries to Theorem 3.1.2
(a) y = cy1 is also a solution if y1 is a solution.
(b) A homogeneous linear DE always possesses the
trivial solution y = 0.
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Ch3_15
Example 4 Superposition – Homogeneous
DE
The function y1 = x2, y2 = x2 ln x are both solutions of
x 3 y  2 xy  4 y  0
Then y = x2 + x2 ln x is also a solution on (0, ).
Definition 3.1.1 Linear Dependence /
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.
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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 3.1.3, neither function is a constant
multiple of the other, then these two functions are
linearly independent.
If two functions are linearly dependent, then one is
simply a constant multiple of the other.
Two functions are linearly independent when neither
is a constant multiple of the other.
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Ch3_17
Fig 3.1.3 The set consisting of f1 and f2 is
linear independent on (-, )
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Ch3_18
Example 5 Linear Dependent Functions
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.
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Ch3_19
Example 6 Linearly Dependent Functions
The functions f1 ( x)  x  5, f 2 ( x)  x  5x, f3 ( x)  x  1,
f 4 ( x)  x 2 are linearly dependent on the interval (0, ),
since
f2 = 1 f1 + 5 f3 + 0 f4
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Ch3_20
Definition 3.1.2 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
f1

f 1( n1)
f2
f 2



f 2( n1) 
fn
f n

f n( n1)
is called the Wronskian of the functions.
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Ch3_21
Theorem 3.1.3 Criterion for Linear Independence
Solutions
Let y1(x), y2(x), …, yn(x) be n solutions of the nth-order
homogeneous DE (6) on an interval I. This set of
solutions is linearly independent on I if and on if
W(y1, y2, …, yn)  0 for every x in the interval.
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Ch3_22
Definition 3.1.3 Fundamental Set of Solutions
Any set y1(x), y2(x),… , yn(x) of n linearly independent
solutions is said to be a fundamental set of solutions.
Theorem 3.1.4 Existence of a Fundamental Set
There exists a fundamental set of solutions for (6) on an
interval I.
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Ch3_23
Theorem 3.1.5 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, i = 1, 2, …, n are arbitrary constants.
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Ch3_24
Example 7 General Solution of a
Homogeneous DE
The functions y1 = e3x, y2 = e-3x are solutions of y” –
9y = 0 on (-, )
Now
3x
e
W (e3 x , e 3 x ) 
3e3 x
e 3 x
 6  0
3 x
 3e
for every x.
So y = c1e3x + c2e-3x is the general solution.
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Ch3_25
Example 8 A Solution Obtained from a
General Solution
The functions y = 4 sinh 3x - 5e3x is a solution of
example 7 (Verify it). Observer
y  2e3 x  2e 3 x  5e 3 x
 e3 x  e 3 x 
 4
  5e 3 x
2


y = 4 sinh 3x – 5e-3x
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Ch3_26
Example 9 General Solution of a
Homogeneous DE
The functions y1 = ex, y2 = e2x , y3 = e3x are solutions
of y’’’ – 6y” + 11y’ – 6y = 0 on (-, ).
Since
ex
W (e x , e 2 x , e 3 x )  e x
ex
e2 x
2e 2 x
4e 2 x
e3 x
3e3 x  2e 6 x  0
9e3 x
for every real value of x.
So y = c1ex + c2 e2x + c3e3x is the general solution on
(-, ).
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Ch3_27
Nonhomogeneous Equations
Theorem 3.1.6 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)
where ci, i = 1, 2, …, n are arbitrary constants.
Complementary Function
y = c1y1 + c2y2 +… + ckyk + yp = yc + yp
= complementary + particular
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Ch3_28
Example 10 General Solution of a
Nonhomogeneous DE
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  c2 e  c3e   x
12 2
x
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2x
3x
Ch3_29
Any Superposition Principles
Theorem 3.1.7 Superposition Principles –
Nonhomogeneous Equations
Given
an ( x) y ( n )  an1 ( x) y ( n1)    a1 ( x) y  a0 ( x) y  g i ( x) (12)
where i = 1, 2, …, k.
If ypi denotes a particular solution corresponding to the
DE (12) with gi(x), then
y p  y p ( x)  y p ( x)    y p ( x)
(13)
is a particular solution of
an ( x) y ( n )  an1 ( x) y ( n1)    a1 ( x) y  a0 ( x) y
(14)
1
2
k
 g1 ( x)  g 2 ( x)    g k ( x)
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Ch3_30
Example 11 Superposition – Nonmogeneous
DE
We find
y p1  4 x 2 is a particular solution of
y"3 y'4 y  16 x 2  24 x  8
y p2  e 2 x is a particular solution of
y"3 y'4 y  2e 2 x
is a particular solution of
y"3 y'4 y  2 xe x  e x
From Theorem 3.1.7, 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 )
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g2 ( x )
g3 ( x )
Ch3_31
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)
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Ch3_32
3.2 Reduction of Order
Introduction
We know the general solution of
a2 ( x) y  a1 ( x) y  a0 ( x) y  0
(1)
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.
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Ch3_33
Example 1 Finding s second Solution
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
y  ue x  e xu , y  ue x  2e xu  e x e
And
y" y  e x (u  2u)  0
Since ex  0, we let w = u’, then
w  c1e2 x  u
1
u   c1e 2 x  c2
2
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Ch3_34
Example 1 (2)
Thus
y  u ( x )e x  
c1  x
e  c2 e x
2
(2)
Choosing c1 = 0, c2 = -2, we have y2 = e-x.
Because W(ex, e-x)  0 for every x, they are independent
on (-, ).
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Ch3_35
General Case
Rewrite (1) as the standard form
y  P( x) y  Q( x) y  0
(3)
Let y1(x) denotes a known solution of (3) and y1(x) 
0 for every x in the interval.
If we define y = uy1, then we have
y  uy1  y1u , y  uy1  2 y1u  y1u
y  Py  Qy
 u[ y1  Py1  Qy1 ]  y1u  (2 y1  Py1 )u  0



zero
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Ch3_36
This implies that
y1u  (2 y1  Py1 )u  0
or
y1w  (2 y1  Py1 )2w  0
(4)
where we let w = u’.
Solving (4), we have
dw
y1
 2 dx  Pdx  0
w
y1
ln | wy12 |   Pdx  c
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or
wy12  c1e  Pdx
Ch3_37
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)
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(5)
Ch3_38
Example 2 A second Solution by Formula (5)
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 y  0
x
x
From (5)
3  dx / x
e
y2  x 2  4 dx  x 2 ln x
x
The general solution is
y  c1 x 2  c2 x 2 ln x
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Ch3_39
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, i = 0, 1, …, n are constants, an  0.
Auxiliary Equation
For n = 2,
ay  by  cy  0
(1)
(2)
Try y = emx, then
e mx (am2  bm  c)  0
am 2  bm  c  0
(3)
is called an auxiliary equation.
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Ch3_40
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.
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Ch3_41
Case 1: Distinct real roots
The general solution is
y  c1e m1x  c2e m2 x
(4)
Case 2: Repeated real roots
y1  e m x and from (5) of Sec 3.2,
1
y2  e
m1x
e 2 m1x
m1x
m1x
dx

e
dx

xe
 e2m1x

(5)
The general solution is
y  c1e m1x  c2 xe m1x
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(6)
Ch3_42
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:
e i x
ei  cos  i sin 
 cos x  i sin x
and e ix  cos x  i sin x
eix  e ix  2 cos x
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(7)
and eix  e ix  2i sin x
Ch3_43
Since y  C1e( i ) x  C2e( i ) x is a solution then set
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  ex (c1 cos x  c2 sin x)
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(8)
Ch3_44
Example 1 Second-Order DEs
Solve the following DEs:
(a) 2 y"5 y'3 y  0
2m2  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)
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Ch3_45
Example 2 An Initial-Value Problem
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 3.3.1.
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Ch3_46
Fig 3.3.1 Graph of a solution of IVP in Ex 2
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Ch3_47
Two Equations worth Knowing
 y  k 2 y  0, y  k 2 y  0, k  0
For the first equation:
y  c1 cos kx  c2 sin kx
(9)
For the second equation:
Let
y  c1e kx  c2e kx
(10)
y1  1/2(e kx  e kx )  cosh kx
Then
y2  1/2(e kx  e kx )  sinh kx
y  c1 cosh kx  c2 sinh kx
Copyright © Jones and Bartlett；滄海書局
(11)
Ch3_48
Higher-Order Equations
Given
an y ( n )  an1 y ( n1)    a2 y  a1 y  a0 y  0
(12)
we have
an m n  an1m n1    a2 m 2  a1m  a0  0
(13)
as an auxiliary equation.
If the roots of (13) are real and distinct, then the
general solution of (12) is
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Ch3_49
Example 3 Third-Order DE
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
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Ch3_50
Example 4 Fourth-Order DE
d4y
d2y
Solve
2 2  y 0
4
dx
dx
Solution:
m4  2m2  1  (m2  1) 2  0
m1  m3  i, m2  m4  i
y  C1eix  C2e ix  C3 xeix  C4 xe ix
 c1 cos x  c2 sin x  c3 x cos x  c4 x sin x
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Ch3_51
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
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Ch3_52
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.
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Ch3_53
Example 1 General Solution Using
Undetermined Coefficients
Solve y  4 y'2 y  2 x 2  3x  6
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 ,
(2)
y p , yp  2 Ax  B, yp  2 A
After substitution,
2A + 8Ax + 4B – 2Ax2 – 2Bx – 2C = 2x2 – 3x + 6
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Ch3_54
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
yp  x  x  9
2
2
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Ch3_55
Example 2 Particular Solution Using
Undetermined Coefficients
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
yp 
6
16
cos 3x  sin 3x
73
73
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Ch3_56
Example 3 Forming yp by Superposition
Solve
y  2 y  3 y  4 x  5  6 xe 2 x
Solution:
We can find yc  c1e  x  c2e3 x
Let y p  Ax  B  Cxe 2 x  Ee2 x
After substitution,
(3)
 3 Ax  2 A  3B  3Cxe 2 x  (2C  3E )e2 x  4 x  5  6 xe2 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
y  c1e  x  c2e3 x  x    2 x   e 2 x
3
9 
3
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Ch3_57
Example 4 A Glitch in the Method
Find yp of y  5 y  4 y  8e x
Solution:
First let yp = Ae2x
After substitution, 0 = 8e2x, (wrong guess)
Let yp = Axex
After substitution, -3Ae2x = 8e2x
Then A = -8/3, yp = (−8/3)xe2x
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Ch3_58
Case I
No function in the assumed yp is part of yc
Table 3.4.1 shows the trial particular solutions.
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Ch3_59
Example 5 Forms if Particular Solution –
Case I
Find the form of yp of
(a) y  8 y  25 y  5x3ex  7ex
Solution:
We have g ( x)  (5x3  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.
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Ch3_60
Rule of Case I
If g(x) consists of a sum of m terms of the kind listed
in the table, then (as in Ex 3) the assumption for a
particular solution yp consists of the sum of the trial
forms
The form of yp is a linear combination of all linearly
independent functions that are generated by repeated
differentiations of g(x).
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Ch3_61
Example 6 Forming yp by Superposition –
Case I
Find the form of yp of
y  9 y  14 y  3x 2  5 sin 2 x  7 xe6 x
Solution:
For 3x2: y p  Ax 2  Bx  C
1
For -5 sin 2x: y p  E cos 2 x  F sin 2 x
2
For 7xe6x: y p  (Gx  H )e6 x
3
No term in y p  y p  y p  y p duplicates a term in yc
1
2
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3
Ch3_62
Example 7 Particular Solution – Case II
Find a particular solution of y  2 y  y  e x
Solution:
The complementary function is yp = c1ex + c2xex.
Assume yp = Aex will fail since it is apparent from yc
that ex is a solution of the associated homogeneous
equation y  2 y  y  0.
And we will not be able to find a particular solution of
the form yp = Axex since the term xex is also duplicated
in yc. We next try yp = Ax2ex, substituting into the given
differential equation yields 2Axex = ex and so A = ½.
The a particular solution is yp = ½x2ex.
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Ch3_63
Multiplication Rule of Case II
If g(x) consists of a sum of m terms of the kind given
in Table 3.4.1, and suppose that the usual assumption
for a particular solution is
y p  y p1  y p2    y pm
where the y pi , i  1, 2, ..., m are the trial particular
solution forms corresponding to these terms.
If any y p consists terms that duplicates terms in yc,
then that y p must be multiplied by xn, where n is the
smallest positive integer that eliminates that
duplication.
i
i
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Ch3_64
Example 8 An Initial-Value Problem
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
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(5)
Ch3_65
Example 9 Using the multiplication Rule
Solve y"6 y'9 y  6 x 2  2  12e3 x
Solution:
yc = c1e3x + c2xe3x
2
3x
yp  
Ax

Bx

C

Ee
 
 
yp
1
yp
2
After substitution and simplification,
A = 2/3, B = 8/9, C = 2/3, E = -6
Then
2
8
2
y  c1e3 x  c2 xe3 x  x 2  x   6 x 2 e3 x
3
9
3
Copyright © Jones and Bartlett；滄海書局
Ch3_66
Example 10 Third-Order-DE CeasI
Solve y  y" e x cos x
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
y  yc  y p  c1  c2 x  c3e  e cos x  e sin x
10
5
x
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Ch3_67
Example 11 Fourth-Order-DE – Case II
Find the form of yp of
y ( 4)  y  1  x 2e  x
Solution:
yc = c1+ c2x + c3x2 + c4e-x
Normal trial:
y  A  Bx 2 e  x  Cxe  x  Ee x
p


yp
yp
1
2
Multiply A by x3 and (Bx2e-x + Cxe-x + Ee-x) by x
Then
yp = Ax3 + Bx3e-x + Cx2e-x + Exe-x
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Ch3_68
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.
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Ch3_69
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)
dx
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(4)
Ch3_70
Making further assumptions:
y1u1’ + y2u2’ = 0, then from (4),
y1’u1’ + y2’u2’ = f(x)
Express the above in terms of determinants
u1 
W1
y2 f ( x)

W
W
and u2 
W2
y1 f ( x)

W
W
(5)
where
y1
W
y1
y2
0
, W1 
y2
f ( x)
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y2
y1
, W2 
y2
y1
0
f ( x)
(6)
Ch3_71
Example 1 General Solution Using Variation
of Parameters
Solve y"4 y'4 y  ( x  1)e2 x
Solution:
m2 – 4m + 4 = 0, m = 2, 2
y1 = e2x, y2 = xe2x,
2x
e
W (e 2 x , 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
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0
4x

(
x

1
)
e
( x  1)e 2 x
Ch3_72
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
y  yc  y p  c1e  c2 xe  x e  x e
6
2
2x
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2x
Ch3_73
Example 2 General Solution Suing Variation
of Parameters
Solve 4 y  36 y  csc 3x
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 sin 3x 3 cos 3x
0
sin 3 x
1
W1 
 ,
1/4 csc 3x 3 cos 3x
4
3
cos 3x
0
1 cos 3x
W2 

 3 sin 3 x 1/4 csc 3x 4 sin 3x
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Ch3_74
Example 2 (2)
W1
1
u1 

W
12
u2 
Then
W2 1 cos 3x

W 12 sin 3x
u1  1/12 x,
u2  1/36 ln | sin 3x |
1
1
y p   x cos 3x  (sin 3x) ln | sin 3x |
And
12
36
y  yc  y p
1
1
 c1 cos 3 x  c2 sin 3 x  x cos 3x  (sin 3x) ln | sin 3x | (7)
12
36
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Ch3_75
Example 3 General Solution Using Variation
pf Paramenters
Solve y  y 
1
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
2
2 0 t
e x (1 / x)
1 x et
u2  
, u2   
dt
x
2
2 0 t
The low and up bounds of the integral are x0 and x,
respectively.
Copyright © Jones and Bartlett；滄海書局
Ch3_76
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
t
t
x e
x e
1
1
y  yc  y p  c1e x  c2 e  x  e x 
dt  e  x 
dt
x0 t
2 x0 t
2
Copyright © Jones and Bartlett；滄海書局
Ch3_77
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)
y1

u1  y2
( n1)
u2    yn
( n1)
un  f ( x)
(9)
and uk’ = Wk/W, k = 1, 2, …, n.
Copyright © Jones and Bartlett；滄海書局
Ch3_78
For the case n = 3,
u1 
0
W1  0
f ( x)
W1
,
W
y2
y2
y2
y1 y2
and W  y1 y2
y1 y2
u2 
W
W2
, u3  3
W
W
y3
y1
y3 , W2  y1
y3
y1
0
0
f ( x)
y3
y1 y2
y3 , W3  y1 y2
y3
y1 y2
(10)
0
0
f ( x)
y3
y3
y3
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Ch3_79
3.6 Cauchy-Eulaer Equation
Form of Cauchy-Euler Equation
n
n1
d
y
d
y
dy
n
n1
an x
 an1 x
   a1 x  a0 y  g ( x)
n
n1
dx
dx
dx
Method of Solution
We try y = xm, since
k
d
y
k
k
m k
ak x
k  ak x m( m  1)(m  2) ( m  k  1) x
dx
 ak m(m  1)(m  2)(m  k  1) x m
Copyright © Jones and Bartlett；滄海書局
Ch3_80
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  c1 x m1  c2 x m2
Copyright © Jones and Bartlett；滄海書局
(2)
Ch3_81
Example 1 Distinct Roots
2
d
y
dy
2
Solve x 2  2 x  4 y  0
dx
dx
Solution:
We have a = 1, b = -2 , c = -4
m2 – 3m – 4 = 0, m = -1, 4,
y = c1x-1 + c2x4
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Ch3_82
Case 2: Repeated Real Roots
Using (5) of Sec 3.2, we have y2  x m ln x
Then
1
y  c1 x m1  c2 x m1 ln x
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(3)
Ch3_83
Example 2 Repeated Roots
d2y
dy
Solve 4 x 2  8 x  y  0
dx
dx
2
Solution:
We have a = 4, b = 8, c = 1
4m2 + 4m + 1 = 0, m = -½ , -½
y  c1 x 1/2  c2 x 1/2 ln x
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Ch3_84
Case 3: Conjugate Complex Roots
Higher-Order: multiplicity
x m1 , x m1 ln x , 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)
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Ch3_85
Example 3 Am Initial-Value Problem
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  x1/ 2 [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 3.6.1.
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Ch3_86
Fig 3.6.1 Graph of solution of IVP in Ex 3
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Ch3_87
Example 4 Third-Order Equation
3
2
d
y
d
Solve x 3 3  5 x 2 y2  7 x dy  8 y  0
dx
dx
dx
Solution:
Let y = xm,
2
dy
d
y
 mx m1 , 2  m(m  1) x m2 ,
dx
dx
d3y
m3

m
(
m

1
)(
m

2
)
x
dx 3
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)
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Ch3_88
Example 5 Variation of Prarmeters
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
y  y  2 y  2 x 2 e x
x
x
Then P = -3/x, Q = 3/x2, f = 2x2ex
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Ch3_89
Example 5 (2)
Thus
x x3
3
W

2
x
,
2
1 3x
0
W1 
2 x 2e x
x
0
x3
5 x
3 x


2
x
e
,
W


2
x
e
2
2 x
2
1 2x e
3x
We find
2 x 5e x
2 x
u1  


x
e ,
3
2x
2 x 5e x
x
u2 

e
2 x3
u1   x e  2 xe  2e ,
2 x
x
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x
u2  e x
Ch3_90
Example 5 (3)
Then
y p  u1 y1  u2 y2  ( x 2 e x  2 xe x  2e x ) x  e x x 3
 2 x 2 e x  2 xe x
y  yc  y p  c1 x  c2 x 3  2 x 2e x  2 xe x
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Ch3_91
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