136
Chapter 4 Higher Order Differential Equations
Highest differentiation:
dny
, n>1
n
dx
Most of the methods in Chapter 4 are applied for the linear DE.
137
附錄四 DE 的分類
Homogeneous
Constant coefficients
Linear
Cauchy-Euler
DE
Nonhomogeneous
Homogeneous
Nonhomogeneous
Others
Nonlinear
Homogeneous
Nonhomogeneous
附錄五 Higher Order DE 解法
reduction of order
linear
homogeneous
part
auxiliary function
Cauchy-Euler
“guess” method
particular
solution
multiple linear DEs
reduction of order
nonlinear
annihilator
variation of parameters
elimination method
Taylor series
numerical method
both
(but mainly linear)
series solution
transform
Laplace transform
Fourier series
Fourier cosine series
Fourier sine series
Fourier transform
138
139
4-1 Linear Differential Equations:
Basic Theory
4.1.1 Initial-Value and Boundary Value Problems
4.1.1.1 The nth Order Initial Value Problem
i.e., nth order linear DE with IVP at the same point
dny
d n1 y
an  x  n  an1  x  n1 
dx
dx
y  x0   y0 ,
y  x0   y1 ,
 a1  x 
y  x0   y2 ,
……………….. y ( n1)  x0   yn1
n initial conditions
dy
 a0  x  y  g  x 
dx
…………..
140
Theorem 4.1.1
For an interval I that contains the point x0
 If a0(x), a1(x), a2(x), ……., an −1(x ), an(x) are continuous at x = x0
 an(x0)  0
(很像沒有 singular point 的條件)
then for the problem on page 138, the solution y(x) exists and is unique
on the interval I that contains the point x0
(Interval I 的範圍，取決於何時 an(x) = 0 以及 何時 ak(x) (k = 0 ~ n)
不為continuous)
Otherwise, the solution is either non-unique or does not exist.
(infinite number of solutions) (no solution)
141
Example 1 (text page 117)
3 y  5 y  y  7 y  0
Example 2 (text page 118)
y  4 y  12 x
2
 x y  2 xy  2 y  6
y (1)  0
y(1)  0
y (0)  4
y(0)  1
y (0)  3
y(0)  1
有無限多組解
y  cx 2  x  3
c 為任意之常數
y(1)  0
142
 比較：
x 2 y  2 xy  2 y  6
y (1)  3
y(1)  1
只有一個解
y  x2  x  3
x  (0, )
 Note:
The initial value can also be the form as:
 y( x0 )   y( x0 )  0
N 1
(n)

y
 n ( x0 )  0 (general initial condition)
n 0
4.1.1.2 nth Order Boundary Value Problem
Boundary conditions are specified at different points
比較：Initial conditions are specified at the same points
例子： a2  x  y  a1  x  y  a0  x   g  x 
subject to
或
或
y (a)  y0 ,
y (b)  y1
y(a)  y0 ,
y(b)  y1
1 y (a)  1 y(a)   1

 2 y (b)   2 y(b)   2
An nth order linear DE with n boundary conditions may have a
unique solution, no solution, or infinite number of solutions.
143
144
Example 3 (text page 119)
y  16 y  0
solution: y  c1 cos  4 x   c2 sin  4 x 
(1) y  0   0
y  / 2   0
y  c2 sin  4 x 
c2 is any constant (infinite number of solutions)
(2) y  0   0
y  /8  0
y0
(unique solution)
145
4.1.2 Homogeneous Equations
4.1.2.1 Definition
dny
d n1 y
an  x  n  an1  x  n1 
dx
dx
dy
 a1  x   a0  x  y  g  x 
dx
g(x) = 0
homogeneous
g(x)  0
nonhomogeneous
 重要名詞：Associated homogeneous equation
The associated homogeneous equation of a nonhomogeneous DE:
Setting g(x) = 0
• Review: Section 2-3, pages 53, 55
146
4.1.2.2 New Notations
n
d
Notation: D y  y
dx n
n
d2y
dy
可改寫成 2
可改寫成

5

6
y
D
y

5
Dy

6
y
( D2  5D  6) y
2
dx
dx
可再改寫成
L( y )
L  D 2  5D  6
147
4.1.2.3 Solution of the Homogeneous Equation
[Theorem 4.1.5]
For an nth order homogeneous linear DE L(y) = 0, if
 y1(t), y2(t), ….., yn(t) are the solutions of L(y) = 0
 y1(t), y2(t), ….., yn(t) are linearly independent
then any solution of the homogeneous linear DE can be
expressed as:
y  c1 y1  c2 y2 
可以和矩陣的概念相比較
 cn yn
148
From Theorem 4.1.5:
An nth order homogeneous linear DE has n linearly independent
solutions.
Find n linearly independent solutions
== Find all the solutions of an nth order homogeneous linear DE
y1(t), y2(t), ….., yn(t): fundamental set of solutions
y  c1 y1  c2 y2 
 cn yn : general solution of the homogenous linear DE
(又稱做 complementary function)
也是重要名詞
149
Definition 4.1 Linear Dependence / Independence
If there is no solution other than c1 = c2 = ……. = cn = 0 for the
following equality
c1 y1  x   c2 y2  x  
 cn yn  x   0
then y1(t), y2(t), ….., yn(t) are said to be linearly independent.
Otherwise, they are linearly dependent.
判斷是否為 linearly independent 的方法: Wronskian
150
Definition 4.2 Wronskian
W  y1 , y2 ,
W  y1 , y2 ,
 y1
 y
, yn   det  1

 y ( n1)
 1
, yn   0
y2
y2
y2( n1)
yn 
yn 


yn( n1) 
linearly independent
151
4.1.2.4 Examples
Example 9 (text page 124)
y  6 y  11y  6 y  0
y1 = ex, y2 = e2x, and y3 = e3x are three of the solutions
Since
 y1 y2
det  y1 y2

 y1 y2
y3  e x

y3   e x

y3 e x
e2 x
2e 2 x
4e 2 x
e3 x 
1 1 1 

3e3 x   e x2 x3 x 1 2 3  2e6 x  0


3x
9e 
1 4 9 
Therefore, y1, y2, and y3 are linear independent for any x
general solution:
y  c1e x  c2e2 x  c3e3 x
x  (−, )
4.1.3 Nonhomogeneous Equations (可和 page 55 相比較)
152
Nonhomogeneous linear DE
an  x  y ( n )  x   an1  x  y ( n1)  x  
Part 1
Associated homogeneous DE
an  x  y ( n )  x   an1  x  y ( n1)  x 

 a1  x  y( x)  a0  x  y  0
 a1  x  y( x )  a0  x  y  g  x 
Part 2
particular solution y p
(any solution of the
nonhomogeneous linear DE)
find n linearly independent solutions g  x   g1  x   g2  x  
y1  x  , y2  x  ,
, yn  x 
y p  x   y p1  x   y p2  x  
general solution of the nonhomogeneous linear DE
y  x   c1 y1  x   c2 y2  x  
 cn yn  x   y p  x 
 gk  x 
 y pk  x 
153
Theorem 4.1.6 general solution of a nonhomogeneous linear DE
y  yc  y p
general solution of the associated
homogeneous function
(complementary function)
particular solution (any solution)
of the nonhomogeneous linear
DE
general solution of the
nonhomogeneous linear DE
Example 10 (text page 125)
y  6 y  11y  6 y  3x
y  6 y  11y  6 y  0
Three linearly independent
solution
e x , e 2 x , e3x
Check by Wronskian (Example 9)
ex
ex
ex
e2 x
2e 2 x
4e 2 x
Particular solution
y p   11  1 x
12 2
e3 x
3e3 x  2e6 x
9e3 x
General solution: y  c1e x  c2e 2 x  c3e3 x  11  1 x
12 2
154
155
Theorem 4.1.7 Superposition Principle
If y p1  x  is the particular solution of
an  x  y ( n )  x   an1  x  y ( n1)  x  
 a1  x  y  x   a0  x  y  x   g1  x 
y p2  x  is the particular solution of
an  x  y ( n )  x   an1  x  y ( n1)  x  
:
y pk  x  is the particular solution of
an  x  y ( n )  x   an1  x  y ( n1)  x  
then y p1  x   y p2  x  
 a1  x  y  x   a0  x  y  x   g 2  x 
 a1  x  y  x   a0  x  y  x   g k  x 
 y pk  x  is the particular solution of
an  x  y ( n )  x   an1  x  y ( n1)  x  
 g1  x   g 2  x  
 gk  x 
 a1  x  y  x   a0  x  y  x 
Example 11 (text page 126)
156
y p1  x   4 x 2 is a particular solution of y  3 y  4 y  16 x 2  24 x  8
y p2  x   e2 x
2x
is a particular solution of y  3 y  4 y  2e
y p3  x   xe x
is a particular solution of y  3 y  4 y  2 xe x  e x
y  y p1  y p2  y p3  4 x 2  e 2 x  xe x is a particular solution of
y  3 y  4 y  16 x 2  24 x  8  2e2 x  2 xe x  e x
157
4.1.4 名詞
 initial conditions, boundary conditions (pages 138, 142)
(重要名詞)
 associated homogeneous equation , complementary function (page 144)
(重要名詞)
 fundamental set of solutions (page 147)
 Wronskian (page 149)
 particular solution (page 151)
 general solution of the homogenous linear DE (page 147)
 general solution of the nonhomogenous linear DE (page 151)
y  x   c1 y1  x   c2 y2  x  
 cn yn  x   y p  x 
158
4.1.5 本節要注意的地方
(1) Most of the theories in Section 4.1 are applied to the linear DE
(2) 注意 initial conditions 和 boundary conditions 之間的不同
(3) 快速判斷 linear independent
159
(補充 1) Theorem 4.1.1 的解釋
dny
d n1 y
an  x  n  an1  x  n1 
dx
dx
 a1  x 
dy
 a0  x  y  g  x 
dx
y  x0   y1 ………………..
y  x0   y0
y ( n1)  x0   yn1
When an(x0)  0
y
(n)
an1  x0  ( n1)
y
 x0  
 x0  
an  x0 
a1  x0 
g  x0 
a0  x 


y  x0  
y  x0  
an  x0 
an  x0 
an  x0 
find y(n)(x0)
y ( n1)  x0     y ( n1)  x0   y ( n )  x0  
(根據 f   t  
f t     f t 
,

find y(n−1)(x0+)
f t     f t   f  t   )
160
以此類推
y ( n2)  x0     y ( n2)  x0   y ( n1)  x0  
find y(n−2)(x0+)
y ( n3)  x0     y ( n3)  x0   y ( n2)  x0  
:
:
y  x0     y  x0   y  x0  
y

(n)
an1  x0    ( n1)
y
 x0    
 x0    
an  x0   
find y(n−3)(x0+)
find y(x0+)
a1  x0   

y  x0   
an  x0   
g  x0   
a0  x   
y  x0    
an  x0   
an  x0   
find y(n)(x0+)
y ( n1)  x0  2   y ( n1)  x0     y ( n )  x0    
y
( n  2)
 x0  2   y
( n  2)
 x0     y
( n 1)
 x0    
find y(n−1)(x0+2)
find y(n−2)(x0+2)
161
:
:
y  x0  2   y  x0     y  x0    
y ( n )  x0  2  
an1  x0  2  ( n1)
y
 x0  2  
an  x0  2 
find y(x0+2)

a1  x0  2 
y  x0  2 
an  x0  2 
g  x0  2 
a0  x  2 

y  x0  2  
an  x0  2 
an  x0  2 
以此類推，可將 y(x0+3), y(x0+4), y(x0+5), ……………
以至於將 y(x) 所有的值都找出來。
(求 y(x) for x > x0 時, 用正的  值，
求 y(x) for x < x0 時, 用負的  值)
162
Requirement 1: a0(x), a1(x), a2(x), ……., an −1(x ), an(x) are continuous
是為了讓 ak(x0+m) 皆可以定義
Requirement 2: an(x)  0 是為了讓 ak(x0+m) /an(x0+m) 不為無限
大
163
4-2 Reduction of Order
4.2.1 適用情形
(1)
(2)
(3)
Suitable for the 2nd order linear homogeneous DE
a2  x  y  a1  x  y  a0  x  y  0
(4) One of the nontrivial solution y1(x) has been known.
164
4.2.2 解法
假設
y2  x   u  x  y1  x 
先將DE 變成 Standard form
y  P  x  y  Q  x  y  0
If y(x) = u(x) y1(x) ,
y  uy1  uy1
(比較 Section 2-3)
y  uy1  2uy1  uy1
uy1  2uy1  uy1  P  x  uy1  P  x  uy1  Q  x  uy1  0
u( y1  P  x  y1  Q  x  y1 )  2uy1  uy1  P  x  uy1  0
zero
165
uy1  u(2 y1  P  x  y1 )  0
set w = u'
dw
dy
y1  w(2 1  P  x  y1 )  0
dx
dx
multiplied by dx/(y1w)
dw
dy
 2 1  P  x  dx  0
w
y1
separable variable
(with 3 variables)
dw
dy1

2
 w  y1   P  x  dx  0
ln w  c3  2ln y1  c4    P  x  dx
ln w  2ln y1  ln w  ln y1  ln w y1  ln wy12
2
ln wy12    P  x  dx  c
2
166
ln wy12    P  x  dx  c
wy  e 
 P x dx c
2
1
w  c1e 
 P x dx
/ y12
e 
u   wdx  c1 
y12
 P x dx
dx  c2
e 
y2  x   y1  x   2
dx
y1  x 
 P x dx
We can setting c1 = 1 and c2 = 0
(因為我們算 u(x) 的目的，只是為
了要算出與 y1(x) 互相independent
的另一個解)
167
4.2.3 例子
Example 1 (text page 130)
y  y  0
We have known that y1 = ex is one of the solution
P(x) = 0
1
y2  x   e x  ce 2 x dx   ce  x
2
Specially, set c = –2, (y2(x) 只要 independent of y1(x) 即可
所以 c 的值可以任意設)
y2  x   e  x
General solution: y  x   c1e x  c2e  x
168
(將課本 x 的範圍做更改)
Example 2 (text page 131)
x 2 y  3xy  4 y  0
when x  (−, 0)
We have known that y1 = x2 is one of the solution
Note: the interval of x
If x  (0, ) (x > 0),
 dx / x  ln x
 dx / x  ln( x)
If x < 0,
3
e3ln(  x )
2 ( x)
y2  x   x  4 dx  x 
dx
4
x
( x)
1
  x 2  dx   x 2 ln x
x
2
y  x   c1 x 2  c2 x 2 ln x
如課本
169
4.2.4 本節需注意的地方

e
(1) 記住公式 y2  x   y1  x 
 y12  x  dx
 P x dx
(2) 若不背公式 (不建議)，在計算過程中別忘了對 w(x) 做積分
(3) 別忘了 P(x) 是 “standard form” 一次微分項的 coefficient
term
(4) 同樣有 singular point 的問題
(5) 因為 y2(x) 是 homogeneous linear DE 的 “任意解”，所以計
算時，常數的選擇以方便為原則
  P x dx
e
dx 的計算較複雜且花時間，所以要多加練習
(6) 由於  2
y1  x 
多算習題
170
附錄六： Hyperbolic Function
x
x
e

e
sinh  x  
2
x
x
e

e
cosh  x  
2
sinh  x  e x  e x
tanh  x  
 x x
cosh  x  e  e
cosh  x  e x  e x
coth  x  
 x x
sinh  x  e  e
sech  x  
1
 x 2 x
cosh  x  e  e
csch  x  
1
 x 2 x
sinh  x  e  e
jx
 jx
e

e
比較： sin  x  
2j
jx
 jx
e

e
cos  x  
2
171
5
5
2
sinh(x)
1
0
0
tanh(x)
0
cosh(x)
-1
-5
-2
0
2
3
2
-5
-2
0
2
2
-2
0
2
3
sech(x)
coth(x)
-2
2
csch(x)
1
1
1
0
0
0
-1
-1
-1
-2
-3
-2
-2
0
2
-2
-2
0
2
-3
-2
0
2
d sinh  ax   a cosh  ax 
dx
sinh  0   0
d cosh  ax   a sinh  ax 
dx
cosh  0   1
d tanh  ax   a sech 2  ax 
dx
sinh  0   1
d coth  ax   a csch 2  ax 
dx
cosh  0   0
d sech  ax   a sech  ax  tanh  ax 
dx
sin  ix   i sinh  x 
d csch  ax   a csch  ax  coth  ax 
dx
cos  ix   cosh  x 
172
 sinh  ax  dx 
173
cosh  ax 
c
a
sinh  ax 
 cosh  ax  dx  a  c
ln cosh  ax 
c
 tanh  ax  dx 
a
ln sinh  ax 
c
 coth  ax  dx 
a


2 tan 1 tanh( a x)
2
sech
ax
dx

c



a
ln tanh( a x)
2 c
csch
ax
dx




a
附錄七 Linear DE 解法的步驟 (參照講義 page 151)
Step 1: Find the general solution (i.e., the complementary function )
of the associated homogeneous DE
(Sections 4-2, 4-3, 4-7)
Step 2: Find the particular solution
(Sections 4-4, 4-5, 4-6)
Step 3: Combine the complementary function and the particular solution
Extra Step: Consider the initial (or boundary) conditions
174
175
4-3 Homogeneous Linear Equations with
Constant Coefficients
本節使用 auxiliary equation 的方法來解 homogeneous DE
KK: [
]
4-3-1 限制條件
限制條件: (1) homogeneous
(2) linear
(3) constant coefficients
an y ( n )  x   an1 y ( n1)  x    a1 y( x )  a0 y  0
a0, a1, a2, …. , an are constants
(the simplest case of the higher order DEs)
176
4-3-2 解法
解法核心：
Suppose that the solutions has the form of
emx
Example: y''(x) 3 y'(x) + 2 y(x) = 0
Set y(x) = emx,
m2 emx  3m emx + 2 emx = 0
m2  3m + 2 = 0
solve m
可以直接把 n 次微分用 mn 取代，變成一個多項式
這個多項式被稱為 auxiliary equation
177
 解法流程
an y ( n )  x   an1 y ( n1)  x  
Step 1-1
auxiliary function an mn  an1mn1 
Step 1-1
 a1 y( x )  a0 y  0
 a1m  a0  0
Find n roots , m1, m2, m3, …., mn
(If m1, m2, m3, …., mn are distinct)
Step 1-2 n linearly independent solutions e
m1x
, em2 x , em3x ,
, emn x
(有三個 Cases)
Step 1-3 Complementary y  c1e
function
m1x
 c2em2 x  c3em3x 
 cnemn x
4-3-3 Three Cases for Roots
(2nd
178
Order DE)
a2 y  x   a1 y  x   a0 y  x   0
a2 m2  a1m  a0  0
 a1  a12  4a2 a0
roots m1 
2a2
 a1  a12  4a2 a0
m2 
2a2
solutions
Case 1 m1  m2, m1, m2 are real
y  c1e
m1x
 c2em2 x
(其實 m1, m2 不必限制為 real)
179
Case 2 m1 = m2 (m1 and m2 are of course real)
First solution: y1  e
m1 x
Second solution: using the method of “Reduction of Order”
e 
y2  x   y1  x   2
dx
y1  x 
 P x dx
e
m1 x
e
m1 x
e
m1 x
e
( 2 m1 
m1x
a1
)x
a2
dx
dx
e
 dx  e  x  c 
m1x
y2  x   xe
y  c1e
e 
2 m1x  a1 / a2 dx
m1 x
 c2 xem1x
m1 
 a1
2a2
Case 3 m1  m2 , m1 and m2 are conjugate and complex
 a1  a12  4a2 a0
m1 
   j
2a2
  a1 / 2a2 ,
m2    j 
  4a2 a0  a12 / 2a2
Solution: y  C1e x j x  C2e x j x
Another form:
y  e x  C1e j x  C2e  j x 
 e x  C1 cos  x  jC1 sin  x  C2 cos  x  jC2 sin  x 
set c1 = C1 + C2 and c2 = jC1 − jC2
y  e x  c1 cos  x  c2 sin  x 
c1 and c2 are some constant
180
181
Example 1 (text page 134)
(a) 2 y  5 y  3 y  0
2m2 − 5m − 3 = 0,
m1 = −1/2, m2 = 3
y  c1e x / 2  c2e3 x
(b) y  10 y  25 y  0
m2 − 10m + 25 = 0,
m1 = 5, m2 = 5
y  c1e5 x  c2 x e5 x
(c)
y  4 y  7 y  0
m2 + 4m + 7 = 0,

m1  2  i 3,
y  e2 x c1 cos 3x  c2 sin 3x

m2  2  i 3
4-3-4 Three + 1 Cases for Roots (Higher Order DE)
For higher order case
an y ( n )  x   an1 y ( n1)  x  
n
n 1
auxiliary function: an m  an1m 
182
 a1 y( x )  a0 y  0
 a1m  a0  0
roots: m1, m2, m3, …., mn
(1) If mp  mq for p = 1, 2, …, n and p  q
(也就是這個多項式在 mq 的地方只有一個根)
then e mq x is a solution of the DE.
重覆次數
(2) If the multiplicities of mq is k (當這個多項式在 mq 的地方有 k 個根),
e
mq x
, xe
mq x
, x 2e
mq x
,
are the solutions of the DE.
, x k 1e
mq x
(3) If both  + j and  − j are the roots of the auxiliary function,
then
183
e x cos   x  , e x sin   x 
are the solutions of the DE.
(4) If the multiplicities of  + j is k and the multiplicities of  − j
is also k，then
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 
are the solutions of the DE.
184
Note: If  + j is a root of a real coefficient polynomial,
then  − j is also a root of the polynomial.
an (  j  )n  an1 (  j  ) n1 
a0, a1, a2, …. , an are real
 a1 (  j  )  a0  0
185
Example 3 (text page 135)
Solve
y  3 y  4 y  0
3
2
Step 1-1 m  3m  4  0
 m  1  m2  4m  4   0
Step 1-2
m1 = 1, m2 = m3 = 2
x
2 x
2 x
3 independent solutions: e , e , xe
x
2 x
2 x
Step 1-3 general solution: y  c1e  c2e  c3 xe
Example 4 (text page 136)
Solve
y (4)  x   2 y  x   y  x   0
4
2
m

2
m
1  0
Step 1-1
(m2  1)2  0
four roots: i, i, i, i
Step 1-2 4 independent solutions: cos x , x cos x , sin x , x sin x
Step 1-3 general solution: y  c1 cos x  c2 x cos x  c3 sin x  c4 x sin x
186
187
4-3-5 How to Find the Roots
(1) Formulas
 a1  a12  4a2 a0
m1 
2a2
a2 m  a1m  a0  0
2
 a1  a12  4a2 a0
m2 
2a2
a3m3  a2 m2  a1m  a0
m1  S  T 
Solutions:
a2
3a3
a
m2   1  S  T   2  i 1 3  S  T 
2
3a3
2
a
m3   1  S  T   2  i 1 3  S  T 
2
3a3 2
S  3 R  Q3  R3
1a1 a22
Q

3a3 9a32
,
, T  3 R  Q3  R3
9a1a2  27a3a0  2a23
R
54a32
(太複雜了)
188
(2) Observing
例如：1 是否為 root
看係數和是否為 0
又如：
3m3  5m 2  10m  4
factor: 1,3
factor: 1,2,4
possible roots: 1, 2, 4, 1/3, 2/3, 4/3
test for each possible root
find that 1/3 is indeed a root
1

3m3  5m 2  10m  4   m    3m 2  6m  12 
3

189
(3) Solving the roots of a polynomial by software
Maple
Mathematica (by the commands of Nsolve and FindRoot)
Matlab ( by the command of roots)
4-3-6 本節需注意的地方
(1) 注意重根和 conjugate complex roots 的情形
(2) 寫解答時，要將 “General solution” 寫出來
y  c1 y1  c2 y2 
 cn yn
(3) 因式分解要熟練
(4) 本節的方法，也適用於 1st order 的情形
190
191
練習題
Sec. 4-1: 3, 7, 8, 10, 13, 20, 26, 29, 33, 36
Sec. 4-2: 2, 4, 9, 13, 14, 16, 18, 19
Sec. 4-3: 7, 16, 20, 24, 28, 33, 39, 41, 52, 54, 56, 59, 61, 63