# 微分方程

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2-4 Exact Equations
2-4-1 方法的條件

M  x, y  dx  N  x, y  dy  0   的型態
              
(1) 當 y M  x, y   N  x , y      成立時，
x

可以用本節的 Exact Equation 的方法來解
              
M  x, y   N  x, y 
(2) 當 y             x          is independent of x
M  x, y 
               
或 y M  x, y   N  x, y 
x                  is independent of y
N  x, y 

可以用Modified Exact Equation Method 來解 (見 2-4-5)
99
2-4-2 方法的來源

• Review the concept of partial differentiation
f    f
df ( x, y )  dx  dy
x    y

• Specially, when f(x, y) = c where c is some constant,

f     f
dx  dy  0
x     y
100

f     f   f
df ( x, y, z )          dx  dy  dz
x     y   z
f        f        f              f
df ( x1 , x2 , x3 ,    , xk )        dx1      dx2      dx3           dxk
x1       x2       x3             xk

(p, q) 的地方，高度會比 (0, 0) 高多少？
ap+bq
(p, q)
f ( x, y )      f ( x, y )
df ( x, y )                dx              dy
x               y
p                q           (0, 0)       (p, 0)
a                  b
101
[Definition 2.4.1] We can express any 1st order DE as
M  x, y  dx  N  x, y  dy  0
 If there exists some function f(x, y) that satisfies
f  x, y                   f  x, y 
 M  x, y  and              N  x, y       ,
x                           y
then we call the 1st order DE as the exact equation.

 The method for checking whether the DE is an exact equation:
M  x, y  N  x, y 

y           x
f  x, y                     f  x, y 
(Proof): If                 M  x, y    and              N  x, y    ,
x                             y
M  x, y                                     N  x, y 
then                      f  x, y         f  x, y  
y         y x              x y                 x
102
For the exact equation,
M  x, y  dx  N  x, y  dy  0

f  x, y       f  x, y 
dx              dy  0
x               y

可改寫成           df  x, y   0,

f  x, y   c
2-4-3 解法                                                                           103

The method for solving the exact equation (A):
f  x, y                                f  x, y 
 M  x, y                                N  x, y 
x                                          y
(2)
(1)                                     (2)
f  x, y    M  x, y  dx  g  y   c     f      
  M  x, y  dx  g   y   N  x, y 
y y
g(y) is a constant for x                                        (2)
(3)          (3) 代入                             
g   y   N  x, y    M  x, y  dx
y
 M  x, y  dx  g  y   c                              (2)

(4)                         g  y    g   y  dy
further
computation
Solution

M  x, y  N  x, y 
Previous Step: Check whether                               is satisfied.
y           x

f  x, y 
Step 1: Solve              M  x, y        f  x, y    M  x, y  dx  g  y 
x
Step 2: 將 f(x, y) 代入 f  x, y   N  x, y  ，以解出 g(y)
y
Step 3: Substitute g(y) into

f  x, y    M  x, y  dx  g  y   c

Step 4: Further computation and obtain the solution

Extra Steps: (a) Consider the initial value problem
105
The method for solving the exact equation (B):
f  x, y                               f  x, y 
 M  x, y                               N  x, y 
x                                       y
(2)
(2)                                   (1)
f 
  N  x, y  dy  h  x   M  x, y  f  x, y   N  x, y  dy  h  x   c
x x                                                       
(2)                                     h(x) is a constant for y
(3) 代入
                                (3)
h  x   M  x, y    N  x, y  dy
x
(2)                       N  x, y  dy  h  x   c
h  x    h  x  dx                        (4)           further
computation
Solution
2-4-4 例子                                                                          106

Example 1 (text page 65)
2 xydx  ( x 2  1)dy  0

M  x, y   2 xy         N  x, y   x 2  1    Step 0: check whether it is exact
M         N
f                        f                                    2x 
 2 xy                     x2 1                        y        x
x                        y
Step 1      Step 2                Step 2
f  x, y   x 2 y  g  y    f
 x2  g  y   x2  1
y
Step 2
Step 3
Step 3 g   y   1
x2 y  y  c                Step 2
Step 4                        g  y  y         思考: 是否有其他的方法可
y  c /( x 2  1)                                以解 Example 1?
107
Example 2 (text page 65)
(e2 y  y cos xy)dx  (2 xe2 y  x cos xy  2 y)dy  0

(a) 自行由另一個方向 f  x, y    M  x, y  dx  g  y  來練習，
看是否得出同樣的結果。
(b) 得出的解 xe2 y  sin xy  y 2  c  0 為 implicit solution

(c) 思考：何時用 f  x, y    M  x, y  dx  g  y 
何時用 f  x, y   N  x, y  dy  h  x 

Example 3 (text page 66)                                        108

dy xy 2  cos x sin x                y  0  2

dx     y 1  x 2 

(a) initial value problem,
(b)何時用 f  x, y    M  x, y  dx  g  y 
何時用 f  x, y    N  x, y  dy  h  x 
(c) 得出的 implicit solution 為 y 1  x   cos x  3 , 範圍：x  (-1, 1)
2      2       2

3  cos 2 x
而 explicit solution 為 y              , 範圍： x  (-1, 1)
1 x 2

為何 y   3  cos 2 x 不為解？
1  x2
109
2-4-5 Modified Exact Equation Method

Technique: Use the integrating factor (x, y) to convert the 1st order
DE into the exact equation.
M  x, y  dx  N  x, y  dy  0

  x, y  M  x, y  dx    x, y  N  x, y  dy  0

such that   x, y  M  x, y     x, y  N  x, y 
y                       x

 y M   M y  x N   N x
x N   y M  (M y  N x )
It is hard to find .
110
x N   y M  (M y  N x )

(1) When ( M y  N x ) / M is a function of y alone:
We can set  to be dependent on y alone.

Therefore,
 y M  (M y  N x )
d  Nx  M y
          
dy       M
用 separable variable 的方法
d       Nx  M y
                   dy
            M
( N x M y )

  y   e
dy
M
x N   y M  (M y  N x )        111

(2) When (M y  N x ) / N is a function of x alone:
We can set  to be dependent on x alone.

Therefore,
x N  (M y  N x )
d  M y  Nx
         
dx     N
用 separable variable 的方法
d       Nx  M y
                 dy
         M
(M y Nx )

  x   e
dx
N
112

Step 0:                Case 1
Yes                                               (小心易背錯)
Use the process of 2-4-3
Case 2 ( N x M y )
Whether                                M y  Nx    Yes     y  e  M dy
M y  Nx
Whether          M                        M y  Nx
No is independent of x                 Whether
No                  N
is independent of y

Case 3 ( M y  N x )
In Cases 2 and 3,                                  Yes                N dx
  x  e
M  x, y  dx  N  x, y  dy  0
Case 4
No
Use other methods
 M  x, y  dx   N  x, y  dy  0

Using the process of 2-4-3, but M(x, y) should be modified as M(x, y)
N(x, y) should be modified as N(x, y)
Example 4 (text page 67)                                                                       113
xydx  (2 x2  3 y 2  20)dy  0

Step 0: M  xy             N  2 x 2  3 y 2  20

M y  N x  x  4 x  3x

M y  Nx              3x                M y  Nx              3           (independent of x)
                                                
N           2 x 2  3 y 2  20            M                 y                (Case 2)

3                           Q: 為何 c 以及
     dy
  y  e       y
e   3ln y
 y3     可省略?

xy 4dx  (2 x2 y3  3 y5  20 y3 )dy  0
double N
1 2 4 1 6
Steps 1~4:      x y  y  5 y4  c
2          2
114
2-4-6 本節需要注意的地方

(1) 使本節方法時，要先將 DE 改成如下的型態
M  x, y  dx  N  x, y  dy  0
並且假設
 f  x, y   M  x, y  ,       f  x, y   N  x , y 
x                               y
(2) 對 x 而言，g(y) 是個常數；對 y 而言，h(x) 是個常數

(3) 本節很少有 singular solution 的問題，
但是可能有 singular point 的問題

(4) 背熟三個判別式，二種情況的 integrating factor (小心勿背錯)
並熟悉解法的流程
115
2-5 Solutions by Substitutions
介紹 3 個特殊解法

Question: 尚有不少的 1st order DE 無法用 Sections 2-2~2-4 的方法
來解

對症下藥
116
2-5-1 特殊解法 1: Homogeneous Equations

If g(tx, ty) = tαg(x, y),
then g(x, y) is a homogeneous function of degree α.

Which one is homogeneous?
g(x, y) = x3 + y3
g(x, y) = x3 + y3 +1

一種是 Section 2-3 的定義 (較常用)
一種是這裡的定義
兩者並不相同
117
 For a   1st   order DE:
M  x, y  dx  N  x, y  dy  0
If M(x, y) and N(x, y) are homogeneous functions of the same degree,
then the 1st order DE is homogeneous.
解法的限制條件

It can by solved by setting
y = xu,               dy = udx + xdu,
and use the separable value method.
118
If M  x, y  dx  N  x, y  dy  0 is homogeneous
M  tx, ty   t  M  x, y    N  tx, ty   t  N  x, y 
then                                                                  以 t = 1/x 得出
M  x, y   x M 1, u        N  x, y   x N 1, u 

where u = y/x, y = xu

dy  udx  xdu

x M 1, u  dx  x N 1, u  (udx  xdu )  0
[M 1, u   uN 1, u ]dx  xN 1, u  du  0

dx        N 1, u  du
                        (separable)
x    M 1, u   uN 1, u 
119
Procedure for solving the homogeneous 1st order DE

Previous Step: Conclude whether the DE is homogeneous
(快速判斷法：看 powers (指數) 之和)

Step 1: Set y = ux, dy = udx + xdu
並化簡
Step 2: Convert into the separable 1st order DE

Step 3: Solve the separable 1st order DE (用 Sec. 2-2 的方法)
Step 4: Substitute u = y/x (別忘了這個步驟)
120
Example 1 (text page 71)
( x 2  y 2 )dx   x 2  xy  dy  0

M(x, y)          N(x, y)                        Previous Step:
Conclude whether the
M(tx, ty) = t2M(x, y)       N(tx, ty) = t2N(x, y)
DE is homogeneous
homogeneous DE

Step 1: Set y = ux,       dy = udx + xdu
原式

( x 2  u 2 x 2 )dx   x 2  ux 2  (udx  xdu )  0

(1  u 2 )dx  1  u  (udx  xdu )  0
1  u      dx
(1  u)dx  1  u  xdu  0    1  u  du   0
            x
Step 2: Convert into the separable 1st order DE
1  u      dx                                                    121
1  u  du   0
            x
Step 3: Solve the separable 1st order DE
      2        dx
  1 u 

1 

du    0
x

u  2ln 1  u  ln x  c1  0

ln[ (1  u ) x ]  u  c1
2

(1  u )2 x  eu c1

(1  u ) x  c2e
2           u            (c2  e c1 )

Step 4 代回 u = y/x
(1  y / x) x  c2e
2              y/x        ( x  y)2 x  c2 xe y / x
122
2-5-2 特殊解法 2: Bernoulli’s Equations

【定義】 Bernoulli’s equation:
dy
 P  x y  f  x yn
dx
n
dy   1 1n du
We can set u = y1–n       ,         u        , and use the method of
dx 1  n     dx
solving the 1st order linear DE to solve Bernoulli’s equation.

1
1 n    n
dy dy du du du     1 1n du
                  u
dx du dx   du dx 1  n     dx
(Chain rule)
123
Procedure for solving Bernoulli’s equation

Previous Step : Conclude whether the DE is Bernoulli’s equation
n
1      dy   1 1n du
Step 1: Set y  u   1n   ,         u
dx 1  n     dx
Step 2: Convert Bernoulli’s equation into the 1st order linear DE

Step 3: Solve the 1st order linear DE (use the method in Sec. 2-3)
Step 4: Substitute u = y1–n (別忘了)
Example 2 (text page 77)                                        124
dy
x       y  x2 y 2
dx

Previous Step: 判斷              (Bernoulli, n = 2)
dy dy du       2 du
Step 1: set u = y–1 (y = u–1)                 u
dx du dx          dx
(Chain rule)
Step 2: Convert into the 1st order linear DE (standard form)
2 du                   du 1
原式         xu           1
u  x u 2 2
 u  x
dx                   dx x
Step 3: Obtain the solution of the 1st order DE
u   x 2  cx
Step 4: 代回 u = y–1 y                 1
 x 2  cx
125
2-5-3 特殊解法 3

If the 1st order DE has the form,
dy
 f  Ax  By  C  (B  0)
dx
(解法的限制條件)

we can set u = Ax + By + C to solve it.
dy 1 du A
    
dx B dx B

Since du  Adx  Bdy      (這式子也許較好記)
126
dy
Procedure for solving       f  Ax  By  C 
dx
Previous Step: Conclude
dy 1 du A
Step 1: Set u  Ax  By  C      du  Adx  Bdy         
dx B dx B
Step 2: Converting (轉化成用其他方法可以解出來的 DE
未必一定是轉化成 separable variable DE)

Step 3: Solving

Step 4: Substitute u  Ax  By  C (別忘了)
Example 3 (text page 77)                                   127
dy
  2 x  y   7 ,       y (0)  0
2

dx
Previous Step: 判斷
dy du
Step 1: Set     u  2 x  y     du  2dx  dy      2
dx dx
Step 2: Converting
du                    du
原式               2  u2  7            dx
dx                   u 9
2

Step 3: Obtain the solution (別忘了在運算過程中，代回 u = Ax + By)
du
注意  2         的算法
u 9
Step 4: 代回 u = Ax + By +C
3(1  e6 x )
y  2x 
1  e6 x
128
2-5-4 本節要注意的地方

(1) 對症下藥，先判斷 DE 符合什麼樣的條件，再決定要什麼

(2) 別忘了，寫出解答時，要將 u 用 y/x, y1-n, 或 Ax + By + C
代回來

(3) 本節方法皆有五大步驟
Previous Step: 判斷用什麼方法
Step 1: Set u = …, du/dx = …
Step 2: Converting，
Step 3: Solving，
Step 4: 將 u 用 x, y 代回來
129
整理：Methods of solving the           1st      order DE

(1)Direct computation            破解法： 直接積分
dy
條件：  f  x 
dx
(2) Separable variable           破解法： x, y 各歸一邊後積分
dy
條件：  g  x  h  y 
破解法：算 e 
dx                                          P ( x ) dx

(3) Linear DE
ye                   P ( x ) dx f  x  dx  ce  P ( x ) dx
dy                                P ( x ) dx
條件：   P  x  y  f  x 
dx                                              e
(4) Exact equation               破解法：double N method
dy     M  x, y 
條件：                                 f
dx      N  x, y          先處理     M  x, y 
x
(或反過來)
M  x, y  N  x, y 
                再處理 f  N  x, y 
y           x               y
( N x M y )
130
  y   e
dy
(4-1) Exact equation 變型                    破解法：                    M

dy     M  x, y 
條件：                                      dy      y  M  x, y 
dx     N  x, y                                           is exact
dx      y  N  x, y 
( M y  N x ) / M independent of x
(M y Nx )

  x   e
dx
(4-2) Exact equation 變型                    破解法：                    N

dy     M  x, y 
條件：  
dx     N  x, y                   dy      x  M  x, y 
                    is exact
( M y  N x ) / N independent of y            dx      x  N  x, y 

(5) Homogeneous equation                   破解法：u = y/x, (y = xu)
dy    M  x, y                             dy  udx  xdu
條件：    
dx    N  x, y 
再用 separable variable method
M  tx, ty   t M  x, y 


N  tx, ty   t  N  x, y 
131
(6) Bernoulli’s Equation           破解法： u = y1–n
n
dy                                 dy    1 1n du
條件：   P  x  y  f  x  y
n                u
dx                                 dx 1  n      dx
再用 linear DE 的方法

(7) Ax + By + C                    破解法： u = Ax + By + c
dy
條件：  f  Ax  By  C                   dy 1 du A
    
dx                               dx B dx B

(b) Exercises in Review 2多練習
(c) 行有餘力，觀察 singular solution 和 singular point
132

Sec. 2-4: 3, 13, 17, 25, 29, 32, 35, 38
Sec. 2-5: 3, 5, 14, 17, 20, 22, 24, 25, 29
Chap. 2 Review: 2, 5 , 8, 9, 10, 11, 14, 15, 16, 19
133
Chapter 4 Higher Order Differential Equations

dny
Highest differentiation:      n
, n>1
dx

Most of the methods in Chapter 4 are applied for the linear DE.
134
附錄三 DE 的分類

Homogeneous
Constant coefficients
Nonhomogeneous
Linear
Homogeneous
Cauchy-Euler
DE                                        Nonhomogeneous

Others                  Homogeneous

Nonhomogeneous
Nonlinear
附錄四 Higher Order DE 解法                             135
reduction of order
homogeneous
auxiliary function
linear            part
Cauchy-Euler
“guess” method
particular
annihilator
solution
variation of parameters
multiple linear DEs       elimination method
reduction of order
nonlinear      Taylor series
numerical method
Laplace transform
series solution
both                                         Fourier series
transform
(but mainly linear)                          Fourier cosine series
Fourier sine series
Fourier transform
136
4-1 Linear Differential Equations:
Basic Theory
4.1.1 Initial-Value and Boundary Value Problems
4.1.1.1 nth order Initial Value Problem

i.e., nth order linear DE with IVP at the same point

dny            d n1 y                    dy
an  x  n  an1  x  n1          a1  x        a0  x  y  g  x 
dx             dx                         dx
y  x0   y0     y  x0   y1     y  x0   y2

……………….. y ( n1)  x0   yn1

initial condition
137
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 136, 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)

Otherwise, the solution is either non-unique or does not exist.

(infinite number of solutions) (no solution)
138
Example 1 (text page 118)
3 y  5 y  y  7 y  0   y (1)  0    y(1)  0      y(1)  0

Example 2 (text page 119)
y  4 y  12 x               y (0)  4       y(0)  1

 x y  2 xy  2 y  6
2
y (0)  3      y(0)  1
有無限多組解
y  cx2  x  3          c 為任意之常數
139
 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
or     n y ( n ) ( x0 )  0
n 0
(general initial condition)
4.1.1.2 nth Order Boundary Value Problem                          140

Boundary conditions are specified at different points

例子： a2  x  y  a1  x  y  a0  x   g  x 

subject to      y (a )  y0 ,   y (b)  y1

或           y(a )  y0 ,    y (b)  y1

1 y (a)  1 y(a)   1
或         
 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.
141
Example 3 (text page 120)
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)
142
4.1.2 Homogeneous Equations
4.1.2.1 Definition
dny            d n1 y               dy
an  x  n  an1  x  n1        a1  x   a0  x  y  g  x 
dx             dx                    dx
g(x) = 0            homogeneous
g(x)  0            nonhomogeneous

The associated homogeneous equation of a nonhomogeneous DE:
setting g(x) = 0

solving the 1st order
• Review:
nonhomogeneous
linear DE
143
4.1.2.2 New Notations

Notation: D y 
n   dny
dx n

d2y     dy    可改寫成 2              可改寫成
2
 5  6y     D y  5Dy  6 y      ( D2  5D  6) y
dx      dx
可再改寫成

L( y )
L  D 2  5D  6
144
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     cn yn

可以和矩陣的概念相比較
145
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)
146
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.

147
Definition 4.2 Wronskian

 y1            y2          yn 
 y             
y2          yn 

W  y1 , y2 ,   , yn   det  1                               
                                 
 y ( n1)     y2n1)
(
ynn1) 
(
 1                               

W  y1 , y2 ,   , yn   0                linearly independent
4.1.2.4 Examples                                                                     148

Example 9 (text page 125)
y  6 y  11y  6 y  0
y1 = ex, y2 = e2x, and y3 = e3x are three of the solutions
Since
 y1 y2      y3  e x      e2 x       e3 x                1 1 1 
                       
det  y1 y2
        y3   e x
             2e 2 x    3e3 x   e x2 x3 x 1 2 3  2e6 x  0
                                                                
  
 y1 y2      y3  e x
          4e 2 x    9e 
3x
               1 4 9 
      
Therefore, y1, y2, and y3 are linear independent for any x

general solution:
y  c1e x  c2e2 x  c3e3 x          x  (−, )
149
4.1.3 Nonhomogeneous Equations (可和 page 53 相比較)
Nonhomogeneous linear DE
an  x  y ( n )  x   an1  x  y ( n1)  x       a1  x  y( x)  a0  x  y  g  x 
Part 1                                                     Part 2
Associated homogeneous DE                                          particular solution y p
an  x  y ( n )  x   an1  x  y ( n1)  x                  (any solution of the
nonhomogeneous linear DE)
      a1  x  y( x)  a0  x  y  0
find n linearly independent solutions                       g  x   g1  x   g2  x            gk  x 
y1  x  , y2  x  ,       , yn  x 
y p  x   y p1  x   y p2  x          y pk  x 

general solution of the nonhomogeneous linear DE
y  x   c1 y1  x   c2 y2  x    cn yn  x   y p  x 
150
Theorem 4.1.6 general solution of a nonhomogeneous linear DE
y  yc  y p

general solution of the associated     particular solution (any solution)
homogeneous function                   of the nonhomogeneous linear
(complementary function)               DE

general solution of the
nonhomogeneous linear DE
Example 10 (text page 126)                                       151
y  6 y  11y  6 y  3x

y  6 y  11y  6 y  0          Particular solution
Three linearly independent                 y p   11  1 x
solution                                           12 2
e x , e 2 x , e3x
Check by Wronskian (Example 9)
ex    e2 x      e3 x
ex    2e 2 x   3e3 x  2e6 x
ex    4e 2 x   9e3 x

General solution: y  c1e x  c2e2 x  c3e3 x  11  1 x
12 2
152
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       a1  x  y  x   a0  x  y  x   g 2  x 
:
y pk  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   g k  x 

then y p1  x   y p2  x           y pk  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   g 2  x               gk  x 
153
Example 11 (text page 127)
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   e 2 x   is a particular solution of y  3 y  4 y  2e
2x

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
154
4.1.4 名詞

 initial conditions, boundary conditions (pages 136, 140)
 associated homogeneous equation , complementary function (page 142)
 fundamental set of solutions (page 145)
 Wronskian (page 147)
 particular solution (page 149)
 general solution of the homogenous linear DE (page 145)
 general solution of the nonhomogenous linear DE (page 149)
an  t  y ( n )  t   an1  t  y ( n1)  t      a1  t  y  t   a0 t  y t   g t 
155
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
156
(補充 1) Theorem 4.1.1 的解釋

dny            d n1 y                             dy
an  x  n  an1  x  n1                   a1  x        a0  x  y  g  x 
dx             dx                                  dx

y  x0   y0       y  x0   y1 ………………..                      y ( n1)  x0   yn1

When an(x0)  0
an1  x0  ( n1)              a1  x0              a  x               g  x0 
y   (n)
 x0              y        x0                y   x0   0        y  x0  
an  x0                       an  x0              an  x0             an  x0 

find y(n)(x0)
y ( n1)  x0     y ( n1)  x0   y ( n )  x0                              find y(n−1)(x0+)
f t     f t 
(根據 f   t                                   ,       f t     f t   f  t   )

157
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                              find y(n−3)(x0+)
:
:
y  x0     y  x0   y  x0                                      find y(x0+)

an1  x0    ( n1)                        a1  x0   
y    (n)
 x0                    y        x0                            y  x0   
an  x0                                   an  x0   
a0  x                    g  x0   
                  y  x0                                                          find y(n)(x0+)
an  x0                   an  x0   

y ( n1)  x0  2   y ( n1)  x0     y ( n )  x0                              find y(n−1)(x0+2)
find y(n−2)(x0+2)
y   ( n2)
 x0  2   y      ( n2)
 x0     y   ( n1)
 x0    
158
:
:
y  x0  2   y  x0     y  x0                                   find y(x0+2)

an1  x0  2  ( n1)                     a1  x0  2 
y ( n )  x0  2                     y        x0  2                       y  x0  2 
an  x0  2                              an  x0  2 
a0  x  2                   g  x0  2 
                y  x0  2  
an  x0  2                  an  x0  2 

(求 y(x) for x > x0 時, 用正的  值，

159

Requirement 1: a0(x), a1(x), a2(x), ……., an −1(x ), an(x) are continuous

Requirement 2: an(x)  0 是為了讓 ak(x0+m) /an(x0+m) 不為無限

160
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.
161
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                               
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
162
uy1  u(2 y1  P  x  y1 )  0
                        set w = u'

dw         dy
y1  w(2 1  P  x  y1 )  0       multiplied by dx/(y1w)
dx         dx
dw    dy
 2 1  P  x  dx  0                 separable variable
w     y1                                 (with 3 variables)

dw      dy
w   2  1   P  x  dx  0
y1

ln w  c3  2ln y1  c4    P  x  dx

ln w  2ln y1  ln w  ln y1  ln w y1  ln wy12
2        2

ln wy12    P  x  dx  c
163
ln wy12    P  x  dx  c

wy  e 
2        P x dx c
1

w  c1e 
 P x dx
/ y12

e 
 P x dx

u   wdx  c1                      dx  c2
y12                        We can setting c1 = 1 and c2 = 0
(因為我們算 u(x) 的目的，只是為
e 
 P x dx

y2  x   y1  x   2        dx            了要算出與 y1(x) 互相independent
y1  x                的另一個解)
164
4.2.3 例子

Example 1 (text page 131)
y  y  0

We have known that y1 = ex is one of the solution
1
P(x) = 0      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
165
Example 2 (text page 132)                                  (將課本 x 的範圍做更改)
x2 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   如課本
If x < 0,        dx / x  ln( x)
e3ln(  x )    2 ( x)
3
y2  x   x  4 dx  x 
2
dx
x              ( x) 4

1
  x 2  dx   x 2 ln x
x
y  x   c1 x 2  c2 x 2 ln x
166
4.2.4 本節需注意的地方

e 
 P x dx
(1) 記住公式 y2  x   y1  x 
 y12  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

  P x dx
e
(6) 由於  2              dx 的計算較複雜且花時間，所以要多加練習
y1  x 
多算習題
167
附錄五： Hyperbolic Function

sinh  x   e x  e x                              e jx  e jx
2                   比較： sin  x  
2j

cosh  x   e x  e x                       cos  x   e jx  e jx
2                                             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
168
5                  5                   2

sinh(x)                                tanh(x)
1

0                  0                   0
cosh(x)
-1

-5                 -5                  -2
-2    0   2        -2   0     2        -2    0   2

3                  2                   3
2                       sech(x)        2    csch(x)
coth(x)       1
1                                      1

0                  0                   0
-1                                     -1
-1
-2                                     -2

-3                 -2                  -3
-2    0   2        -2   0     2        -2    0   2
169
d sinh  ax   a cosh  ax                 sinh  0   0
dx
d cosh  ax   a sinh  ax                 cosh  0   1
dx
d tanh  ax   a sech 2  ax               sinh  0   1
dx
d coth  ax   a csch 2  ax              cosh  0   0
dx
d sech  ax   a sech  ax  tanh  ax 
dx                                           sin  ix   i sinh  x 
d csch  ax   a csch  ax  coth  ax    cos  ix   cosh  x 
dx
170
cosh  ax 
 sinh  ax  dx  | a |  c
sinh  ax 
 cosh  ax  dx       |a|
c

ln cosh  ax 
 tanh  ax  dx  | a |  c
ln sinh  ax 
 coth  ax  dx  | a |  c

sech  ax  dx 

2 tan 1 tanh( a x)

2    c
|a|

ln tanh( a x)
 csch  ax  dx           2 c
|a|
171
附錄六 Linear DE 解法的步驟 (參照講義 page 149)

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
172
4-3 Homogeneous Linear Equations with
Constant Coefficients

KK: [                    ]

4-3-1 限制條件

(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)
173
4-3-2 解法
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

174
 解法流程
an y ( n )  x   an1 y ( n1)  x        a1 y( x)  a0 y  0

Step 1-1
auxiliary function an mn  an1mn1                      a1m  a0  0

Step 1-1                 Find n roots , m1, m2, m3, …., mn

(If m1, m2, m3, …., mn are distinct)
m1 x
Step 1-2 n linearly independent solutions e                           , em2 x , em3x ,       , emn x
(有三個 Cases)

Step 1-3 Complementary y  c1e                          c2em2 x  c3e m3 x              cne mn x
m1 x

function
175
4-3-3 Three Cases for Roots               (2nd     Order DE)

a2 y  x   a1 y  x   a0 y  x   0

a2 m2  a1m  a0  0

a1  a  4a2 a0
2                             a1  a12  4a2 a0
roots m1              1                        m2 
2a2                                     2a2

solutions
(或 m1, m2 雖為 complex
Case 1 m1  m2, m1, m2 are real                    但不為 conjugate)
y  c1e           c2e m2 x
m1 x
176
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 
 P x dx

y2  x   y1  x   2        dx
y1  x 

e 
2 m1x  a1 / a2 dx
e           e
m1 x
dx
( 2 m1 
a1
)x                  a1
m1 
e           e
m1 x                      a2
dx
2 a2
e            dx  e  x  c 
m1 x                      m1x

y2  x   xe
m1 x

y  c1e            c2 xe m1x
m1 x
177
Case 3 m1  m2 , m1 and m2 are conjugate and complex

a1  a12  4a2 a0
m1                        j                   m2    j 
2a2

   a1 / 2a2 ,              4a2a0  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 = C1 − jC2
y  e x  c1 cos  x  c2 sin  x    c1 and c2 are some constant
178
Example 1 (text page 135)
(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 xe5 x

(c)   y  4 y  7 y  0
m2 + 4m + 7 = 0,          m1  2  i 3,    m2  2  i 3


y  e2 x c1 cos 3x  c2 sin 3x   
179
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      a1 y( x)  a0 y  0

n1
auxiliary function: an m  an1m                               a1m  a0  0
n

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.

(2) If the multiplicities of mq is k (當這個多項式在 mq 的地方有 k 個根),
, x k 1e
mq x          mq x            mq x                      mq x
e          , xe          , x 2e          ,
are the solutions of the DE.
180
(3) If both  + j and  − j are the roots of the auxiliary function,
then
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.
181

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     a1 (  j  )  a0  0
a0, a1, a2, …. , an are real
182
Example 3 (text page 136)
Solve      y  3 y  4 y  0

Step 1-1 m  3m  4  0
3       2

 m  1  m 2  4m  4   0     m1 = 1,   m2 = m3 = 2

x   2 x  2 x
Step 1-2   3 independent solutions: e , e , xe
2 x     2 x
Step 1-3 general solution: y  c1e  c2e  c3 xe
x
183
Example 4 (text page 137)
Solve       y (4)  x   2 y  x   y  x   0

Step 1-1 m 4  2m 2  1  0
(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
184
4-3-5 How to Find the Roots
(1) Formulas
a1  a12  4a2 a0         a1  a12  4a2 a0
a2 m  a1m  a0  0
2
m1                        m2 
2a2                        2a2

a3m3  a2m2  a1m  a0
a2
m1  S  T 
3a3
a
m2   1  S  T   2  i 1 3  S  T 
2            3a3    2
Solutions:
a
m3   1  S  T   2  i 1 3  S  T 
2            3a3 2

S  3 R  Q3  R3          , T  3 R  Q3  R3

1a1 a22
9a1a2  27a3a0  2a2
3
Q     2            ,   R            2
3a3 9a3                          54a3
185
(2) Observing
例如：1 是否為 root                  看係數和是否為 0

又如：
3m3  5m2  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
186
(3) Solving the roots of a polynomial by software

Maple
Mathematica (by the commands of Nsolve and FindRoot)
Matlab ( by the command of roots)
187
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
188
Exercise for practice

Sec. 4-1: 3, 7, 8, 13, 20, 24, 29, 33, 36
Sec. 4-2: 2, 4, 9, 13, 14, 18, 19
Sec. 4-3: 7, 16, 20, 24, 28, 33, 39, 41, 50, 51
189
4-4 Undetermined Coefficients –
Superposition Approach
This section introduces some method of “guessing” the particular
solution.
4-4-1 方法適用條件
(1)         (2)
Suitable for linear and constant coefficient DE.

an y ( n )  x   an1 y ( n1)  x      a1 y( x)  a0 y  g  x 

(3) g(x), g'(x), g'' (x), g'''(x), g(4)(x), g(5)(x), ………contain
finite number of terms.
190
4-4-2 方法

g(x) 長什麼樣子，particular solution 就應該是什麼樣子.

(計算時要把 A, B, C, … 這些 unknowns 解出來)
191
(from text page 144)

It comes from the “form rule”. See page 196.
192

g  x   e2 x  xe3 x                  yp = ?

g  x   cos  x   x 2 sin  2 x    yp = ?

g  x   cosh  2 x                   yp = ?
193
4-4-3 Examples
Example 2      y  y  y  2sin 3 x   (text page 142)

Step 1: find the solution of the associated homogeneous equation
Guess
Step 2: particular solution
y p  A cos3x  B sin 3x
yp  3 A sin 3x  3B cos3x
y  9 A cos3x  9 B sin 3x
p

y  yp  y p  (8 A  3B)cos3x  (3 A  8B)sin 3x  2sin 3x
p

8 A  3B  0
                              A = 6/73, B = 16/73
 3 A  8B  2
y p  6 cos3x  16 sin 3x
73         73
Step 3: General solution:
                         
y  e x / 2  c1 cos 3 x  c2 sin 3 x   6 cos3 x  16 sin 3 x
        2            2  73             73
194
Example 3      y  2 y  3 y  4 x  5  6 xe   2x
(text page 143)

Step 1: Find the solution of
y  2 y  3 y  0.
yc  c1e3 x  c2e x
Step 2: Particular solution
y  2 y  3 y  4 x  5            y  2 y  3 y  6 xe2 x
guess                                   guess
y p1  Ax  B                              y p2  Cxe 2 x  Ee 2 x
yp1  A                                  yp2  2Cxe 2 x  Ce 2 x  2 Ee 2 x
y1  0
p                                        y2  4Cxe 2 x  4Ce 2 x  4 Ee 2 x
p
3 Ax  2 A  3B  4 x  5              3Cxe2 x   2C  3E  e2 x  6 xe2 x
A   4 , B  23                          C  2, E   4
3         9                                           3
y p1   4 x  23                       y p2  (2 x  4)e2 x
3     9                                       3
195
Particular solution
y p  y p1  y p2   4 x  23  (2 x  4)e x
3     9           3

Step 3: General solution
y  yc  y p
y  c1e3 x  c2e x  4 x  23  (2 x  4)e2 x
3     9           3
196
4-4-4 方法的解釋

Form Rule: yp should be a linear combination of g(x), g'(x),
g'' (x), g'''(x), g(4)(x), g(5)(x), …………….

Why? 如此一來，在比較係數時才不會出現多餘的項
197
When g(x) =   xn
x n  x n1  x n2  x n3                1 0

y p  An x n  An1 x n1  An2 x n2       A0

When g(x) = cos kx
cos kx  sin kx

y p  A1 cos kx  A2 sin kx

When g(x) = exp(kx)
e kx

y p  A exp(kx)
198
When g(x) =   xnexp(kx)
g   x   nx n1ekx  kx nekx
g   x   n(n  1) x n2ekx  2nkx n1ekx  k 2 x ne kx
g   x   n(n  1)(n  2) x n3e kx  3kn(n  1) x n2e kx
3k 2 nx n1e kx  k 3 x ne kx
:
:
會發現 g(x) 不管多少次微分，永遠只出現

x nekx , x n1ekx , x n2ekx , x n3ekx ,          , ekx

y p  cn x n e kx  cn1 x n1e kx  cn2 x n2e kx             c0e kx
199
4-4-5 Glitch of the method:

Example 4      y  5 y  4 y  8e x       (text page 143)

Particular solution guessed by Form Rule:
y p  Ae x
y  5 yp  4 y p  Ae x  5 Ae x  4 Ae x  8e x
p

0  8e x   (no solution)

Why?
200
Glitch condition 1: The particular solution we guess belongs to the
complementary function.

For Example 4       y  5 y  4 y  8e x
Complementary function yc  c1e  c2e                 Ae x  yc
x      4x

解決方法：再乘一個 x
y p  Axe   x     yp  Axe x  Ae x
y  Axe x  2 Ae x
p

y  5 yp  4 y p  3 Ae x  8e x
p
A  8/ 3

y p   8 xe x
3
y  c1e x  c2e4 x  8 xe x
3
201
Example 7      y  2 y  y  e   x
(text page 145)
yc  c1e x  c2 xe x
From Form Rule, the particular solution is Aex
Ae x  yc
Axe x  yc          如果乘一個 x 不夠，則再乘一個 x
y p  Ax 2e x      yp  ( Ax 2  2 Ax)e x
y  ( Ax 2  4 Ax  2 A)e x
p

y  2 yp  y p  2 Ae x  e x
p                                            A  1/ 2
y p  x 2e x / 2

y  c1e x  c2 xe x  x 2e x / 2
202
Example 8 (text page 146)
y  y  4 x  10sin x        y    0      y    2

Step 1       yc  c1 cos x  c2 sin x

Step 2       y p  Ax  B  Cx sin x  Ex cos x

y p  4 x  5 x cos x
Step 3       y  c1 cos x  c2 sin x  4 x  5 x cos x

Step 4       Solving c1 and c2 by initial conditions        (最後才解 IVP)
y    c1  4  5  0              c1  9
y  c1 sin x  c2 cos x  4  5cos x  5 x sin x
y    c  9  2              c2  7
2

y  9 cos x  7sin x  4 x  5 x cos x
203
Example 11 (text page 147)

y (4)  y  1  x 2e x
yc  c1  c2 x  c3 x 2  c4e x

From Form Rule
y p  A  Bx 2e  x  Cxe  x  Ee  x            yp 只要有一部分和 yc 相
同就作修正
修正
y p  Ax3  Bx3e  x  Cx 2e  x  Exe  x           乘上 x
乘上 x3
If we choose    y p  A  Bx 2e  x  Cxe  x  Ee  x
y (4)  y( )  2 Bxe  x  (6 B  C )e  x  1  x 2e  x
p         p

沒有 1, x2ex 兩項，不能比較係數，無解
204
If we choose y p  Ax3  Bx 2e  x  Cxe  x  Ee  x
y (4)  y( )  6 A  2 Bxe  x  (6 B  C )e  x  1  x 2e  x
p         p

沒有 x2ex 這一項，不能比較係數，無解
If we choose y p  Ax3  Bx3e  x  Cx 2e  x  Exe  x
y (4)  y( )
p         p

 6 A  3Bx 2e  x  18B  2C  xe  x  (18B  6C  E )e  x
 1  x 2e x
A = 1/6, B = 1/3, C = 3, E = 12
y p  1 x3  1 x3e x  3x 2e x  12 xe x
6      3
y  c1  c2 x  c3 x 2  c4e x  1 x3  1 x3e x  3x 2e  x  12 xe x
6      3
205
Glitch condition 2: g(x), g'(x), g'' (x), g'''(x), g(4)(x), g(5)(x), ……………
contain infinite number of terms.

If g(x) = ln x
1   1  1
ln x       2 3
x  x  x
If g(x) = exp(x2)
g   x   2 xe x
2

g   x   (4 x  2)e
2        x2

g   x   (8 x 3  12 x)e x
2

:
:
206
4-4-6 本節需要注意的地方

(1) 記住 Table 4.1 的 particular solution 的假設方法
(其實和 “form rule” 有相密切的關聯)
(2) 注意 “glitch condition”
另外， “同一類” 的 term 要乘上相同的東西 (參考 Example
11)
(3) 所以要先算 complementary function，再算 particular solution
(4) 同樣的方法，也可以用在 1st order 的情形
(5) 本方法只適用於 linear, constant coefficient DE

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