1

工程數學--微分方程
Differential Equations (DE)
授課者：丁建均
教學網頁：http://djj.ee.ntu.edu.tw/DE.htm
(請上課前來這個網站將講義印好)

歡迎大家來修課！
 2
授課者：丁建均
Office： 明達館723室, TEL： 33669652
Office hour： 週一至週四的下午皆可來找我
個人網頁：http://disp.ee.ntu.edu.tw/
E-mail: jjding@ntu.edu.tw

上課時間： 星期三 第 3, 4 節 (AM 10:20~12:10)

上課地點： 明達205
課本： "Differential Equations-with Boundary-Value Problem,"
Dennis G. Zill and Michael R. Cullen, 9th edition, 2017.
(metric version, international version)
評分方式：四次作業 15%, 期中考 42.5%, 期末考 42.5%
教學網頁：http://djj.ee.ntu.edu.tw/DE.htm
共同教學網頁： http://cc.ee.ntu.edu.tw/~tomme/DE/DE.html
 3
注意事項：

(1) 本課程採行雙軌制，同學們可以來現場上課，或是可觀看
NTUCool 的影片

(2)請上課前，來這個網頁，將上課資料印好。

http://djj.ee.ntu.edu.tw/DE.htm

(3) 請各位同學踴躍出席 。

(4) 作業不可以抄襲。作業若寫錯但有用心寫仍可以有
40%~90% 的分數，但抄襲或借人抄襲不給分。

(5) 每次作業有11題

(6) 我週一至週四下午都在辦公室，有什麼問題 ，歡迎同學們

來找我
 上課日期 4
Week Number Date (Wednesday) Remark
1. 9/6
2. 9/13
3. 9/20
4. 9/27: HW1
5. 10/4
6. 10/11
7. 10/18: HW2
8. 10/25: Midterms 範圍： (Sections 2-2 ~ 4-5)
9. 11/1
10. 11/8
11. 11/15
12. 11/22: HW3
13. 11/29
14. 12/6
15. 12/13: HW4
16. 12/20: Finals 範圍： (Sections 4-6 ~ 12-4)
 課程大綱 5

Introduction (Chap. 1)
解法 (Chap. 2)
First Order DE 應用 (Chap. 3)
矩陣解 (Chap. 8，範圍外)
解法 (Chap. 4)
Higher Order DE 應用 (Chap. 5，範圍外)
非線性 (Sections 4-10, 5-3, 工數特論)
多項式解法 (Chap. 6)
解法 (Sections 12-1, 12-4)
Partial DE
直角座標 (Chapter 12，工數特論)
圓座標 (Chapter 13，工數特論)

Laplace Transform (Chap. 7 ，範圍外)

Transforms Fourier Series (Chap. 11)
Fourier Transform (Chap. 14，工數特論)
 6
授課範圍

期中考範圍 Sections 1-1, 1-2, 1-3

Sections 2-1, 2-2, 2-3, 2-4, 2-5, 2-6
Sections 3-1, 3-2
Sections 4-1, 4-2, 4-3, 4-4, 4-5

期末考範圍 Sections 4-6, 4-7

Sections 6-1, 6-2, 6-3
Sections 11-1, 11-2, 11-3
Sections 12-1, 12-4

blue colors: 要考的章節

 7
Chapter 1 Introduction to Differential Equations
1.1 Definitions and Terminology (術語)
(1)Differential Equation (DE): any equation containing derivation
(text page 3, definition 1.1)

(2)
dy ( x) x: independent variable 自變數
1
dx y(x): dependent variable 應變數

x d 3 f ( x)
0 sin(t ) f ( x  t )dt  dx3  cos  x 
 8

• Note: In the text book, f(x) is often simplified as f

• notations of differentiation
df d2 f d3 f d4 f
dx , dx 2 , dx 3 , dx 4 , ………. Leibniz notation
f , f  , f  , f ( 4) , ………. prime notation
f , 
f , 
f , 
f , ………. dot notation
fx , f xx , f xxx , f xxxx , ………. subscript notation
 9
(3) Ordinary Differential Equation (ODE):
differentiation with respect to one independent variable

d 3u d 2u du dx dy dz
3
 2  cos(6 x )u  0    2 xy  z
dx dx dx dt dt dt

(4) Partial Differential Equation (PDE):

differentiation with respect to two or more independent variables
 2u  2u x y
 2 0 
x 2
y  t 
 10

(5) Order of a Differentiation Equation: the order of the highest

derivative in the equation
d 7u d 6u d 5u d 4u 7th order
7
2 6 2 5 4 4 0
dx dx dx dx

d2y dy 2nd order

 4  5 y  e x

dx 2 dx
 11
(6) Linear Differentiation Equation:
dny d n1 y dy
an  x  n  an1  x  n1    a1  x   a0  x  y  g  x 
dx dx dx
dy d n1 y d n y
(i) For y, only the terms y, dx ,  , n1 , n appear.
dx dx
(ii) All of the coefficient terms am(x) m = 1, 2, …, n are independent of y.
Property of linear differentiation equations:
d n y1 d n1 y1 dy1
If an  x  n  an1  x  n1    a1  x   a0  x  y1  g1  x 
dx dx dx
d n y2 d n1 y2 dy
an  x  n  an1  x  n1    a1  x  2  a0  x  y2  g 2  x 
dx dx dx
and y3 = by1 + cy2, then
d n y3 d n1 y3 dy
an  x  n  an1  x  n1    a1  x  3  a0  x  y3  bg1  x   cg 2  x 
dx dx dx
(if y(x) is treated as the input and g(x) is the output)
 12
(7) Non-Linear Differentiation Equation

d 2 y dy
( y  3) 2   2 y  x
dx dx
d 2 y dy
  y 2
 e x

dx 2 dx

d 2 y dy
  e y
 e x

dx 2 dx
[Example 1.1.2] Linear and Nonlinear ODEs
(a) The equations

d3y dy
( y  x ) dx  4 xd y  0, y "  2 y  y  0, x 3
 x 3
 5 y  ex
dx dx

are, in turn, linear first-, second-, and third-order ordinary

differential equations. We have just demonstrated that the first
equation is linear in the variable y by writing it in the alternative
form 4xy’ + y = x.
(b) The equations
nonlinear term: nonlinear term: nonlinear term:
coefficient depends on y nonlinear function of y power not 1
d2y d4y
(1  y ) y ' 2 y  e ,
x
2
 si n y  0, an d 4
 y 2
0
dx dx
are examples of nonlinear first-, second-, and fourth-order ordinary
differential equations, respectively.
 14
(8) Explicit Solution (text page 8)
The solution is expressed as y = (x)
(9) Implicit Solution (text page 8)

dy 2
Example:  x ,
dx

Solution: 1 x2  y2  c (implicit solution)

2

y  c  x2 / 2
or (explicit solution)
y   c  x2 / 2
 15

1.2 Initial Value Problem (IVP)

A differentiation equation always has more than one solution.
dy
for 1 ,
dx
y = x, y = x+1 , y = x+2 … are all the solutions of the above
differentiation equation.
General form of the solution: y = x+ c, where c is any constant.

The initial value (未必在 x = 0) is helpful for obtain the unique solution.
dy
 1 and y(0) = 2 y = x+2
dx
dy
 1 and y(2) =3.5 y = x+1.5
dx
 16

The kth order linear differential equation usually requires k independent

initial conditions (or k independent boundary conditions) to obtain the
unique solution.
d2y
2
1
dx solution: y = x2/2 + bx + c,
b and c can be any constant
y(1) = 2 and y(2) = 3 (boundary conditions，在不同點)
y(0) = 1 and y'(0) =5 (initial conditions ，在相同點)
y(0) = 1 and y'(3) =2 (boundary conditions，在不同點)

For the kth order differential equation, the initial conditions can be 0th ~
(k–1)th derivatives at some points.
 17
1.3 Differential Equations as Mathematical
Model
Physical meaning of differentiation:
the variation at certain time or certain place

dx  t  dv  t  d 2 x  t 
[Example 1]: v t   , a t   
dt dt dt 2

F   v  ma dx(t ) d 2 x(t )
F  m
dt dt 2

x(t): location, v(t): velocity, a(t): acceleration

F: force, β: coefficient of friction, m: mass
 18
[Example 2]: 人口隨著時間而增加的模型

dA  t  A: population
 kA  t 
dt 人口增加量和人口呈正比
 19
[Example 3]: 開水溫度隨著時間會變冷的模型
dT T: 熱開水溫度,
 k (T  Tm )
dt
Tm: 環境溫度
t: 時間
 20
大一微積分所學的：

例如：  1 dt  ln t  c
 f  t  dt 的解
t
dA  t 
 f  t   A  t    f  t  dt  c
dt
dA  t  1
Example:  A  t   ln t  c
dt t
dA  t  1 1
 2  At    2 dt  c  ?
dt t 4 t 4
Problems
(1) 若等號兩邊都出現 dependent variable (如 pages 18, 19 的例子)

(2) 若 order of DE 大於 1 (如 page 17 的例子)

該如何解？
 21

Review
• dependent variable and independent variable
• DE
• PDE and ODE
• Order of DE
• linear DE and nonlinear DE
• explicit solution and implicit solution
• initial value; boundary value
• IVP
 22

Chapter 2 First Order Differential Equation

2-1 Solution Curves without a Solution

Instead of using analytic methods, the DE can be solved by graphs (圖解)

dy
slopes and the field directions:  f  x, y 
dx
y-axis
the slope is f(x0, y0)
(x0, y0)

x-axis
 23
Example 1 dy/dx = 0.2xy

From： Fig. 2-1-3(a) in “Differential Equations-with Boundary-Value

Problem”, 9th ed., Dennis G. Zill and Michael R. Cullen.
 24
Example 2 dy/dx = sin(y), y(0) = –3/2

From： Fig. 2-1-4 in “Differential Equations-with Boundary-Value Problem”,

9th ed., Dennis G. Zill and Michael R. Cullen.

With initial conditions, one curve can be obtained

 25
Advantage:
It can solve some 1st order DEs that cannot be solved by
mathematics.

Disadvantage:
It can only be used for the case of the 1st order DE.
It requires a lot of time
 26
Section 2-6 A Numerical Method

• Another way to solve the DE without analytic methods

sampling(取樣)
• independent variable x x0, x1, x2, …………

dy ( x)
• Find the solution of  f  x, y 
dx
Since dy  x   f  x, y  approximation y  xn1   y  xn 
 f  xn , y ( xn ) 
dx xn1  xn

y  xn1   y  xn   f  xn , y ( xn )  xn1  xn 

前一點的值 取樣間格
dy  x  27
 f  x, y  y  xn1   y  xn   f  xn , y ( xn )  xn1  xn 
dx

If 𝑦 𝑥 is known
y  x1   y  x0   f  x0 , y ( x0 )  x1  x0 

y  x2   y  x1   f  x1 , y ( x1 )  x2  x1 

y  x3   y  x2   f  x2 , y ( x2 )   x3  x2 

:
:
:
:
 dy  x  28
 f  x, y  y  xn1   y  xn   f  xn , y ( xn )  xn1  xn 
dx

Example:

• dy(x)/dx = 0.2xy y(xn+1) = y(xn) + 0.2xn y(xn )*(xn+1 –xn).

• dy/dx = sin(x) y(xn+1) = y(xn) + sin(xn)*(xn+1 –xn).

後頁為 dy/dx = sin(x), y(0) = –1,

(a) xn+1 –xn = 0.01, (b) xn+1 –xn = 0.1,
(c) xn+1 –xn = 1, (d) xn+1 –xn = 0.1, dy/dx = 10sin(10x) 的例子

Constraint for obtaining accurate results:

(1) small sampling interval (2) small variation of f(x, y)
 29
(a) 1 (b) 1
0.5 0.5

0 0

-0.5 -0.5

-1 -1

-1.5 -1.5
0 5 10 0 5 10

(c) (d)
1 1

0.5 0.5

0 0

-0.5 -0.5

-1 -1

-1.5 -1.5
0 5 10 0 5 10
 30

Advantages
-- It can solve some 1st order DEs that cannot be solved by mathematics.
-- can be used for solving a complicated DE (not constrained for the 1st
order case)
-- suitable for computer simulation

Disadvantages
-- numerical error (數值方法的課程對此有詳細探討)
附錄一 Table of Integration 31

1/x ln|x| + c
cos(x) sin(x) + c
sin(x) –cos(x) + c
tan(x) –ln|cos(x)| + c
cot(x) ln|sin(x)| + c
ax ax/ln(a) + c
1 1 1 x
x2  a2 tan c
a a

1/ a  x2 2 sin 1 ( x / a )  c

1 / a  x 2 2 cos 1 ( x / a )  c
e ax  1
x eax  x  c
a  a
e ax  2 2 x 2 
x2 eax x   2 c
a  a a 
 32

Exercises for Practicing

(not homework, but are encouraged to practice)
1-1: 1, 13, 19, 23, 37
1-2: 3, 13, 21, 33
1-3: 2, 7, 28
2-1: 1, 13, 25, 33
2-6: 1, 3