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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： 星期三下午 1:00~5:00
個人網頁：http://disp.ee.ntu.edu.tw/
E-mail: djj@cc.ee.ntu.edu.tw, djj1@ms63.hinet.net
上課時間： 星期三 第 3, 4 節 (AM 10:20~12:10)
星期五 第 2 節 (AM 9:10~10:00)
上課地點： 電二143
課本： "Differential Equations-with Boundary-Value Problem",
7th edition, Dennis G. Zill and Michael R. Cullen
評分方式：四次作業一次小考 10%, 期中考 45%, 期末考 45%
3
注意事項：
(1)請上課前，來這個網頁，將上課資料印好。
http://djj.ee.ntu.edu.tw/DE.htm
(2) 請各位同學踴躍出席 。
(3) 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。
(4) 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, Friday) Remark
1. 9/14, 9/16
2. 9/21, 9/23
3. 9/28, 9/30
4. 10/5, 10/7
5. 10/12, 10/14
6. 10/19, 10/21
7. 10/26, 10/28
8. 11/2, 11/4
9. 11/9: Midterm; (Chaps.1-5), 11/11 範圍： (Chaps.1-5)
10. 11/16, 11/18
11. 11/23, 11/25
12. 11/30, 12/2
13. 12/7, 12/9
14. 12/14, 12/16
15. 12/21, 12/23
16. 12/28, 12/30
17. 1/4, 1/6
18. 1/11 Finals 範圍： (Chaps. 6, 7, 8, 11, 12, 14)
課程大綱 5
Introduction (Chap. 1) 解法 (Chap. 2)
First Order DE 應用 (Chap. 3)
矩陣解法 (Chap. 8)
解法 (Chap. 4)
Higher Order DE 應用 (Chap. 5)
多項式解法 (Chap. 6)
Partial DE (Chap. 12)
Laplace Transform (Chap. 7)
Transforms Fourier Series (Chap. 11)
Fourier Transform (Chap. 14)
6
Chapter 1 Introduction to Differential Equations
1.1 Definitions and Terminology (術語)
(1)Differential Equation (DE): any equation containing derivation
(page 2, definition 1.1)
dy ( x) x: independent variable 自變數
1
dx y(x): dependent variable 應變數
d 3 f (t )
0 sin(2 t ) f (t )dt dt 3 g t
t
7
• 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
8
(2) 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 d x dx dt dt dt
(3) Partial Differential Equation (PDE):
differentiation with respect to two or more independent variables
2u 2u x y
2 0
x y
2
t
9
(4) Order of a Differentiation Equation: the order of the highest
derivative in the equation
d 7u d 6u du
2 6 3 u 0 7th order
dx 7 dx dx
d2y dy
5 4 y ex 2nd order
dx 2 dx
10
(5) Linear Differentiation Equation:
dny d n1 y dy
an x n an1 x n1 a1 x a0 x y g x
dx dx dx
All the coefficient terms are independent of y.
Property of linear differentiation equations:
d n y1 d n1 y1 dy
If an x n an1 x n1 a1 x 1 a0 x y1 g1 x
dx dx dx
d n y2 d n1 y2 dy2
an x n an1 x n1 a1 x a0 x y2 g 2 x
dx dx dx
and y3 = by1 + cy2, then
d n y3 d n1 y3 dy3
an x n an1 x n1 a1 x a0 x y3 bg1 x cg 2 x
dx dx dx
11
(6) Non-Linear Differentiation Equation
d 2 y dy
( y 3) 2 2 y x
dx dx
d 2 y dy
y2 ex
dx 2 dx
d 2 y dy
e y ex
dx 2 dx
12
(7) Explicit Solution (page 6)
The solution is expressed as y = (x)
(8) Implicit Solution (page 7)
dy 2
Example: x ,
dx
Solution: 1 x2 y 2 c (implicit solution)
2
y c x2 / 2
or (explicit solution)
y c x2 / 2
13
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
14
The kth order differential equation usually requires k initial conditions (or
k boundary conditions) to obtain the unique solution.
d2y
2
1 solution: y = x2/2 + bx + c,
dx
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)
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.
15
1.3 Differential Equations as Mathematical
Model
Physical meaning of differentiation:
the variation at certain time or certain place
Example 1:
dA t A: population
kA t
dt 人口增加量和人口呈正比
16
Example 2:
dT T: 熱開水溫度,
k (T Tm )
dt
Tm: 環境溫度
t: 時間
17
大一微積分所學的：
例如： 1 dt ln t c
t
f t dt 的解
1
t 2 4 dt ?
dA t 1
A t ln t c
dt t
問題：
(1) 若等號兩邊都出現 dependent variable (如 pages 15, 16 的例子)
(2) 若order of DE 大於 1
d 2 A(t ) dA(t )
2
2 1
dt dt
18
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
• IVP
19
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
20
Example 1 dy/dx = 0.2xy
資料來源： Fig. 2-1-3(a) of “Differential Equations-with Boundary-
Value Problem”, 7th ed., Dennis G. Zill and Michael R. Cullen.
21
Example 2 dy/dx = sin(y), y(0) = –3/2
資料來源： Fig. 2-1-4 of “Differential Equations-with Boundary-Value
Problem”, 7th ed., Dennis G. Zill and Michael R. Cullen.
With initial conditions, one curve can be obtained
22
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
23
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 xn1 y xn
f xn , y ( xn )
dx xn1 xn
y xn1 y xn f xn , y ( xn ) xn1 xn
前一點的值 取樣間格
24
• 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)
25
(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
26
Advantages
-- can be used for solving a complicated DE (not constrain for the 1st
order case)
-- suitable for computer simulation
Disadvantages
-- more time for computation
-- numerical error (數值方法的課程對此有詳細探討)
27
Exercises for Practicing
(not homework, but are encouraged to practice)
1-1: 1, 13, 19, 23, 33
1-2: 3, 13, 21, 33
1-3: 2, 7, 28
2-1: 1, 13, 20, 25, 33
2-6: 1, 3
28
附錄一 Methods of Solving the First Order Differential Equation
graphic method
direct integration
numerical method
separable variable
analytic method
method for linear equation
method for exact equation
homogeneous equation method
Bernoulli’s equation method
method for Ax + By + c
series solution
Laplace transform
transform Fourier series
Fourier cosine series
Fourier sine series
Fourier transform
29
Simplest method for solving the 1st order DE:
Direct Integration
dy(x)/dx = f(x)
y x f ( x )dx
dF ( x)
F ( x) c where f ( x)
dx
Table of Integration 30
1/x ln|x| + c
cos(x) sin(x) + c
sin(x) –cos(x) + c
tan(x) ln|sec(x)| + c
cot(x) ln|sin(x)| + c
ax ax/ln(a) + c
1 1 1 x
tan c
x2 a2 a a
1 x
sin 1 c
a2 x2 a
e ax 1
x eax x c
a a
eax 2 2 x 2
x2 eax x 2 c
a a a
31
2-2 Separable Variables
2-2-1 方法的限制條件
1st order DE 的一般型態: dy(x)/dx = f(x, y)
[Definition 2.2.1] (text page 45)
If dy(x)/dx = f(x, y) and f(x, y) can be separate as
f(x, y) = g(x)h(y)
i.e., dy(x)/dx = g(x)h(y)
then the 1st order DE is separable (or have separable variable).
32
條件： dy(x)/dx = g(x)h(y)
dy
cos( x)e x2 y
dx
dy
x y
dx
2-2-2 解法 33
dy
If g ( x ) h( y ) , then
dx
dy
Step 1 h( y ) g ( x )dx 分離變數
p( y )dy g ( x)dx where p(y) = 1/h(y)
Step 2 p( y)dy g ( x)dx 個別積分
P ( y ) c1 G ( x) c2 dP ( y ) dG ( x)
where p( y) g ( x)
dy dx
P( y ) G ( x) c
Extra Step: (a) Initial conditions
(b) Check the singular solution
34
Extra Step (b) Check the singular solution:
Suppose that y is a constant r
dy
g ( x ) h( y )
dx
0 g ( x ) h( r )
h( r ) 0
solution for r
See whether the solution is a special case of the general solution.
35
2-2-3 Examples
Example 1 (text page 46)
(1 + x) dy – y dx = 0 Extra Step (b)
dy y check the singular
dy dx dx 1 x solution
Step 1
y 1 x set y = r ,
0 = r/(1+x)
Step 2 ln y ln 1 x c1
r = 0,
y eln 1 x ec1 y ec1 eln 1 x y=0
(a special case of the
y ec1 1 x ec1 (1 x) general solution)
y c(1 x) c ec1
36
Example 練習小技巧
遮住解答和筆記，自行重新算一次
(任何和解題有關的提示皆遮住)
Exercise 練習小技巧
初學者，先針對有解答的題目作練習
累積一定的程度和經驗後，再多練習沒有解答的題目
將題目依類型分類，多綀習解題正確率較低的題型
動筆自己算，就對了
37
Example 2 (with initial condition and implicit solution, text page 46)
dy x, y(4) = –3
dx y
Extra Step (b)
ydy xdx check the singular solution
Step 1
Step 2 y / 2 x / 2 c
2 2
Extra Step (a)
4.5 8 c, c 12.5
x 2 y 2 25 (implicit solution)
y 25 x 2 invalid
y 25 x 2 valid
(explicit solution)
38
Example 3 (with singular solution, text page 47)
dy
y2 4 Extra Step (b)
dx
check the singular solution
dy dy
Step 1 dx y2 4
y2 4 dx
set y = r ,
1 dy 1 dy
dx
4 y2 4 y2 0 = r2 – 4
Step 2
r = 2,
1 1
ln y 2 ln y 2 x c1
4 4 y = 2
y2
ln 4 x 4c1
y2
y2 1 ce4 x
e4 x 4 c1 ce4 x y2 or y=2
y2 1 ce4 x
c e4c1
39
Example 4 (text page 47)
自修
sin(2 x)
注意如何計算 cos x dx , ye y dy
40
Example in the bottom of page 48
dy
xy1/2, y(0) = 0
dx
Extra Step (b)
Step 1
Check the singular solution
Step 2
Extra Step (a)
Solution: y 1 x 4 or y 0 其實，還有更多的解
16
dy 41
xy1/2 , y(0) = 0
dx
solutions: (1) y 1 x 4 (2) y 0
16
1 x 2 b 2 2 for x b
16
(3) y 0
for b x a b0a
1 2
16 x a
2 2
for x a
42
2-2-4 IVP 是否有唯一解？
dy
f x, y y x0 y0
dx
這個問題有唯一解的條件：(Theorem 1.2.1, text page 15)
如果 f(x, y), f x, y 在 x = x0, y = y0 的地方為 continuous
y
則必定存在一個 h，使得 IVP 在 x0−h < x < x0 +h 的區間當中
有唯一解
hint: 用「圖解」的角度來思考
43
2-2-5 Solutions Defined by Integral
d x
(1)
dx x0 g t dt g x
(2) If dy/dx = g(x) and y(x0) = y0, then
y x y0 g t dt
x
x0
積分 (integral, antiderivative) 難以計算的 function，
被稱作是 nonelementary
x2
如 e , sin x 2
此時，solution 就可以寫成 y x y0 x g t dt
x
的型態
0
dy 44
Example 5 e x2
y 3 5
dx
Solution y x 5 3 e dt
x
t 2
或者可以表示成 complementary error function
y x 5 erfc 3 erfc x
2
45
error function (useful in probability)
2 x t 2
erf x e dt
0
complementary error function
2
erfc x e dt 1 erf x
t 2
x
See text page 59 in Section 2.3
46
2-2-6 本節要注意的地方
(1) 複習並背熟幾個重要公式的積分
(2) 別忘了加 c
並且熟悉什麼情況下 c 可以合併和簡化
(3) 若時間允許，別忘了計算 singular solution
(4) 多練習，加快運算速度
附錄二 微分方程查詢 47
http://integrals.wolfram.com/index.jsp
輸入數學式，就可以查到積分的結果
範例：
(a) 先到integrals.wolfram.com/index.jsp 這個網站
(b) 在右方的空格中輸入數學式，例如
數學式
(c) 接著按 “Compute Online with Mathematica” 48
就可以算出積分的結果
按
結果
(d) 有時，對於一些較複雜的數學式，下方還有連結，點進去就可49
以看到相關的解說
連結
其他有用的網站 50
http://mathworld.wolfram.com/
對微分方程的定理和名詞作介紹的百科網站
http://www.sosmath.com/tables/tables.html
眾多數學式的 mathematical table (不限於微分方程)
http://www.seminaire-sherbrooke.qc.ca/math/Pierre/Tables.pdf
眾多數學式的 mathematical table，包括 convolution, Fourier
transform, Laplace transform, Z transform
軟體當中， Maple, Mathematica, Matlab 皆有微積分結果查詢的
功能
51
2-3 Linear Equations
“friendly” form of DEs
2-3-1 方法的適用條件
[Definition 2.3.1] The first-order DE is a linear equation if it has
the following form:
dy
a1 x a0 x y g x
dx
g(x) = 0: homogeneous
g(x) 0: nonhomogeneous
52
dy
Standard form: P x y f x
dx
dy dy a0 x g x
a1 x a0 x y g x y
dx dx a1 x a1 x
許多自然界的現象，皆可以表示成 linear first order DE
53
2-3-2 解法的推導
dy
P x y f x
dx
子問題 1 子問題 2
dyc dy p ( x)
P x yc 0 P x y p ( x) f x
dx dx
Find the general solution yc(x) Find any solution yp(x)
(homogeneous solution) (particular solution)
Solution of the DE
y x yc x y p ( x)
54
yc + yp is a solution of the linear first order DE, since
d ( yc y p )
P x ( yc y p )
dx
dy dy
c P x yc p P x y p
dx dx
0 f x f x
Any solution of the linear first order DE should have the form yc + yp .
The proof is as follows. If y is a solution of the DE, then
dy dy
P x y p P x yp f x f x 0
dx dx
d ( y yp )
P x( y yp ) 0
dx
dyc
Thus, y − yp should be the solution of P x yc 0
dx
y should have the form of y = yc + yp
55
Solving the homogeneous solution yc(x) (子問題一）
dyc
P x yc 0
dx
separable variable
dyc
P x dx
yc
ln yc P x dx c1
yc ce
P ( x ) dx
y1 e
P ( x ) dx
Set , then yc cy1
56
Solving the particular solution yp(x) (子問題二）
dy p ( x)
P x y p ( x) f x
dx
Set yp(x) = u(x) y1(x) (猜測 particular solution 和 homogeneous
solution 有類似的關係)
dy1 ( x) du ( x)
u ( x) y1 ( x) P x u ( x) y1 ( x) f x
dx dx
du ( x) dy ( x)
y1 ( x) u ( x) 1 P x y1 ( x) f x
dx dx
equal to zero
du ( x)
y1 ( x) f x
dx
f x f x f x
du ( x) dx u ( x) dx y p ( x) y1 ( x) dx
y1 ( x) y1 ( x) y1 ( x)
57
yp x e [e
P ( x ) dx
yc ce
P ( x ) dx P ( x ) dx
f ( x)]dx
solution of the linear 1st order DE:
y x c e e [e
P ( x ) dx P ( x ) dx
P ( x ) dx
f ( x)]dx
where c is any constant
e : integrating factor
P ( x ) dx
2-3-3 解法 58
(Step 1) Obtain the standard form and find P(x)
(Step 2) Calculate e P ( x ) dx
(Step 3) The standard form of the linear 1st order DE can be rewritten as:
d P ( x ) dx P ( x ) dx
e y e f x
dx
remember it
(Step 4) Integrate both sides of the above equation
e y e f x dx c,
P ( x ) dx P ( x ) dx
ye P ( x ) dx f x dx ce P ( x ) dx
P ( x ) dx
e
or remember it, skip Step 3
(Extra Step) (a) Initial value
(c) Check the Singular Point
59
dy dy
a1 x a0 x y g x P x y f x
dx dx
Singular points: the locations where a1(x) = 0
i.e., P(x)
More generally, even if a1(x) 0 but P(x) or f(x) , then
the location is also treated as a singular point.
(a) Sometimes, the solution may not be defined on the interval
including the singular points. (such as Example 4)
(b) Sometimes the solution can be defined at the singular points,
such as Example 3
60
More generally, even if a1(x) 0 but P(x) or f(x) , then the
location is also treated as a singular point.
Exercise 29
dy
( x 1) y ln x
dx
61
2-3-4 例子
Example 2 (text page 55)
dy
3y 6
dx
Step 1 P( x) 3 Extra Step (c)
check the singular point
Step 2 e P ( x ) dx e 3 x
為何在此時可以將
–3x+c 簡化成 –3x?
d 3 x
e y 6e3 x
dx
Step 3
或著，跳過 Step 3，直接代公式
Step 4 e3 x y 2e3 x c
ye P ( x ) dx f x dx ce P ( x ) dx
P ( x ) dx
e
y 2 ce3 x
Example 3 (text page 56) 62
dy
x 4 y x 6e x
dx
Extra Step (c)
dy y 4
Step 1 4 x 5e x , P x check the singular point
dx x x x=0
e
4
e4ln x x
P ( x ) dx
Step 2
若只考慮 x > 0 的情形, e P ( x ) dx x 4 考慮 x < 0 的情形
d 4
Step 3 x y xe x
dx
Step 4 x4 y ( x 1)e x c
y ( x5 x 4 )e x cx 4 修正： x (, )
x 的範圍: (0, )
Example 4 (text page 57) 63
x 2 9 xy 0
dy
dx Extra Step (c)
check the singular point
dy x
2 y0
dx x 9
x
P x 2
x 9
x
1
dx ln x 2 9
e x 2 9
e 2
| x2 9 |
d
| x2 9 | y 0
dx
| x2 9 | y c
c defined for x (–, –3), (–3, 3), (3, )
y
| x2 9 | not includes the points of x = –3, 3
Example 6 (text, page 58) 64
1, 0 x 1
dy
y f x y 0 0 f x
dx 0, x 1
e P ( x ) dx e x
d x check the singular point
(e y ) e x f x
dx
0x1 x>1
d x d x
(e y ) e x (e y ) 0
dx dx
e x y e x c1 e x y c2
y 1 c1e x y c2e x
要求 y(x) 在 x = 1 的地方
from initial condition 為 continuous
y 1 e x y (e 1)e x
65
2-3-5 名詞和定義
(1) transient term, stable term
Example 5 (text page 58) 的解為 y x 1 5e x
5e x : transient term 當 x 很大時會消失
x 1: stable term
10
8
6
4
y
2
0
x1
-2
0 2 4 6 8 10
x-axis
66
(2) piecewise continuous
A function g(x) is piecewise continuous in the region of [x1, x2] if
g'(x) exists for any x [x1, x2].
In Example 6, f(x) is piecewise continuous in the region of [0, 1)
or (1, )
(3) Integral (積分) 有時又被稱作 antiderivative
(4) error function
2
erf x
x
t 2
e dt
0
complementary error function
2
erfc x e dt 1 erf x
t 2
x
67
(5) sine integral function
x sin(t )
Si x dt
0 t
Fresnel integral function
S x sin t 2 / 2 dt
x
0
(6) dy P x y f x
dx
f(x) 常被稱作 input 或 deriving function
Solution y(x) 常被稱作 output 或 response
68
2-3-6 小技巧
dy
When is not easy to calculate:
dx
dx
Try to calculate
dy
dy 1
Example: (not linear, not separable)
dx x y 2
dx
x y2 (linear)
dy
x y 2 2 y 2 ce y (implicit solution)
69
2-3-6 本節要注意的地方
(1) 要先將 linear 1st order DE 變成 standard form
(2) 別忘了 singular point
注意：singular point 和 Section 2-2 提到的 singular solution 不同
(3) 記熟公式
d P ( x ) dx P ( x ) dx
e y e f x
dx
或
ye P ( x ) dx f x dx ce P ( x ) dx
P ( x ) dx
e
(4) 計算時， e P ( x ) dx 的常數項可以忽略
70
太多公式和算法，怎麼辦？
最上策： realize + remember it
上策： realize it
中策： remember it
下策： read it without realization and remembrance
最下策： rest z…..z..…z……
71
Chapter 3 Modeling with First-Order
Differential Equations
應用題
(1) Convert a question into a 1st order DE.
將問題翻譯成數學式
(2) Many of the DEs can be solved by
Separable variable method or
Linear equation method
(with integration table remembrance)
72
3-1 Linear Models
Growth and Decay (Examples 1~3)
Change the Temperature (Example 4)
Mixtures (Example 5)
Series Circuit (Example 6)
可以用 Section 2-3 的方法來解
Example 1 (an example of growth and decay, text page 83) 73
Initial: A culture (培養皿) initially has P0 number of bacteria.
翻譯 A(0) = P0
The other initial condition: At t = 1 h, the number of bacteria is
measured to be 3P0/2.
翻譯 A(1) = 3P0/2
關鍵句: If the rate of growth is proportional to the number of
bacteria A(t) presented at time t,
dA
翻譯 kA k is a constant
dt
Question: determine the time necessary for the number of bacteria to
triple
翻譯 find t such that A(t) = 3P0
這裡將課本的 P(t) 改成 A(t)
dA 74
kA A(0) = P0, A(1) = 3P0/2 可以用 什麼方法解？
dt
Extra Step (b)
dA check singular solution
Step 1 kdt
A
Step 2 ln A kt c1
A ekt c1
A cekt c ec1
Extra (1) P0 c 1 c = P0
Step (a)
(2) 3P0 / 2 cek k = ln(3/2) = 0.4055
A P0e0.4055t
針對這一題的問題
3P0 P0e0.4055t t ln(3) / 0.4055 2.71h
75
課本用 linear (Section 2.3) 的方法來解 Example 1
思考：為什麼此時需要兩個 initial values 才可以算出唯一解？
Example 4 (an example of temperature change, text page 85) 76
Initial: When a cake is removed from an oven, its temperature is measured at
300 F.
翻譯 T(0) = 300
The other initial condition: Three minutes later its temperature is 200 F.
翻譯 T(3) = 200
question: Suppose that the room temperature is 70 F. How long will it take
for the cake to cool off to 75 F? (註：這裡將課本的問題做一些修改)
翻譯 find t such that T(t) = 75.
另外，根據題意，了解這是一個物體溫度和周圍環境的溫度交互作用的
問題，所以 T(t) 所對應的 DE 可以寫成
dT k T 70 k is a constant
dt
77
dT k T 70 T(0) = 300 T(3) = 200
dt
課本用 separable variable 的方法解
如何用 linear 的方法來解？
78
Example 5 (an example for mixture, text page 86)
Concentration:
2 lb/gal
300 gallons
3 gal/min 3 gal/min
A: the amount of salt in the tank
dA
(input rate of salt) (output rate of salt)
dt
3A
3 2
300
79
80
Figure 3.1.5 LR series circuit
From Kirchhoff’s second law
di
L Ri E t
dt
81
Figure 3.1.6 RC series circuit
q
Ri E t
C
q dq
R E t
C dt
82
C
How about an LRC series circuit?
q dq d 2q
R L 2 E t
C dt dt
83
Example 6 (text page 88) LR series circuit
E(t): 12 volt, inductance: 1/2 henry,
resistance: 10 ohms, initial current: 0
e
1 di di
P(t ) 20
P ( t ) dt
10i 12 20i 24 e 20t c1
2 dt dt
這裡 + c1 可省略
6 d 20t
i (t ) ce 20t 20 t6 20t
e i e c e i 24e20t
5 5 dt
i (0) 0
6
0 c
5
6 6
i(t ) e20t
5 5
84
Circuit problem for t is small and t
For the LR circuit: L R
transient stable
For the RC circuit: R C
transient stable
85
3-2 Nonlinear Models
可以用 separable variable 或其他的方法來解
3-2-1 Logistic Equation
used for describing the growth of population
dP b
aP(1 P) P(a bP)
dt a
The solution of a logistic equation is called the logistic function.
a
Two stable conditions: P 0 and P .
b
86
Figure 3.2.2 Logistic curves for differential initial conditions
87
Solving the logistic equation
dP
P(a bP)
dt
dP separable
dt variable
P(a bP)
1/ a b / a
dP dt
P a bP
b dP ( a bP )
d
註： dP
1 1 dP ln a bP c0
ln P ln a bP t c a bP a bP
a a
P
ln at ac
a bP (with initial condition P(0) = P0)
P ac1 aP0
c1e at P t P t
a bP bc1 e at bP0 (a bP0 )e at
c1 eac logistic function
Example 1 (text page 96) There are 1000 students. 88
Suppose a student carrying a flu virus returns to an isolate college
campus of 1000 students.
翻譯 x(0) = 1
If it is assumed that the rate at which the virus spreads is proportional
not only to the number x of infected students but also to the number of
students not infected,
dx t
翻譯 kx 1000 x k is a constant
dt
determine the number of infected students after 6 days
翻譯 find x(6)
if it is further observed that after 4 days x(4) = 50
整個問題翻譯成 89
dx t
kx 1000 x Initial: x(0) = 1, x(4) = 50
dt
find x(6)
可以用separable variable 的方法
dx t (c2e1000 kt 1) x c21000e1000 kt
90
kx 1000 x
dt
x
1000 (c c2 1 )
dx t 1 ce 1000 kt
kdt
x 1000 x
x 0 1
1000
1 dx dx 1
kdt 1 c
1000 x 1000 x
c 999
dx dx
1000kdt
x x 1000 x
1000 x 4 50
1 999e 1000 kt
ln x ln x 1000 1000kt c1 1000
50
1 999e 4000 k
x
e1000 kt c1
x 1000 1000k 0.9906
x 1000
c2 e1000 kt (c ec1 ) x x 6 276
x 1000 2
1 999e 0.9906t
91
Logistic equation 的變形
dP
(1) P(a bP) h 人口有遷移的情形
dt
dP
(2) P(a bP) cP 遷出的人口和人口量呈正比
dt
dP
(3) P(a bP) ce kP 人口越多，遷入的人口越少
dt
dP
(4) P(a b ln P) Gompertz DE
dt
bP(a / b ln P ) 飽合人口為 ea /b
飽合人口
人口增加量，和 ln
P
呈正比
92
3-2-2 化學反應的速度
A+ B C
• Use compounds A and B to for compound C
• x(t): the amount of C
• To form a unit of C requires s1 units of A and s2 units of B
• a: the original amount of A
• b: the original amount of B
• The rate of generating C is proportional to the product of the
amount of A and the amount of B
dx t
k a s1 x b s2 x
dt
See Example 2
93
3-3 Modeling with Systems of DEs
Some Systems are hard to model by one dependent variable
but can be modeled by the 1st order ordinary differential equation
dx t
g1 t , x, y
dt
dy t
g 2 t , x, y
dt
They should be solved by the Laplace Transform and other
methods
94
from Kirchhoff’s 1st law
i1 t i2 t i3 t
from Kirchhoff’s 2nd law
di t
(1) E t i1R1 L1 2 i2 R2
dt
di3 t
(2) E t i1R1 L2
dt
Three dependent variable
Fig. 3-3-3 in the textbook. We can only simplify it into two
dependent variable
95
from Kirchhoff’s 1st law
i1 t i2 t i3 t
from Kirchhoff’s 2nd law
di1 t
(1) E t L i2 t R
dt
(2) q3 t i2 t R
C
Fig. 3-3-4 in the textbook.
i1 t i2 t R i2 t
1 d
C dt
96
Chapter 3: 訓練大家將和 variation 有關的問題寫成 DE 的能力
……. the variation is proportional to………………
97
練習題
Section 2-2: 4, 7, 12, 13, 18, 21, 25, 28, 32, 46
Section 2-3: 7, 9, 13, 15, 21, 27, 29, 47, 49(a), 50(a)
Section 3-1: 4, 5, 10, 15, 20, 29, 32
Section 3-2: 2, 5, 14, 15
Section 3-3: 12, 13
Review 3: 3, 4, 11, 12
Homework 1 (due: 10/12)
(1)Sec. 2-2 8, (2) Sec. 2-2 20, (3) Sec. 2-2 27, (4) Sec. 2-3 18,
(5) Sec. 2-3 25, (6) Sec. 2-3 30, (7) Sec. 2-3 37 , (8) Sec. 3-1 3,
(9) Sec. 3-1 33, (10) Sec. 3-2 3
