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# 投影片1

VIEWS: 18 PAGES: 70

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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
x2  a2              tan     c
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 46)
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

dy
 cos( x)e x2 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 (i.e., the constant 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 47)
(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

(任何和解題有關的提示皆遮住)

Exercise 練習小技巧

37
Example 2 (with initial condition and implicit solution, text page 48)
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 48)
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 y2 4 y2                                         0 = r2 – 4
Step 2
r = 2,
1           1
ln y  2  ln y  2  x  c1
4           4                                             y = 2

y2
ln        4 x  4c1
y2

y2                                1  ce4 x
 e4 x  4 c1  ce4 x     y2              or   y = 2
y2                                1  ce4 x
c  e4c1
39
Example 4 (text page 49)

sin(2 x)

40
Example in the top of page 50
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   b0a
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


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
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
44
Example 5 (text page 50)

dy
 e x          y  3  5
2

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

http://integrals.wolfram.com/index.jsp
輸入數學式，就可以查到積分的結果

(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

(c) 接著按 “Compute Online with Mathematica”   48

就可以算出積分的結果

按

(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

連結

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

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 

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 3a) 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 3b) Integrate both sides of the above equation
e                y   e                f  x  dx  c,
P ( x ) dx              P ( x ) dx

ye                       P ( x ) dx f  x  dx  ce  P ( x ) dx
 P ( x ) dx
e
or remember it, skip Step 3a
(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 33

dy
( x  1)  y  ln x
dx
61
2-3-4 例子
Example 2 (text page 56)
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   6e3 x
dx     
Step 3

或著，跳過 Step 3，直接代公式
Step 4 e3 x y  2e3 x  c
ye                   P ( x ) dx f  x  dx  ce  P ( x ) dx
 P ( x ) dx
e
y  2  ce3 x
Example 3 (text page 57)                                                    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
 e4ln x  x
P ( x ) dx
Step 2
若只考慮 x > 0 的情形, e  P ( x ) dx  x 4            思考： x < 0 的情形

d 4
Step 3     x y   xe x
dx      

Step 4    x4 y  ( x  1)e x  c
y  ( x5  x 4 )e x  cx 4
x 的範圍: (0, )
Example 4 (text page 58)                                                                       63

 x 2  9   xy  0
dy
dx                                                   Extra Step (c)
check the singular point
dy      x
 2      y0
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 59)                                           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
0x1                   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
x1

-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
2-3-7 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
ye                     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？

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)

Example 1     (an example of growth and decay, text page 84)           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

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

3P0  P0e0.4055t          t  ln(3) / 0.4055  2.71h
75

Example 4 (an example of temperature change, text page 86)                     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.

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 87)

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

LR series circuit

From Kirchhoff’s second law
di
L       Ri  E  t 
dt
81

RC series circuit
q
 Ri  E  t    q: 電荷
C
q    dq
 R  E t 
C    dt
82

How about an LRC series circuit?

q    dq  d 2q
 R  L 2  E t 
C    dt  dt
83
Example 7 (text page 89) 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 )   e20t
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

3-2-1 Logistic Equation
used for describing the growth of population
dP                  a
 P(a  bP)  bP(  P)
dt                  b
The solution of a logistic equation is called the logistic function.
a
Two stable conditions: P  0 and P        .
b
86

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
1       1
ln P  ln a  bP  t  c       註：         dP                 dP  ln a  bP  c0
a       a                             a  bP           a  bP

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 97) 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

dx  t 
 kx 1000  x    Initial: x(0) = 1, x(4) = 50
dt
find x(6)

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
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

 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

Section 2-2:   4, 7, 12, 13, 18, 21, 25, 28, 30, 36, 46, 50, 54(a)
Section 2-3:   7, 9, 13, 15, 21, 29, 33, 36, 40, 53, 55(a), 56(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

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

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