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投影片1

VIEWS: 18 PAGES: 70

  • pg 1
									                                                               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
 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
條件: dy(x)/dx = g(x)h(y)


       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
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 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)
自修


                   sin(2 x)
注意如何計算             cos x dx ,    ye y dy
                                                                        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
這個問題有唯一解的條件:(Theorem 1.2.1, text page 16)
                 
 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
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有
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 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?


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 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
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 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.

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 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
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                  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
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
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
練習題
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
第七版練習題
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

								
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