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

Simplest method for solving the 1st order DE: Direct Integration dy(x)/dx = f(x) where

Table of Integration 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 x eax x2 eax

2-2 Separable Variables 2-2-1 方法的限制條件 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).

條件： dy(x)/dx = g(x)h(y)

2-2-2 解法 If , then where p(y) = 1/h(y) 分離變數 Step 1 where 個別積分 Step 2 2-2-2 解法 If , then where p(y) = 1/h(y) where 分離變數 Step 1 個別積分 Step 2 Extra Step: (a) Initial conditions (b) Check the singular solution (i.e., the constant solution)

Extra Step (b) Check the singular solution: Suppose that y is a constant r solution for r See whether the solution is a special case of the general solution.

2-2-3 Examples Example 1 (text page 46) (1 + x) dy – y dx = 0 Extra Step (b) check the singular solution Step 1 set y = r , 0 = r/(1+x) r = 0, y = 0 Step 2 (a special case of the general solution)

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

Example 2 (with initial condition and implicit solution, text page 46) , y(4) = –3 Extra Step (b) Step 1 check the singular solution Step 2 Extra Step (a) (implicit solution) invalid valid (explicit solution)

Example 3 (with singular solution, text page 47) Extra Step (b) check the singular solution Step 1 set y = r , 0 = r2 – 4 r = 2, y = 2 Step 2 or y = 2

Example 4 (text page 47) 自修 注意如何計算 ,

Example in the bottom of page 48 , y(0) = 0 Step 1 Extra Step (b) Check the singular solution Step 2 Extra Step (a) Solution: or 其實，還有更多的解

, y(0) = 0 solutions: (1) (2) (3) b  0  a

2-2-4 IVP 是否有唯一解？ 這個問題有唯一解的條件：(Theorem 1.2.1, text page 15) 如果 f(x, y), 　　　　　 在 x = x0, y = y0 的地方為 continuous 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

2-2-5 Solutions Defined by Integral (1) (2) If dy/dx = g(x) and y(x0) = y0, then 積分 (integral, antiderivative) 難以計算的 function， 被稱作是 nonelementary 如 , 此時，solution 就可以寫成 的型態

Example 5 Solution 或者可以表示成 complementary error function

 error function (useful in probability)  complementary error function See text page 59 in Section 2.3

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 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 http://integrals.wolfram.com/index.jsp 輸入數學式，就可以查到積分的結果 範例： 附錄二 微分方程查詢 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” 就可以算出積分的結果

(d) 有時，對於一些較複雜的數學式，下方還有連結，點進去就可以看到相關的解說

其他有用的網站 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

2-3 Linear Equations 2-3-1 方法的適用條件 “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: g(x) = 0: homogeneous g(x)  0: nonhomogeneous

Standard form: 許多自然界的現象，皆可以表示成 linear first order DE

2-3-2 解法的推導 子問題 2 子問題 1 Find the general solution yc(x) 2-3-2 解法的推導 子問題 2 子問題 1 Find the general solution yc(x) (homogeneous solution) Find any solution yp(x) (particular solution) Solution of the DE

 yc + yp is a solution of the linear first order DE, since 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 Thus, y − yp should be the solution of y should have the form of y = yc + yp

Solving the homogeneous solution yc(x) (子問題一） separable variable Set , then

Solving the particular solution yp(x) (子問題二） Set yp(x) = u(x) y1(x) (猜測 particular solution 和 homogeneous solution 有類似的關係) equal to zero

solution of the linear 1st order DE: where c is any constant : integrating factor

2-3-3 解法 (Step 1) Obtain the standard form and find P(x) 2-3-3 解法 (Step 1) Obtain the standard form and find P(x) (Step 2) Calculate (Step 3) The standard form of the linear 1st order DE can be rewritten as: (Step 4) Integrate both sides of the above equation remember it (Extra Step) (a) Initial value (c) Check the Singular Point or remember it, skip Step 3

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

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

2-3-4 例子 Example 2 (text page 55) Step 1 Extra Step (c) check the singular point Step 2 為何在此時可以將 –3x+c 簡化成 –3x? Step 3 Step 4 或著，跳過 Step 3，直接代公式

Example 3 (text page 56) Extra Step (c) check the singular point Step 1 x = 0 Step 2 思考： x < 0 的情形 若只考慮 x > 0 的情形, Step 3 Step 4 x 的範圍: (0, )

Example 4 (text page 57) Extra Step (c) check the singular point defined for x  (–, –3), (–3, 3), (3, ) not includes the points of x = –3, 3

Example 6 (text, page 58) check the singular point 0  x  1 x > 1 要求 y(x) 在 x = 1 的地方為 continuous from initial condition

2-3-5 名詞和定義 (1) transient term, stable term 2-3-5 名詞和定義 (1) transient term, stable term Example 5 (text page 58) 的解為 : transient term 當 x 很大時會消失 x 1: stable term y x1 x-axis

(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 complementary error function

(5) sine integral function Fresnel integral function (6) f(x) 常被稱作 input 或 deriving function Solution y(x) 常被稱作 output 或 response

2-3-6 小技巧 When is not easy to calculate: Try to calculate Example: 2-3-6 小技巧 When is not easy to calculate: Try to calculate Example: (not linear, not separable) (linear) (implicit solution)

2-3-7 本節要注意的地方 (1) 要先將 linear 1st order DE 變成 standard form 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 或 (4) 計算時， 的常數項可以忽略

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……

Chapter 3 Modeling with First-Order Differential Equations 應用題 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)

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) 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, 翻譯  k is a constant 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)

A(0) = P0, A(1) = 3P0/2 可以用 什麼方法解？ Extra Step (b) Step 1 check singular solution Step 2 Extra Step (a) (1) c = P0 (2) k = ln(3/2) = 0.4055 針對這一題的問題

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) 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 k is a constant

T(0) = 300 T(3) = 200 課本用 separable variable 的方法解 如何用 linear 的方法來解？

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

LR series circuit From Kirchhoff’s second law

RC series circuit q: 電荷

How about an LRC series circuit?

Example 6 (text page 88) LR series circuit  E(t): 12 volt,  inductance: 1/2 henry,  resistance: 10 ohms,  initial current: 0 這裡 + c1 可省略

Circuit problem for t is small and t  For the LR circuit: L R transient stable For the RC circuit: R C transient stable

3-2 Nonlinear Models 3-2-1 Logistic Equation 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 The solution of a logistic equation is called the logistic function. Two stable conditions: and .

Logistic curves for differential initial conditions (b) Logistic curves for differential initial conditions

Solving the logistic equation separable variable 註： (with initial condition P(0) = P0) logistic function

Example 1 (text page 96) There are 1000 students.  Suppose a student carrying a flu virus returns to an isolate college campus of 1000 students.  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, 翻譯  x(0) = 1 翻譯  k is a constant  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 Initial: x(0) = 1, x(4) = 50 find x(6) 可以用separable variable 的方法

Logistic equation 的變形 (1) 人口有遷移的情形 遷出的人口和人口量呈正比 (2) 人口越多，遷入的人口越少 (3) (4) Gompertz DE 飽合人口為 人口增加量，和 呈正比

3-2-2 化學反應的速度 A + B  C Use compounds A and B to for compound C 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 See Example 2

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 They should be solved by the Laplace Transform and other methods

from Kirchhoff’s 1st law from Kirchhoff’s 2nd law (1) (2) Three dependent variable We can only simplify it into two dependent variable

from Kirchhoff’s 1st law from Kirchhoff’s 2nd law (1) (2)

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………………

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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頁碼 作品 版權圖示 來源/作者 台灣大學 電信工程研究所 丁建均教授 以創用CC「姓名標示－非商業性－相同方式分享」臺灣3.0版授權釋出。 86 台灣大學 電信工程研究所 丁建均教授 以創用CC「姓名標示－非商業性－相同方式分享」臺灣3.0版授權釋出。 94 95