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, 30, 36, 46, 48, 50, 54(a) Section 2-3: 7, 9, 13, 15, 21, 29, 30, 33, 36, 40, 42, 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

2-4 Exact Equations 2-4-1 方法的條件 任何 first order DE 皆可改寫成 的型態 (1) 當 成立時， (2) 當 is independent of x 或 is independent of y 可以用Modified Exact Equation Method 來解 (見講義 2-4-5)

2-4-2 方法的來源 Review the concept of partial differentiation Specially, when f(x, y) = c where c is some constant,

補充： 思考：假設一個人在山坡的某處。若往東走，每走一公尺，高度會增加 a 公尺。若往北走，每走一公尺，高度會增加 b 公尺。假設這人現在所在的位置是(0, 0)。那麼這人的東北方，座標為 (p, q) 的地方，高度會比 (0, 0) 高多少？ a  p + b  q (p, q) p q (p, 0) (0, 0) a b

[Definition 2.4.1] We can express any 1st order DE as  If there exists some function f(x, y) that satisfies and , then we call the 1st order DE the exact equation.  The method for checking whether the DE is an exact equation: (Proof): If and , then

For the exact equation, 可改寫成

2-4-3 解法 The method for solving the exact equation (A): (2) (1) (2) g(y) is a constant for x (2) (3) 代入 (3) (2) (4) further computation Solution

整理 Previous Step: Check whether is satisfied. Step 1: Solve Step 2: 將 f(x, y) 代入 ，以解出 g(y) Step 3: Substitute g(y) into Step 4: Further computation and obtain the solution Extra Steps: (a) Consider the initial value problem

The method for solving the exact equation (B): (2) (2) (1) (2) h(x) is a constant for y (3) 代入 (3) (2) (4) further computation Solution

2-4-4 例子 Example 1 (text page 65) Step 0: check whether it is exact

Example 2 (text page 65) Step 0: check for exact Step 1 Step 2 Step 3 Step 3 Step 2 Step 4

要注意 (a) 自行由另一個方向 來練習， 看是否得出同樣的結果。 (b) 得出的解 為 implicit solution (c) 思考：何時用 何時用

Example 3 (text page 66) 自修，但注意 initial value problem, 何時用 (c) 得出的 implicit solution 為 , 範圍：x  (-1, 1) 而 explicit solution 為 , 範圍： x  (-1, 1) 為何 不為解？

2-4-5 Modified Exact Equation Method Technique: Use the integrating factor (x, y) to convert the 1st order DE into the exact equation. such that It is hard to find .

(1) When is a function of y alone: We can set  to be dependent on y alone. Therefore, 用 separable variable 的方法

(2) When is a function of x alone: We can set  to be dependent on x alone. Therefore, 用 separable variable 的方法

前面 2-4-3 (pages 105~107) 的解法流程再加一個步驟： Step 0: Case 1 Yes (小心易背錯) Use the process of 2-4-3 Case 2 Whether Yes Whether Whether No is independent of x No is independent of y Case 3 In Cases 2 and 3, Yes Case 4 No Use other methods Using the process of 2-4-3, but M(x, y) should be modified as M(x, y) N(x, y) should be modified as N(x, y)

(independent of x) (Case 2) Example 4 (text page 67) Step 0: (independent of x) (Case 2) Q: 為何 c 以及  可省略? double N Steps 1~4:

2-4-6 本節需要注意的地方 (1) 使本節方法時，要先將 DE 改成如下的型態 並且假設 2-4-6 本節需要注意的地方 (1) 使本節方法時，要先將 DE 改成如下的型態 並且假設 (2) 對 x 而言，g(y) 是個常數；對 y 而言，h(x) 是個常數 (3) 本節很少有 singular solution 的問題， 但是可能有 singular point 的問題 (4) 背熟三個判別式，二種情況的 integrating factor (小心勿背錯) 並熟悉解法的流程

2-5 Solutions by Substitutions 介紹 3 個特殊解法 Question: 尚有不少的 1st order DE 無法用 Sections 2-2~2-4 的方法來解 本節所提到的特殊解法的共通點： 用新的變數 u 來取代 y 對 症 下 藥

2-5-1 特殊解法 1: Homogeneous Equations If g(tx, ty) = tαg(x, y), then g(x, y) is a homogeneous function of degree α. Which one is homogeneous? g(x, y) = x3 + y3 g(x, y) = x3 + y3 +1 注意：課本中，homogeneous 有兩種定義 一種是 Section 2-3 的定義 (較常用) 一種是這裡的定義 兩者並不相同

If M(x, y) and N(x, y) are homogeneous functions of the same degree,  For a 1st order DE: If M(x, y) and N(x, y) are homogeneous functions of the same degree, then the 1st order DE is homogeneous. It can by solved by setting y = xu, dy = udx + xdu, and use the separable value method. 解法的限制條件 (from page 102)

If is homogeneous then where u = y/x, y = xu 以 t = 1/x 得出 (separable)

Procedure for solving the homogeneous 1st order DE Previous Step: Conclude whether the DE is homogeneous (快速判斷法：看 powers (指數) 之和) Step 1: Set y = ux, dy = udx + xdu 並化簡 Step 2: Convert into the separable 1st order DE Step 3: Solve the separable 1st order DE (用 Sec. 2-2 的方法) Step 4: Substitute u = y/x (別忘了這個步驟)

Example 1 (text page 71) M(x, y) N(x, y) Previous Step: Conclude whether the DE is homogeneous M(tx, ty) = t2M(x, y) N(tx, ty) = t2N(x, y) homogeneous DE Step 1: Set y = ux, dy = udx + xdu 原式 Step 2: Convert into the separable 1st order DE

Step 3: Solve the separable 1st order DE Step 4 代回 u = y/x

2-5-2 特殊解法 2: Bernoulli’s Equations We can set u = y1–n , , and the method of solving the 1st order linear DE to solve the Bernoulli’s equation. so (Chain rule)

Procedure for solving the Bernoulli’s equation Previous Step : Conclude whether the DE is a Bernoulli’s equation Step 1: Set , Step 2: Convert the Bernoulli’s equation into the 1st order linear DE Step 3: Solve the 1st order linear DE (use the method in Sec. 2-3) Step 4: Substitute u = y1–n (別忘了)

Example 2 (text page 77) Previous Step: 判斷 (Bernoulli, n = 2) Step 1: set u = y–1 (y = u–1) (Chain rule) Step 2: Convert into the 1st order linear DE (standard form) 原式 Step 3: Obtain the solution of the 1st order DE Step 4: 代回 u = y–1

2-5-3 特殊解法 3 If the 1st order DE has the form, (B  0) 2-5-3 特殊解法 3 If the 1st order DE has the form, we can set u = Ax + By + C to solve it. (B  0) (解法的限制條件) Since (這式子也許較好記)

Procedure for solving Previous Step: Conclude Step 1: Set Step 2: Converting (轉化成用其他方法可以解出來的 DE 未必一定是轉化成 separable variable DE) Step 3: Solving Step 4: Substitute (別忘了)

Example 3 (text page 77) , Previous Step: 判斷 Step 1: Set Step 2: Converting 原式 Step 3: Obtain the solution (別忘了在運算過程中，代回 u = Ax + By) 注意 的算法 Step 4: 代回 u = Ax + By +C

2-5-4 本節要注意的地方 (1) 對症下藥，先判斷 DE 符合什麼樣的條件，再決定要什麼方法來解 (部分的 DE 可以用兩個以上的方法來解) (2) 別忘了，寫出解答時，要將 u 用 y/x, y1-n, 或 Ax + By + C 代回來 (3) 本節方法皆有五大步驟 Previous Step: 判斷用什麼方法 Step 1: Set u = …, du/dx = … Step 2: Converting， Step 3: Solving， Step 4: 將 u 用 x, y 代回來

整理：Methods of solving the 1st order DE Direct computation (2) Separable variable (3) Linear DE (4) Exact equation 破解法： 直接積分 條件： 破解法： x, y 各歸一邊後積分 條件： 破解法：算 條件： 破解法：double N method 先處理 再處理 條件： (或反過來)

(4-1) Exact equation 變型 (4-2) Exact equation 變型 (5) Homogeneous equation 破解法： 條件： is exact independent of x 破解法： 條件： is exact independent of y 破解法：u = y/x, (y = xu) 再用 separable variable method 條件：

(6) Bernoulli’s Equation (7) Ax + By + C 破解法： u = y1–n 再用 linear DE 的方法 條件： 破解法： u = Ax + By + c 條件： 注意 (a) 速度的訓練 (b) Exercises in Review 2多練習 (c) 行有餘力，觀察 singular solution 和 singular point

練習題 Sec. 2-4: 3, 8, 13, 17, 25, 29, 32, 34, 35, 38, 42 Sec. 2-5: 3, 5, 10, 13, 14, 17, 20, 22, 24, 25, 29, 31 Chap. 2 Review: 2, 13 , 16, 17, 18, 19, 22, 23, 24, 26, 27

Chapter 4 Higher Order Differential Equations Highest differentiation: , n > 1 Most of the methods in Chapter 4 are applied for the linear DE.

附錄四 DE 的分類 Homogeneous Constant coefficients Nonhomogeneous Linear Cauchy-Euler DE Nonhomogeneous Homogeneous Others Nonhomogeneous Nonlinear

附錄五 Higher Order DE 解法 reduction of order homogeneous part auxiliary function linear Cauchy-Euler “guess” method particular solution annihilator variation of parameters multiple linear DEs elimination method reduction of order nonlinear Taylor series numerical method Laplace transform series solution both (but mainly linear) Fourier series transform Fourier cosine series Fourier sine series Fourier transform

4-1 Linear Differential Equations: Basic Theory 4.1.1 Initial-Value and Boundary Value Problems 4.1.1.1 The nth Order Initial Value Problem i.e., the nth order linear DE with the constraints at the same point ………….. ……………….. n initial conditions

Theorem 4.1.1 For an interval I that contains the point x0  If a0(x), a1(x), a2(x), ……., an −1(x ), an(x) are continuous at x = x0  an(x0)  0 (很像沒有 singular point 的條件) then for the problem on page 139, the solution y(x) exists and is unique on the interval I that contains the point x0 (Interval I 的範圍，取決於何時 an(x) = 0 以及 何時 ak(x) (k = 0 ~ n) 不為continuous) Otherwise, the solution is either non-unique or does not exist. (infinite number of solutions) (no solution)

Example 1 (text page 117) Example 2 (text page 118)  有無限多組解 c 為任意之常數

 比較： There is only one solution x  (0, )  Note: The initial value can also be the form as: (general initial condition)

4.1.1.2 nth Order Boundary Value Problem Boundary conditions are specified at different points 比較：Initial conditions are specified at the same points 例子： subject to 或 或 An nth order linear DE with n boundary conditions may have a unique solution, no solution, or infinite number of solutions.

Example 3 (text page 119) solution: (1) c2 is any constant (infinite number of solutions) (2) (unique solution)

4.1.2 Homogeneous Equations 4.1.2.1 Definition g(x) = 0 homogeneous g(x)  0 nonhomogeneous  重要名詞：Associated homogeneous equation The associated homogeneous equation of a nonhomogeneous DE: Setting g(x) = 0 Review: Section 2-3, pages 53, 55

4.1.2.2 New Notations Notation: 可改寫成 可改寫成 可再改寫成

4.1.2.3 Solution of the Homogeneous Equation [Theorem 4.1.5] For an nth order homogeneous linear DE L(y) = 0, if  y1(t), y2(t), ….., yn(t) are the solutions of L(y) = 0  y1(t), y2(t), ….., yn(t) are linearly independent then any solution of the homogeneous linear DE can be expressed as: 可以和矩陣的概念相比較

From Theorem 4.1.5: An nth order homogeneous linear DE has n linearly independent solutions. Find n linearly independent solutions == Find all the solutions of an nth order homogeneous linear DE y1(t), y2(t), ….., yn(t): fundamental set of solutions : general solution of the homogenous linear DE (又稱做 complementary function) 也是重要名詞

Definition 4.1 Linear Dependence / Independence If there is no solution other than c1 = c2 = ……. = cn = 0 for the following equality then y1(t), y2(t), ….., yn(t) are said to be linearly independent. Otherwise, they are linearly dependent. 判斷是否為 linearly independent 的方法: Wronskian

Definition 4.2 Wronskian linearly independent

4.1.2.4 Examples Example 9 (text page 124) y1 = ex, y2 = e2x, and y3 = e3x are three of the solutions Since Therefore, y1, y2, and y3 are linear independent for any x general solution: x  (−, )