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Section 2-3 當中沒有 singular point 的條件) then for the problem on page 140, 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 119) Example 2 (text page 120)  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 120) 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 54, 56

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 127) 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  (−, )

4.1.3 Nonhomogeneous Equations (可和 page 56 相比較) Nonhomogeneous linear DE Part 1 Part 2 Associated homogeneous DE particular solution (any solution of the nonhomogeneous linear DE) find n linearly independent solutions general solution of the nonhomogeneous linear DE

Theorem 4.1.6 general solution of a nonhomogeneous linear DE general solution of the associated homogeneous function (complementary function) particular solution (any solution) of the nonhomogeneous linear DE general solution of the nonhomogeneous linear DE

Example 10 (text page 128) Particular solution Three linearly independent solution , , Check by Wronskian (Example 9) General solution:

Theorem 4.1.7 Superposition Principle If is the particular solution of is the particular solution of : then is the particular solution of

Example 11 (text page 129) is a particular solution of

4.1.4 名詞 initial conditions, boundary conditions (pages 140, 144) 4.1.4 名詞 initial conditions, boundary conditions (pages 140, 144) (重要名詞) associated homogeneous equation , complementary function (page 146) (重要名詞) fundamental set of solutions (page 149) Wronskian (page 151) particular solution (page 153) general solution of the homogenous linear DE (page 149) general solution of the nonhomogenous linear DE (page 153)

4.1.5 本節要注意的地方 (1) Most of the theories in Section 4.1 are applied to the linear DE (2) 注意 initial conditions 和 boundary conditions 之間的不同 (3) 快速判斷 linear independent

(補充 1) Theorem 4.1.1 的解釋 ……………….. When an(x0)  0 find y(n)(x0) find y(n−1)(x0+) (根據 , )

以此類推 find y(n−2)(x0+) find y(n−3)(x0+) : find y(x0+) find y(n)(x0+) find y(n−1)(x0+2) find y(n−2)(x0+2)

: find y(x0+2) 以此類推，可將 y(x0+3), y(x0+4), y(x0+5), …………… 以至於將 y(x) 所有的值都找出來。 (求 y(x) for x > x0 時, 用正的  值， 求 y(x) for x < x0 時, 用負的  值)

Requirement 1: a0(x), a1(x), a2(x), …… Requirement 1: a0(x), a1(x), a2(x), ……., an −1(x ), an(x) are continuous 是為了讓 ak(x0+m) 皆可以定義 Requirement 2: an(x)  0 是為了讓 ak(x0+m) /an(x0+m) 不為無限大

4-2 Reduction of Order 4.2.1 適用情形 (1) (2) (3) 4.2.1 適用情形 (1) (2) (3) Suitable for the 2nd order linear homogeneous DE (4) One of the nontrivial solution y1(x) has been known.

4.2.2 解法 假設 先將DE 變成 Standard form If y(x) = u(x) y1(x) , 4.2.2 解法 假設 先將DE 變成 Standard form If y(x) = u(x) y1(x) , (比較 Section 2-3) zero

separable variable (with 3 variables) set w = u' multiplied by dx/(y1w) separable variable (with 3 variables)

We can set c1 = 1 and c2 = 0 (因為我們算 u(x) 的目的，只是為了要算出與 y1(x) 互相independent 的另一個解)

4.2.3 例子 Example 1 (text page 132) We have known that y1 = ex is one of the solution P(x) = 0 Specially, set c = –2, (y2(x) 只要 independent of y1(x) 即可 所以 c 的值可以任意設) General solution:

Example 2 (text page 133) (將課本 x 的範圍做更改) when x  (−, 0) We have known that y1 = x2 is one of the solution Note: the interval of x If x  (0, ) (x > 0), 如課本 If x < 0,

4.2.4 本節需注意的地方 (1) 記住公式 (2) 若不背公式 (不建議)，在計算過程中別忘了對 w(x) 做積分 4.2.4 本節需注意的地方 (1) 記住公式 (2) 若不背公式 (不建議)，在計算過程中別忘了對 w(x) 做積分 (3) 別忘了 P(x) 是 “standard form” 一次微分項的 coefficient term (4) 同樣有 singular point 的問題 (5) 因為 y2(x) 是 homogeneous linear DE 的 “任意解”，所以計算時，常數的選擇以方便為原則 (6) 由於 的計算較複雜且花時間，所以要多加練習 多算習題

附錄六： Hyperbolic Function 比較：

附錄七 Linear DE 解法的步驟 (參照講義 page 153) Step 1: Find the general solution (i.e., the complementary function ) of the associated homogeneous DE (Sections 4-2, 4-3, 4-7) Step 2: Find the particular solution (Sections 4-4, 4-5, 4-6) Step 3: Combine the complementary function and the particular solution Extra Step: Consider the initial (or boundary) conditions

4-3 Homogeneous Linear Equations with Constant Coefficients 本節使用 auxiliary equation 的方法來解 homogeneous DE KK: [             ] 4-3-1 限制條件 限制條件: (1) homogeneous (2) linear (3) constant coefficients a0, a1, a2, …. , an are constants (the simplest case of the higher order DEs)

4-3-2 解法 解法核心： Suppose that the solutions has the form of emx 4-3-2 解法 解法核心： Suppose that the solutions has the form of emx Example: y''(x) 3 y'(x) + 2 y(x) = 0 Set y(x) = emx, m2 emx  3m emx + 2 emx = 0 m2  3m + 2 = 0 solve m 可以直接把 n 次微分用 mn 取代，變成一個多項式 這個多項式被稱為 auxiliary equation

 解法流程 Step 1-1 auxiliary function Step 1-1 Find n roots , m1, m2, m3, …., mn (If m1, m2, m3, …., mn are distinct) Step 1-2 n linearly independent solutions (有三個 Cases) Step 1-3 Complementary function

4-3-3 Three Cases for Roots (2nd Order DE) solutions Case 1 m1  m2, m1, m2 are real (其實 m1, m2 不必限制為 real)

Case 2 m1 = m2 (m1 and m2 are of course real) First solution: Second solution: using the method of “Reduction of Order”

Case 3 m1  m2 , m1 and m2 are conjugate and complex Solution: Another form: set c1 = C1 + C2 and c2 = jC1 − jC2 c1 and c2 are some constant

Example 1 (text page 137) (a) 2m2 − 5m − 3 = 0, m1 = −1/2, m2 = 3 (b) m2 − 10m + 25 = 0, m1 = 5, m2 = 5 (c) m2 + 4m + 7 = 0,

4-3-4 Three + 1 Cases for Roots (Higher Order DE) For higher order case auxiliary function: roots: m1, m2, m3, …., mn (1) If mp  mq for p = 1, 2, …, n and p  q (也就是這個多項式在 mq 的地方只有一個根) then is a solution of the DE. (2) If the multiplicities of mq is k (當這個多項式在 mq 的地方有 k 個根), are the solutions of the DE. 重覆次數

(3) If both  + j and  − j are the roots of the auxiliary function, then are the solutions of the DE. (4) If the multiplicities of  + j is k and the multiplicities of  − j is also k，then

Note: If  + j is a root of a real coefficient polynomial, then  − j is also a root of the polynomial. a0, a1, a2, …. , an are real

Example 3 (text page 138) Solve Step 1-1 m1 = 1, m2 = m3 = 2 Step 1-2 3 independent solutions: Step 1-3 general solution:

Example 4 (text page 138) Solve Step 1-1 four roots: i, i, i, i Step 1-2 4 independent solutions: Step 1-3 general solution:

4-3-5 How to Find the Roots (1) Formulas Solutions: (太複雜了)

(2) Observing 例如：1 是否為 root 看係數和是否為 0 又如： factor: 1,3 factor: 1,2,4 possible roots: 1, 2, 4, 1/3, 2/3, 4/3 test for each possible root find that 1/3 is indeed a root

(3) Solving the roots of a polynomial by software Maple Mathematica (by the commands of Nsolve and FindRoot) Matlab ( by the command of roots)

4-3-6 本節需注意的地方 (1) 注意重根和 conjugate complex roots 的情形 4-3-6 本節需注意的地方 (1) 注意重根和 conjugate complex roots 的情形 (2) 寫解答時，要將 “General solution” 寫出來 (3) 因式分解要熟練 (4) 本節的方法，也適用於 1st order 的情形

4-4 Undetermined Coefficients – Superposition Approach This section introduces some method of “guessing” the particular solution. 4-4-1 方法適用條件 (1) (2) Suitable for linear and constant coefficient DE. (3) g(x), g'(x), g'' (x), g'''(x), g(4)(x), g(5)(x), ………contain finite number of terms.

4-4-2 方法 把握一個原則： g(x) 長什麼樣子，particular solution 就應該是什麼樣子. 記熟下一頁的規則 4-4-2 方法 把握一個原則： g(x) 長什麼樣子，particular solution 就應該是什麼樣子. 記熟下一頁的規則 (計算時要把 A, B, C, … 這些 unknowns 解出來)

Trial Particular Solutions (from text page 146) g(x) Form of yp 1 (any constant) A 5x + 7 Ax + B 3x2 – 2 Ax2 + Bx + C x3 – x + 1 Ax3 + Bx2 + Cx + E sin4x Acos4x + Bsin4x cos4x e5x Ae5x (9x – 2)e5x (Ax + B)e5x x2e5x (Ax2 + Bx + C)e5x e3xsin4x Ae3xcos4x + Be3xsin4x 5x2sin4x (Ax2 + Bx + C)cos4x + (Ex2 + Fx + G)sin4x xe3xcos4x (Ax + B)e3xcos4x + (Cx + E)e3xsin4x It comes from the “form rule”. See page 199.

yp = ? yp = ? yp = ?

4-4-3 Examples Example 2 (text page 144) Step 1: find the solution of the associated homogeneous equation Guess Step 2: particular solution A = 6/73, B = 16/73 Step 3: General solution:

Example 3 (text page 145) Step 1: Find the solution of Step 2: Particular solution guess guess

Particular solution Step 3: General solution

4-4-4 方法的解釋：Form Rule Form Rule: yp should be a linear combination of g(x), g'(x), g'' (x), g'''(x), g(4)(x), g(5)(x), ……………. Why? 如此一來，在比較係數時才不會出現多餘的項

When g(x) = xn When g(x) = cos kx When g(x) = exp(kx)

When g(x) = xnexp(kx) : 會發現 g(x) 不管多少次微分，永遠只出現

4-4-5 Glitch of the method: Example 4 (text page 146) Particular solution guessed by Form Rule: (no solution) Why?

Glitch condition 1: The particular solution we guess belongs to the complementary function. For Example 4 Complementary function 解決方法：再乘一個 x

Example 7 (text page 148) From Form Rule, the particular solution is Aex 如果乘一個 x 不夠，則再乘一個 x

Example 8 (text page 148) Step 1 注意： sinx, cosx 都要 乘上 x Step 2 Step 3 Step 4 Solving c1 and c2 by initial conditions (最後才解 IVP)

Example 11 (text page 149) From Form Rule yp 只要有一部分和 yc 相同就作修正 修正 乘上 x 乘上 x3 If we choose 沒有 1, x2ex 兩項，不能比較係數，無解

If we choose 沒有 x2ex 這一項，不能比較係數，無解 If we choose A = 1/6, B = 1/3, C = 3, E = 12

Glitch condition 2: g(x), g'(x), g'' (x), g'''(x), g(4)(x), g(5)(x), …………… contain infinite number of terms. If g(x) = ln x If g(x) = exp(x2) :

4-4-6 本節需要注意的地方 (1) 記住 Table 4.1 的 particular solution 的假設方法 (其實和 “form rule” 有相密切的關聯) (2) 注意 “glitch condition” 另外，“同一類” 的 term 要乘上相同的東西 (參考 Example 11) (3) 所以要先算 complementary function，再算 particular solution (4) 同樣的方法，也可以用在 1st order 的情形 (5) 本方法只適用於 linear, constant coefficient DE

4-5 Undetermined Coefficients – Annihilator Approach For a linear DE: Annihilator Operator: 能夠「殲滅」 g(x) 的 operator 4-5-1 方法適用條件 (1) Linear， (2) Constant coefficients (3) g(x), g'(x), g'' (x), g'''(x), g(4)(x), g(5)(x), ………contain finite number of terms.

4-5-2 Find the Annihilator Example 1: (text page 153) annihilator: D4 annihilator: D + 3

annihilator: (D − 2)2 (D − 2)2 = D2 − 4D + 4 註：當各個微分項的 coefficients 皆為 constants 時，function of D 的計算方式和 function of x 的計算方式相同 (x − 2)2 = x2 − 4x + 4  (D − 2)2 = D2 − 4D + 4

General rule 1: If then the annihilator is 注意： annihilator 和 a0, a1, …… , an 無關 只和 , n 有關 The annihilator is independent of the constant multiplied in the front of each term.

General rule 2: If b1  0 or b2  0 then the annihilator is Example 2: (text page 154) annihilator Example 5: (text page 156) annihilator Example 6: (text page 157) annihilator

General rule 3: If g(x) = g1(x) + g2(x) + …… + gk(x) Lh[gh(x)] = 0 but Lh[gm(x)]  0 if m  h, then the annihilator of g(x) is the product of Lh (h = 1 ~ k) Proof: (因為 L1, L2 為 linear DE with constant coefficient, L1L2 = L2L1 )

Similarly, : Therefore,

Example 7 (text page 157) annihilator: D3 annihilator: D − 5 annihilator: (D − 2)3 annihilator of g(x): D3 (D − 2)3 (D − 5)

4-5-3 Using the Annihilator to Find the Particular Solution Step 2-1 Find the annihilator L1 of g(x) Step 2-2 如果原來的 linear & constant coefficient DE 是 那麼將 DE 變成如下的型態： (homogeneous linear & constant coefficient DE) 註： If then

Step 2-3 Use the method in Section 4-3 to find the solution of Step 2-4 Find the particular solution. The particular solution yp is a solution of but not a solution of (Proof): Since , if g(x)  0, should be nonzero. Moreover, . Step 2-5 Solve the unknowns

solutions of particular solution yp solutions of particular solution yp  solutions of  solutions of 本節核心概念

4-5-4 Examples Example 3 (text page 155) Step 1: Complementary function (solution of the associated homogeneous function) Step 2-1: Annihilation: D3 Step 2-2: Step 2-3: auxiliary function roots: m1 = m2 = m3 = 0, m4 = −1, m5 = −2 移除和 complementary function 相同的部分 Solution for :

Step 2-4: particular solution

Example 4 (text page 156) Step 1: Complementary function From auxiliary function, m2 − 3m = 0, roots: 0, 3 Step 2-1: Find the annihilator D − 3 annihilate but cannot annihilate (D2 + 1) annihilate but cannot annihilate (D − 3)(D2 + 1) is the annihilator of Step 2-2:

Step 2-3: auxiliary function: 易犯錯的地方 solution of : Step 2-4: particular solution 代回原式 並比較係數 Step 2-5: Step 3: general solution

4-5-5 本節要注意的地方 (1) 所以要先算 complementary function，再算 particular solution 4-5-5 本節要注意的地方 (1) 所以要先算 complementary function，再算 particular solution (2) 若有兩個以上的 annihilator，選其中較簡單的即可 (3) 計算 auxiliary function 時有時容易犯錯 (4) 的解和 的解不一樣。 (5) 這方法，只適用於 constant coefficient linear DE (因為，還需借助 auxiliary function) The thing that can be done by the annihilator approach can always be done by the “guessing” method in Section 4-4, too.

練習題 Sec. 4-1: 3, 7, 8, 10, 13, 20, 24, 29, 33, 36 Sec. 4-2: 2, 4, 9, 13, 14, 16, 18, 19 Sec. 4-3: 7, 16, 20, 22, 24, 28, 33, 39, 41, 52, 54, 56, 59, 61, 63 Section 4-4 5, 6, 14, 17, 18, 24, 26, 32, 33, 39, 42 Section 4-5 2, 7, 8, 13, 18, 31, 45, 60, 62, 69, 70 Review 4 2, 21, 22, 25, 33, 34, 37