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: 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.

4-6 Variation of Parameters 4-6-1 方法的限制 The method can solve the particular solution for any linear DE (1) May not have constant coefficients (2) g(x) may not be of the special forms

4-6-2 Case of the 2nd order linear DE associated homogeneous equation: Suppose that the solution of the associated homogeneous equation is Then the particular solution is assumed as: (方法的基本精神)

代入原式後，總是可以簡化 代入 zero zero

簡化 進一步簡化： 假設 聯立方程式

where | |: determinant 可以和 1st order case (page 58) 相比較

4-6-3 Process for the 2nd Order Case Step 2-1 變成 standard form Step 2-2 Step 2-3 Step 2-4 Step 2-5

4-6-4 Examples Example 1 (text page 162) Step 1: solution of Step 2-2:

Step 2-4: Step 2-5: Step 3:

Example 2 (text page 163) Step 1: solution of Step 2-1: standard form: Step 2-2: Step 2-3: Step 2-4: (未完待續) 注意 算法

Step 2-5: Step 3: Note: 課本 Interval (0, /6) 應該改為(0, /3)

Example 3 (text page 164) Note: 沒有 analytic 的解 所以直接表示成 (複習 page 45)

4-6-5 Case of the Higher Order Linear DE Solution of the associated homogeneous equation: The particular solution is assumed as:

Wk: replace the kth column of W by For example, when n = 3,

4-6-6 Process of the Higher Order Case Step 2-1 變成 standard form Step 2-2 Calculate W, W1, W2, …., Wn (see page 239) Step 2-3 ……… Step 2-4 ……. Step 2-5

Exercise 26 Complementary function:

for -/4 < x < /4 Note: -/4 , /4 are singular points

4-6-7 本節需注意的地方 養成先解 associated homogeneous equation 的習慣 記熟幾個重要公式 4-6-7 本節需注意的地方 養成先解 associated homogeneous equation 的習慣 記熟幾個重要公式 這裡 | | 指的是 determinant (4) 算出 u1(x) 和 u2(x) 後別忘了作積分 (5) f(x) = g(x)/an(x) (和 1st order 的情形一樣，使用 standard form) (6) 計算 u1'(x) 和 u2'(x) 的積分時，+ c 可忽略 因為 我們的目的是算particular solution yp yp 是任何一個能滿足原式的解 (7) 這方法解的範圍，不包含 an(x) = 0 的地方 特別要小心

4-7 Cauchy-Euler Equation 4-7-1 解法限制條件 k not constant coefficients but the coefficients of y(k)(x) have the form of ak is some constant associated homogeneous equation particular solution

4-7-2 解法 Associated homogeneous equation of the Cauchy-Euler equation 4-7-2 解法 Associated homogeneous equation of the Cauchy-Euler equation Guess the solution as y(x) = xm , then

auxiliary function 比較: 和 constant coefficient 時有何不同？ 規則：把 變成

4-7-3 For the 2nd Order Case auxiliary function: roots [Case 1]: m1  m2 and m1, m2 are real two independent solution of the homogeneous part:

[Case 2]: m1 = m2 Use the method of reduction of order Note 1: 原式 Note 2: 此時

If y2(x) is a solution of a homogeneous DE then c y2(x) is also a solution of the homogeneous DE If we constrain that x > 0, then

[Case 3]: m1  m2 and m1, m2 are the form of two independent solution of the homogeneous part: 同理

Example 1 (text page 167) Example 2 (text page 168)

Example 3 (text page 169)

solution of the nth order associated homogeneous equation 4-7-4 For the Higher Order Case Process: auxiliary function Step 1-1 roots n independent solutions Step 1-2 solution of the nth order associated homogeneous equation Step 1-3

(1) 若 auxiliary function 在 m0 的地方只有一個根 是 associated homogeneous equation 的其中一個解 (2) 若 auxiliary function 在 m0 的地方有 k 個重根 皆為 associated homogeneous equation 的解

(3) 若 auxiliary function 在  + j 和  − j 的地方各有一個根 (未出現重根) 是 associated homogeneous equation 的其中二個解 (4) 若 auxiliary function 在  + j 和  − j 的地方皆有 k 個重根 是 associated homogeneous equation 的其中2k 個解

Example 4 (text page 169) auxiliary function

4-7-5 Nonhomogeneous Case To solve the nonhomogeneous Cauchy-Euler equation: Method 1: (See Example 5) (1) Find the complementary function (general solutions of the associated homogeneous equation) from the rules on pages 248-251, 255-256. (2) Use the method in Sec.4-6 (Variation of Parameters) to find the particular solution. (3) Solution = complementary function + particular solution Method 2: See Example 6，很重要 Set x = et, t = ln x

Example 5 (text page 169, illustration for method 1) Step 1 solution of the associated homogeneous equation auxiliary function Step 2-2 Particular solution Step 2-3

Step 2-4 Step 2-5 Step 3

Example 6 (text page 170, illustration for method 2) Set x = et, t = ln x (chain rule) Therefore, the original equation is changed into

(別忘了 t = ln x 要代回來) Note 1: 以此類推 Note 2: 簡化計算的小技巧：結合兩種解 nonhomogeneous Cauchy-Euler equation 的長處

4-7-6 本節要注意的地方 (1) 本節公式記憶的方法： 把 Section 4-3 的 ex 改成 x，x 改成 ln(x) 4-7-6 本節要注意的地方 (1) 本節公式記憶的方法： 把 Section 4-3 的 ex 改成 x，x 改成 ln(x) 把 auxiliary function 的 mn 改成 (2) 如何解 particular solution? Variation of Parameters 的方法 (3) 解的範圍將不包括 x = 0 的地方 (Why?)

還有很多 linear DE 沒有辦法解，怎麼辦 (1) numerical approach (Section 4-9-3) (2) using special function (Chap. 6) (3) Laplace transform and Fourier transform (Chaps. 7, 11, 14) (4) 查表 (table lookup)

(1) 即使用了 Section 4-7 的方法，大部分的 DE還是沒有辦法解 (2) 所幸，自然界真的有不少的例子是 linear DE 甚至是 constant coefficient linear DE

Exercise for practice 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 Section 4-6 4, 5, 8, 13, 14, 17, 18, 21, 28, 29, 34 Section 4-7 11, 17, 18, 20, 21, 24, 32, 34, 36, 37, 40, 42 Review 4 2, 21, 22, 25, 27, 28, 29, 30, 32, 33, 34, 37, 42