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4-4 Undetermined Coefficients – Superposition Approach

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.

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4-4 Undetermined Coefficients – Superposition Approach

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  1. 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.

  2. 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)

  3. Trial Particular Solutions (from text page 143) It comes from the “form rule”. See page 198.

  4. yp = ? yp = ? yp = ?

  5. 4-4-3 Examples Example 2 (text page 141) 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:

  6. Example 3 (text page 142) Step 1: Find the solution of Step 2: Particular solution guess guess

  7. Particular solution Step 3: General solution

  8. 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

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

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

  11. 4-4-5 Glitch of the method: Example 4 (text page 142) Particular solution guessed by Form Rule: (no solution) Why?

  12. 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

  13. Example 7 (text page 144) From Form Rule, the particular solution is Aex 如果乘一個 x不夠，則再乘一個 x

  14. Example 8 (text page 145) 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)

  15. Example 11 (text page 146) 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

  16. 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

  17. 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) : :

  18. 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

  19. 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.

  20. 4-5-2 Find the Annihilator Example 1: (text page 150) annihilator: D4 annihilator: D + 3

  21. annihilator: (D − 2)2 (D − 2)2 = D2 − 4D + 4 註：當 coefficient 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

  22. General rule 1: If then the annihilator is 注意： annihilator 和 a0, a1, …… , an無關 只和 , n有關

  23. General rule 2: If b1 0 or b2 0 then the annihilator is Example 2: (text page 151) annihilator Example 5: (text page 154) annihilator Example 6: (text page 155) annihilator

  24. 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 )

  25. Similarly, : : Therefore,

  26. Example 7 (text page 154) annihilator: D3 annihilator: D − 5 annihilator: (D− 2)3 annihilator of g(x): D3 (D− 2)3 (D − 5)

  27. 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： (homogeneouslinear & constant coefficient DE) 註： If then

  28. 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

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

  30. 4-5-4 Examples Example 3 (text page 152) 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 :

  31. Step 2-4: particular solution Step 2-5: Step 3:

  32. Example 4 (text page 153) 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:

  33. 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

  34. 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)

  35. The thing that can be done by the annihilator approach can always be done by the “guessing” method in Section 4-4, too.

  36. 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

  37. 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)

  38. 代入原式後，總是可以簡化 代入 zero zero

  39. 簡化 進一步簡化： 假設 聯立方程式

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

  41. 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

  42. 4-6-4 Examples Example 1 (text page 159) Step 1: solution of Step 2-2: Step 2-3:

  43. Step 2-4: Step 2-5: Step 3:

  44. Example 2 (text page 159) 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

  45. Step 2-5: Step 3: Note: 課本 Interval (0, /6) 改為(0, /3) Example 3 (text page 160) Note: 沒有 analytic 的解 所以直接表示成 (複習 page 45)

  46. 4-6-5 Case of the Higher Order Linear DE Solution of the associated homogeneous equation: The particular solution is assumed as:

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

  48. 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 237) Step 2-3 ……… Step 2-4 ……. Step 2-5

  49. 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

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