Microsoft PowerPoint - DE3 [相容模式]

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1 -4-1 方法的條件任何 first order DE 皆可改寫成 -4 Eact Equations, d N, d 0 (1) 當, N, 成立時, 的型態 100 () 當可以用本節的 Eact Equation 的方法來解,, N, 或, N, N, is independent of is independent of 可以用 odified Eact Equation ethod 來解 ( 見講義 -4-5)

2 -4- 方法的來源 101 Review the concept of partial differentiation f f df (, ) d d Speciall, when f(, ) = c where c is some constant, f d f d 0

3 補充 : f f f df (,, z) d d dz z f f f f df (,,,, ) d d d d 1 3 k k k 10 思考 : 假設一個人在山坡的某處若往東走, 每走 1 公尺, 高度會增加 a 公尺若往北走, 每走 1 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 f (, ) f(, ) df (, ) d d p q (0, 0) a b (p, q) (p, 0)

4 [Definition.4.1] We can epress an 1 st order DE as, d N, d 0 If there eists some function f(, ) that satisfies f,, and, then we call the 1 st order DE the eact equation. f, N, 103 The method for checking whether the DE is an eact equation: (Proof): If and, then f, N,,, f, N,, N, f, f,

5 For the eact equation,, d N, d 0 104,, f f d d 0 可改寫成 df, 0, f, c

6 -4-3 解法 105 The method for solving the eact equation (A): f,, f, N, () (1) () f,, d g c f, d g N, g() is a constant for () (3) (3) 代入 g N,, d, d g c () (4) further computation Solution g g d

7 整理 Previous Step: Check whether f Step 1: Solve,, Step : 將 f(, ) 代入 f, Step 3: Substitute g() into,, g N, f d c, N, 106,, d f g, 以解出 g() Step 4: Further computation and obtain the solution Etra Steps: (a) Consider the initial value problem is satisfied.

8 The method for solving the eact equation (B): f,, () () f N, d h, () (3) 代入 h, N, d () h h d f, N,,, 107 f Ndh c (3) (1) h() is a constant for, N d h c (4) Solution further computation

9 -4-4 例子 Eample 1 (tet page 67) d ( 1) d 0 108,, 1 Step 0: check whether it is eact N f f 1 Step 1 Step Step N, f g Step 3 c Step 4 c /( 1) Step 3 f g g 1 g Step Step 1 思考 : 是否有其他的方法可以解 Eample 1?

10 Eample (tet page 67) 109 ( e cos ) d (e cos ) d 0, cos e f Step e cos N, e cos f e cos Step 1 f(, ) sin h e cos h e cos h Step h 0 c1 Step 3 e Step 3 f (, ) e sin c1 c e sin c Step 4 e sin c 0 1 Step 0: check for eact cos e sin N

11 110 要注意 (a) 自行由另一個方向 f,, d g 來練習, 看是否得出同樣的結果 (b) 得出的解 e c sin 0 為 implicit solution (c) 思考 : 何時用何時用,, f d g,, f N d h

12 Eample 3 (tet page 68) 111 d cos sin d 1 0 自修, 但注意 (a)initial value problem, (b) 何時用 f,, d g 何時用 f, N, d h (c) 得出的 implicit solution 為而 eplicit solution 為 1 cos 3, 範圍 : (-1, 1) 3 cos 1, 範圍 : (-1, 1) 為何 3 cos 不為解? 1

13 -4-5 odified Eact Equation ethod 11 Technique: Use the integrating factor (, ) to convert the 1 st order DE into the eact equation., d N, d 0 d,, N, d, 0 such that,,, N, It is hard to find. N N N ( N )

14 N ( N ) N N 113 (1) When ( N )/ is a function of alone: Therefore, We can set to be dependent on alone. d N d 用 separable variable 的方法 d N d N e ( N) d

15 () When ( N )/ N is a function of alone: Therefore, N ( N ) We can set to be dependent on alone. d d N N 用 separable variable 的方法 d N N N N d N N N 114 e ( N) d N

16 前面 -4-3 (pages 105~107) 的解法流程再加一個步驟 : 115 Step 0: Whether N Yes No Case 1 Use the process of -4-3 Whether N is independent of Yes No Case e ( 小心易背錯 ) ( N ) d Whether N N is independent of In Cases and 3,, d N, d 0, d N, d 0 Yes No Case 3 e ( N) d N Case 4 Use other methods Using the process of -4-3, but (, ) should be modified as (, ) N(, ) should be modified as N(, )

17 Eample 4 (tet page 69) d d ( 3 0) Step 0: N 3 0 N 43 N 3 N 3 0 N 3 (independent of ) (Case ) 3 d e e 3ln 3 Q: 為何 c 以及可省略? Steps 1~4: d d ( 3 0 ) 0 double N c

18 -4-6 本節需要注意的地方 117 (1) 使本節方法時, 要先將 DE 改成如下的型態並且假設, d N, d 0 f,,,,, f N () 對而言,g() 是個常數 ; 對而言,h() 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

19 -5 Solutions b Substitutions 介紹 3 個特殊解法 118 Question: 尚有不少的 1 st order DE 無法用 Sections -~-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 來取代對症下藥

20 -5-1 特殊解法 1: Homogeneous Equations 119 If g(t, t) = t α g(, ), then g(, ) is a homogeneous function of degree α. Which one is homogeneous? g(, ) = g(, ) = 注意 : 課本中,homogeneous 有兩種定義一種是 Section -3 的定義 ( 較常用 ) 一種是這裡的定義兩者並不相同

21 For a 1 st order DE:, d N, d 0 10 If (, ) and N(, ) are homogeneous functions of the same degree, then the 1 st order DE is homogeneous. 解法的限制條件 It can b solved b setting = u, d = ud + du, and use the separable value method. (from page 10)

22 If, d N, d 0 is homogeneous then, t, Nt, t t N, t t, 1, N, N1, u u 以 t = 1/ 得出 11 where u = /, = u d ud du 1, u d N 1, u ( uddu) 0 [ 1, u un 1, u ] dn 1, u du 0 N1, udu 1, 1, d u un u (separable)

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

24 Eample 1 (tet page 73) ( ) d d0 (, ) N(, ) (t, t) = t (, ) N(t, t) = t N(, ) homogeneous DE 13 Previous Step: Conclude whether the DE is homogeneous Step 1: Set = u, d = ud + du 原式 ( u ) d u ( ud du) 0 (1 u ) d 1 u ( ud du) 0 1 u (1 ud ) 1 u du0 du 1 u Step : Convert into the separable 1 st order DE d 0

25 1 u du 1 u d 0 14 Step 3: Solve the separable 1 st order DE d 1 du 0 1 u uln1u ln c 0 1 ln[ (1 u) ] uc 1 (1 ) u c u e 1 (1 ) Step 4 代回 u = / u u ce (1 / ) c e / 1 ( c e c ) ( ) ce /

26 -5- 特殊解法 : Bernoulli s Equations 15 定義 Bernoulli s equation: d P f d n so d d n 1n P f n d 1 du We can set u = 1 n 1n, u, and the method of solving d 1 n d the 1 st order linear DE to solve the Bernoulli s equation. 1 1n n d d du du du 1 du 1n u d du d du d 1 n d (Chain rule)

27 Procedure for solving the Bernoulli s equation 16 Previous Step : Conclude whether the DE is a Bernoulli s equation n 1 d 1 du 1n Step 1: Set 1 n u, u d 1 n d Step : Convert the Bernoulli s equation into the 1st order linear DE Step 3: Solve the 1st order linear DE (use the method in Sec. -3) Step 4: Substitute u = 1 n ( 別忘了 )

28 Eample (tet page 74) d d 17 Previous Step: 判斷 (Bernoulli, n = ) Step 1: set u = 1 ( = u 1 d d du du ) u d du d d (Chain rule) Step : Convert into the 1 st order linear DE (standard form) 原式 du u u u d 1 Step 3: Obtain the solution of the 1 st order DE u c Step 4: 代回 u = 1 1 c du 1 u d

29 -5-3 特殊解法 3 18 If the 1 st order DE has the form, d f A B C d (B 0) ( 解法的限制條件 ) we can set u = A + B + C to solve it. d 1 du A d B d B Since du Ad Bd ( 這式子也許較好記 )

30 d d Procedure for solving f A B C Previous Step: Conclude Step 1: Set u AB C du Ad Bd d 1 du A d B d B 19 Step : 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 u AB C ( 別忘了 )

31 Eample 3 (tet page 74) d 7, d (0) 0 Previous Step: 判斷 Step 1: Set u du d d Step : Converting d d du d du 原式 u 7 d du d u 9 Step 3: Obtain the solution ( 別忘了在運算過程中, 代回 u = A + B) du 注意的算法 u 9 Step 4: 代回 u = A + B +C 1 e 6 3(1 e ) 6 130

32 -5-4 本節要注意的地方 131 (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 ) () 別忘了, 寫出解答時, 要將 u 用 /, 1-n, 或 A + B + 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/d = Step : Converting, Step 3: Solving, Step 4: 將 u 用, 代回來

33 整理 :ethods of solving the 1 st order DE 13 (1)Direct computation d 條件 : f d () Separable variable d 條件 : d g h (3) Linear DE 條件 : d P f d (4) Eact equation d, 條件 : d N,, N, 破解法 : 直接積分破解法 :, 各歸一邊後積分破解法 : 算 ( ) e P d e e f dce 破解法 :double N method 先處理再處理 f f, P( ) d P( ) d P( ) d ( 或反過來 ) N,

34 (4-1) Eact equation 變型 d, 條件 : d N, ( N )/ (4-) Eact equation 變型 d, 條件 : d N, ( N )/ N (5) Homogeneous equation d, 條件 : d N, independent of independent of, t, t t, t N, N t t 破解法 : d d 破解法 : d d e ( N ) d, N, e ( N) d N, N, 破解法 :u = /, ( = u) d ud du 133 is eact is eact 再用 separable variable method

35 (6) Bernoulli s Equation d P f d 條件 : (7) A + B + C d f A B C d 條件 : n 破解法 : u = 1 n n d 1 du 1n u d 1 n d 再用 linear DE 的方法破解法 : u = A + B + c d 1 du A d B d B 134 注意 (a) 速度的訓練 (b) Eercises in Review 多練習 (c) 行有餘力, 觀察 singular solution 和 singular point

36 135 練習題 Sec. -4: 3, 8, 13, 17, 5, 9, 3, 34, 35, 38, 4 Sec. -5: 3, 5, 10, 13, 14, 17, 0,, 4, 5, 9, 31 Chap. Review:, 13, 16, 17, 18, 19,, 3, 4, 6, 33

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Untitled-3 SEC.. Separable Equations In each of problems 1 through 8 solve the given differential equation : ü 1. y ' x y x y, y 0 fl y - x 0 fl y - x 0 fl y - x3 3 c, y 0 ü. y ' x ^ y 1 + x 3 x y 1 + x 3, y 0 fl