方法限制條件ppt課件

1、附錄一附錄一 Methods of Solving the First Order Differential Equationgraphic methodnumerical methodanalytic methodseparable variablemethod for linear equationmethod for exact equationhomogeneous equation methodtransform Laplace transform Fourier transform direct integrationseries solution Bernoullis equat

2、ion methodmethod for Ax + By + cFourier seriesFourier sine seriesFourier cosine seriesSimplest method for solving the 1st order DE: Direct Integration dy(x)/dx = f(x) where ( )( )y xf x dxF xc( )( )dF xf xdxTable of Integration 1/x ln|x| + ccos(x) sin(x) + csin(x)cos(x) + ctan(x)ln sec(x) + ccot(x)l

3、n sin(x) + caxax/ln(a) + cx eaxx2 eax11tanxcaa221xa2222axexxcaaa221ax1sinxca1axexcaa2-2 Separable Variables2-2-1 方法的限制條件方法的限制條件1st order DE 的普通型態(tài): dy(x)/dx = f(x, y)Definition 2.2.1 (text page 45) If dy(x)/dx = f(x, y) and f(x, y) can be separate as f(x, y) = g(x)h(y) i.e., dy(x)/dx = g(x)h(y)then t

4、he 1st order DE is separable (or have separable variable). dyxydx222cos( )xydyx edxdy(x)/dx = g(x)h(y)條件：If , thenStep 1 where p(y) = 1/h(y)Step 2 where( ) ( )dyg x h ydx( )( )dyg x dxh y( )( )p y dyg x dx( )( )p y dyg x dx12( )( )P ycG xc( )( )P yG xc( )( )dP yp ydy( )( )dG xg xdx2-2-2 解法解法Step 4 C

5、heck the singular solution 分離變數(shù)個別積分Step 3 Initial conditionsStep 4 Check the singular solution: Suppose that y is a constant r( ) ( )dyg x h ydx0( ) ( )g x h r( )0h r solution for rSee whether the solution is a special case of the general solution.Example 1 (text, page 46) (1 + x) dy y dx = 0 1dydxy

6、x1lnln1yxc1ln 1 xcyee1ln 1 xcye e 111(1)ccyexex (1)ycx1cce 1dyydxxcheck the singular solution set y = r , 0 = r/(1+x) r = 0, y = 02-2-3 例子例子(a special case of the general solution)Step 4Step 1Step 2Example 練習(xí)小技巧遮住解答和筆記，自行重新算一次(任何和解題有關(guān)的提示皆遮住)Exercise 練習(xí)小技巧初學(xué)者，先針對有解答的題目作練習(xí)累積一定的程度和經(jīng)驗後，再多練習(xí)沒有解答的題目將題目依類型

7、分類，多綀習(xí)解題正確率較低的題型動筆本人算，就對了Example 2, with initial condition and implicit solution (text, page 46) , y(4) = 3 dyxdxy check the singular solution ydyxdx 22/2/2yxc 4.58,12.5cc 2225xy(implicit solution)225yx225yx (explicit solution)validinvalidStep 1Step 2Step 3Step 4Example 3, with singular solution (te

8、xt page 47) 24dyydx24dydxy114242dydydxyy111ln2ln244yyxc12ln442yxcy144422xcxyecey 14cce 44121xxceycecheck the singular solution 24dyydxset y = r , 0 = r2 4 r = 2, y = 2Step 4Step 1Step 2ory = 2Example 4 (text page 47)自修留意如何計算 , yye dysin(2 )cosxdxxExample in the bottom of page 481/2dyxydx, y(0) = 0St

9、ep 1Step 2Step 34116yxStep 4 check the singular solution Solution: or 0y 其實，還有更多的解1/2dyxydx, y(0) = 0solutions: (1) (2) (3) 4116yx0y 2222221160116xbfor xbyfor bxaxafor xab 0 a2-2-4 IVP 能否有獨(dú)一解？能否有獨(dú)一解？,dyf x ydx00y xy這個問題有獨(dú)一解的條件：(Theorem 1.2.1, text page 15)假設(shè) f(x, y), 在 x = x0, y = y0 的地方為 continuous

10、 ,f x yy則必定存在一個 h，使得 IVP 在 x0 h x 0 的情形,( )4P x dxexx = 0Step 44(1)xx yxec544()xyxx ecxx 的範(fàn)圍: (0, )考慮 x 1()xxde yedx1xxe yec11xyce 1xye from initial condition()0 xde ydx2xe yc2xyc e(1)xyee要求 y(x) 在 x = 1 的地方為 continuous 2-3-4 名詞和定義名詞和定義(1) transient term, stable termExample 5 (text page 58) 的解為 : tr

11、ansient term 當(dāng) x 很大時會消逝x 1: stable term1 5xyxe 5xe0246810-20246810yx1x-axis(2) piecewise continuous A function g(x) is piecewise continuous in the region of x1, x2 if g(x) exists for any x x1, x2. In Example 6, f(x) is piecewise continuous in the region of 0, 1) or (1, )(3) Integral (積分) 有時又被稱作 anti

12、derivative(4) error function 202erfxtxedtcomplementary error function 22erfc1 erftxxedtx (5) sine integral function 0Sisin/xxt t dtFresnel integral function 20sin/ 2xS xtdt(6) dyP x yf xdxf(x) 常被稱作 input 或 deriving function Solution y(x) 常被稱作 output 或 response When is not easy to calculate: Try to c

13、alculate dydxdxdy2-3-5 小技巧小技巧Example: 21dydxxy(not linear, not separable)2dxxydy(linear)222yxyyce (implicit solution)2-3-6 本節(jié)要留意的地方本節(jié)要留意的地方(1) 要先將 linear 1st order DE 變成 standard form(2) 別忘了 singular point留意：singular point 和 Section 2-2 提到的 singular solution 不同 (3) 記熟公式 ( )( )( )P x dxP x dxP x dxye

14、ef x dxce ( )( )P x dxP x dxdeyef xdx或( )P x dxe(4) 計算時， 的常數(shù)項可以忽略最上策： realize + remember it上策： realize it中策： remember it 下策： read it without realization and remembrance最下策： rest z.z.z太多公式和算法，怎麼辦？Chapter 3 Modeling with First-Order Differential Equations應(yīng)用題 Convert a question into a 1st order DE. 將問題

15、翻譯成數(shù)學(xué)式 (2) Many of the DEs can be solved by Separable variable method or Linear equation method (with integration table remembrance) 3-1 Linear ModelsGrowth and Decay (Examples 13)Change the Temperature (Example 4)Mixtures (Example 5)Series Circuit (Example 6)可以用 Section 2-3 的方法來解翻譯 A(0) = P0翻譯 A(1)

16、 = 3P0/2翻譯 k is a constant dAkAdt這裡將課本的 P(t) 改成 A(t) 翻譯 find t such that A(t) = 3P0Example 1 (an example of growth and decay, text page 83) Initial: A culture (培養(yǎng)皿培養(yǎng)皿) initially has P0 number of bacteria. The other initial condition: At t = 1 h, the number of bacteria is measured to be 3P0/2. 關(guān)鍵句關(guān)鍵句

17、: If the rate of growth is proportional to the number of bacteria P(t) presented at time t, Question: determine the time necessary for the number of bacteria to tripledAkdtA0.40550tAPeln(3)/0.40552.71thdAkAdtA(0) = P0, A(1) = 3P0/2可以用 什麼方法解？1ln Aktc1kt cAektAce1cce check singular solutionStep 1Step

18、2Step 4Step 3 (1) c = P0 (2) k = ln(3/2) = 0.405501Pc03/ 2kPce針對這一題的問題針對這一題的問題0.4055003tPPe思索：為什麼此時需求兩個 initial values 才可以算出獨(dú)一解？課本用 linear (Section 2.3) 的方法來解 Example 1Example 4 (an example of temperature change, text page 85) Initial: When a cake is removed from an oven, its temperature is measured

19、 at 300 F.翻譯 T(0) = 300The other initial condition: Three minutes later its temperature is 200 F.翻譯 T(3) = 200question: Suppose that the room temperature is 70 F. How long will it take for the cake to cool off to 75 F? (註：這裡將課本的問題做一些修正)翻譯 find t such that T(t) = 70. 另外，根據(jù)題意，了解這是一個物體溫度和周圍環(huán)境的溫度交互作用的問題

20、，所以 T(t) 所對應(yīng)的 DE 可以寫成70dTk Tdtk is a constant 70dTk TdtT(0) = 300T(3) = 200課本用 separable variable 的方法解如何用 linear 的方法來解？Example 5 (an example for mixture, text page 86)300 gallons3 gal/min3 gal/minConcentration: 2 lb/galA: the amount of salt in the tank(input rate of salt)(output rate of salt)33 2300

21、dAdtA diLRiE tdtFrom Kirchhoffs second lawFigure 3.1.5 LR series circuit qRiE tC qdqRE tCdtFigure 3.1.6 RC series circuitCHow about an LRC series circuit? 22qdqd qRLE tCdtdtExample 6 (text page 88) LR series circuit E(t): 12 volt, inductance: 1/2 henry, resistance: 10 ohms, initial current: 0110122d

22、iidt2024diidt( )20P t 1( )20P t dtt cee這裡 + c1 可省略202024ttdeiedt202065tteiec206( )5ti tce(0)0i605c2066( )55ti teCircuit problem for t is small and t For the LR circuit: L RFor the RC circuit: R C transient stabletransient stable3-2 Nonlinear Models3-2-1 Logistic Equationused for describing the growt

23、h of population(1)()dPbaPPP abPdtaThe solution of a logistic equation is called the logistic function. Two stable conditions: and . 可以用 separable variable 或其他的方法來解0P aPbFigure 3.2.2 Logistic curves for differential initial conditionsSolving the logistic equation()dPP abPdt()dPdtP abP1/ab adPdtPabP0(

24、)lnddPabPbdPdPabPcabPabP註：11lnlnPabPtcaa lnPatacabPseparable variable1atPc eabP1acce 11atacP tbce 000()ataPP tbPabP e(with initial condition P(0) = P0)logistic functionExample 1 (text page 96) There are 1000 students. Suppose a student carrying a flu virus returns to an isolate college campus of 100

25、0 students. If it is assumed that the rate at which the virus spreads is proportional not only to the number x of infected students but also to the number of students not infected, 翻譯 x(0) = 1翻譯 1000dx tkxxdtk is a constant determine the number of infected students after 6 days翻譯 find x(6) if it is

26、further observed that after 4 days x(4) = 50整個問題翻譯成 1000dx tkxxdtInitial: x(0) = 1, x(4) = 50find x(6)可以用separable variable 的方法 1000dx tkxxdt 1000dx tkdtxx110001000dxdxkdtxx10001000dxdxkdtxx1lnln10001000 xxktc110001000kt cxex100021000ktxc ex12()cce 1000100022(1)1000ktktc exce100010001ktxce12()cc1000

27、10001999ktxe 01x100011c999c 450 x40001000501999ke10000.9906k 0.990610001999txe 6276xLogistic equation 的變形()dPP abPhdt人口有遷移的情形()dPP abPcPdt遷出的人口和人口量呈正比()kPdPP abPcedt人口越多，遷入的人口越少(ln)( /ln)dPP abPdtbP a bPGompertz DE飽合人口為 人口添加量，和呈正比 /a belnP飽合人口(1)(2)(3)(4)3-2-2 化學(xué)反應(yīng)的速度化學(xué)反應(yīng)的速度 12dx tk as xbs xdt Use c

28、ompounds A and B to for compound C x(t): the amount of C To form a unit of C requires s1 units of A and s2 units of B a: the original amount of A b: the original amount of B The rate of generating C is proportional to the product of the amount of A and the amount of BA + B CSee Example 23-3 Modeling with Systems of DEsSome Systems are hard to model by one dependent variable but can be modeled by the 1st order ord