方法的限制条件课件.ppt

1、27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 equ

2、ation methodmethod for Ax+By+cFourier seriesFourier sine seriesFourier cosine series28Simplest method for solving the 1st order DE:Direct Integration dy(x)/dx=f(x)where()()y xf x dxF xc()()dF xf xdx29Table of Integration 1/x ln|x|+ccos(x)sin(x)+csin(x)cos(x)+ctan(x)ln|sec(x)|+ccot(x)ln|sin(x)|+caxax

3、/ln(a)+cx eaxx2 eax11tanxcaa221xa2222axexxcaaa221ax1sinxca1axexcaa302-2 Separable Variables2-2-1 方法的限制條件方法的限制條件1st order DE 的一般型態: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 the 1st order DE is separable(or ha

4、ve separable variable).31dyxydx222cos()xydyx edxdy(x)/dx=g(x)h(y)條件：32If ,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 Check the singular solution 分離變數個別積分Step 3 Initial condi

5、tions33Step 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.34Example 1(text,page 46)(1+x)dy y dx=0 1dydxyx1lnln1yxc1ln 1 xcyee1ln 1 xcye e 111(1)ccyexex (1)ycx1cce 1dyydxxc

6、heck 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 235Example 練習小技巧遮住解答和筆記，自行重新算一次(任何和解題有關的提示皆遮住)Exercise 練習小技巧初學者，先針對有解答的題目作練習累積一定的程度和經驗後，再多練習沒有解答的題目將題目依類型分類，多綀習解題正確率較低的題型動筆自己算，就對了36Example 2,with initial condition and implicit solu

7、tion(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 437Example 3,with singular solution(text page 47)24dyydx24dydxy114242dydydxyy111ln2ln244yyxc12ln442yxcy144422xcxyecey 14c

8、ce 44121xxceycecheck the singular solution 24dyydxset y=r,0=r2 4 r=2,y=2Step 4Step 1Step 2ory=238Example 4(text page 47)自修注意如何計算 ,yye dysin(2)cosxdxx39Example in the bottom of page 481/2dyxydx,y(0)=0Step 1Step 2Step 34116yxStep 4 check the singular solution Solution:or 0y 其實，還有更多的解401/2dyxydx,y(0)=0

9、solutions:(1)(2)(3)4116yx0y 2222221160116xbfor xbyfor bxaxafor xab 0 a412-2-4 IVP 是否有唯一解？是否有唯一解？,dyf x ydx00y xy這個問題有唯一解的條件：(Theorem 1.2.1,text page 15)如果 f(x,y),在 x=x0,y=y0 的地方為 continuous,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 的範圍:(0,)考慮 x 1()xxde yedx1

10、xxe yec11xyc e 1xye from initial condition()0 xde ydx2xe yc2xyc e(1)xyee要求 y(x)在 x=1 的地方為 continuous 622-3-4 名詞和定義名詞和定義(1)transient term,stable termExample 5 (text page 58)的解為 :transient term 當 x 很大時會消失x 1:stable term1 5xyxe 5xe0246810-20246810yx1x-axis63(2)piecewise continuous A function g(x)is pie

11、cewise 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 antiderivative(4)error function 202erfxtxedtcomplementary error function 22erfc1 erftxxedtx 64(5)sine integral function 0Sisin/xxt t dtFres

12、nel 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 65When is not easy to calculate:Try to calculate dydxdxdy2-3-5 小技巧小技巧Example:21dydxxy(not linear,not separable)2dxxydy(linear)222yxyyce(implicit solution)662-3-6 本節要注意的地方本節要注意的地方

13、(1)要先將 linear 1st order DE 變成 standard form(2)別忘了 singular point注意：singular point 和 Section 2-2 提到的 singular solution 不同(3)記熟公式()()()P x dxP x dxP x dxyeef x dxce()()P x dxP x dxdeyef xdx或()P x dxe(4)計算時，的常數項可以忽略67最上策：realize+remember it上策：realize it中策：remember it 下策：read it without realization and

14、remembrance最下策：rest z.z.z太多公式和算法，怎麼辦？68Chapter 3 Modeling with First-Order Differential Equations應用題(1)Convert a question into a 1st order DE.將問題翻譯成數學式(2)Many of the DEs can be solved by Separable variable method or Linear equation method (with integration table remembrance)693-1 Linear ModelsGrowth

15、 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70翻譯 A(0)=P0翻譯 A(1)=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(培養皿)initia

16、lly has P0 number of bacteria.The other initial condition:At t=1 h,the number of bacteria is measured to be 3P0/2.關鍵句關鍵句: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 triple71dAkdtA0.40550tA

17、Peln(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 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0.4055003tPPe72思考：為什麼此時需要兩個 initial values 才可以算出唯一解？課本用 linear(Section 2.3)的方法來解 Example 173Example 4 (an example of tem

18、perature change,text page 85)Initial:When a cake is removed from an oven,its temperature is measured 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 o

19、ff to 75 F?(註：這裡將課本的問題做一些修改)翻譯 find t such that T(t)=70.另外，根據題意，了解這是一個物體溫度和周圍環境的溫度交互作用的問題，所以 T(t)所對應的 DE 可以寫成70dTk Tdtk is a constant 7470dTk TdtT(0)=300T(3)=200課本用 separable variable 的方法解如何用 linear 的方法來解？75Example 5 (an example for mixture,text page 86)300 gallons3 gal/min3 gal/minConcentration:2 l

20、b/galA:the amount of salt in the tank(input rate of salt)(output rate of salt)33 2300dAdtA 7677 diLRiE tdtFrom Kirchhoffs second lawFigure 3.1.5 LR series circuit78 qRiE tC qdqRE tCdtFigure 3.1.6 RC series circuit79CHow about an LRC series circuit?22qdqd qRLE tCdtdt80Example 6 (text page 88)LR serie

21、s circuit E(t):12 volt,inductance:1/2 henry,resistance:10 ohms,initial current:0110122diidt2024diidt()20P t 1()20P t dtt cee這裡+c1 可省略202024ttdeiedt202065tteiec206()5ti tce(0)0i605c2066()55ti te81Circuit problem for t is small and t For the LR circuit:L RFor the RC circuit:R C transient stabletransie

22、nt stable823-2 Nonlinear Models3-2-1 Logistic Equationused for describing the growth 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 aPb83Figure 3.2.2 Logistic curves for differential ini

23、tial conditions84Solving the logistic equation()dPP abPdt()dPdtP abP1/ab adPdtPabP0()lnddPabPbdPdPabPcabPabP註：11lnlnPabPtcaa lnPatacabPseparable variable1atPc eabP1acce 11atacP tbce 000()ataPP tbPabP e(with initial condition P(0)=P0)logistic function85Example 1 (text page 96)There are 1000 students.

24、Suppose a student carrying a flu virus returns to an isolate college campus of 1000 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 constan

25、t determine the number of infected students after 6 days翻譯 find x(6)if it is further observed that after 4 days x(4)=5086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 87 1000dx tkxxdt 1000dx tkdtxx110001000dxdxkdtxx10001000dxdxkdtxx1lnln10001000 xxktc110001000kt cxex1

26、00021000ktxc ex12()cce 1000100022(1)1000ktktc exce100010001ktxce12()cc100010001999ktxe 01x100011c999c 450 x40001000501999ke10000.9906k 0.990610001999txe 6276x88Logistic 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 人口增加量

27、，和呈正比/a belnP飽合人口(1)(2)(3)(4)893-2-2 化學反應的速度化學反應的速度 12dx tk as xbs xdt Use compounds 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 produ

28、ct of the amount of A and the amount of BA+B CSee Example 2903-3 Modeling with Systems of DEsSome Systems are hard to model by one dependent variable but can be modeled by the 1st order ordinary differential equation 1,dx tg t x ydt 2,dy tgt x ydtThey should be solved by the Laplace Transform and ot

29、her methods91from Kirchhoffs 1st law 123i titi tfrom Kirchhoffs 2nd law 211122ditE ti RLi Rdt 3112di tE ti RLdtThree dependent variable We can only simplify it into two dependent variable Fig.3-3-3 in the textbook.(1)(2)92from Kirchhoffs 1st law 123i titi tfrom Kirchhoffs 2nd law 12di tE tLit Rdt 32

30、qtit RCFig.3-3-4 in the textbook.1221di titRitCdt(1)(2)93Chapter 3:訓練大家將和 variation 有關的問題寫成 DE 的能力.the variation is proportional to94練習題Section 2-2:4,7,12,13,18,19,21,25,28,32 Section 2-3:7,9,13,15,21,27,29,37,47,49(a),50(a)Section 3-1:4,5,10,15,20,29,32Section 3-2:2,5,14,15Section 3-3:12,13Review 3:3,4,11,12 Homework 1(due:10/13)(1)Sec.2-2 7,(2)Sec.2-2 18,(3)Sec.2-2 28,(4)Sec.2-3 13,(5)Sec.2-3 21,(6)Sec.2-3 29,(7)Sec.2-3 37,(8)Sec.3-1 5,(9)Sec.3-1 32,(10)Sec.3-2 15