Basics Concepts and Ideas First Order Differential Equations. Dr. Omar R. Daoud

Transcription

1 Basics Concepts and Ideas First Order Differential Equations Dr. Omar R. Daoud

2 Differential Equations Man Phsical laws and relations appear mathematicall in the form of Differentia Equations The are one of the important fundamentals in engineering mathematics. An ordinar differential equation is an equation that contains one or several derivatives of unknown functions A differential equation is a relationship between an independent variable, a dependent variable and one or more derivatives of with respect to. If the unknown function depends onl on one independent variable, then its called b Ordinar D.E O.D.E., while its denoted b Partial D.E. P.D.E if the function depends on two or more independent variables. 11/13/013 Part I

3 Differential Equations The order of a differential equation is given b the highest derivative involved. A function which satisfies the equation is called a solution to the differential equation. Solving a differential equation is the reverse process to the one just considered. To solve a differential equation a function has to be found for which the equation holds true. The solution will contain a number of arbitrar constants the number equalling the order of the differential equation. d sin d 3 d d 3 d d d d 4 + e 0 is an equation of the 1st order 0 is an equation of the nd order 0 is an equation of the 3rd order 11/13/013 Part I 3

4 Differential Equations Linear vs. nonlinear differential equations A linear differential equation contains onl terms that are linear in the dependent variable or its derivatives. d + d 4 A nonlinear differential equation contains nonlinear function of the dependent variable. d + d d d d d d d d d [ ] 0, + 0, + sin 11/13/013 Part I 4

5 ordinar differential equations Definition: A differential equation is an equation containing an unknown function and its derivatives. Eamples: d d d + 3 d a d d 3 4 d d d d is dependent variable and is independent variable, and these are ordinar differential equations 0

6 Partial Differential Equation Eamples: 0 + u u t u u t u t u u u is dependent variable and and are independent variables, and is partial differential equation. u is dependent variable and and t are independent variables

7 Order of Differential Equation The order of the differential equation is order of the highest derivative in the differential equation. Differential Equation d d + 3 ORDER 1 d d d d 0 d 3 d 3 4 d d 3 3

8 Degree of Differential Equation The degree of a differential equation is power of the highest order derivative term in the differential equation. Differential Equation Degree d d d 3 d 3 d a 0 d 4 d d d d 3 + d d

9 Linear Differential Equation A differential equation is linear, if 1. dependent variable and its derivatives are of degree one,. coefficients of a term does not depend upon dependent variable. Eample: 1. d d d d 0. is linear. Eample:. d 3 d 3 4 d d 3 is non - linear because in nd term is not of degree one.

10 Eample: 3. d d d + d 3 is non - linear because in nd term coefficient depends on. Eample: 4. d d sin is non - linear because sin 3 3! + is non linear

11 Differential Equations Homogeneous vs. Inhomogeneous differential equations A linear differential equation is homogeneous if ever term contains the dependent variable or its derivatives. d d d d A homogeneous differential equation can be written as where L is a linear differential operator. d e.g., L d L 0, A homogeneous differential equation alwas has a trivial solution 0. d d 11/13/013 Part I 11

12 Differential Equations An inhomogeneous differential equation has at least one term that contains no dependent variable. d + 4 d The general solution to a linear inhomogeneous differential equation can be written as the sum of two parts: h + p Here h is the general solution of the corresponding homogeneous equation, and p is an particular solution of the inhomogeneous equation. 11/13/013 Part I 1

13 Differential Equations Thus, the first order differential equitation contains and ma contains and given function of ; F,, ' 0 or ' f, A solution of a given F.O.D.E. on some open interval a<<b is a function h that has a derivative which satisfies Equation 1 for all in that interval. When a differential equation is of 1 d d f d f d the form d d f, We can then integrate both sides. This will obtain the general solution. 11/13/013 Part I 13

14 Differential Equations Modelling is the steps that lead from the phsical situation to a mathematical formulation and solution and to the Phsical interpretation of the result; setting up a mathematical model Differential Equations of the phsical process. Solving the D.Es. Determination of a particular solution from an initial conditions to transform the general solution to a particular one. An initial value problem is a differential equation together with an initial condition. Checking. 11/13/013 Part I 14

15 Differential Equations Formation of differential equations Differential equations ma be formed from a consideration of the phsical problems to which the refer. Mathematicall, the can occur when arbitrar constants are eliminated from a given function. For eample, let: d Asin + Bcos so that Acos Bsin therefore d d Asin Bcos d d That is + 0 d 11/13/013 Part I 15

16 Differential Equations Formation of differential equations Here the given function had two arbitrar constants: Asin + Bcos and the end result was a second order differential equation: d 0 d + In general an nth order differential equation will result from consideration of a function with n arbitrar constants. 11/13/013 Part I 16

17 Differential Equations Various methods of solving F.O.D.E. will be discussed. These include: Variables separable Homogeneous equations Eact equations Equations that can be made eact b multipling b an integrating factor 11/13/013 Part I 17

18 f, Linear Non-Linear Integrating Factor Separable Homogeneous Eact Integrating Factor Change to Separable Change to Eact 11/13/013 Part I 18

19 Differential Equations Solution of linear differential equations d 1. When a differential equation is of the form f d 1 d d We can then integrate both sides. f d f d This will obtain the general solution. 11/13/013 Part I 19

20 Differential Equations Solution of linear differential equations. When the equation is of the form 1 d f d g d g f d d d f g, then 11/13/013 Part I 0

21 Differential Equations Solution of linear differential equations 3. A first order linear differential equation is an equation of the form d P Q d + 1 To find a method for solving this equation, lets consider the simpler equation d P 0 d + Which can be solved b separating the variables. 11/13/013 Part I 1

22 d P 0 d + d P d d P d ln P d + c e + P d c P d e e Ce P d c or e P d C Using the product rule to differentiate the LHS we get: 11/13/013 Part I

23 d e d P d d e d P e P d P d + d + d P e P d d Returning to equation 1, P Q d + If we multipl both sides b d P d P d + P e Qe d d P d P d e Q e d Now integrate both sides. e P d P d P d e Q e d P d P d and Q e d For this to work we need to be able to find 11/13/013 Part I 3

24 Solve the differential equation d + d Step 1: Comparing with equation 1, we have P P d d ln ln ln P d ln Step : e e e P d is called the integrating factor Step 3: Multipl both sides b the integrating factor. d d 3 + d d 3 d 4 + c c 4 11/13/013 Part I 4

25 d Solve the differential equation 3 d Step 1: Comparing with equation 1, we have P and Q 3 P d d Step : Integrating Factor e P d e Step 3: Multipl both sides b the integrating factor. d e e 3 d d e d 3 e 11/13/013 Part I 5

26 d e d 3 e e 3e d To solve this integration we need to use substitution. Let t dt 3 t 3e d e dt d 3 e t 3 e 3 e e + c 3 + ce 11/13/013 Part I 6

27 d 3 Solve the differential equation 0 d + 1 rewriting in standard form: d + d 1 Step 1: Comparing with equation 1, we have P and Q d P d ln Step : Integrating Factor P d ln e e Step 3: Multipl both sides b the integrating factor. d d note the shortcut I have taken here d 1 + c 3 1 The modulus vanishes as we will have either both positive on either side or both negative. Their effect is cancelled. 11/13/013 Part I 7

28 d Solve the differential equation cos sin cos d + d rewriting in standard form: tan cos d + Step 1: Comparing with equation 1, we have P tan and Q cos P d tan d ln sec Step : Integrating Factor P d ln sec e e sec Step 3: Multipl both sides b the integrating factor. d sec 1 note the shortcut I have taken here d sec 1d cos + ccos The modulus vanishes as we will have either both positive on either side or both negative. Their effect is cancelled. 11/13/013 Part I 8

29 d d Solve the differential equation given 1 when 0 1 rewriting in standard form: d d Step 1: Comparing with equation 1, we have P P d 1 d ln Step : Integrating Factor e e P d ln 1 Step 3: Multipl both sides b the integrating factor. d d /13/013 Part I 9

30 1d + c + c given 1 when c c 1 Hence the particular solution is 11/13/013 Part I 30

31 Differential Equations Solution of Separable differential equations F.O.D.E. can be reduced to the form of g ' f 1.1 since ', then g f 1. This equation is called separable because that the variables and could be separated, so that appears onl on one side while on the other one. 11/13/013 Part I 31

32 Differential Equations Solution of Separable differential equations Direct integration If the differential equation to be solved can be arranged in the form: d d f the solution can be found b direct integration. That is: f d 11/13/013 Part I 3

33 Differential Equations Solution of Separable differential equations Direct integration For eample: so that: d d d C + + This is the general solution or primitive of the differential equation. If a value of is given for a specific value of then a value for C can be found. This would then be a particular solution of the differential equation. 11/13/013 Part I 33

34 Differential Equations Solution of Separable differential equations Separating the variables If a differential equation is of the form: d d f F Then, after some manipulation, the solution can be found b direct integration. F d f d so F d f d 11/13/013 Part I 34

35 Differential Equations Solution of Separable differential equations Separating the variables For eample: so that: That is: Finall: d d d d so + 1 d d + + C + C C 11/13/013 Part I 35

36 Differential Equations Solution of Separable differential equations Separating the variables For eample: so that: 3 Let z; that is: d d dz z + d d d 0 3 z 3 z z 3 z 11/13/013 Part I 36

37 Differential Equations Solution of Separable differential equations Separating the variables For eample: 3 ln d + d 0 Separating the variables, we get 1 3 d d ln Integrating we get the solution as ln ln 3ln + k or ln 3 3 c 11/13/013 Part I 37

38 11/13/ Part I Differential Equations Solution of Inseparable differential equations Certain Des are inseparable, however the could be transferred to separable DE b the introduction of a new unknown function,. g For eample: Divide b, then ' u u u u u u u u u u ' 1 1 ' ' assume that ' u

39 Differential Equations Solution of Inseparable differential equations Cont. For eample: ln 1 + u 1+ u Finall, use u u u Integrate u + 1 ln + C, take the eponential C + ' C 11/13/013 Part I 39

40 Differential Equations Solution of Inseparable differential equations For eample: 3 d d 0 so that: d d 3 z Let z; that is: dz z z + or d 3 z 3 z dz d z 3 z z 3 z + z 3 z 11/13/013 Part I 40

41 Differential Equations Solution of Inseparable differential equations Cont. For eample: 3 d d 0 Separating the variables, we get 3 z 3 z z dz d, Integrating we get 3 z z z dz 3 d We epress the LHS integral b partial fractions. We get 3 z z dz z 1 d + ln c or 3 z z + 1 z 1 c 11/13/013 Part I 41

42 Differential Equations Solution of Inseparable differential equations Cont. For eample: 3 d d 0 Noting z /, the solution is: 3 + c or c 3 11/13/013 Part I 4

43 Differential Equations Homogenous differential equations The general form of the L.D.E. is ' + p r 0 r If then the L.D.E. is called Homogenous otherwise it is not. The solution of Homogenous D.E. could be attained b using the separable method. The Integrating Factor IF will be used to change the Inhomogeneous D.E. to a homogenous D.E. 11/13/013 Part I 43

44 Differential Equations Solution of Homogeneous differential equations For eample: ' + 3 Solution: Integrate 0 3 Ce 3 3 ln C, take the eponential 3 11/13/013 Part I 44

45 Differential Equations Solution of Homogeneous differential equations For eample: Solution: + sin Let z dz dz z + z + sin z or sin z d d Separating the variables, we get 1 d dz sin z Integrating cosec z-cot z C cosec -cot C 11/13/013 Part I 45

46 Differential Equations Solution of Inhomogeneous differential equations ' + p r or a1 ' + a0 b Two Different was to solve it a 0 a 1 a 0 a 1 11/13/013 Part I 46

47 11/13/ Part I Differential Equations Solution of Inhomogeneous differential equations C b a b a b a b a a a a b a a a a 1 integrate with respect to, Then ' But ' 1 1 ' 1 ' 1 ' 1 ' 1 1 ' 1 1 ' 1 0

48 Differential Equations Solution of Inhomogeneous differential equations a 0 a Multipl the inhomogenous D.E. b the IF to get a 0 IF ' 1 ' 1 a Let IF ' ' + p r IF ' + IF p IF r 1 IF p then 1 could be rewritten as IF ' IF r 11/13/013 Part I 48

49 11/13/ Part I Differential Equations Solution of Inhomogeneous differential equations ' substitute in 1 then ' since ' ' 1 0 r e e r e p e e e IF p IF IF p IF IF a a p p p p p p +

50 11/13/ Part I Differential Equations Solution of Inhomogeneous differential equations h h h p p p p Ce r e e h p C r e e r e e a a + + b Denote 1 Integrate with respect to ' ' 1 0

51 11/13/ Part I Differential Equations Solution of Inhomogeneous differential equations For eample: Ce e Ce e e e h e r p e + + 1, 1, Solution:

52 Differential Equations Solution of Inhomogeneous differential equations For eample: Solution: + p, r e e ke e make use of a e e 3sin + cos + Ce sin b 3sin + cos 3sin + cos, h a ke a + b asin b bcos b a a ke ke cos b acos b + bsin b + C 11/13/013 Part I a + b 5 + C, and

53 11/13/ Part I Differential Equations Solution of Inhomogeneous differential equations For eample: e C e Ce C e e Ce C e e e sin sin sin cos 3sin Solution:

54 Differential Equations Solution of Inhomogeneous differential equations For eample: Solution: + tan sin, for 0 1 p e cos C 3 tan, r sin, h tan ln cos lncos e sin + Ce lncos lncos sin cos + C cos cos make use of sin sin cos cos sin + C cos + C cos, at 0, /13/013 Part I 54

55 Differential Equations Eact differential equations M, d + N, d A F.O.D.E. is called an Eact if there eits a function f, such that df M, d + N, d Here df is the total differential of f, and equals f d + f d Hence the given DE becomes df 0 Integrating, we get the solution as f, C 0 11/13/013 Part I 55

56 11/13/013 Part I 56

57 Eactness Test f N, 11/13/013 Part I ۵

58 STEP:1 STEP: 11/13/013 Part I ۵۸

59 STEP:3 Find k STEP:4 STEP:5 11/13/013 STEP:6 Part I ۵۹

60 f, + e + h 11/13/013 Part I ٦۰

61 f h' h 0 0 Constant + e + h' f, + e But for the general solution f, f, + e Constant + e The integral of zero is ZERO, simple. Although derivative of a constant would be zero, but integral of zero would alwas be zero. One thing to note: Integral is NOT antiderivative in strict sense. Its an area under graph f in Cartesian sstem, where it is ofcourse in a two dimensional plane. 11/13/013 Part I ٦۱

62 Eample 6 Test whether the following DE is eact. If eact, solve it. Here M, N + M N 1 + d + d 0 Hence eact. Now f, M d d g + Differentiating partiall w.r.t., we get Hence f + g N + g 11/13/013 Part I 6

63 Eample 6 Test whether the following DE is eact. If eact, solve it. + d + d 0 Integrating, we get g ln Note that we have NOT put the arb constant Hence f, + g + ln Thus the solution of the given D.E. is f, c or + ln c, 11/13/013 Part I 63

64 Eample 7 Test whether the following DE is eact. If eact, solve it sin d cos d 0 Here 4 3 M + sin, N 4 + cos M 3 N 8 + cos Hence eact. f, M d sin d + Now 4 4 g + sin + Differentiating partiall w.r.t., we get f + 3 N 4 cos cos g Hence g 0 11/13/013 Part I 64

65 Integrating, we get Hence g f, + sin + g + sin Thus the solution of the given d.e. is f, c or 4 + c sin, c an arb const. 11/13/013 Part I 65

66 In the above problems, we found f, b integrating M partiall w.r.t. and then equated f to N. We can reverse the roles of and. That is we can find f, b integrating w.r.t. N partiall and then equate f to M. 11/13/013 Part I 66

67 Eample 8 Test whether the following DE is eact. If eact, solve it. Here M M Now sin ; 1+ sin d cos d 0 1+ sin, f, N d co s d N N g cos + cos 4cos sin sin Hence eact. 11/13/013 Part I 67

68 Differentiating partiall w.r.t., we get f cos sin + g + M 1 sin Integrating, we get g sin + g gives g 1 Hence f, cos + g cos + Thus the solution of the given D.E. is or f, c cos c, c an arb const. 11/13/013 Part I 68

69 IF M N, DE is not eact. Making Eact 11/13/013 Part I ٦۹

70 Integrating Factor To make DE Eact 11/13/013 Part I 70

71 Integrating Factor To make DE Eact 11/13/013 Part I ۱

72 11/13/013 Part I ۲

73 11/13/013 Part I ۳

74 11/13/013 Part I

75 Eample 3 Find an I.F. for the following DE and hence solve it. + 3 d + d 0 Here Now M M + 3 ; N M N 6 µ e N N g d 6 e d Hence the given DE is not eact. g, is an integrating factor of the given DE a function of alone. Hence 11/13/013 Part I 75

76 Multipling b, the given DE becomes d + d 0 which is of the form M d + N d 0 Note that now 3 3 M + 3 ; N Integrating, we easil see that the solution is c, 11/13/013 Part I 76

77 Eample 4 Find an I.F. for the following DE and hence solve it. Here e d + e cot + csc d 0 M e ; N e cot + csc M N 0 e cot Hence the given DE is not eact. Now M N M 0 e cot e cot h, a function of alone. Hence 11/13/013 Part I 77

78 h d µ e e cot d sin is an integrating factor of the given DE Multipling b sin, the given DE becomes e sin d + e cos + d 0 which is of the form Note that now M d + N d M e sin ; N e cos + Integrating, we easil see that the solution is e c sin +, 0 c an arbitrar constant. 11/13/013 Part I 78

79 Eample 5 Find an I.F. for the following DE and hence solve it. Here Now d + d 3 M ; N M 3 N M N N M a function of z alone. Hence Hence the given DE is not eact. z gz, 11/13/013 Part I 79

80 µ e g z dz e dz z 1 1 z is an integrating factor of the given DE Multipling b 1, which is of the form M d the given DE becomes + N d 1 1 d+ d 0 Integrating, we easil see that the solution is 1 + c, 0 c an arbitrar constant. 11/13/013 Part I 80

81 Differential Equations Bernoulli differential equations A Bernoulli equation is a differential equation of the form: This is solved b: d P Q d + n a Divide both sides b n to give: n d 1 n + P Q d b Let z 1 n so that: dz n d 1 n d d 11/13/013 Part I 81

82 Differential Equations Bernoulli differential equations Substitution ields: then: becomes: dz d 1 n d n + P n Q d n 1 n 1 1 Which can be solved using the integrating factor method. n dz Pz Q d d d 11/13/013 Part I 8

83 Eample 1 Solve the following D.E. ' A B, where A and B are positive constants Solution a Divide both sides b n to give: ' A b Let z 1 n so that: z z' 1 n z' ' A 1 + B Az + B z' + Az B B n A 11/13/013 Part I 83 B

84 Eample 1 Solve the following D.E. ' A B, where A and B are positive constants Solution p z e e A A A, r z 1 A e B B A e A B, h + B A Ce 1 A + Ce A + Ce A A B A A + Ce A 11/13/013 Part I 84