1. Differential equations 机动 目录 上页 下页 返回 结束 15.2 First-order linear equations 15.3 Exact equations 15.4 Strategy for solving first-order equations Chapter 1 5 15.1 Basic concepts, separable and homogeneous equations

2. A second-order linear differential equation has the form (1) where P, Q, R, and G are continuous functions. 15.5 Second-Order Linear Equations If G(x) = 0 for all x, such equations are called second- order homogeneous linear equations. (This use of the word homogeneous has nothing to do with the meaning given in Section 15.1.) (2)

3. If for some x, Equation 1 is nonhomogeneous. Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions and of such an equation, then the linear combination is also a solution. (3)Theorem If and are both solutions of the linear equation (2) and and are any constants, then the function is also a solution of Equation 2.

4. Proof Since and are solutions of Equation 2, we have and Therefore Thus is a solution of Equation 2.

5. Let x and y are two variables, if neither x nor y is a constant multiple of the other, we say x and y are two linearly independent variables. For instance, the function and are linearly dependent, but and are linearly independent.

6. The second theorem says that the general solution of a homogeneous linear equation is a linear combination of two linearly independent solutions. (4)Theorem If and are linearly independent solutions of Equation 2, then the general solution is given by where and are arbitrary constants.

7. In general, it is not easy to discover particular solutions to a second-order linear equation. But it is always possible to do so if the coefficient functions P, Q and R are constant functions, that is, if the differential equation has the form (5) It is not hard to think of some likely candidates for particular solutions of Equation 5. For example, the exponential function y because its derivatives are constants multiple of itself:. Substitute these expression into Equation 5

8. Notice is never 0 so is a solution of Equation 5 if r is a root of the equation (6) which is called the auxiliary equation(or characteristic equation) of Equation 5. Using the quadratic formula, the root and of the auxiliary equation can be found: (7) We distinguish three cases according to the sign of the discriminant.

9. In this case the roots and of the auxiliary equation are real and distinct, so and are two linearly independent solutions of Equation 5. (8) If the roots and of the auxiliary equation ` are real and unequal, then the general solution of is Example 1 Solve the equation Case  Example 2 Solve the equation

10. In this case ; that is, the root of the auxiliary equation are real and equal. Denote r as the common value of and, we have (9) We know that is one solution of Equation 5. We now verify that is also a solution: Case 

11. (10) If the auxiliary equation has only one real root r, then the general solution of ` is Since and are linearly independent solutions, Theorem 4 provides us with the general solution: Example 3 Solve the equation

12. In this case the roots and of the auxiliary equation are complex numbers, we can write Case  Using Euler’s equation we write the solution of the differential equation as

14. We summarize the discussion as follows: (11) If the roots of auxiliary equation are the complex numbers, then the general solution of is Example 4 Solve the equation

15. An initial-value problem for the second-order Equation 1 or 2 consists of finding a solution y of the differential equation that also satisfies initial conditions of the form where and are given constants. If P, Q, R, and G are continuous on an interval and there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem. Initial-value and boundary-value problems

16. A boundary-value problem for Equation 1 consists of finding a solution of the differential equation that also satisfies boundary conditions of the form In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution.

17. Example 5 Solve the initial-value problem Example 6 Solve the initial-value problem Example 7 Solve the boundary-value problem

18. 1 5.1 Basic concepts, separable and homogeneous equations 机动 目录 上页 下页 返回 结束

19. 1 0 Two kinds of equations An ordinary differential equation: Basic concepts We have known the concept of the differential equations. In Sections 8 ， Such as involves an unknown function of a single variable and some of its derivatives. Def:

20. A partial differential equation : involves an unknown function of two or more variables and some of its partial derivatives. Such as

21. 机动 目录 上页 下页 返回 结束 The order of a differential equation is the order of the highest derivative that appears in the equation. 2 0 The order of the equation Is an ordinary differential equation of order 1 Is a third-order differential equation Is a second-order partial differential equation

22. 机动 目录 上页 下页 返回 结束 1 0 Define the separable equations 12.1.2 Separable equations Note: We study only ordinary differential equations mainly In general,a first-order differential equation has the form : where F is some function of the two variables x and y When F can be factored as a function of x times a function Is called a separable equation then

23. 机动 目录 上页 下页 返回 结束 Form (1) Also can be written as: (2) We integrate both sides ： This defines y implicitly as a function of x. thus, y=f(x) are called general solutions of the equation Sometimes,we can solve for y in terms of x ，

24. 机动 目录 上页 下页 返回 结束 2 0 An initial-value problem Has a solution satisfies an initial condition of the form The separable equation We say this is an initial problem 。 In general, there is a unique solution to the initial problem given by equations (2),(3) (3)

25. 机动 目录 上页 下页 返回 结束 Example 1 Solution (a) Write it as This is the general solution, involves an arbitrary constant C Solution (b)y(0) =1 tell us x=o,y=1 So (b) Solve the initial-value problem (a) Solve the differential equation

26. 机动 目录 上页 下页 返回 结束 Example 2 Solution: Write it as We integrate both sides So Solve the differential equation

27. 机动 目录 上页 下页 返回 结束 Example 3 Solution ： We integrate both sides So Write it as Solve the differential equation

28. 机动 目录 上页 下页 返回 结束 Homogeneous equations A first-order differential equation If F(x,y) can be written as The form (4) is called homogeneous equations For instance: Turn it as :

29. 机动 目录 上页 下页 返回 结束 We see make the change of variable then thus This is a separable differential equation, So, the original differential equation has the general solution it’s general solution or

30. 机动 目录 上页 下页 返回 结束 Example 5 Solution:write it as Solve the differential equation

31. 机动 目录 上页 下页 返回 结束 Example 6 Solve the homogeneous differential equation

32. 15.1Homework P 955 1.3.5.

33. 15.2 First-order linear equations 机动 目录 上页 下页 返回 结束

34. The general solutions of the first-order linear equations It’s solution is (1) (2)

35. 机动 目录 上页 下页 返回 结束 Is the solution of the equation form (3), This is a separate equations,solve it (3) When Q(x)=0, form (1) become Thus (4) C is an arbitrary constant

36. 机动 目录 上页 下页 返回 结束 The solution is written as (3) Equation Let us replace the constant C in it by arbitrary We look for the solution of the equation of the form (5)

37. 机动 目录 上页 下页 返回 结束 Differentiating Equation (5),we get Put it into (5), (6) Another method see page 997 thus: The solution of the form (1) is

38. 机动 目录 上页 下页 返回 结束 Example 1 Solution: By the general solution form (6) First step: The second step: Last step: this is a linear equation Solve the differential equation

39. 机动 目录 上页 下页 返回 结束 Example 2 Solution : First: Second: Last: Since y(1)=2, Therefore, We must first put the differential equation into standard form: We have the solution to the initial-value problem is Find the solution of the initial-value problem

40. 机动 目录 上页 下页 返回 结束 Example 3 find the solution of the initial-value problem

41. 机动 目录 上页 下页 返回 结束 A Bernoulli differential equation Observe that,if n=0 or 1,it is linear. Example 1 transforms the Bernoulli equation into the linear equation Solve the differential equation the substitution

42. 15.2 homework

43. 15.3 Exact equations

44. 机动 目录 上页 下页 返回 结束 Definition of the exact Suppose the equation Then y=f(x) satisfies a first-order differential equation obtained by using the Chain Rule to differentiate both sides of equation with respect to x: (1) (2) A differential equation of the form of Equation (2)is called exact 。

45. 机动 目录 上页 下页 返回 结束 Definition ： is called exact if there is a function f(x,y) such that If f(x,y) is known,thus the solution is given implicitly by (3) (4) We may be able to solve Equation 4 for y as an explicit function of x. A first –order differential equation of the form

46. 机动 目录 上页 下页 返回 结束 Theorem We have the following convenient method for the exactness of a differential equation. (5) is exact if and only if The the differential equation have continuous partial derivatives on a simply –connected domain. Theorem: Suppose and

47. 机动 目录 上页 下页 返回 结束 Example 1 Solve the differential equation Solution Here and have continuous partial derivatives on So the differential equation is exact by theorem. (6) (7) Find solutions of exact equation Thus there exists a function f such that Also

48. 机动 目录 上页 下页 返回 结束 To determine f we first integrate 6 with respect to x : Now we differentiate 8 with respect to y: (8) (9) Comparing 7 and 9,we see that We do not need the arbitrary constant here Thus So is the solution. and so

49. 机动 目录 上页 下页 返回 结束 Example 2 Solution Here and have continuous partial derivatives on So the differential equation is exact by theorem. thus there exists a function f such that Also Solve the differential equation

50. 机动 目录 上页 下页 返回 结束 Comparing them,we see We do not need the arbitrary constant here Thus So is the solution. To determine f we first integrate Now we differentiate it with respect to y: then with respect to x : that

51. 机动 目录 上页 下页 返回 结束 Integrating factors If the differential equation The differential equation (10 ) is not exact. the resulting equation is exact. Such that,after multiplication by I(x,y), we look for an integrating factor I(x,y) is not exact,

52. 机动 目录 上页 下页 返回 结束 Equation (10) is exact if That is or (11) In general,it is harder to solve this partial differential equation than the original differential equation. But it is sometimes possible to find I that is a function of x or y alone.

53. 机动 目录 上页 下页 返回 结束 Suppose I is a function of x alone. Then So equation 11 becomes(12) if Then Equation 12 is a first –order linear (and separable ) ordinary differential equation. It can be solved for I(x) Then Equation 10 is exact and can be solved In Example 1 or 2. is a function of x alone,

54. 机动 目录 上页 下页 返回 结束 Example 3 Solution :Here since the given equation is not exact. But is a function of x alone. So by Equation 12 there is an integrating factor I that satisfies we get Solve the differential equation

55. 机动 目录 上页 下页 返回 结束 Multiplying the original equation by x,we get Comparison then gives (13) so If we let Then So Equation 13 is now exact.thus there is a function f such that Integrating the first of these equations,we get (which we can take to be 0),so g is a constant

56. 机动 目录 上页 下页 返回 结束 The solution is therefore

57. 15.3homework

58. 15.4 Strategy for solving first- order equations

59. 机动 目录 上页 下页 返回 结束 In solving first-order differential equations we used the technique for separable equations in Sections 8.1 and the method for linear equations in Section 15.2.(on the text). In this section we present a miscellaneous collection of first-order differential equations and part of the problem is to recognize which technique should be used on each equation. We also develops methods for solving homogeneous equations in Section 15.1 (on the text) and exact equations in Section 15.3 (on the text) 。

60. 机动 目录 上页 下页 返回 结束 Here, however, the important thing is not so much the form of the functions involved e as it is the form of the equation itself. As with the strategy of integration (Section7.6) and the strategy of testing series (Section 10.7),the main idea is to classify the equation according to its form.

61. 机动 目录 上页 下页 返回 结束 A linear equation can be put into the form that is,the expression for dy/dx can be factored and a product of a function of x and a function of y. (1) Recall that a separable equation can be written in the form (2)

62. 机动 目录 上页 下页 返回 结束 A homogeneous equation can be expressed in the form (3) If an equation has none of these forms,we can,as a last resort, attempt to find and integrating factor and thus equation exact. where (4) An exact equation has the form

63. 机动 目录 上页 下页 返回 结束 (This step is analogous to Step l of the strategy for integration: algebraic implification. ) In each of these cases,some preliminary algebra may be required in order to put a given equation into one of the preceding forms.

64. 机动 目录 上页 下页 返回 结束 (compare with Equation 2) It is also linear,since we can write the equation as (compare with Equation 1) is separable because dy/dx can be written as For instance, the equation It may happen that a given equation is of more than one type.

65. 机动 目录 上页 下页 返回 结束 In such a case we could solve the equation using any one of the corresponding methods,although one of the methods might be easier than the others. (compare with Equation 2). Furthermore,it is also homogeneous because we can write it as

66. 机动 目录 上页 下页 返回 结束 Example1 We now recognize the equation ad being separable and we can solve it using the methods of Section 15.1. Initially,this equation may not appear to be in any of the forms of Equations 1,2,3,or 4,but observe that we can factor the right side and therefore write the equation as In the following examples we identify the type of each equation without working our the details of the solutions.

67. 机动 目录 上页 下页 返回 结束 Example 2 The change of variable v=y\x converts the equation into a separable equation. (We could have anticipated this because the expressions x2,y2,and 2xy are all of degree 2.) Which shows that y’ is a function of y/x and the equation is homogeneous (see Equation 3). it is not exact. Since The equation is clearly mot separable, nor separable, nor is it linear. But if we solve for y’, we get

68. 机动 目录 上页 下页 返回 结束 Example3 Therefore,the equation is indeed exact and can be dolled by the methods of Section 15.3. then If We suspect is might be exact, so we write it in the form This equation is not separable,linear,or homogeneous.

69. 15.4homework

70. It is therefore linear and can be solved using the integrating factor We recognize it as having the form of Equation 2. If we put the equation in the form Example4

71. In this section we deal with second-order nonhomogeneous linear differential equations with constant coefficient : (1) where a, b, c are constants and G is a continuous function. 15.6 Nonhomogeneous Linear Equations The related homogeneous equation (2) is called the complementary equation and plays an important role in the solution of the original nonhomogeneous equation (1).

72. (3)Theorem The general solution of the nonhomogeneous differential equation (1) can be written as where is a particular solution of Equation 1 and is the general solution of the complementary Equation 2.

73. Proof We want to verify that if y is any solution of Equation 1, then is a solution of the complementary Equation 2. Indeed

74. We know from Section 15.5 that the solution of the complementary equation is, where and are linearly independent solutions of Equation 2. There are two methods for finding a particular solution. The method of undetermined coefficients is straightforward but works only for a restricted class of functions G. The method of variation of parameters works for every function G but is usually more difficult to apply in practice. Therefore, Theorem 3 says that we know the general solution of the nonhomogeneous equation as soon as we know a particular solution.

75. It is reasonable to guess that the particular solution is a polynomial of the same degree as G because if y is a polynomial, then is also a polynomial. We therefore substitute a polynomial (of the same degree as G ) into the differential equation and determine the coefficients. Example 1 Solve the equation The method of undetermined coefficients We first illustrate the method of undetermined coefficients for the equation Case 1 Case 1 G(x) is a polynomial

76. Case 2 Case 2 G(x) is of the form, where C and k are constants. We take as a trial solution a function of the same form `, because the derivatives of are constant multiples of. Example 2 Solve the equation Because of the rules for differentiating the sine and cosine functions, we take as a trial particular solution a function of the form Example 3 Solve the equation Case 3 Case 3 G(x) is either Ccoskx or Csinkx

77. We take the trial solution to be a product of functions of the same type. For instance, in solving the differential equation we would try We use the easily verified principle of superposition, which says if and are solutions of respectively, then is a solution of Case 4 Case 4 G(x) is a product of functions of the preceding types Case 5 Case 5 G(x) is a sum of functions of these types

78. Example 4 Solve Example 5 Solve Finally we note that the recommended trial solution sometimes turns out to be a solution of the complementary equation and therefore cannot be a solution of the nonhomogeneous equation. In such cases we multiply the recommended trial solution by x (or if necessary) so that no term in is a solution of the complementary equation.

79. G(x) = First try Modification: If any term of is a solution of the complementary equation, multiply by x (or if necessary). Summary of undetermined coefficient

80. Suppose we have already solved the homogeneous equation and written the solution as (4) where and are linearly independent solutions. Let us replace the constants and in Equation 4 by arbitrary functions and. We look for a particular solution of the nonhomogeneous equation of the form (5) The method of variation of parameters

81. Since and are arbitrary functions, we can impose two conditions on them. One condition is that is a solution of the differential equation; we can choose the other condition so as to simplify our calculations. In view of the expression in Equation 6, let us impose the condition that (7) Differentiating Equation 5, we get (6) Then

82. Substituting in the differential equation, we get or (8) But and are solutions of the complementary equation, so and Equation 8 simplifies to (9)

83. Equation 7 and 9 form a system of two equations in the unknown functions and. After solving this system we may be able to integrate to find and then the particular solution is given by Equation 5. Example 6 Solve the equation

84. is not easy to solve. But it is important to be able to solve equations such as Equation 1 since they arise from physical problems and, in particular, in connection with the Schrödinger equation( 薛定谔方程 ) in quantum mechanics. Even a simple-looking equation like (1) 15.8 Series Solutions So far, the only second-order differential equations that we have been able to solve are linear equations with constant coefficients.

85. In such a case, we use the method of power series; that is, we look for a solution of the form The method is to substitute this expression into the differential equation and determine the values of the coefficients Example 1 Use power series to solve the equation Solution We assume there is a solution of the form (2)

86. Rewrite as In order to compare the expression for y and more easily, we substituting these expressions into the differential equation, we obtain By Theorem 10.9.2 we can differentiate power series term by term. So (3)

87. If two power series are equal, then the corresponding coefficients must be equal (from Section 10.10). Therefore, the coefficients of in Equation 5 must be 0: Equation 6 is called a recursion relation. If and are known, this equation allows us to determine the remaining coefficients recursively by putting n = 0, 1, 2, 3, …in succession. or (5) (6)

89. By now we see the pattern: For the even coefficients, For the odd coefficients, Putting these values back into Equation 2, we write the solution as

90. Notice that there are two arbitrary constants and, which is to be expected. Note 1 : We recognize the series obtained in Example 1 as being the Maclaurin series for cosx and sinx. (See Equation 10.10.16 and 10.10.15.) Therefore, we could write the solution as But we are not usually able to express power series solutions of differential equations in terms of known functions.

91. Solution We assume there is a solution of the form Example 2 Solve Then and Substituting in the differential equation, we get

92. We solve this recursion relation by putting n = 0, 1, 2, 3, … successively in Equation 7: This is true if the coefficient of is 0: (7)

93. and the odd coefficients are given by In general, the even coefficients are given by

94. Or (8) The solution is

95. Note 2 : In Example 2 we had to assume that the differential equation had a series solution. But now we could verify directly that the function given by Equation 8 is indeed a solution.

96. Note 3 : Unlike the situation of Example 1, the power series that arise in the solution of Example 2 do not define elementary functions. The functions and are perfectly good functions but they cannot be expressed in terms of familiar functions. We can use these power series expressions for and to compute approximate values of the functions and even to graph them.

97. Note 4 : If we are asked to solve the initial-value problem we would observe from Theorem 10.10.5 that This would simplify the calculations in Example 2, since all of the even coefficients would be 0. The solution to the initial-value problem is