1. Chapter 1 First-Order Differential Equations Shurong Sun University of Jinan Semester 1, 2010-2011

2. 1.1 for the unknown variable x. In this course one of our tasks will be to solve differential equations such as for the unknown function y=y(t). Analogous to a course in algebra and trigonometry, where a good amount of time is spent solving equations such as

3. 1.2 1.1 Separable Differential Equations Many first order differential equations can be written as: We can solve these equations by integrating:

4. 1.3 Separable Differential Equation Setand multiply byto obtain Integrating, we have A first order differential equation of the form is called a separable differential equation. I.

5. 1.4 II.then is one solution to the equation

6. 1.5 Example Solve the equation Solution: Separate variables and rewrite the equation in the form Integrating, we have or Solving for y, we obtain the solution in explicit form as Separable Equation Example This is a separable equation.

7. 1.6 Separable Equation Example Solve the differential equation: Solution:We first rewrite the DE in fractional form Separate variables and rewrite the equation in the form Integrating, we get

8. 1.7 Example Solve the IVP Solution: Separate variables and rewrite the equation in the form Integrating, we have or Solving for y, we obtain the solution in explicit form as

9. 1.8 Note that y=1 is also a solution to the equation. So the general solution to the equation is is an arbitrary constant. Applying the initial condition directly, we have or Thus

10. 1.9 Example Solve the equation Solution: Separate variables and rewrite the equation in the form Integrating, we have or

11. 1.10 Example- Newton ’ s Law of Cooling A copper sphere is heated to then placed in water that is maintained at At time t=0 it is Derive an equation for the temperature T of the ball as a function of time t. After 3 minutes the sphere ’ s temperature is Newton ’ s law of cooling implies:

12. 1.11 Example- Newton ’ s Law of Cooling Step 1 - Separate variables Step 2 - Integrate

13. 1.12 Example- Newton ’ s Law of Cooling Step 3 - Determine a particular solution Step 4 - Determine k

14. 1.13 Example A hemispherical bowl has top radius 4 ft and at time t=0 is full of water. At that moment a circular hole with diameter 1 in. is opened in the bottom of the tank. How long will it take for all the water to drain from the tank? Solution

15. 1.14 1.2 Reduction to Separable Form Sometimes first order differential equations can be made separable by a simple change of variable. Consider equations of the form: (1) Homogeneous Equation is homogeneous iff Remark:

16. 1.15 Writing the differential equation in terms of u gives: This suggests we set:

17. 1.16 Separating variables gives:

18. 1.17 Example: Solve Solution: We express (6) in the derivative form (6) then we see that the equation (6) is homogeneous. Now letand recall that With these substitution, equation (6) becomes

19. 1.18 The above equation is separable, and, on separating the variables and integrating, we obtain Hence, Also note that x=0 is a solution.

20. 1.19 (2) Equations of the Form SetThen the equation is transformed into a separable one. Example: Solve Solution:Set Then and so (8)

21. 1.20 Substituting into (8) yields Solving this separable equation, we obtain from which it follows that Finally, replacing z by x-y yields

22. 1.21 where theare constants. I. In this case, the equations can be reduced to the form (3)

23. 1.22 II. Then the system of equations has a unique solution The above DE can be written in the form

24. 1.23 which yields the DE after the translation of axes of the form Homogenous Equation

25. 1.24 Example: Solve Solution: From we obtain Hence, we let

26. 1.25 The differential equation for v is or The above equation is homogenous, so we let Thenor

27. 1.26 Separating variables gives from which it follows that When we substitute back in for z, u, and v, we find

28. 1.27 1.2 Linear First-Order DE (1) Definition Linear Equation A first-order differential equation of the form is said to be linear equation. When the linear equation is said to be homogeneous; (1) or inhomogenous. otherwise, it is non-homogenous

29. 1.28 (2) Standard Form By dividing both sides of (1) by the lead coefficient we obtain a more useful form, the stand form, of a linear equation: We seek a solution of (2) on an interval I for which (2) both function P and f are continuous. (3) Variation of parameters

30. 1.29 I. The Homogeneous Equation This is a separable equation. (3) Writing (3) asand integrating. Solving for y gives II. The Nonhomogeneous Equation(2) Method of variation of constants

31. 1.30 II. The Nonhomogeneous Equation (2) Method of variation of constants The basic idea is the constant C in the general solution of the homogeneous equation (3) is replaced by a function C(x). The calculation of an appropriate choice of C(x) gives a solution of the nonhomogeneous equation (2). Substitutinginto (2) gives

32. 1.31 so Hence Thus if (2) has a solution, it must be of form (4). and (4) that is a general solution of equation (2). Conversely, it is easy to verify that (4) constitutes one-parameter family of solutions of equation (2),

33. 1.32 ExampleFinding the general solution to The associated homogeneous equation is Solution: We write the differential equation in standard form Separating variable, we find

34. 1.33 substituting So the general solution to the homogeneous equation is By the method of variation of parameter, into the equation gives So Thus the general solution to

35. 1.34 is

36. 1.35 An integrating factor Method of Integrating factor

37. 1.36 Find the general solution to Solution: We write the differential equation in the standard form: Here So the integrating factor is: Multiplying the original differential equation by we get:or Example :

38. 1.37 We now integrate both sides and solve for y to find or

39. 1.38 Find the general solution to Solution: Here So

40. 1.39 Solve Solution: Here

41. 1.40 Bernoulli Equations Remark: when n=0,1, equation (9) is also a (9) linear equation. Fordividing equation (9) by yields Taking we find via the chain rule that (10)

42. 1.41 and so equation (10) becomes Linear equation Example : Solve Solution: This is a Bernoulli equation with n=3, (11)

43. 1.42 We make the substitution Sincethe transformed equation is This is a linear equation, its solution is

44. 1.43 Substitutinggive the solution Not included in the last equation is the solution that was lost in the process of dividing (11) by

45. 1.3 Exact Differential Equations and Integrating factors

46. 1.45 Exact Differential Equations Recall that the total or exact differential of a function u(x,y) is: We will use the concept of exactness to study differential equations of the form:

47. 1.46 Compare the following We say an ODE is exact if there exists a function Note this means that we can now write our ODE as: That is such that

48. 1.47 In this case, its solution (in implicit form) is given by Remember we need: Note this means that we can now write our ODE as:

49. 1.48 Testing for Exactness In practice we often don ’ t know about u, only about M and N and it is hard to check that We want a technique of testing for exactness based on knowing M and N, not u

50. 1.49 Let ’ s check the derivatives of M and N: A necessary and sufficient condition for exactness is: If M(x,y) and N(x,y) are continuous functions and have continuous first partial derivatives on some simply connected region of xy-plane, then

51. 1.50 Solving Exact Equations Once we have checked the equation is exact it can be readily solved by evaluating either:

52. 1.51

53. 1.52 Exact Equations: Example Example Solve Solution the equation given is exact. Fromwe have that

54. 1.53 Next we take the partial derivative of u with respect to y and substitute for N: Here we drop the constant of integration that technically should be present in g(y) since it will just get absorbed into the constant we pick up in the next step. Hence So, the implicit solution to the differential equation is

55. 1.54 Exact Equations: Example

56. 1.55 Exact Equations: Example Solve for k(x) and then u :

57. 1.56 Example Find the solution for the following IVP. Solution Let ’ s identify M and N and check that it ’ s exact. So, it ’ s exact.

58. 1.57 With the proper simplification integrating the second one isn ’ t too bad. However, the first is already set up for easy integration so let ’ s do that one. Differentiate with respect to y and compare to N. So, we get Recall that actually h(y) = k, but we drop the k because it will get absorbed in the next step. That gives us h(y) = 0.

59. 1.58 Therefore, we get The implicit solution is then Applying the initial condition gives 1 = C The implicit solution is then

60. 1.59 Remember Exact Equations Alternate method: we can often rearrange linear first order ODEs into the form: (Necessary and sufficient) We can check whether an equation in this form is exact by checking if

61. 1.60 Solve the following DE Solution:Here it ’ s exact. We first check to see if we have an exact equation. Since

62. 1.61 Note that So the general solution is

63. 1.62

64. 1.63 Example. Find the general solution to Step 1: Check to see So, it is exact.

65. 1.64 Then, So, we get as the general solution.

66. 1.65 Example. Solve the homogeneous DE Solution: This equation can be written in the form which is an exact equation. In this case, the solution in implicit form is i.e.,

67. 1.66 Integrating Factors If the equation is not exact we can consider such that the new equation is exact: multiplying it by a function

68. 1.67 Finding Integrating Factors For exactness we require: This equation can be simplified in special cases, two. of which we treat next

69. 1.68 whereis just a function of x. Let If we choose

70. 1.69 Now solve for

71. 1.70 Conversely, ifis just a function of x. Then Let is an integrating factor for Equation.

72. 1.71 In a similar fashion, if equation has an integrating is just a function of y. Conversely, ifis just a function of y. Let is an integrating factor for Equation. Then factor that depends only y, then

73. 1.72 Integrating Factors: Example Example: Solve Solution: The equation is not exact.We compute

74. 1.73 So an integrating factor for the equation is given by When we multiplying by we get the exact equation Solving this equation, we obtain the implicit solution

75. 1.74 Hence,and are solutions to the given equation. Note that the solution x=0 is lost in multiplying by

76. 1.75 Linear Differential Equations Recall a differential equation is linear if it can be written: If q(x)=0 the equation is homogeneous, otherwise the equation is nonhomogeneous.

77. 1.76 Homogeneous Linear Case Separating variables and integrating:

78. 1.77 Nonhomogeneous Linear Case Nonhomogeneous linear equations can be solved using the integrating factor method. First rewrite the differential equation as: i.e.

79. 1.78 Finding the integrating factor To find the integrating factor we first compute

80. 1.79 Nonhomogeneous Linear Case We can now find an integrating factor Now multiply the differential equation by the integrating factor:

81. 1.80 Nonhomogeneous Linear Case

82. 1.81 Integrating Factors: Example Example: Solve Solution: The equation is not exact. We compute Here

83. 1.82 So an integrating factor for is given by When we multiplying by we get the exact equation Solving this equation, we obtain as the solution in implicit form.

84. 1.83 In general, integrating factors are difficult to uncover. If a differential equation does not have one of the forms given above, then a search for an integrating factor likely will not be successful, and other methods of solution are recommended.

85. 1.84 Example : Solve This differential equation is not exact, and no integrating factor is immediately apparent. Note however, that if terms are strategically regrouped, the DE can be rewritten as Solution: Rewriting this equation in differential form, we have

86. 1.85 The first group of terms has many integrating factors (see Table 2). One of these factors, namelysee Table 2 is an integrating factor for the entire equation. Multiplying (1) by (1) we find (2)

87. 1.86 Since (2) is exact, it can be solved using the steps described previously. Alternatively, we note from Table 1, so that (2) can be rewrite as Integrating both sides of this last equation, we find

88. 1.87 or equivalently,

89. 1.88 Solve Solution: Here and, since The differential equation is not exact. Since is a function of y alone. we have an integrating factor

90. 1.89 Multiplying the given DE by we obtain the exact equation or equivalently,

91. 1.90 Integrating both sides of this last equation, we find (ii) Note that the differential equation can be rewritten as The first group of terms has many integrating factors (see Table 1). One of these factors, namelysee Table 1

92. 1.91 is an integrating factor for the entire equation. Multiplying (1) by we find Integrating both sides of this last equation, we find

93. 1.92 Now letand recall that With these substitution, equation (3) becomes (iii) Rewriting this equation in the derivative form, then we see that the equation (3) is homogeneous. (3)

94. 1.93 The above equation is separable, and, by separating the variables and integrating, we obtain

95. 1.94 (iv) We rewrite the differential equation in the form which is linear equation. Its solution is Also note that x=0 is a solution.

96. 1.95 Table 1

97. 1.96 Table 2

98. 1.97

99. 1.98 Exercise Solve

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