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Unformatted text preview: 5E-18(pp 1176-1185) 1/19/06 3:42 PM Page 1176 CHAPTER 18 The charge in an electric circuit is governed by the differential equations that we solve in Section 18.3. Second-Order Differential Equations 5E-18(pp 1176-1185) 1/19/06 3:42 PM Page 1177 The basic ideas of differential equations were explained in Chapter 10; there we concentrated on first-order equations. In this chapter we study second-order linear differential equations and learn how they can be applied to solve problems concerning the vibrations of springs and electric circuits. We will also see how infinite series can be used to solve differential equations. |||| 18.1 Second-Order Linear Equations A second-order linear differential equation has the form 1 P͑x͒ d 2y dy ϩ R͑x͒y ෇ G͑x͒ ϩ Q͑x͒ dx 2 dx where P, Q, R, and G are continuous functions. We saw in Section 10.1 that equations of this type arise in the study of the motion of a spring. In Section 18.3 we will further pursue this application as well as the application to electric circuits. In this section we study the case where G͑x͒ ෇ 0, for all x, in Equation 1. Such equations are called homogeneous linear equations. Thus, the form of a second-order linear homogeneous differential equation is 2 P͑x͒ d 2y dy ϩ Q͑x͒ ϩ R͑x͒y ෇ 0 2 dx dx If G͑x͒ 0 for some x, Equation 1 is nonhomogeneous and is discussed in Section 18.2. Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions y1 and y2 of such an equation, then the linear combination y ෇ c1 y1 ϩ c2 y2 is also a solution. 3 Theorem If y1͑x͒ and y2͑x͒ are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the function y͑x͒ ෇ c1 y1͑x͒ ϩ c2 y2͑x͒ is also a solution of Equation 2. Proof Since y1 and y2 are solutions of Equation 2, we have P͑x͒y1Љ ϩ Q͑x͒y1 ϩ R͑x͒y1 ෇ 0 Ј and P͑x͒y2Љ ϩ Q͑x͒yЈ ϩ R͑x͒y2 ෇ 0 2 1177 5E-18(pp 1176-1185) 1178 ❙❙❙❙ 1/19/06 3:42 PM Page 1178 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS Therefore, using the basic rules for differentiation, we have P͑x͒yЉ ϩ Q͑x͒yЈ ϩ R͑x͒y ෇ P͑x͒͑c1 y1 ϩ c2 y2͒Љ ϩ Q͑x͒͑c1 y1 ϩ c2 y2͒Ј ϩ R͑x͒͑c1 y1 ϩ c2 y2͒ ෇ P͑x͒͑c1 y1Љ ϩ c2 y2Љ͒ ϩ Q͑x͒͑c1 yЈ ϩ c2 yЈ͒ ϩ R͑x͒͑c1 y1 ϩ c2 y2͒ 1 2 ෇ c1͓P͑x͒y1Љ ϩ Q͑x͒y1 ϩ R͑x͒y1͔ ϩ c2 ͓P͑x͒y2Љ ϩ Q͑x͒y2 ϩ R͑x͒y2͔ Ј Ј ෇ c1͑0͒ ϩ c2͑0͒ ෇ 0 Thus, y ෇ c1 y1 ϩ c2 y2 is a solution of Equation 2. The other fact we need is given by the following theorem, which is proved in more advanced courses. It says that the general solution is a linear combination of two linearly independent solutions y1 and y2. This means that neither y1 nor y2 is a constant multiple of the other. For instance, the functions f ͑x͒ ෇ x 2 and t͑x͒ ෇ 5x 2 are linearly dependent, but f ͑x͒ ෇ e x and t͑x͒ ෇ xe x are linearly independent. 4 Theorem If y1 and y2 are linearly independent solutions of Equation 2, and P͑x͒ is never 0, then the general solution is given by y͑x͒ ෇ c1 y1͑x͒ ϩ c2 y2͑x͒ where c1 and c2 are arbitrary constants. Theorem 4 is very useful because it says that if we know two particular linearly independent solutions, then we know every solution. 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 ayЉ ϩ byЈ ϩ cy ෇ 0 where a, b, and c are constants and a 0. It’s not hard to think of some likely candidates for particular solutions of Equation 5 if we state the equation verbally. We are looking for a function y such that a constant times its second derivative yЉ plus another constant times yЈ plus a third constant times y is equal to 0. We know that the exponential function y ෇ e rx (where r is a constant) has the property that its derivative is a constant multiple of itself: yЈ ෇ re rx. Furthermore, yЉ ෇ r 2e rx. If we substitute these expressions into Equation 5, we see that y ෇ e rx is a solution if ar 2e rx ϩ bre rx ϩ ce rx ෇ 0 ͑ar 2 ϩ br ϩ c͒e rx ෇ 0 or But e rx is never 0. Thus, y ෇ e rx is a solution of Equation 5 if r is a root of the equation 6 ar 2 ϩ br ϩ c ෇ 0 Equation 6 is called the auxiliary equation (or characteristic equation) of the differential equation ayЉ ϩ byЈ ϩ cy ෇ 0. Notice that it is an algebraic equation that is obtained from the differential equation by replacing yЉ by r 2, yЈ by r, and y by 1. 5E-18(pp 1176-1185) 1/19/06 3:42 PM Page 1179 SECTION 18.1 SECOND-ORDER LINEAR EQUATIONS ❙❙❙❙ 1179 Sometimes the roots r1 and r 2 of the auxiliary equation can be found by factoring. In other cases they are found by using the quadratic formula: r1 ෇ 7 Ϫb ϩ sb 2 Ϫ 4ac 2a r2 ෇ Ϫb Ϫ sb 2 Ϫ 4ac 2a We distinguish three cases according to the sign of the discriminant b 2 Ϫ 4ac. b2 Ϫ 4ac Ͼ 0 In this case the roots r1 and r 2 of the auxiliary equation are real and distinct, so y1 ෇ e r 1 x and y2 ෇ e r 2 x are two linearly independent solutions of Equation 5. (Note that e r 2 x is not a constant multiple of e r 1 x.) Therefore, by Theorem 4, we have the following fact. CASE I ■ If the roots r1 and r 2 of the auxiliary equation ar 2 ϩ br ϩ c ෇ 0 are real and unequal, then the general solution of ayЉ ϩ byЈ ϩ cy ෇ 0 is 8 y ෇ c1 e r 1 x ϩ c2 e r 2 x |||| In Figure 1 the graphs of the basic solutions f ͑x͒ ෇ e 2 x and t͑x͒ ෇ eϪ3 x of the differential equation in Example 1 are shown in black and red, respectively. Some of the other solutions, linear combinations of f and t , are shown in blue. 8 EXAMPLE 1 Solve the equation yЉ ϩ yЈ Ϫ 6y ෇ 0. SOLUTION The auxiliary equation is r 2 ϩ r Ϫ 6 ෇ ͑r Ϫ 2͒͑r ϩ 3͒ ෇ 0 whose roots are r ෇ 2, Ϫ3. Therefore, by (8) the general solution of the given differential equation is 5f+g y ෇ c1 e 2x ϩ c2 eϪ3x f+5g f+g f g _1 g-f f-g _5 1 We could verify that this is indeed a solution by differentiating and substituting into the differential equation. EXAMPLE 2 Solve 3 FIGURE 1 d 2y dy ϩ Ϫ y ෇ 0. dx 2 dx SOLUTION To solve the auxiliary equation 3r 2 ϩ r Ϫ 1 ෇ 0 we use the quadratic formula: r෇ Ϫ1 Ϯ s13 6 Since the roots are real and distinct, the general solution is y ෇ c1 e (Ϫ1ϩs13 ) x͞6 ϩ c2 e (Ϫ1Ϫs13 ) x͞6 b 2 Ϫ 4 ac ෇ 0 In this case r1 ෇ r2 ; that is, the roots of the auxiliary equation are real and equal. Let’s denote by r the common value of r1 and r 2. Then, from Equations 7, we have CASE II 9 ■ r෇Ϫ b 2a so 2ar ϩ b ෇ 0 5E-18(pp 1176-1185) 1180 ❙❙❙❙ 1/19/06 3:42 PM Page 1180 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS We know that y1 ෇ e rx is one solution of Equation 5. We now verify that y2 ෇ xe rx is also a solution: ay2Љ ϩ byЈ ϩ cy2 ෇ a͑2re rx ϩ r 2xe rx ͒ ϩ b͑e rx ϩ rxe rx ͒ ϩ cxe rx 2 ෇ ͑2ar ϩ b͒e rx ϩ ͑ar 2 ϩ br ϩ c͒xe rx ෇ 0͑e rx ͒ ϩ 0͑xe rx ͒ ෇ 0 The first term is 0 by Equations 9; the second term is 0 because r is a root of the auxiliary equation. Since y1 ෇ e rx and y2 ෇ xe rx are linearly independent solutions, Theorem 4 provides us with the general solution. If the auxiliary equation ar 2 ϩ br ϩ c ෇ 0 has only one real root r, then the general solution of ayЉ ϩ byЈ ϩ cy ෇ 0 is 10 y ෇ c1 e rx ϩ c2 xe rx |||| Figure 2 shows the basic solutions f ͑x͒ ෇ eϪ3x͞2 and t͑x͒ ෇ xeϪ3x͞2 in Example 3 and some other members of the family of solutions. Notice that all of them approach 0 as x l ϱ. f-g 8 5f+g f+g g-f g _5 FIGURE 2 SOLUTION The auxiliary equation 4r 2 ϩ 12r ϩ 9 ෇ 0 can be factored as ͑2r ϩ 3͒2 ෇ 0 so the only root is r ෇ Ϫ 3 . By (10) the general solution is 2 f _2 EXAMPLE 3 Solve the equation 4yЉ ϩ 12yЈ ϩ 9y ෇ 0. y ෇ c1 eϪ3x͞2 ϩ c2 xeϪ3x͞2 f+5g 2 b 2 Ϫ 4ac Ͻ 0 In this case the roots r1 and r 2 of the auxiliary equation are complex numbers. (See Appendix G for information about complex numbers.) We can write CASE III ■ r1 ෇ ␣ ϩ i␤ r 2 ෇ ␣ Ϫ i␤ where ␣ and ␤ are real numbers. [In fact, ␣ ෇ Ϫb͑͞2a͒, ␤ ෇ s4ac Ϫ b 2͑͞2a͒.] Then, using Euler’s equation e i␪ ෇ cos ␪ ϩ i sin ␪ from Appendix G, we write the solution of the differential equation as y ෇ C1 e r 1 x ϩ C2 e r 2 x ෇ C1 e ͑␣ϩi␤͒x ϩ C2 e ͑␣Ϫi␤͒x ෇ C1 e ␣ x͑cos ␤ x ϩ i sin ␤ x͒ ϩ C2 e ␣ x͑cos ␤ x Ϫ i sin ␤ x͒ ෇ e ␣ x ͓͑C1 ϩ C2 ͒ cos ␤ x ϩ i͑C1 Ϫ C2 ͒ sin ␤ x͔ ෇ e ␣ x͑c1 cos ␤ x ϩ c2 sin ␤ x͒ where c1 ෇ C1 ϩ C2 , c2 ෇ i͑C1 Ϫ C2͒. This gives all solutions (real or complex) of the differential equation. The solutions are real when the constants c1 and c2 are real. We summarize the discussion as follows. 5E-18(pp 1176-1185) 1/19/06 3:42 PM Page 1181 SECTION 18.1 SECOND-ORDER LINEAR EQUATIONS ❙❙❙❙ 1181 If the roots of the auxiliary equation ar 2 ϩ br ϩ c ෇ 0 are the complex numbers r1 ෇ ␣ ϩ i␤, r 2 ෇ ␣ Ϫ i␤, then the general solution of ayЉ ϩ byЈ ϩ cy ෇ 0 is 11 y ෇ e ␣ x͑c1 cos ␤ x ϩ c2 sin ␤ x͒ |||| Figure 3 shows the graphs of the solutions in Example 4, f ͑x͒ ෇ e 3 x cos 2x and t͑x͒ ෇ e 3 x sin 2x, together with some linear combinations. All solutions approach 0 as x l Ϫϱ. EXAMPLE 4 Solve the equation yЉ Ϫ 6yЈ ϩ 13y ෇ 0. SOLUTION The auxiliary equation is r 2 Ϫ 6r ϩ 13 ෇ 0. By the quadratic formula, the roots are r෇ 3 f+g g f-g 6 Ϯ s36 Ϫ 52 6 Ϯ sϪ16 ෇ ෇ 3 Ϯ 2i 2 2 By (11) the general solution of the differential equation is f _3 y ෇ e 3x͑c1 cos 2x ϩ c2 sin 2x͒ 2 Initial-Value and Boundary-Value Problems _3 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 FIGURE 3 y͑x 0 ͒ ෇ y0 yЈ͑x 0 ͒ ෇ y1 where y0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and P͑x͒ 0 there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initial-value problem. Examples 5 and 6 illustrate the technique for solving such a problem. EXAMPLE 5 Solve the initial-value problem yЉ ϩ yЈ Ϫ 6y ෇ 0 |||| Figure 4 shows the graph of the solution of the initial-value problem in Example 5. Compare with Figure 1. y͑0͒ ෇ 1 yЈ͑0͒ ෇ 0 SOLUTION From Example 1 we know that the general solution of the differential equa- tion is y͑x͒ ෇ c1 e 2x ϩ c2 eϪ3x 20 Differentiating this solution, we get yЈ͑x͒ ෇ 2c1 e 2x Ϫ 3c2 eϪ3x To satisfy the initial conditions we require that _2 FIGURE 4 0 2 12 y͑0͒ ෇ c1 ϩ c2 ෇ 1 13 yЈ͑0͒ ෇ 2c1 Ϫ 3c2 ෇ 0 From (13) we have c2 ෇ 2 c1 and so (12) gives 3 c1 ϩ 2 c1 ෇ 1 3 c1 ෇ 3 5 Thus, the required solution of the initial-value problem is y ෇ 3 e 2x ϩ 2 eϪ3x 5 5 c2 ෇ 2 5 5E-18(pp 1176-1185) 1182 ❙❙❙❙ 1/19/06 3:43 PM Page 1182 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ■ The solution to Example 6 is graphed in Figure 5. It appears to be a shifted sine curve and, indeed, you can verify that another way of writing the solution is y ෇ s13 sin͑x ϩ ␾͒ where tan ␾ ෇ 2 3 EXAMPLE 6 Solve the initial-value problem yЉ ϩ y ෇ 0 y͑0͒ ෇ 2 yЈ͑0͒ ෇ 3 SOLUTION The auxiliary equation is r 2 ϩ 1 ෇ 0, or r 2 ෇ Ϫ1, whose roots are Ϯi. Thus ␣ ෇ 0, ␤ ෇ 1, and since e 0x ෇ 1, the general solution is 5 y͑x͒ ෇ c1 cos x ϩ c2 sin x _2π 2π yЈ͑x͒ ෇ Ϫc1 sin x ϩ c2 cos x Since the initial conditions become y͑0͒ ෇ c1 ෇ 2 _5 FIGURE 5 yЈ͑0͒ ෇ c2 ෇ 3 Therefore, the solution of the initial-value problem is y͑x͒ ෇ 2 cos x ϩ 3 sin x A boundary-value problem for Equation 1 consists of finding a solution y of the differential equation that also satisfies boundary conditions of the form y͑x 0 ͒ ෇ y0 y͑x 1 ͒ ෇ y1 In contrast with the situation for initial-value problems, a boundary-value problem does not always have a solution. EXAMPLE 7 Solve the boundary-value problem yЉ ϩ 2yЈ ϩ y ෇ 0 y͑0͒ ෇ 1 y͑1͒ ෇ 3 SOLUTION The auxiliary equation is r 2 ϩ 2r ϩ 1 ෇ 0 ͑r ϩ 1͒2 ෇ 0 or whose only root is r ෇ Ϫ1. Therefore, the general solution is y͑x͒ ෇ c1 eϪx ϩ c2 xeϪx The boundary conditions are satisfied if y͑0͒ ෇ c1 ෇ 1 |||| Figure 6 shows the graph of the solution of the boundary-value problem in Example 7. y͑1͒ ෇ c1 eϪ1 ϩ c2 eϪ1 ෇ 3 The first condition gives c1 ෇ 1, so the second condition becomes 5 eϪ1 ϩ c2 eϪ1 ෇ 3 _1 5 Solving this equation for c2 by first multiplying through by e, we get 1 ϩ c2 ෇ 3e _5 FIGURE 6 so c2 ෇ 3e Ϫ 1 Thus, the solution of the boundary-value problem is y ෇ eϪx ϩ ͑3e Ϫ 1͒xeϪx 5E-18(pp 1176-1185) 1/19/06 3:43 PM Page 1183 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ❙❙❙❙ 1183 Ј Ј Summary: Solutions of ayЈЈ ϩ byЈ ϩ c ϭ 0 Roots of ar 2 ϩ br ϩ c ෇ 0 General solution y ෇ c1 e r 1 x ϩ c2 e r 2 x y ෇ c1 e rx ϩ c2 xe rx y ෇ e ␣ x͑c1 cos ␤ x ϩ c2 sin ␤ x͒ r1, r2 real and distinct r1 ෇ r2 ෇ r r1, r2 complex: ␣ Ϯ i␤ |||| 18.1 1–13 Exercises 20. 2yЉ ϩ 5yЈ Ϫ 3y ෇ 0, Solve the differential equation. |||| y͑0͒ ෇ 1, y͑␲͞4͒ ෇ Ϫ3, yЈ͑0͒ ෇ 4 yЈ͑␲͞4͒ ෇ 4 1. yЉ Ϫ 6yЈ ϩ 8y ෇ 0 2. yЉ Ϫ 4yЈ ϩ 8y ෇ 0 21. yЉ ϩ 16y ෇ 0, 3. yЉ ϩ 8yЈ ϩ 41y ෇ 0 4. 2yЉ Ϫ yЈ Ϫ y ෇ 0 22. yЉ Ϫ 2yЈ ϩ 5y ෇ 0, y͑␲͒ ෇ 0, yЈ͑␲͒ ෇ 2 5. yЉ Ϫ 2yЈ ϩ y ෇ 0 6. 3yЉ ෇ 5yЈ 23. yЉ ϩ 2yЈ ϩ 2y ෇ 0, y͑0͒ ෇ 2, yЈ͑0͒ ෇ 1 7. 4yЉ ϩ y ෇ 0 8. 16yЉ ϩ 24yЈ ϩ 9y ෇ 0 24. yЉ ϩ 12yЈ ϩ 36y ෇ 0, ■ 9. 4yЉ ϩ yЈ ෇ 0 10. 9yЉ ϩ 4y ෇ 0 11. d 2y dy Ϫ2 Ϫy෇0 dt 2 dt 13. 25–32 d 2y dy ϩ ϩy෇0 dt 2 dt ■ ■ 14–16 ■ ■ 12. d 2y dy Ϫ6 ϩ 4y ෇ 0 dt 2 dt ■ ■ ■ ■ ■ ■ ■ |||| 17–24 |||| ■ ■ ■ ■ ■ ■ ■ ■ Solve the initial-value problem. 17. 2yЉ ϩ 5yЈ ϩ 3y ෇ 0, 18. yЉ ϩ 3y ෇ 0, 19. 4 yЉ Ϫ 4 yЈ ϩ y ෇ 0, |||| 18.2 y͑0͒ ෇ 3, y͑0͒ ෇ 1, ■ ■ ■ ■ ■ ■ y͑␲͒ ෇ Ϫ4 y͑0͒ ෇ 1, 30. yЉ Ϫ 6yЈ ϩ 9y ෇ 0, ■ ■ y͑1͒ ෇ 2 y͑0͒ ෇ 1, y͑0͒ ෇ 2, 29. yЉ Ϫ 6yЈ ϩ 25y ෇ 0, ■ ■ ■ y͑3͒ ෇ 0 y͑␲͒ ෇ 5 y͑0͒ ෇ 1, y͑0͒ ෇ 1, y͑0͒ ෇ 2, y͑␲͒ ෇ 2 y͑1͒ ෇ 0 y͑␲ ͞2͒ ෇ 1 y͑␲͒ ෇ 1 y͑0͒ ෇ 0, ■ ■ ■ ■ ■ 33. Let L be a nonzero real number. d y dy Ϫ2 ϩ 5y ෇ 0 dx 2 dx ■ 28. yЉ ϩ 100 y ෇ 0, ■ yЈ͑1͒ ෇ 1 ■ y͑0͒ ෇ 3, 32. 9yЉ Ϫ 18yЈ ϩ 10 y ෇ 0, 2 ■ ■ Solve the boundary-value problem, if possible. 31. yЉ ϩ 4yЈ ϩ 13y ෇ 0, d y dy Ϫ Ϫ 2y ෇ 0 dx 2 dx y͑1͒ ෇ 0, ■ 27. yЉ Ϫ 3yЈ ϩ 2y ෇ 0, ■ d 2y dy Ϫ8 ϩ 16y ෇ 0 15. dx 2 dx ■ |||| ■ 26. yЉ ϩ 2yЈ ෇ 0, 2 16. ■ 25. 4 yЉ ϩ y ෇ 0, Graph the two basic solutions of the differential equation ; and several other solutions. What features do the solutions have in common? 14. 6 ■ yЈ͑0͒ ෇ Ϫ4 34. If a, b, and c are all positive constants and y͑x͒ is a solution yЈ͑0͒ ෇ 3 y͑0͒ ෇ 1, ■ (a) Show that the boundary-value problem yЉ ϩ ␭y ෇ 0, y͑0͒ ෇ 0, y͑L͒ ෇ 0 has only the trivial solution y ෇ 0 for the cases ␭ ෇ 0 and ␭ Ͻ 0. (b) For the case ␭ Ͼ 0, find the values of ␭ for which this problem has a nontrivial solution and give the corresponding solution. yЈ͑0͒ ෇ Ϫ1.5 of the differential equation ayЉ ϩ byЈ ϩ cy ෇ 0, show that lim x l ϱ y͑x͒ ෇ 0. Nonhomogeneous Linear Equations In this section we learn how to solve second-order nonhomogeneous linear differential equations with constant coefficients, that is, equations of the form 1 ayЉ ϩ byЈ ϩ cy ෇ G͑x͒ 5E-18(pp 1176-1185) 1184 ❙❙❙❙ 1/19/06 3:43 PM Page 1184 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS where a, b, and c are constants and G is a continuous function. The related homogeneous equation 2 ayЉ ϩ byЈ ϩ cy ෇ 0 is called the complementary equation and plays an important role in the solution of the original nonhomogeneous equation (1). 3 Theorem The general solution of the nonhomogeneous differential equation (1) can be written as y͑x͒ ෇ yp͑x͒ ϩ yc͑x͒ where yp is a particular solution of Equation 1 and yc is the general solution of the complementary Equation 2. Proof All we have to do is verify that if y is any solution of Equation 1, then y Ϫ yp is a solution of the complementary Equation 2. Indeed a͑y Ϫ yp ͒Љ ϩ b͑y Ϫ yp ͒Ј ϩ c͑y Ϫ yp ͒ ෇ ayЉ Ϫ aypЉ ϩ byЈ Ϫ byЈ ϩ cy Ϫ cyp p ෇ ͑ayЉ ϩ byЈ ϩ cy͒ Ϫ ͑aypЉ ϩ byp ϩ cyp ͒ Ј ෇ t͑x͒ Ϫ t͑x͒ ෇ 0 We know from Section 18.1 how to solve the complementary equation. (Recall that the solution is yc ෇ c1 y1 ϩ c2 y2 , where y1 and y2 are linearly independent solutions of Equation 2.) Therefore, Theorem 3 says that we know the general solution of the nonhomogeneous equation as soon as we know a particular solution yp . 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. The Method of Undetermined Coefficients We first illustrate the method of undetermined coefficients for the equation ayЉ ϩ byЈ ϩ cy ෇ G͑x͒ where G͑x) is a polynomial. It is reasonable to guess that there is a particular solution yp that is a polynomial of the same degree as G because if y is a polynomial, then ayЉ ϩ byЈ ϩ cy is also a polynomial. We therefore substitute yp͑x͒ ෇ a polynomial (of the same degree as G ) into the differential equation and determine the coefficients. EXAMPLE 1 Solve the equation yЉ ϩ yЈ Ϫ 2y ෇ x 2. SOLUTION The auxiliary equation of yЉ ϩ yЈ Ϫ 2y ෇ 0 is r 2 ϩ r Ϫ 2 ෇ ͑r Ϫ 1͒͑r ϩ 2͒ ෇ 0 with roots r ෇ 1, Ϫ2. So the solution of the complementary equation is yc ෇ c1 e x ϩ c2 eϪ2x 5E-18(pp 1176-1185) 1/19/06 3:43 PM Page 1185 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ❙❙❙❙ 1185 Since G͑x͒ ෇ x 2 is a polynomial of degree 2, we seek a particular solution of the form yp͑x͒ ෇ Ax 2 ϩ Bx ϩ C Then yp ෇ 2Ax ϩ B and ypЉ ෇ 2A so, substituting into the given differential equation, we Ј have ͑2A͒ ϩ ͑2Ax ϩ B͒ Ϫ 2͑Ax 2 ϩ Bx ϩ C͒ ෇ x 2 or Ϫ2Ax 2 ϩ ͑2A Ϫ 2B͒x ϩ ͑2A ϩ B Ϫ 2C͒ ෇ x 2 Polynomials are equal when their coefficients are equal. Thus |||| Figure 1 shows four solutions of the differential equation in Example 1 in terms of the particular solution yp and the functions f ͑x͒ ෇ e x and t͑x͒ ෇ eϪ2 x. 8 2A Ϫ 2B ෇ 0 2A ϩ B Ϫ 2C ෇ 0 The solution of this system of equations is A ෇ Ϫ1 2 yp+2f+3g yp+3g Ϫ2A ෇ 1 B ෇ Ϫ1 2 C ෇ Ϫ3 4 A particular solution is therefore yp+2f _3 3 yp͑x͒ ෇ Ϫ 1 x 2 Ϫ 1 x Ϫ 3 2 2 4 yp and, by Theorem 3, the general solution is _5 y ෇ yc ϩ yp ෇ c1 e x ϩ c2 eϪ2x Ϫ 1 x 2 Ϫ 1 x Ϫ 3 2 2 4 FIGURE 1 If G͑x͒ (the right side of Equation 1) is of the form Ce k x, where C and k are constants, then we take as a trial solution a function of the same form, yp͑x͒ ෇ Ae k x, because the derivatives of e k x are constant multiples of e k x. EXAMPLE 2 Solve yЉ ϩ 4y ෇ e 3x. |||| Figure 2 shows solutions of the differential equation in Example 2 in terms of yp and the functions f ͑x͒ ෇ cos 2x and t͑x͒ ෇ sin 2x. Notice that all solutions approach ϱ as x l ϱ and all solutions resemble sine functions when x is negative. 4 SOLUTION The auxiliary equation is r 2 ϩ 4 ෇ 0 with roots Ϯ2i, so the solution of the complementary equation is yc͑x͒ ෇ c1 cos 2x ϩ c2 sin 2x For a particular solution we try yp͑x͒ ෇ Ae 3x. Then yp ෇ 3Ae 3x and ypЉ ෇ 9Ae 3x. SubstiЈ tuting into the differential equation, we have 9Ae 3x ϩ 4͑Ae 3x ͒ ෇ e 3x yp+f+g yp+g yp _4 yp+f _2 FIGURE 2 1 so 13Ae 3x ෇ e 3x and A ෇ 13 . Thus, a particular solution is 2 1 yp͑x͒ ෇ 13 e 3x and the general solution is 1 y͑x͒ ෇ c1 cos 2x ϩ c2 sin 2x ϩ 13 e 3x If G͑x͒ is either C cos kx or C sin kx, then, because of the rules for differentiating the sine and cosine functions, we take as a trial particular solution a function of the form yp͑x͒ ෇ A cos kx ϩ B sin kx 5E-18(pp 1186-1195) 1186 ❙❙❙❙ 1/19/06 3:44 PM Page 1186 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 3 Solve yЉ ϩ yЈ Ϫ 2y ෇ sin x. SOLUTION We try a particular solution yp͑x͒ ෇ A cos x ϩ B sin x yp ෇ ϪA sin x ϩ B cos x Ј Then ypЉ ෇ ϪA cos x Ϫ B sin x so substitution in the differential equation gives ͑ϪA cos x Ϫ B sin x͒ ϩ ͑ϪA sin x ϩ B cos x͒ Ϫ 2͑A cos x ϩ B sin x͒ ෇ sin x ͑Ϫ3A ϩ B͒ cos x ϩ ͑ϪA Ϫ 3B͒ sin x ෇ sin x or This is true if Ϫ3A ϩ B ෇ 0 and ϪA Ϫ 3B ෇ 1 The solution of this system is 1 A ෇ Ϫ 10 3 B ෇ Ϫ 10 so a particular solution is 1 3 yp͑x͒ ෇ Ϫ 10 cos x Ϫ 10 sin x In Example 1 we determined that the solution of the complementary equation is yc ෇ c1 e x ϩ c2 eϪ2x. Thus, the general solution of the given equation is 1 y͑x͒ ෇ c1 e x ϩ c2 eϪ2x Ϫ 10 ͑cos x ϩ 3 sin x͒ If G͑x͒ is a product of functions of the preceding types, then we take the trial solution to be a product of functions of the same type. For instance, in solving the differential equation yЉ ϩ 2yЈ ϩ 4y ෇ x cos 3x we would try yp͑x͒ ෇ ͑Ax ϩ B͒ cos 3x ϩ ͑Cx ϩ D͒ sin 3x If G͑x͒ is a sum of functions of these types, we use the easily verified principle of superposition, which says that if yp1 and yp2 are solutions of ayЉ ϩ byЈ ϩ cy ෇ G1͑x͒ ayЉ ϩ byЈ ϩ cy ෇ G2͑x͒ respectively, then yp1 ϩ yp2 is a solution of ayЉ ϩ byЈ ϩ cy ෇ G1͑x͒ ϩ G2͑x͒ EXAMPLE 4 Solve yЉ Ϫ 4y ෇ xe x ϩ cos 2x. SOLUTION The auxiliary equation is r 2 Ϫ 4 ෇ 0 with roots Ϯ2, so the solution of the com- plementary equation is yc͑x͒ ෇ c1 e 2x ϩ c2 eϪ2x. For the equation yЉ Ϫ 4y ෇ xe x we try yp1͑x͒ ෇ ͑Ax ϩ B͒e x 5E-18(pp 1186-1195) 1/19/06 3:44 PM Page 1187 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ❙❙❙❙ 1187 Ј Љ Then yp1 ෇ ͑Ax ϩ A ϩ B͒e x, yp1 ෇ ͑Ax ϩ 2A ϩ B͒e x, so substitution in the equation gives ͑Ax ϩ 2A ϩ B͒e x Ϫ 4͑Ax ϩ B͒e x ෇ xe x ͑Ϫ3Ax ϩ 2A Ϫ 3B͒e x ෇ xe x or Thus, Ϫ3A ෇ 1 and 2A Ϫ 3B ෇ 0, so A ෇ Ϫ 1 , B ෇ Ϫ 2 , and 3 9 yp1͑x͒ ෇ (Ϫ 1 x Ϫ 2 )e x 3 9 For the equation yЉ Ϫ 4y ෇ cos 2x, we try yp2͑x͒ ෇ C cos 2x ϩ D sin 2x |||| In Figure 3 we show the particular solution yp ෇ yp1 ϩ yp 2 of the differential equation in Example 4. The other solutions are given in terms of f ͑x͒ ෇ e 2 x and t͑x͒ ෇ eϪ2 x. 5 Substitution gives Ϫ4C cos 2x Ϫ 4D sin 2x Ϫ 4͑C cos 2x ϩ D sin 2x͒ ෇ cos 2x Ϫ8C cos 2x Ϫ 8D sin 2x ෇ cos 2x or yp+2f+g Therefore, Ϫ8C ෇ 1, Ϫ8D ෇ 0, and yp+g yp+f _4 yp2͑x͒ ෇ Ϫ 1 cos 2x 8 1 yp By the superposition principle, the general solution is _2 y ෇ yc ϩ yp1 ϩ yp2 ෇ c1 e 2x ϩ c2 eϪ2x Ϫ ( 1 x ϩ 2 )e x Ϫ 1 cos 2x 3 9 8 FIGURE 3 Finally we note that the recommended trial solution yp sometimes turns out to be a solution of the complementary equation and therefore can’t be a solution of the nonhomogeneous equation. In such cases we multiply the recommended trial solution by x (or by x 2 if necessary) so that no term in yp͑x͒ is a solution of the complementary equation. EXAMPLE 5 Solve yЉ ϩ y ෇ sin x. SOLUTION The auxiliary equation is r 2 ϩ 1 ෇ 0 with roots Ϯi, so the solution of the com- plementary equation is yc͑x͒ ෇ c1 cos x ϩ c2 sin x Ordinarily, we would use the trial solution yp͑x͒ ෇ A cos x ϩ B sin x but we observe that it is a solution of the complementary equation, so instead we try yp͑x͒ ෇ Ax cos x ϩ Bx sin x Then yp͑x͒ ෇ A cos x Ϫ Ax sin x ϩ B sin x ϩ Bx cos x Ј ypЉ͑x͒ ෇ Ϫ2A sin x Ϫ Ax cos x ϩ 2B cos x Ϫ Bx sin x 5E-18(pp 1186-1195) 1188 ❙❙❙❙ 1/19/06 3:44 PM Page 1188 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS |||| The graphs of four solutions of the differential equation in Example 5 are shown in Figure 4. Substitution in the differential equation gives ypЉ ϩ yp ෇ Ϫ2A sin x ϩ 2B cos x ෇ sin x 4 so A ෇ Ϫ 1 , B ෇ 0, and 2 _2π 2π yp͑x͒ ෇ Ϫ 1 x cos x 2 yp The general solution is _4 y͑x͒ ෇ c1 cos x ϩ c2 sin x Ϫ 1 x cos x 2 FIGURE 4 We summarize the method of undetermined coefficients as follows: 1. If G͑x͒ ෇ e kxP͑x͒, where P is a polynomial of degree n, then try yp͑x͒ ෇ e kxQ͑x͒, where Q͑x͒ is an nth-degree polynomial (whose coefficients are determined by substituting in the differential equation.) 2. If G͑x͒ ෇ e kxP͑x͒ cos mx or G͑x͒ ෇ e kxP͑x͒ sin mx, where P is an nth-degree polynomial, then try yp͑x͒ ෇ e kxQ͑x͒ cos mx ϩ e kxR͑x͒ sin mx where Q and R are nth-degree polynomials. Modification: If any term of yp is a solution of the complementary equation, multiply yp by x (or by x 2 if necessary). EXAMPLE 6 Determine the form of the trial solution for the differential equation yЉ Ϫ 4yЈ ϩ 13y ෇ e 2x cos 3x. SOLUTION Here G͑x͒ has the form of part 2 of the summary, where k ෇ 2, m ෇ 3, and P͑x͒ ෇ 1. So, at first glance, the form of the trial solution would be yp͑x͒ ෇ e 2x͑A cos 3x ϩ B sin 3x͒ But the auxiliary equation is r 2 Ϫ 4r ϩ 13 ෇ 0, with roots r ෇ 2 Ϯ 3i, so the solution of the complementary equation is yc͑x͒ ෇ e 2x͑c1 cos 3x ϩ c2 sin 3x͒ This means that we have to multiply the suggested trial solution by x. So, instead, we use yp͑x͒ ෇ xe 2x͑A cos 3x ϩ B sin 3x͒ The Method of Variation of Parameters Suppose we have already solved the homogeneous equation ayЉ ϩ byЈ ϩ cy ෇ 0 and written the solution as 4 y͑x͒ ෇ c1 y1͑x͒ ϩ c2 y2͑x͒ where y1 and y2 are linearly independent solutions. Let’s replace the constants (or parame- 5E-18(pp 1186-1195) 1/19/06 3:44 PM Page 1189 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ❙❙❙❙ 1189 ters) c1 and c2 in Equation 4 by arbitrary functions u1͑x͒ and u2͑x͒. We look for a particular solution of the nonhomogeneous equation ayЉ ϩ byЈ ϩ cy ෇ G͑x͒ of the form yp͑x͒ ෇ u1͑x͒y1͑x͒ ϩ u2͑x͒y2͑x͒ 5 (This method is called variation of parameters because we have varied the parameters c1 and c2 to make them functions.) Differentiating Equation 5, we get 6 yp ෇ ͑u1 y1 ϩ u2 y2 ͒ ϩ ͑u1 y1 ϩ u2 y2 ͒ Ј Ј Ј Ј Ј Since u1 and u2 are arbitrary functions, we can impose two conditions on them. One condition is that yp 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’s impose the condition that u1 y1 ϩ u2 y2 ෇ 0 Ј Ј 7 Then ypЉ ෇ u1 y1 ϩ u2 y2 ϩ u1 y1Љ ϩ u2 y2Љ Ј Ј Ј Ј Substituting in the differential equation, we get a͑u1 y1 ϩ u2 y2 ϩ u1 y1Љ ϩ u2 y2Љ͒ ϩ b͑u1 y1 ϩ u2 y2 ͒ ϩ c͑u1 y1 ϩ u2 y2 ͒ ෇ G Ј Ј Ј Ј Ј Ј or 8 u1͑ay1Љ ϩ by1 ϩ cy1 ͒ ϩ u2͑ay2Љ ϩ by2 ϩ cy2 ͒ ϩ a͑u1 y1 ϩ u2 y2 ͒ ෇ G Ј Ј Ј Ј Ј Ј But y1 and y2 are solutions of the complementary equation, so ay1Љ ϩ by1 ϩ cy1 ෇ 0 Ј and ay2Љ ϩ by2 ϩ cy2 ෇ 0 Ј and Equation 8 simplifies to 9 a͑u1 y1 ϩ u2 y2 ͒ ෇ G Ј Ј Ј Ј Equations 7 and 9 form a system of two equations in the unknown functions u1 and u2 . Ј Ј After solving this system we may be able to integrate to find u1 and u2 and then the particular solution is given by Equation 5. EXAMPLE 7 Solve the equation yЉ ϩ y ෇ tan x, 0 Ͻ x Ͻ ␲ ͞2. SOLUTION The auxiliary equation is r 2 ϩ 1 ෇ 0 with roots Ϯi, so the solution of yЉ ϩ y ෇ 0 is c1 sin x ϩ c2 cos x. Using variation of parameters, we seek a solution of the form yp͑x͒ ෇ u1͑x͒ sin x ϩ u2͑x͒ cos x Then yp ෇ ͑u1 sin x ϩ u2 cos x͒ ϩ ͑u1 cos x Ϫ u2 sin x͒ Ј Ј Ј Set 10 uЈ sin x ϩ uЈ cos x ෇ 0 1 2 5E-18(pp 1186-1195) ❙❙❙❙ 1190 1/19/06 3:45 PM Page 1190 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ypЉ ෇ u1 cos x Ϫ u2 sin x Ϫ u1 sin x Ϫ u2 cos x Ј Ј Then For yp to be a solution we must have ypЉ ϩ yp ෇ uЈ cos x Ϫ uЈ sin x ෇ tan x 1 2 11 Solving Equations 10 and 11, we get u1͑sin 2x ϩ cos 2x͒ ෇ cos x tan x Ј u1 ෇ sin x Ј u1͑x͒ ෇ Ϫcos x (We seek a particular solution, so we don’t need a constant of integration here.) Then, from Equation 10, we obtain u2 ෇ Ϫ Ј sin x sin 2x cos 2x Ϫ 1 u1 ෇ Ϫ Ј ෇ ෇ cos x Ϫ sec x cos x cos x cos x u2͑x͒ ෇ sin x Ϫ ln͑sec x ϩ tan x͒ |||| Figure 5 shows four solutions of the differential equation in Example 7. So 2.5 (Note that sec x ϩ tan x Ͼ 0 for 0 Ͻ x Ͻ ␲ ͞2.) Therefore yp͑x͒ ෇ Ϫcos x sin x ϩ ͓sin x Ϫ ln͑sec x ϩ tan x͔͒ cos x ෇ Ϫcos x ln͑sec x ϩ tan x͒ π 2 0 yp and the general solution is _1 y͑x͒ ෇ c1 sin x ϩ c2 cos x Ϫ cos x ln͑sec x ϩ tan x͒ FIGURE 5 |||| 18.2 Exercises 12. 2yЉ ϩ 3yЈ ϩ y ෇ 1 ϩ cos 2x 1–10 |||| Solve the differential equation or initial-value problem using the method of undetermined coefficients. 1. yЉ ϩ 3yЈ ϩ 2y ෇ x 2 2. yЉ ϩ 9y ෇ e 3x 3. yЉ Ϫ 2yЈ ෇ sin 4x ■ 4. yЉ ϩ 6yЈ ϩ 9y ෇ 1 ϩ x Ϫx 7. yЉ ϩ y ෇ e x ϩ x 3, y͑0͒ ෇ 2, 8. yЉ Ϫ 4y ෇ e x cos x, 9. yЉ Ϫ yЈ ෇ xe x, y͑0͒ ෇ 2, 10. yЉ ϩ yЈ Ϫ 2y ෇ x ϩ sin 2x, ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 15. yЉ ϩ 9 yЈ ෇ 1 ϩ xe 9x 16. yЉ ϩ 3yЈ Ϫ 4 y ෇ ͑x 3 ϩ x͒e x 17. yЉ ϩ 2 yЈ ϩ 10 y ෇ x 2eϪx cos 3x yЈ͑0͒ ෇ 0 ■ ■ 18. yЉ ϩ 4y ෇ e 3x ϩ x sin 2x ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ |||| Graph the particular solution and several other solutions. What characteristics do these solutions have in common? 19–22 |||| Solve the differential equation using (a) undetermined coefficients and (b) variation of parameters. 11. 4yЉ ϩ 5yЈ ϩ y ෇ e x 19. yЉ ϩ 4y ෇ x ; 11–12 ■ 14. yЉ ϩ 9 yЈ ෇ xeϪx cos ␲ x yЈ͑0͒ ෇ 1 ■ ■ 13. yЉ ϩ 9 y ෇ e 2x ϩ x 2 sin x yЈ͑0͒ ෇ 2 y͑0͒ ෇ 1, ■ |||| Write a trial solution for the method of undetermined coefficients. Do not determine the coefficients. yЈ͑0͒ ෇ 0 y͑0͒ ෇ 1, ■ 13–18 6. yЉ ϩ 2yЈ ϩ y ෇ xeϪx 5. yЉ Ϫ 4yЈ ϩ 5y ෇ e ■ 20. yЉ Ϫ 3yЈ ϩ 2y ෇ sin x ■ 5E-18(pp 1186-1195) 1/19/06 3:45 PM Page 1191 SECTION 18.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS 21. yЉ Ϫ 2yЈ ϩ y ෇ e 2x 22. yЉ Ϫ yЈ ෇ e ■ ■ ■ 25. yЉ Ϫ 3yЈ ϩ 2y ෇ x ■ ■ ■ ■ ■ ■ ■ ■ 1191 1 1 ϩ eϪx 26. yЉ ϩ 3yЈ ϩ 2y ෇ sin͑e x ͒ ■ 1 x 23–28 |||| Solve the differential equation using the method of variation of parameters. 27. yЉ Ϫ y ෇ 23. yЉ ϩ y ෇ sec x, 0 Ͻ x Ͻ ␲͞2 28. yЉ ϩ 4yЈ ϩ 4y ෇ 24. yЉ ϩ y ෇ cot x, 0 Ͻ x Ͻ ␲͞2 |||| 18.3 ❙❙❙❙ ■ ■ ■ ■ eϪ2x x3 ■ ■ ■ ■ ■ ■ ■ ■ Applications of Second-Order Differential Equations Second-order linear differential equations have a variety of applications in science and engineering. In this section we explore two of them: the vibration of springs and electric circuits. Vibrating Springs We consider the motion of an object with mass m at the end of a spring that is either vertical (as in Figure 1) or horizontal on a level surface (as in Figure 2). In Section 6.4 we discussed Hooke’s Law, which says that if the spring is stretched (or compressed) x units from its natural length, then it exerts a force that is proportional to x : 0 equilibrium position restoring force ෇ Ϫkx x m where k is a positive constant (called the spring constant). If we ignore any external resisting forces (due to air resistance or friction) then, by Newton’s Second Law (force equals mass times acceleration), we have m x FIGURE 1 1 equilibrium position m 0 FIGURE 2 x m d 2x ෇ Ϫkx dt 2 or m d 2x ϩ kx ෇ 0 dt 2 This is a second-order linear differential equation. Its auxiliary equation is mr 2 ϩ k ෇ 0 with roots r ෇ Ϯ␻ i, where ␻ ෇ sk͞m. Thus, the general solution is x͑t͒ ෇ c1 cos ␻ t ϩ c2 sin ␻ t x which can also be written as x͑t͒ ෇ A cos͑␻ t ϩ ␦͒ where ␻ ෇ sk͞m (frequency) A ෇ sc 2 ϩ c 2 1 2 cos ␦ ෇ c1 A (amplitude) sin ␦ ෇ Ϫ c2 A ͑␦ is the phase angle͒ (See Exercise 17.) This type of motion is called simple harmonic motion. 5E-18(pp 1186-1195) 1192 ❙❙❙❙ 1/19/06 3:45 PM Page 1192 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 1 A spring with a mass of 2 kg has natural length 0.5 m. A force of 25.6 N is required to maintain it stretched to a length of 0.7 m. If the spring is stretched to a length of 0.7 m and then released with initial velocity 0, find the position of the mass at any time t . SOLUTION From Hooke’s Law, the force required to stretch the spring is k͑0.2͒ ෇ 25.6 so k ෇ 25.6͞0.2 ෇ 128. Using this value of the spring constant k, together with m ෇ 2 in Equation 1, we have 2 d 2x ϩ 128x ෇ 0 dt 2 As in the earlier general discussion, the solution of this equation is x͑t͒ ෇ c1 cos 8t ϩ c2 sin 8t 2 We are given the initial condition that x͑0͒ ෇ 0.2. But, from Equation 2, x͑0͒ ෇ c1. Therefore, c1 ෇ 0.2. Differentiating Equation 2, we get xЈ͑t͒ ෇ Ϫ8c1 sin 8t ϩ 8c2 cos 8t Since the initial velocity is given as xЈ͑0͒ ෇ 0, we have c2 ෇ 0 and so the solution is x͑t͒ ෇ 1 cos 8t 5 Damped Vibrations m We next consider the motion of a spring that is subject to a frictional force (in the case of the horizontal spring of Figure 2) or a damping force (in the case where a vertical spring moves through a fluid as in Figure 3). An example is the damping force supplied by a shock absorber in a car or a bicycle. We assume that the damping force is proportional to the velocity of the mass and acts in the direction opposite to the motion. (This has been confirmed, at least approximately, by some physical experiments.) Thus FIGURE 3 damping force ෇ Ϫc dx dt where c is a positive constant, called the damping constant. Thus, in this case, Newton’s Second Law gives m d 2x dx ෇ restoring force ϩ damping force ෇ Ϫkx Ϫ c dt 2 dt or 3 m d 2x dx ϩc ϩ kx ෇ 0 dt 2 dt 5E-18(pp 1186-1195) 1/19/06 3:45 PM Page 1193 SECTION 18.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS ❙❙❙❙ 1193 Equation 3 is a second-order linear differential equation and its auxiliary equation is mr 2 ϩ cr ϩ k ෇ 0. The roots are r1 ෇ 4 Ϫc ϩ sc 2 Ϫ 4mk 2m r2 ෇ Ϫc Ϫ sc 2 Ϫ 4mk 2m According to Section 18.1 we need to discuss three cases. x CASE I c 2 Ϫ 4 mk Ͼ 0 (overdamping) ■ In this case r1 and r 2 are distinct real roots and 0 x ෇ c1 e r1 t ϩ c2 e r2 t t x 0 t Since c, m, and k are all positive, we have sc 2 Ϫ 4mk Ͻ c, so the roots r1 and r 2 given by Equations 4 must both be negative. This shows that x l 0 as t l ϱ. Typical graphs of x as a function of t are shown in Figure 4. Notice that oscillations do not occur. (It’s possible for the mass to pass through the equilibrium position once, but only once.) This is because c 2 Ͼ 4mk means that there is a strong damping force (high-viscosity oil or grease) compared with a weak spring or small mass. c 2 Ϫ 4mk ෇ 0 (critical damping) This case corresponds to equal roots CASE II FIGURE 4 Overdamping ■ r1 ෇ r 2 ෇ Ϫ c 2m and the solution is given by x ෇ ͑c1 ϩ c2 t͒eϪ͑c͞2m͒t It is similar to Case I, and typical graphs resemble those in Figure 4 (see Exercise 12), but the damping is just sufficient to suppress vibrations. Any decrease in the viscosity of the fluid leads to the vibrations of the following case. c 2 Ϫ 4mk Ͻ 0 (underdamping) Here the roots are complex: CASE III ■ ͮ r1 c ෇Ϫ Ϯ ␻i r2 2m x x=Ae– (c/ 2m)t where 0 t x=_Ae– (c/ 2m)t FIGURE 5 Underdamping ␻෇ s4mk Ϫ c 2 2m The solution is given by x ෇ eϪ͑c͞2m͒t͑c1 cos ␻ t ϩ c2 sin ␻ t͒ We see that there are oscillations that are damped by the factor eϪ͑c͞2m͒t. Since c Ͼ 0 and m Ͼ 0, we have Ϫ͑c͞2m͒ Ͻ 0 so eϪ͑c͞2m͒t l 0 as t l ϱ. This implies that x l 0 as t l ϱ; that is, the motion decays to 0 as time increases. A typical graph is shown in Figure 5. 5E-18(pp 1186-1195) 1194 ❙❙❙❙ 1/19/06 3:45 PM Page 1194 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 2 Suppose that the spring of Example 1 is immersed in a fluid with damping constant c ෇ 40. Find the position of the mass at any time t if it starts from the equilibrium position and is given a push to start it with an initial velocity of 0.6 m͞s. SOLUTION From Example 1 the mass is m ෇ 2 and the spring constant is k ෇ 128, so the differential equation (3) becomes 2 d 2x dx ϩ 40 ϩ 128x ෇ 0 dt 2 dt d 2x dx ϩ 64x ෇ 0 2 ϩ 20 dt dt or The auxiliary equation is r 2 ϩ 20r ϩ 64 ෇ ͑r ϩ 4͒͑r ϩ 16͒ ෇ 0 with roots Ϫ4 and Ϫ16, so the motion is overdamped and the solution is x͑t͒ ෇ c1 eϪ4t ϩ c2 eϪ16t |||| Figure 6 shows the graph of the position function for the overdamped motion in Example 2. We are given that x͑0͒ ෇ 0, so c1 ϩ c2 ෇ 0. Differentiating, we get 0.03 xЈ͑t͒ ෇ Ϫ4c1 eϪ4t Ϫ 16c2 eϪ16t xЈ͑0͒ ෇ Ϫ4c1 Ϫ 16c2 ෇ 0.6 so 0 1.5 Since c2 ෇ Ϫc1 , this gives 12c1 ෇ 0.6 or c1 ෇ 0.05. Therefore x ෇ 0.05͑eϪ4t Ϫ eϪ16t ͒ FIGURE 6 Forced Vibrations Suppose that, in addition to the restoring force and the damping force, the motion of the spring is affected by an external force F͑t͒. Then Newton’s Second Law gives m d 2x ෇ restoring force ϩ damping force ϩ external force dt 2 ෇ Ϫkx Ϫ c dx ϩ F͑t͒ dt Thus, instead of the homogeneous equation (3), the motion of the spring is now governed by the following nonhomogeneous differential equation: 5 m d 2x dx ϩc ϩ kx ෇ F͑t͒ dt 2 dt The motion of the spring can be determined by the methods of Section 18.2. 5E-18(pp 1186-1195) 1/19/06 3:45 PM Page 1195 SECTION 18.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS ❙❙❙❙ 1195 A commonly occurring type of external force is a periodic force function F͑t͒ ෇ F0 cos ␻ 0 t where ␻ 0 ␻ ෇ sk͞m In this case, and in the absence of a damping force (c ෇ 0), you are asked in Exercise 9 to use the method of undetermined coefficients to show that 6 x͑t͒ ෇ c1 cos ␻ t ϩ c2 sin ␻ t ϩ F0 2 cos ␻ 0 t m͑␻ Ϫ ␻ 0 ͒ 2 If ␻ 0 ෇ ␻, then the applied frequency reinforces the natural frequency and the result is vibrations of large amplitude. This is the phenomenon of resonance (see Exercise 10). Electric Circuits R switch L E C In Sections 10.3 and 10.6 we were able to use first-order separable and linear equations to analyze electric circuits that contain a resistor and inductor (see Figure 5 on page 639 or Figure 4 on page 671) or a resistor and capacitor (see Exercise 29 on page 673). Now that we know how to solve second-order linear equations, we are in a position to analyze the circuit shown in Figure 7. It contains an electromotive force E (supplied by a battery or generator), a resistor R, an inductor L, and a capacitor C, in series. If the charge on the capacitor at time t is Q ෇ Q͑t͒, then the current is the rate of change of Q with respect to t : I ෇ dQ͞dt. As in Section 10.6, it is known from physics that the voltage drops across the resistor, inductor, and capacitor are FIGURE 7 RI L dI dt Q C respectively. Kirchhoff’s voltage law says that the sum of these voltage drops is equal to the supplied voltage: L dI Q ϩ RI ϩ ෇ E͑t͒ dt C Since I ෇ dQ͞dt, this equation becomes 7 L d 2Q dQ 1 ϩR ϩ Q ෇ E͑t͒ dt 2 dt C which is a second-order linear differential equation with constant coefficients. If the charge Q0 and the current I 0 are known at time 0, then we have the initial conditions Q͑0͒ ෇ Q0 QЈ͑0͒ ෇ I͑0͒ ෇ I 0 and the initial-value problem can be solved by the methods of Section 18.2. 5E-18(pp 1196-1204) 1196 ❙❙❙❙ 1/19/06 3:27 PM Page 1196 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS A differential equation for the current can be obtained by differentiating Equation 7 with respect to t and remembering that I ෇ dQ͞dt : L d 2I dI 1 ϩ I ෇ EЈ͑t͒ 2 ϩ R dt dt C EXAMPLE 3 Find the charge and current at time t in the circuit of Figure 7 if R ෇ 40 ⍀, L ෇ 1 H, C ෇ 16 ϫ 10Ϫ4 F, E͑t͒ ෇ 100 cos 10t, and the initial charge and current are both 0. SOLUTION With the given values of L, R, C, and E͑t͒, Equation 7 becomes 8 d 2Q dQ ϩ 40 ϩ 625Q ෇ 100 cos 10t dt 2 dt The auxiliary equation is r 2 ϩ 40r ϩ 625 ෇ 0 with roots r෇ Ϫ40 Ϯ sϪ900 ෇ Ϫ20 Ϯ 15i 2 so the solution of the complementary equation is Qc͑t͒ ෇ eϪ20t ͑c1 cos 15t ϩ c2 sin 15t͒ For the method of undetermined coefficients we try the particular solution Qp͑t͒ ෇ A cos 10t ϩ B sin 10t Then Qp͑t͒ ෇ Ϫ10A sin 10t ϩ 10B cos 10t Ј QpЉ͑t͒ ෇ Ϫ100A cos 10t Ϫ 100B sin 10t Substituting into Equation 8, we have ͑Ϫ100A cos 10t Ϫ 100B sin 10t͒ ϩ 40͑Ϫ10A sin 10t ϩ 10B cos 10t͒ ϩ 625͑A cos 10t ϩ B sin 10t͒ ෇ 100 cos 10t or ͑525A ϩ 400B͒ cos 10t ϩ ͑Ϫ400A ϩ 525B͒ sin 10t ෇ 100 cos 10t Equating coefficients, we have 525A ϩ 400B ෇ 100 Ϫ400A ϩ 525B ෇ 0 21A ϩ 16B ෇ 4 or or Ϫ16A ϩ 21B ෇ 0 84 64 The solution of this system is A ෇ 697 and B ෇ 697 , so a particular solution is 1 Qp͑t͒ ෇ 697 ͑84 cos 10t ϩ 64 sin 10t͒ and the general solution is 4 Q͑t͒ ෇ Qc͑t͒ ϩ Qp͑t͒ ෇ eϪ20t ͑c1 cos 15t ϩ c2 sin 15t͒ ϩ 697 ͑21 cos 10t ϩ 16 sin 10t͒ 5E-18(pp 1196-1204) 1/19/06 3:28 PM Page 1197 SECTION 18.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS ❙❙❙❙ 1197 Imposing the initial condition Q͑0͒ ෇ 0, we get Q͑0͒ ෇ c1 ϩ 697 ෇ 0 84 84 c1 ෇ Ϫ 697 To impose the other initial condition we first differentiate to find the current: I෇ dQ ෇ eϪ20t ͓͑Ϫ20c1 ϩ 15c2 ͒ cos 15t ϩ ͑Ϫ15c1 Ϫ 20c2 ͒ sin 15t͔ dt 40 ϩ 697 ͑Ϫ21 sin 10t ϩ 16 cos 10t͒ I͑0͒ ෇ Ϫ20c1 ϩ 15c2 ϩ 640 ෇ 0 697 464 c2 ෇ Ϫ 2091 Thus, the formula for the charge is 4 697 Q͑t͒ ෇ ͫ ͬ eϪ20t ͑Ϫ63 cos 15t Ϫ 116 sin 15t͒ ϩ ͑21 cos 10t ϩ 16 sin 10t͒ 3 and the expression for the current is 1 I͑t͒ ෇ 2091 ͓eϪ20t͑Ϫ1920 cos 15t ϩ 13,060 sin 15t͒ ϩ 120͑Ϫ21 sin 10t ϩ 16 cos 10t͔͒ In Example 3 the solution for Q͑t͒ consists of two parts. Since eϪ20t l 0 as t l ϱ and both cos 15t and sin 15t are bounded functions, NOTE 1 ■ 0.2 4 Qc͑t͒ ෇ 2091 eϪ20t͑Ϫ63 cos 15t Ϫ 116 sin 15t͒ l 0 Qp 0 Q 1.2 as t l ϱ So, for large values of t , 4 Q͑t͒ Ϸ Qp͑t͒ ෇ 697 ͑21 cos 10t ϩ 16 sin 10t͒ _0.2 FIGURE 8 and, for this reason, Qp͑t͒ is called the steady state solution. Figure 8 shows how the graph of the steady state solution compares with the graph of Q in this case. NOTE 2 Comparing Equations 5 and 7, we see that mathematically they are identical. This suggests the analogies given in the following chart between physical situations that, at first glance, are very different. ■ 5 7 d 2x dx ϩ c ϩ kx ෇ F͑t͒ dt 2 dt d 2Q dQ 1 L ϩR ϩ Q ෇ E͑t͒ dt 2 dt C m Spring system x dx͞dt m c k F͑t͒ displacement velocity mass damping constant spring constant external force Electric circuit Q I ෇ dQ͞dt L R 1͞C E͑t͒ charge current inductance resistance elastance electromotive force We can also transfer other ideas from one situation to the other. For instance, the steady state solution discussed in Note 1 makes sense in the spring system. And the phenomenon of resonance in the spring system can be usefully carried over to electric circuits as electrical resonance. 5E-18(pp 1196-1204) 1198 ❙❙❙❙ 1/19/06 3:28 PM Page 1198 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS |||| 18.3 Exercises 12. Consider a spring subject to a frictional or damping force. 1. A spring with a 3-kg mass is held stretched 0.6 m beyond its (a) In the critically damped case, the motion is given by x ෇ c1 ert ϩ c2 tert. Show that the graph of x crosses the t-axis whenever c1 and c2 have opposite signs. (b) In the overdamped case, the motion is given by x ෇ c1e r t ϩ c2 e r t, where r1 Ͼ r2. Determine a condition on the relative magnitudes of c1 and c2 under which the graph of x crosses the t-axis at a positive value of t. natural length by a force of 20 N. If the spring begins at its equilibrium position but a push gives it an initial velocity of 1.2 m͞s, find the position of the mass after t seconds. 2. A spring with a 4-kg mass has natural length 1 m and is main- 1 tained stretched to a length of 1.3 m by a force of 24.3 N. If the spring is compressed to a length of 0.8 m and then released with zero velocity, find the position of the mass at any time t. 13. A series circuit consists of a resistor with R ෇ 20 ⍀, an induc- tor with L ෇ 1 H, a capacitor with C ෇ 0.002 F, and a 12-V battery. If the initial charge and current are both 0, find the charge and current at time t. 3. A spring with a mass of 2 kg has damping constant 14, and a force of 6 N is required to keep the spring stretched 0.5 m beyond its natural length. The spring is stretched 1 m beyond its natural length and then released with zero velocity. Find the position of the mass at any time t. 14. A series circuit contains a resistor with R ෇ 24 ⍀, an inductor 4. A spring with a mass of 3 kg has damping constant 30 and ; spring constant 123. (a) Find the position of the mass at time t if it starts at the equilibrium position with a velocity of 2 m͞s. (b) Graph the position function of the mass. ; a voltage of E͑t͒ ෇ 12 sin 10t. Find the charge at time t. 16. The battery in Exercise 14 is replaced by a generator producing critical damping. 6. For the spring in Exercise 4, find the damping constant that ; 7. A spring has a mass of 1 kg and its spring constant is k ෇ 100. The spring is released at a point 0.1 m above its equilibrium position. Graph the position function for the following values of the damping constant c: 10, 15, 20, 25, 30. What type of damping occurs in each case? ; 8. A spring has a mass of 1 kg and its damping constant is c ෇ 10. The spring starts from its equilibrium position with a velocity of 1 m͞s. Graph the position function for the following values of the spring constant k: 10, 20, 25, 30, 40. What type of damping occurs in each case? 9. Suppose a spring has mass m and spring constant k and let ␻ ෇ sk͞m. Suppose that the damping constant is so small that the damping force is negligible. If an external force F͑t͒ ෇ F0 cos ␻ 0 t is applied, where ␻ 0 ␻, use the method of undetermined coefficients to show that the motion of the mass is described by Equation 6. 10. As in Exercise 9, consider a spring with mass m, spring con- stant k, and damping constant c ෇ 0, and let ␻ ෇ sk͞m. If an external force F͑t͒ ෇ F0 cos ␻ t is applied (the applied frequency equals the natural frequency), use the method of undetermined coefficients to show that the motion of the mass is given by x͑t͒ ෇ c1 cos ␻ t ϩ c2 sin ␻ t ϩ ͑F0 ͑͞2m␻͒͒t sin ␻ t. 11. Show that if ␻ 0 ␻, but ␻͞␻ 0 is a rational number, then the motion described by Equation 6 is periodic. with L ෇ 2 H, a capacitor with C ෇ 0.005 F, and a 12-V battery. The initial charge is Q ෇ 0.001 C and the initial current is 0. (a) Find the charge and current at time t. (b) Graph the charge and current functions. 15. The battery in Exercise 13 is replaced by a generator producing 5. For the spring in Exercise 3, find the mass that would produce would produce critical damping. 2 ; a voltage of E͑t͒ ෇ 12 sin 10t. (a) Find the charge at time t. (b) Graph the charge function. 17. Verify that the solution to Equation 1 can be written in the form x͑t͒ ෇ A cos͑␻ t ϩ ␦͒. 18. The figure shows a pendulum with length L and the angle ␪ from the vertical to the pendulum. It can be shown that ␪, as a function of time, satisfies the nonlinear differential equation t d 2␪ ϩ sin ␪ ෇ 0 dt 2 L where t is the acceleration due to gravity. For small values of ␪ we can use the linear approximation sin ␪ Ϸ ␪ and then the differential equation becomes linear. (a) Find the equation of motion of a pendulum with length 1 m if ␪ is initially 0.2 rad and the initial angular velocity is d␪͞dt ෇ 1 rad͞s. (b) What is the maximum angle from the vertical? (c) What is the period of the pendulum (that is, the time to complete one back-and-forth swing)? (d) When will the pendulum first be vertical? (e) What is the angular velocity when the pendulum is vertical? ¨ L 5E-18(pp 1196-1204) 1/19/06 3:29 PM Page 1199 SECTION 18.4 SERIES SOLUTIONS |||| 18.4 ❙❙❙❙ 1199 Series Solutions Many differential equations can’t be solved explicitly in terms of finite combinations of simple familiar functions. This is true even for a simple-looking equation like yЉ Ϫ 2xyЈ ϩ y ෇ 0 1 But it is important to be able to solve equations such as Equation 1 because they arise from physical problems and, in particular, in connection with the Schrödinger equation in quantum mechanics. In such a case we use the method of power series; that is, we look for a solution of the form y ෇ f ͑x͒ ෇ ϱ ͚cx n n ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ и и и n෇0 The method is to substitute this expression into the differential equation and determine the values of the coefficients c0 , c1, c2 , . . . . This technique resembles the method of undetermined coefficients discussed in Section 18.2. Before using power series to solve Equation 1, we illustrate the method on the simpler equation yЉ ϩ y ෇ 0 in Example 1. It’s true that we already know how to solve this equation by the techniques of Section 18.1, but it’s easier to understand the power series method when it is applied to this simpler equation. EXAMPLE 1 Use power series to solve the equation yЉ ϩ y ෇ 0. SOLUTION We assume there is a solution of the form 2 y ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ и и и ෇ ϱ ͚cx n n n෇0 We can differentiate power series term by term, so ϱ ͚ nc x yЈ ෇ c1 ϩ 2c2 x ϩ 3c3 x 2 ϩ и и и ෇ n nϪ1 n෇1 3 yЉ ෇ 2c2 ϩ 2 и 3c3 x ϩ и и и ෇ ϱ ͚ n͑n Ϫ 1͒c x n nϪ2 n෇2 In order to compare the expressions for y and yЉ more easily, we rewrite yЉ as follows: |||| By writing out the first few terms of (4), you can see that it is the same as (3). To obtain (4) we replaced n by n ϩ 2 and began the summation at 0 instead of 2. yЉ ෇ 4 ϱ ͚ ͑n ϩ 2͒͑n ϩ 1͒c nϩ2 xn n෇0 Substituting the expressions in Equations 2 and 4 into the differential equation, we obtain ϱ ͚ ͑n ϩ 2͒͑n ϩ 1͒c xn ϩ nϩ2 n෇0 ϱ ͚cx n n ෇0 n෇0 or ϱ 5 ͚ ͓͑n ϩ 2͒͑n ϩ 1͒c nϩ2 n෇0 ϩ cn ͔x n ෇ 0 5E-18(pp 1196-1204) 1200 ❙❙❙❙ 1/19/06 3:30 PM Page 1200 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS If two power series are equal, then the corresponding coefficients must be equal. Therefore, the coefficients of x n in Equation 5 must be 0: ͑n ϩ 2͒͑n ϩ 1͒cnϩ2 ϩ cn ෇ 0 cnϩ2 ෇ Ϫ 6 cn ͑n ϩ 1͒͑n ϩ 2͒ n ෇ 0, 1, 2, 3, . . . Equation 6 is called a recursion relation. If c0 and c1 are known, this equation allows us to determine the remaining coefficients recursively by putting n ෇ 0, 1, 2, 3, . . . in succession. Put n ෇ 0: c2 ෇ Ϫ c0 1ؒ2 Put n ෇ 1: c3 ෇ Ϫ c1 2ؒ3 Put n ෇ 2: c4 ෇ Ϫ c2 c0 c0 ෇ ෇ 3ؒ4 1ؒ2ؒ3ؒ4 4! Put n ෇ 3: c5 ෇ Ϫ c3 c1 c1 ෇ ෇ 4ؒ5 2ؒ3ؒ4ؒ5 5! Put n ෇ 4: c6 ෇ Ϫ c4 c0 c0 ෇Ϫ ෇Ϫ 5ؒ6 4! 5 ؒ 6 6! Put n ෇ 5: c7 ෇ Ϫ c5 c1 c1 ෇Ϫ ෇Ϫ 6ؒ7 5! 6 ؒ 7 7! By now we see the pattern: For the even coefficients, c2n ෇ ͑Ϫ1͒n c0 ͑2n͒! For the odd coefficients, c2nϩ1 ෇ ͑Ϫ1͒n c1 ͑2n ϩ 1͒! Putting these values back into Equation 2, we write the solution as y ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ c4 x 4 ϩ c5 x 5 ϩ и и и ͩ ෇ c0 1 Ϫ ෇ ෇ c0 ͩ ϩ c1 x Ϫ ϱ ͚ ͑Ϫ1͒ n n෇0 ͪ x2 x4 x6 x 2n ϩ Ϫ ϩ и и и ϩ ͑Ϫ1͒n ϩ иии 2! 4! 6! ͑2n͒! ͪ x3 x5 x7 x 2nϩ1 ϩ Ϫ ϩ и и и ϩ ͑Ϫ1͒n ϩ иии 3! 5! 7! ͑2n ϩ 1͒! ϱ x 2n x 2nϩ1 ϩ c1 ͚ ͑Ϫ1͒n ͑2n͒! ͑2n ϩ 1͒! n෇0 Notice that there are two arbitrary constants, c0 and c1. 5E-18(pp 1196-1204) 1/19/06 3:31 PM Page 1201 SECTION 18.4 SERIES SOLUTIONS ❙❙❙❙ 1201 NOTE 1 We recognize the series obtained in Example 1 as being the Maclaurin series for cos x and sin x. (See Equations 12.10.16 and 12.10.15.) Therefore, we could write the solution as ■ y͑x͒ ෇ c0 cos x ϩ c1 sin x But we are not usually able to express power series solutions of differential equations in terms of known functions. EXAMPLE 2 Solve yЉ Ϫ 2xyЈ ϩ y ෇ 0. SOLUTION We assume there is a solution of the form y෇ ϱ ͚cx n n n෇0 yЈ ෇ Then ϱ ͚ nc x n nϪ1 n෇1 yЉ ෇ and ϱ ͚ n͑n Ϫ 1͒c x n nϪ2 n෇2 ϱ ෇ ͚ ͑n ϩ 2͒͑n ϩ 1͒c nϩ2 xn n෇0 as in Example 1. Substituting in the differential equation, we get ϱ ͚ ͑n ϩ 2͒͑n ϩ 1͒c x n Ϫ 2x nϩ2 n෇0 ϱ nϩ2 xn Ϫ n෇0 ͚ n෇1 2ncn x n ෇ n n෇1 nϩ2 n෇0 ϩ ϱ ͚cx n n ෇0 n ෇0 n෇0 ϱ ͚ ͓͑n ϩ 2͒͑n ϩ 1͒c 2ncn x n nϪ1 ͚ 2nc x ϱ ϱ ͚ n n෇1 ͚ ͑n ϩ 2͒͑n ϩ 1͒c ϱ ϱ ͚ nc x n ϩ ϱ ͚cx n n෇0 Ϫ ͑2n Ϫ 1͒cn ͔x n ෇ 0 n෇0 This equation is true if the coefficient of x n is 0: ͑n ϩ 2͒͑n ϩ 1͒cnϩ2 Ϫ ͑2n Ϫ 1͒cn ෇ 0 7 cnϩ2 ෇ 2n Ϫ 1 cn ͑n ϩ 1͒͑n ϩ 2͒ n ෇ 0, 1, 2, 3, . . . We solve this recursion relation by putting n ෇ 0, 1, 2, 3, . . . successively in Equation 7: Put n ෇ 0: c2 ෇ Ϫ1 c0 1ؒ2 Put n ෇ 1: c3 ෇ 1 c1 2ؒ3 Put n ෇ 2: c4 ෇ 3 3 3 c2 ෇ Ϫ c0 ෇ Ϫ c0 3ؒ4 1ؒ2ؒ3ؒ4 4! Put n ෇ 3: c5 ෇ 5 1ؒ5 1ؒ5 c3 ෇ c1 ෇ c1 4ؒ5 2ؒ3ؒ4ؒ5 5! 5E-18(pp 1196-1204) 1202 ❙❙❙❙ 1/19/06 3:32 PM Page 1202 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS Put n ෇ 4: c6 ෇ 7 3ؒ7 3ؒ7 c4 ෇ Ϫ c0 ෇ Ϫ c0 5ؒ6 4! 5 ؒ 6 6! Put n ෇ 5: c7 ෇ 9 1ؒ5ؒ9 1ؒ5ؒ9 c5 ෇ c1 ෇ c1 6ؒ7 5! 6 ؒ 7 7! Put n ෇ 6: c8 ෇ 11 3 ؒ 7 ؒ 11 c6 ෇ Ϫ c0 7ؒ8 8! Put n ෇ 7: c9 ෇ 13 1 ؒ 5 ؒ 9 ؒ 13 c7 ෇ c1 8ؒ9 9! In general, the even coefficients are given by c2n ෇ Ϫ 3 ؒ 7 ؒ 11 ؒ и и и ؒ ͑4n Ϫ 5͒ c0 ͑2n͒! and the odd coefficients are given by c2nϩ1 ෇ 1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ c1 ͑2n ϩ 1͒! The solution is y ෇ c0 ϩ c1 x ϩ c2 x 2 ϩ c3 x 3 ϩ c4 x 4 ϩ и и и ͩ ෇ c0 1 Ϫ ෇ ͪ 1 2 3 4 3ؒ7 6 3 ؒ 7 ؒ 11 8 x Ϫ x Ϫ x Ϫ x Ϫ иии 2! 4! 6! 8! ͩ ϩ c1 x ϩ ͪ 1 3 1ؒ5 5 1ؒ5ؒ9 7 1 ؒ 5 ؒ 9 ؒ 13 9 x ϩ x ϩ x ϩ x ϩ иии 3! 5! 7! 9! or ͩ y ෇ c0 1 Ϫ 8 ෇ ϱ 1 2 3 ؒ 7 ؒ и и и ؒ ͑4n Ϫ 5͒ 2n x Ϫ ͚ x 2! ͑2n͒! n෇2 ͩ ϩ c1 x ϩ ϱ ͚ n෇1 ͪ 1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ 2nϩ1 x ͑2n ϩ 1͒! ͪ 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. 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 ■ ■ y1͑x͒ ෇ 1 Ϫ and ϱ 1 2 3 ؒ 7 ؒ и и и ؒ ͑4n Ϫ 5͒ 2n x Ϫ ͚ x 2! ͑2n͒! n෇2 y2͑x͒ ෇ x ϩ ͚ ϱ n෇1 1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ 2nϩ1 x ͑2n ϩ 1͒! 5E-18(pp 1196-1204) 1/19/06 3:32 PM Page 1203 CHAPTER 18 REVIEW 2 1203 are perfectly good functions but they can’t be expressed in terms of familiar functions. We can use these power series expressions for y1 and y2 to compute approximate values of the functions and even to graph them. Figure 1 shows the first few partial sums T0, T2, T4, . . . (Taylor polynomials) for y1͑x͒, and we see how they converge to y1 . In this way we can graph both y1 and y2 in Figure 2. T¸ 2 _2 ❙❙❙ T¡¸ NOTE 4 ■ If we were asked to solve the initial-value problem _8 yЉ Ϫ 2xyЈ ϩ y ෇ 0 FIGURE 1 c0 ෇ y͑0͒ ෇ 0 fi _2.5 c1 ෇ yЈ͑0͒ ෇ 1 2.5 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 › _15 |||| 18.4 |||| ϱ ͚ y͑x͒ ෇ x ϩ FIGURE 2 n෇1 1 ؒ 5 ؒ 9 ؒ и и и ؒ ͑4n Ϫ 3͒ 2nϩ1 x ͑2n ϩ 1͒! Exercises 11. yЉ ϩ x 2 yЈ ϩ x y ෇ 0, Use power series to solve the differential equation. 1. yЈ Ϫ y ෇ 0 2. yЈ ෇ x y 3. yЈ ෇ x 2 y 4. ͑x Ϫ 3͒yЈ ϩ 2y ෇ 0 5. yЉ ϩ x yЈ ϩ y ෇ 0 6. yЉ ෇ y ■ ■ 10. yЉ ϩ x 2 y ෇ 0, y͑0͒ ෇ 1, y͑0͒ ෇ 1, 18 Review yЈ͑0͒ ෇ 0 ■ CONCEPT CHECK 1. (a) Write the general form of a second-order homogeneous linear differential equation with constant coefficients. (b) Write the auxiliary equation. (c) How do you use the roots of the auxiliary equation to solve the differential equation? Write the form of the solution for each of the three cases that can occur. 2. (a) What is an initial-value problem for a second-order differ- ential equation? (b) What is a boundary-value problem for such an equation? 3. (a) Write the general form of a second-order nonhomogeneous linear differential equation with constant coefficients. ■ y͑0͒ ෇ 0, ■ ■ yЈ͑0͒ ෇ 1 ■ ■ ■ ■ y͑0͒ ෇ 1 yЈ͑0͒ ෇ 0 is called a Bessel function of order 0. (a) Solve the initial-value problem to find a power series expansion for the Bessel function. (b) Graph several Taylor polynomials until you reach one that looks like a good approximation to the Bessel function on the interval ͓Ϫ5, 5͔. ; yЈ͑0͒ ෇ 0 ■ x 2 yЉ ϩ x yЈ ϩ x 2 y ෇ 0 8. yЉ ෇ x y 9. yЉ Ϫ x yЈ Ϫ y ෇ 0, ■ 12. The solution of the initial-value problem 7. ͑x 2 ϩ 1͒yЉ ϩ x yЈ Ϫ y ෇ 0 |||| yЈ͑0͒ ෇ 1 we would observe from Theorem 11.10.5 that 15 1–11 y͑0͒ ෇ 0 ■ (b) What is the complementary equation? How does it help solve the original differential equation? (c) Explain how the method of undetermined coefficients works. (d) Explain how the method of variation of parameters works. 4. Discuss two applications of second-order linear differential equations. 5. How do you use power series to solve a differential equation? ■ 5E-18(pp 1196-1204) 1204 ❙❙❙❙ 1/19/06 3:33 PM Page 1204 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ■ TRUE-FALSE QUIZ 3. The general solution of yЉ Ϫ y ෇ 0 can be written as Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement. y ෇ c1 cosh x ϩ c2 sinh x 1. If y1 and y2 are solutions of yЉ ϩ y ෇ 0, then y1 ϩ y2 is also a solution of the equation. 4. The equation yЉ Ϫ y ෇ e x has a particular solution of the form 2. If y1 and y2 are solutions of yЉ ϩ 6yЈ ϩ 5y ෇ x, then yp ෇ Ae x c1 y1 ϩ c2 y2 is also a solution of the equation. ■ 1–10 |||| EXERCISES yЉ Ϫ x yЈ Ϫ 2y ෇ 0 2. yЉ ϩ 4yЈ ϩ 13y ෇ 0 17. A series circuit contains a resistor with R ෇ 40 ⍀, an inductor with L ෇ 2 H, a capacitor with C ෇ 0.0025 F, and a 12-V battery. The initial charge is Q ෇ 0.01 C and the initial current is 0. Find the charge at time t. 3. yЉ ϩ 3y ෇ 0 4. 4yЉ ϩ 4yЈ ϩ y ෇ 0 d 2y dy Ϫ4 ϩ 5y ෇ e 2x dx 2 dx 18. A spring with a mass of 2 kg has damping constant 16, and a force of 12.8 N keeps the spring stretched 0.2 m beyond its natural length. Find the position of the mass at time t if it starts at the equilibrium position with a velocity of 2.4 m͞s. d 2y dy Ϫ 2y ෇ x 2 6. 2 ϩ dx dx 7. 8. 19. Assume that the earth is a solid sphere of uniform density with d 2y dy Ϫ2 ϩ y ෇ x cos x dx 2 dx d 2y ϩ 4 y ෇ sin 2 x dx 2 mass M and radius R ෇ 3960 mi. For a particle of mass m within the earth at a distance r from the earth’s center, the gravitational force attracting the particle to the center is Fr ෇ d 2y dy Ϫ Ϫ 6y ෇ 1 ϩ eϪ2x 9. dx 2 dx d 2y ϩ y ෇ csc x, 10. dx 2 ■ ■ 11–14 ■ |||| ■ 0 Ͻ x Ͻ ␲ ͞2 ■ y͑1͒ ෇ 3, 12. yЉ Ϫ 6yЈ ϩ 25y ෇ 0, 13. yЉ Ϫ 5yЈ ϩ 4y ෇ 0, 14. 9yЉ ϩ y ෇ 3x ϩ e Ϫx, ■ ■ ■ ■ ■ ■ ■ ■ ■ Solve the initial-value problem. 11. yЉ ϩ 6yЈ ෇ 0, ■ ■ ■ 16. Use power series to solve the equation Solve the differential equation. 1. yЉ Ϫ 2yЈ Ϫ 15y ෇ 0 5. ■ yЈ͑1͒ ෇ 12 y͑0͒ ෇ 2, y͑0͒ ෇ 0, y͑0͒ ෇ 1, ■ ■ ■ yЈ͑0͒ ෇ 1 yЈ͑0͒ ෇ 2 ■ ■ 15. Use power series to solve the initial-value problem yЉ ϩ x yЈ ϩ y ෇ 0 y͑0͒ ෇ 0 where G is the gravitational constant and Mr is the mass of the earth within the sphere of radius r . ϪGMm (a) Show that Fr ෇ r. R3 (b) Suppose a hole is drilled through the earth along a diameter. Show that if a particle of mass m is dropped from rest, at the surface, into the hole, then the distance y ෇ y͑t͒ of the particle from the center of the earth at time t is given by yЉ͑t͒ ෇ Ϫk 2 y͑t͒ yЈ͑0͒ ෇ 1 ■ ϪGMr m r2 yЈ͑0͒ ෇ 1 ■ ■ where k 2 ෇ GM͞R 3 ෇ t͞R. (c) Conclude from part (b) that the particle undergoes simple harmonic motion. Find the period T. (d) With what speed does the particle pass through the center of the earth? ...
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This note was uploaded on 02/08/2010 for the course M 340L taught by Professor Lay during the Spring '10 term at École Normale Supérieure.

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