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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. S econd-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 d 2y dx 2 Px Qx dy dx Rxy Gx 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. 0, for all x, in Equation 1. Such equaIn this section we study the case where G x tions are called homogeneous linear equations. Thus, the form of a second-order linear homogeneous differential equation is 2 Px d 2y dx 2 Qx dy dx Rxy 0 0 for some x, Equation 1 is nonhomogeneous and is discussed in Section 18.2. If G x 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 c2 y2 is also a solution. 3 Theorem If y1 x and y2 x are both solutions of the linear homogeneous equation (2) and c1 and c2 are any constants, then the function yx c1 y1 x c2 y2 x is also a solution of Equation 2. Proof Since y1 and y2 are solutions of Equation 2, we have P x y1 and Q x y1 R x y1 0 P x y2 Q x y2 R x y2 0 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 Pxy Qxy Rxy P x c1 y1 c2 y2 Q x c1 y1 c2 y2 R x c1 y1 c2 y2 P x c1 y1 c2 y2 Q x c1 y1 c2 y2 R x c1 y1 c2 y2 c1 P x y1 Q x y1 c1 0 Thus, y c1 y1 c2 0 R x y1 c2 P x y2 Q x y2 R x y2 0 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 x x 2 and t x 5x 2 are linearly dependent, x x but f x e and t x xe are linearly independent. 4 Theorem If y1 and y2 are linearly independent solutions of Equation 2, and P x is never 0, then the general solution is given by yx c1 y1 x c2 y2 x 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 ay by 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 y plus another constant times y 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: y re rx. Furthermore, y r 2e rx. rx If we substitute these expressions into Equation 5, we see that y e is a solution if ar 2e rx ar 2 or But e rx is never 0. Thus, y 6 bre rx ce rx 0 br c e rx 0 e rx is a solution of Equation 5 if r is a root of the equation ar 2 br c 0 Equation 6 is called the auxiliary equation (or characteristic equation) of the differential equation a y by c y 0. Notice that it is an algebraic equation that is obtained from the differential equation by replacing y by r 2, y 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: sb 2 2a b r1 7 4 ac b r2 sb 2 2a 4 ac We distinguish three cases according to the sign of the discriminant b 2 4 ac. b2 4 ac 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 unequal, then the general solution of ay by cy 8 c1 e r 1 x y |||| In Figure 1 the graphs of the basic solutions fx e 2 x and t x e 3 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. EXAMPLE 1 Solve the equation y c 0 are real and c2 e r 2 x 6y 0. SOLUTION The auxiliary equation is r2 whose roots are r tial equation is 8 y br 0 is 2, r 6 r 2r 3 0 3. Therefore, by (8) the general solution of the given differen- 5f+g f+5g f+g f g _1 g-f f-g _5 c1 e 2 x y 1 3x c2 e 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 dx 2 dy dx y 0. SOLUTION To solve the auxiliary equation 3r 2 r 1 0 we use the quadratic formula: 1 r s13 6 Since the roots are real and distinct, the general solution is y c1 e ( 1 s13 ) x 6 c2 e ( 1 s13 ) x 6 b 2 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 b 2a so 2 ar b 0 5E-18(pp 1176-1185) 1180 ❙❙❙❙ 1/19/06 3:42 PM Page 1180 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS e rx is one solution of Equation 5. We now verify that y2 We know that y1 a solution: a y2 b y2 a 2re rx c y2 r 2xe rx b e rx 2 ar 0 e rx b e rx ar 2 0 xe rx r xe rx xe rx is also c xe rx c xe rx br 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 br c general solution of ay by c y 0 is 0 has only one real root r, then the 10 c1 e rx y |||| Figure 2 shows the basic solutions fx e 3x 2 and t x x e 3x 2 in Example 3 and some other members of the family of solutions. Notice that all of them approach 0 as x l . EXAMPLE 3 Solve the equation 4 y 12 y SOLUTION The auxiliary equation 4r 2 c2 xe rx 9y 0. 9 0 can be factored as 2 0 12r 2r 3 f-g 8 5f+g _2 f+g 3 2 so the only root is r f g-f g _5 . By (10) the general solution is f+5g y c1 e 3x 2 3x 2 c2 xe 2 b 2 4 ac 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 ■ FIGURE 2 r1 i r2 where and are real numbers. [In fact, using Euler’s equation ei i b 2a , cos s4ac b 2 2a .] Then, i sin from Appendix G, we write the solution of the differential equation as y C1 e r 1 x C1 e e x e x x C2 e r 2 x cos x C1 i sin x C2 cos x c1 cos x i C1 e x C2 e C2 e i C1 x i x cos x i sin x C2 sin x c2 sin x where c1 C1 C2 , c2 i C1 C2 . 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 br c 0 are the complex numbers r1 i , r2 i , then the general solution of a y by cy 0 is 11 y |||| Figure 3 shows the graphs of the solue 3 x cos 2 x and tions in Example 4, f x 3x tx e sin 2 x, together with some linear combinations. All solutions approach 0 as x l . 6y SOLUTION The auxiliary equation is r 2 f-g c2 sin x 13 y 6r 0. 13 0. By the quadratic formula, the roots are 6 r g c1 cos x EXAMPLE 4 Solve the equation y 3 f+g x e s36 2 52 6 s 16 2 3 2i By (11) the general solution of the differential equation is f _3 2 e 3x c1 cos 2 x y c2 sin 2 x 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 y x0 y0 y x0 y1 where y0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and Px 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 y |||| Figure 4 shows the graph of the solution of the initial-value problem in Example 5. Compare with Figure 1. y 6y 0 y0 1 y0 SOLUTION From Example 1 we know that the general solution of the differential equa- tion is yx 20 c1 e 2 x c2 e 3x Differentiating this solution, we get 2c1 e 2 x yx 3c2 e 3x To satisfy the initial conditions we require that _2 FIGURE 4 0 2 0 12 y0 c1 13 y0 2c1 From (13) we have c2 c2 1 3c2 0 2 31 c and so (12) gives c1 2 31 c 1 3 5 c1 c2 Thus, the required solution of the initial-value problem is y 3 5 e 2x 2 5 e 3x 2 5 5E-18(pp 1176-1185) 1182 ❙❙❙❙ 1/19/06 3:43 PM 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 Page 1182 s13 sin x 2 3 where tan EXAMPLE 6 Solve the initial-value problem y 0 y0 2π y0 3 1, and since e 1 0, or r 2 1, whose roots are 1, the general solution is 0x yx _2π 2 SOLUTION The auxiliary equation is r 2 0, 5 y Since c1 cos x yx i. Thus c2 sin x c1 sin x c2 cos x the initial conditions become y0 _5 FIGURE 5 c1 2 y0 c2 3 Therefore, the solution of the initial-value problem is yx 2 cos x 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 y x0 y0 y x1 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 y 2y y 0 y0 1 y1 SOLUTION The auxiliary equation is r2 whose only root is r 2r 1 0 or r 1 2 0 1. Therefore, the general solution is yx c1 e x c2 xe x The boundary conditions are satisfied if y0 y1 |||| Figure 6 shows the graph of the solution of the boundary-value problem in Example 7. The first condition gives c1 5 c1 c1 e 5 FIGURE 6 1 c2 e 3 1 c2 e 1 3 Solving this equation for c2 by first multiplying through by e, we get 1 _5 1 1, so the second condition becomes e _1 1 c2 3e so c2 3e Thus, the solution of the boundary-value problem is y e x 3e 1 xe x 1 3 5E-18(pp 1176-1185) 1/19/06 3:43 PM Page 1183 S ECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS Summary: Solutions of ay Roots of ar 2 by c br c 1–13 General solution 0 1. y 6y 8y 3. y 8y 41y 5. y 2y y 7. 4 y y 20. 2 y 2. y 0 21. y 8. 16 y 0 y 0 0 24 y 9y 0, 22. y 5y 8y 16 y 2y 5y 0, y 0, y 23. y y 6. 3 y 0 4y 4. 2 y 0 5y 2y 2y 0, y0 2, y0 24. y 0 ■ 9. 4 y y 10. 9 y 0 11. d 2y dt 2 2 13. d 2y dt 2 dy dt dy dt y 12. 0 d 2y dt 2 4y 6 0 dy dt 25–32 4y 0 25. 4 y 0 16. ■ ■ ■ ■ ■ ■ 2 17. 2 y 18. y 19. 4 y |||| 5y ■ 0 ■ ■ ■ ■ ■ ■ Solve the initial-value problem. 5y 3y 4y |||| 18.2 3y 0, 0, y0 y 0, ■ ■ 2y 0, y0 1, y3 0, y0 6y 25 y 0, 30. y 6y 9y 4y 13 y ■ 18 y ■ 0, 0, y0 y0 ■ y ■ ■ 5 1, 1, y0 0, 10 y ■ 2, 0 2, 2 y1 0 y 2 0, y0 ■ y ■ 1 y 1 ■ ■ ■ 33. Let L be a nonzero real number. 0 ■ ■ 2 32. 9 y 16 y dy dx ■ 0 ■ 4 y1 ■ dy 8 dx ■ 17–24 2y 1 ■ y 3, y0 31. y dy dx dy dx 2 ■ ■ 1 y1 ■ 1, 29. y 2 ■ y0 100 y d 2y 15. dx 2 0, y1 4 2 0, 28. y ■ |||| dy dx 2 0, 4 4 2y Graph the two basic solutions of the differential equation ; and several other solutions. What features do the solutions have in common? 14. 6 y 3, ■ y0 26. y y ■ 2 4 ■ 0, y 3y 14–16 1, y0 Solve the boundary-value problem, if possible. 27. y ■ y 36 y ■ |||| 0, 3y 12 y ■ ■ ■ c1 e r 1 x c2 e r 2 x c1 e rx c2 xe rx e x c1 cos x c2 sin x y y y Exercises Solve the differential equation. |||| 1183 0 r1, r2 real and distinct r1 r2 r r1, r2 complex: i |||| 18.1 ❙❙❙❙ y0 1, y0 3, y0 1, y0 4 ■ (a) Show that the boundary-value problem y y 0, y0 0, y L 0 has only the trivial solution y 0 for the cases 0 and 0. (b) For the case 0, find the values of for which this problem has a nontrivial solution and give the corresponding solution. 34. If a, b, and c are all positive constants and y x is a solution 3 y0 1.5 of the differential equation ay lim x l y x 0. by cy 0, show that 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 ay by cy Gx 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 ay 2 by 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 yx yp x yc x 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 yp is a solution of the complementary Equation 2. Indeed ay yp by yp cy yp ay a yp by ay by cy tx tx b yp cy 0 a yp b yp c yp c yp We know from Section 18.1 how to solve the complementary equation. (Recall that the solution is yc c1 y1 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 ay by cy Gx where G x) 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 ay by c y is also a polynomial. We therefore substitute yp x a polynomial (of the same degree as G ) into the differential equation and determine the coefficients. EXAMPLE 1 Solve the equation y y 2y SOLUTION The auxiliary equation of y r2 with roots r 1, r y 2 x 2. 2y r 0 is 1r 2 0 2. So the solution of the complementary equation is yc c1 e x c2 e 2x 5E-18(pp 1176-1185) 1/19/06 3:43 PM Page 1185 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS Since G x 1185 x 2 is a polynomial of degree 2, we seek a particular solution of the form Ax 2 yp x Then yp have ❙❙❙❙ B and yp 2 Ax C 2 A so, substituting into the given differential equation, we 2 Ax 2 2A 2 Ax B 2 Ax 2 or Bx 2A 2B x Bx x2 B 2A C 2C x2 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 x ex and t x e 2 x. 2A 1 2B 0 2A B 2C 0 The solution of this system of equations is 8 1 2 A yp+2f+3g yp+3g 2A 1 2 B 3 4 C A particular solution is therefore yp+2f _3 3 1 2 yp x yp 1 2 x2 3 4 x and, by Theorem 3, the general solution is _5 y FIGURE 1 yc c1 e x yp 1 2 2x c2 e 1 2 x2 x 3 4 If G x (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, yp x Ae k x, because the kx kx derivatives of e are constant multiples of e . EXAMPLE 2 Solve y e 3 x. 4y SOLUTION The auxiliary equation is r 2 |||| Figure 2 shows solutions of the differential equation in Example 2 in terms of yp and the functions f x cos 2 x and t x sin 2 x. Notice that all solutions approach as x l and all solutions resemble sine functions when x is negative. 4 yc x c1 cos 2 x yp+g yp _4 _2 FIGURE 2 so 13Ae 3x e 3x and A 1 13 4 Ae 3x 3Ae 3x and yp 9Ae 3x. Substi- e 3x . Thus, a particular solution is 2 yp x yp+f 2i, so the solution of the c2 sin 2 x For a particular solution we try yp x Ae 3x. Then yp tuting into the differential equation, we have 9Ae 3x yp+f+g 0 with roots 4 complementary equation is 1 13 e 3x and the general solution is yx c1 cos 2 x c2 sin 2 x 1 13 e 3x If G x is either C cos k x or C sin k x, then, because of the rules for differentiating the sine and cosine functions, we take as a trial particular solution a function of the form yp x A cos k x B sin k x 5E-18(pp 1186-1195) 1186 ❙❙❙❙ 1/19/06 3:44 PM Page 1186 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS EXAMPLE 3 Solve y y 2y sin x. SOLUTION We try a particular solution yp x yp Then A sin x A cos x B sin x B cos x yp A cos x B sin x so substitution in the differential equation gives A cos x B sin x A sin x B cos x 3A or 2 A cos x B cos x B sin x 3B sin x A sin x sin x This is true if 3A B 0 and A 3B 1 The solution of this system is 1 10 A 3 10 B so a particular solution is 1 10 yp x 3 10 cos x sin x In Example 1 we determined that the solution of the complementary equation is yc c1 e x c2 e 2 x. Thus, the general solution of the given equation is c1 e x yx 1 10 2x c2 e cos x 3 sin x If G x 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 y 2y 4y x cos 3x we would try yp x Ax B cos 3x Cx D sin 3x If G x 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 ay respectively, then yp1 by cy ay by cy G2 x yp2 is a solution of ay EXAMPLE 4 Solve y G1 x 4y xe x by cy G1 x G2 x cos 2 x. 2 SOLUTION The auxiliary equation is r plementary equation is yc x c1 e 2 x yp1 x 4 0 with roots 2, so the solution of the comc2 e 2 x. For the equation y 4 y xe x we try Ax B ex 5E-18(pp 1186-1195) 1/19/06 3:44 PM Page 1187 SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS Then yp1 gives Ax B e x, yp1 A Ax Ax or Thus, 4 Ax 3A 1 and 2 A 3B 4y 5 ( 4 C cos 2 x 4 D sin 2 x 8C 1, 8D )e x D sin 2 x 4 C cos 2 x D sin 2 x cos 2 x 8 D sin 2 x cos 2 x 0, and 1 8 yp2 x 1 yp 2 9 x , and 8C cos 2 x yp+f _4 1 3 2 9 ,B Substitution gives Therefore, yp+g xe x C cos 2 x or yp+2f+g 3B e x cos 2 x, we try yp2 x |||| In Figure 3 we show the particular solution yp yp1 yp 2 of the differential equation in Example 4. The other solutions are given in terms e 2 x and t x of f x e 2 x. xe x 1 3 0, so A yp1 x For the equation y B ex 2A 3Ax 1187 B e x, so substitution in the equation 2A B ex 2A ❙❙❙❙ cos 2 x By the superposition principle, the general solution is _2 y FIGURE 3 yc yp1 c1 e 2 x yp2 c2 e 2x (1x 3 2 9 )e x 1 8 cos 2 x 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 yp x is a solution of the complementary equation. EXAMPLE 5 Solve y y sin x. SOLUTION The auxiliary equation is r 2 1 0 with roots i, so the solution of the com- plementary equation is yc x c1 cos x c2 sin x Ordinarily, we would use the trial solution yp x A cos x B sin x but we observe that it is a solution of the complementary equation, so instead we try yp x Then yp x yp x A cos x 2 A sin x A x cos x A x sin x A x cos x B x sin x B sin x B x cos x 2 B cos x B x 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 yp 4 1 2 so A _2π ,B yp 2 A sin x 2B cos x sin x 0, and 2π 1 2 yp x x cos x yp The general solution is _4 yx FIGURE 4 c1 cos x c2 sin x 1 2 x cos x We summarize the method of undetermined coefficients as follows: e kxP x , where P is a polynomial of degree n, then try yp x e kxQ x , where Q x is an nth-degree polynomial (whose coefficients are determined by substituting in the differential equation.) 1. If G x e kxP x cos m x or G x polynomial, then try e kxP x sin m x, where P is an nth-degree 2. If G x e kxQ x cos m x yp x e kxR x sin m x 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 y 4y 13y e 2 x cos 3x. SOLUTION Here G x has the form of part 2 of the summary, where k Px 2, m 3, and 1. So, at first glance, the form of the trial solution would be e 2 x A cos 3x yp x But the auxiliary equation is r 2 4r of the complementary equation is 13 0, with roots r e 2 x c1 cos 3x yc x B sin 3x 2 3i, so the solution c2 sin 3x This means that we have to multiply the suggested trial solution by x. So, instead, we use xe 2 x A cos 3x yp x B sin 3x The Method of Variation of Parameters Suppose we have already solved the homogeneous equation ay ten the solution as 4 yx c1 y1 x by cy 0 and writ- c2 y2 x 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 u1 x and u2 x . We look for a particular solution of the nonhomogeneous equation ay by c y G x of the form yp x 5 u1 x y1 x u2 x y2 x (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 yp 6 u1 y1 u2 y2 u1 y1 u2 y2 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 7 Then yp u1 y1 u2 y2 u2 y2 0 u1 y1 u2 y2 Substituting in the differential equation, we get a u1 y1 u2 y2 u1 y1 u2 y2 b u1 y1 u2 y2 c u1 y1 u2 y2 G or 8 u1 a y1 b y1 c y1 u2 a y2 b y2 c y2 a u1 y1 u2 y2 G But y1 and y2 are solutions of the complementary equation, so ay1 b y1 c y1 0 and ay2 b y2 c y2 0 and Equation 8 simplifies to a u1 y1 9 u2 y2 G Equations 7 and 9 form a system of two equations in the unknown functions u1 and u2 . 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 y y SOLUTION The auxiliary equation is r 2 y y 0 is c1 sin x of the form tan x, 0 2. x 1 0 with roots i, so the solution of c2 cos x. Using variation of parameters, we seek a solution yp x Then yp u1 x sin x u2 x cos x u1 sin x u2 cos x u1 cos x Set 10 u1 sin x u2 cos x 0 u2 sin x 5E-18(pp 1186-1195) ❙❙❙❙ 1190 1/19/06 3:45 PM Page 1190 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS yp Then u1 cos x u2 sin x u1 sin x u2 cos x For yp to be a solution we must have yp 11 yp u1 cos x u2 sin x tan x Solving Equations 10 and 11, we get u1 sin 2x u1 cos 2x sin x cos x tan x u1 x cos x (We seek a particular solution, so we don’t need a constant of integration here.) Then, from Equation 10, we obtain |||| Figure 5 shows four solutions of the differential equation in Example 7. So 2.5 (Note that sec x u2 x tan x ln sec x x cos x sec x tan x 2.) Therefore cos x sin x sin x cos x ln sec x π 2 yp cos 2x 1 cos x sin x 0 for 0 yp x 0 sin 2x cos x sin x u1 cos x u2 ln sec x tan x cos x tan x and the general solution is _1 yx FIGURE 5 |||| 18.2 x2 1. y 3y 2y 3. y 2y 2. y x 5. y 4y 7. y y 8. y 4y e x cos x, 9. y y x e x, y0 10. y y 2y x 5y ex ■ e x 3, ■ 12. 2 y 2, 9y 2, sin 2 x, ■ 2y y0 1, y0 ; 11–12 6y y0 y0 y 1 5y tan x x y ex 1 ■ 13. y cos 2 x ■ 9y e 2x 9y xe x cos 0 ■ 9y 1 3y 4y ■ ■ ■ ■ ■ 2y 10 y x 2e 4y 3x 18. y ■ ■ x x3 17. y ■ ■ 9x 16. y y0 ■ x 2 sin x 15. y 1 ■ y ■ 14. y xe 2 1, 3y ■ |||| Write a trial solution for the method of undetermined coefficients. Do not determine the coefficients. x |||| Graph the particular solution and several other solutions. What characteristics do these solutions have in common? 11. 4 y cos x ln sec x 13–18 0 y0 ■ ■ e 3x 6. y y0 ■ 9y 4. y sin 4 x ■ c2 cos x Exercises 1–10 |||| Solve the differential equation or initial-value problem using the method of undetermined coefficients. ■ c1 sin x ■ ■ e ■ xe ■ x ex x cos 3 x x sin 2 x ■ ■ ■ ■ ■ ■ ■ 19–22 |||| Solve the differential equation using (a) undetermined coefficients and (b) variation of parameters. 19. y 4y x 20. y 3y 2y sin x ■ 5E-18(pp 1186-1195) 1/19/06 3:45 PM Page 1191 S ECTION 18.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS 21. y 2y y e2x y x ❙❙❙❙ 1191 1 25. y 3y 2y 26. y 3y 2y sin e x 23–28 |||| Solve the differential equation using the method of variation of parameters. 27. y y 23. y y sec x, 0 x 2 28. y 4y 4y e 2x x3 24. y y cot x, 0 x 2 22. y ■ ■ e ■ ■ |||| 18.3 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 1 e x 1 x ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 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 : m equilibrium position 0 restoring force m x kx 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 x FIGURE 1 equilibrium position m 0 FIGURE 2 m 1 x d 2x dt 2 kx or m d 2x dt 2 kx 0 This is a second-order linear differential equation. Its auxiliary equation is mr 2 with roots r i, where sk m. Thus, the general solution is x c1 cos t xt c2 sin t which can also be written as xt where A cos sk m A cos (frequency) sc 2 1 c2 2 t c1 A (amplitude) sin c2 A is the phase angle (See Exercise 17.) This type of motion is called simple harmonic motion. k 0 5E-18(pp 1186-1195) 1192 ❙❙❙❙ 1/19/06 3:45 PM Page 1192 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS E XAMPLE 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 k 0.2 25.6 so k 25.6 0.2 128. Using this value of the spring constant k, together with m in Equation 1, we have 2 d 2x dt 2 128 x 2 0 As in the earlier general discussion, the solution of this equation is xt 2 c1 cos 8t c2 sin 8t We are given the initial condition that x 0 0.2. But, from Equation 2, x 0 Therefore, c1 0.2. Differentiating Equation 2, we get xt 8c1 sin 8t Since the initial velocity is given as x 0 8c2 cos 8t 0, we have c2 1 5 xt c1. 0 and so the solution is cos 8t 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 c dx dt where c is a positive constant, called the damping constant. Thus, in this case, Newton’s Second Law gives m d 2x dt 2 restoring force damping force or 3 m d 2x dt 2 c dx dt kx 0 kx c dx dt 5E-18(pp 1186-1195) 1/19/06 3:45 PM Page 1193 S ECTION 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 cr k 0. The roots are r1 4 c sc 2 2m 4 mk c r2 sc 2 2m 4 mk According to Section 18.1 we need to discuss three cases. x CASE I c2 4 mk 0 (overdamping) In this case r1 and r 2 are distinct real roots and 0 ■ x t x 0 t c1 e r1 t c2 e r2 t Since c, m, and k are all positive, we have sc 2 4 mk 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 . 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 4 mk means that there is a strong damping force (high-viscosity oil or grease) compared with a weak spring or small mass. c 2 4 mk 0 (critical damping) This case corresponds to equal roots CASE II FIGURE 4 Overdamping ■ r1 c 2m r2 and the solution is given by x c1 c2 t e c 2m t 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 4 mk 0 (underdamping) Here the roots are complex: CASE III ■ r1 r2 x x=Ae– (c/ 2m)t t x=_Ae– (c/ 2m)t FIGURE 5 Underdamping i s4 mk c 2 2m where 0 c 2m The solution is given by x e c 2m t c1 cos t c2 sin t We see that there are oscillations that are damped by the factor e c 2 m t. Since c 0 and m 0, we have c 2 m 0 so e c 2 m t l 0 as t l . This implies that x l 0 as t l ; 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 E XAMPLE 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 m s. SOLUTION From Example 1 the mass is m differential equation (3) becomes 2 d 2x dt 2 40 d 2x dt 2 or 2 and the spring constant is k dx dt 20 128 x dx dt 0 64 x 0 The auxiliary equation is r 2 20 r 64 r 4 r 16 and 16, so the motion is overdamped and the solution is xt |||| Figure 6 shows the graph of the position function for the overdamped motion in Example 2. We are given that x 0 0.03 0, so c1 c1 e c2 0 1.5 4 c1 e x0 Since c2 4 c1 c1 , this gives 12c1 FIGURE 6 0.6 or c1 x c2 e 0 with roots 4 16 t 0. Differentiating, we get xt so 4t 128, so the 0.05 e 4t 16 c2 e 16 c2 16 t 0.6 0.05. Therefore 4t e 16 t 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 F t . Then Newton’s Second Law gives m d 2x dt 2 restoring force kx c dx dt damping force external force Ft 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 dt 2 c dx dt kx Ft 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 Ft F0 cos 0 t where sk m 0 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 xt 6 c1 cos t F0 c2 sin t 2 m 2 0 cos 0 t 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 Q t , then the current is the rate of change of Q with respect to t : I dQ dt. 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 Since I 7 dI dt RI Q C Et dQ dt, this equation becomes L d 2Q dt 2 R dQ dt 1 Q C Et 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 Q0 Q0 Q0 I0 I0 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 dQ dt : L d 2I dt 2 R dI dt 1 I C Et EXAMPLE 3 Find the charge and current at time t in the circuit of Figure 7 if R L 1 H, C both 0. 16 10 4 40 , 100 cos 10 t, and the initial charge and current are F, E t SOLUTION With the given values of L, R, C, and E t , Equation 7 becomes d 2Q dt 2 8 40 The auxiliary equation is r 2 dQ dt 40r 625 40 r 625Q 100 cos 10 t 0 with roots s 900 2 20 15i so the solution of the complementary equation is Qc t e 20 t c1 cos 15t c2 sin 15t For the method of undetermined coefficients we try the particular solution Qp t Then A cos 10 t B sin 10 t Qp t 10 A sin 10 t 10 B cos 10 t Qp t 100 A cos 10 t 100 B sin 10 t Substituting into Equation 8, we have 100 A cos 10 t 100 B sin 10 t 40 10 A sin 10 t 10 B cos 10 t 625 A cos 10 t or 525 A 400 B cos 10 t 400 A B sin 10 t 525B sin 10 t 100 cos 10 t 100 cos 10 t Equating coefficients, we have 525 A 400 A 400 B 525B The solution of this system is A Qp t 100 21 A 0 84 697 and B 64 697 1 697 84 cos 10 t 16B 4 16 A or or 21B 0 , so a particular solution is 64 sin 10 t and the general solution is Qt Qc t Qp t e 20t c1 cos 15t c2 sin 15t 4 697 21 cos 10 t 16 sin 10 t 5E-18(pp 1196-1204) 1/19/06 3:28 PM Page 1197 S ECTION 18.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS Imposing the initial condition Q 0 Q0 ❙❙❙❙ 1197 0, we get 84 697 c1 84 697 c1 0 To impose the other initial condition we first differentiate to find the current: dQ dt I e 20t 20c1 40 697 I0 20c1 15c2 cos 15t 21 sin 10 t 640 697 15c2 0 15c1 20c2 sin 15t 16 cos 10 t c2 464 2091 Thus, the formula for the charge is 4 697 Qt e 20 t 63 cos 15t 3 116 sin 15t 21 cos 10 t 16 sin 10 t and the expression for the current is 1 2091 It NOTE 1 tl e 20 t 1920 cos 15t 13,060 sin 15t 120 21 sin 10 t 16 cos 10 t In Example 3 the solution for Q t consists of two parts. Since e and both cos 15t and sin 15t are bounded functions, ■ 20 t l 0 as 0.2 Qp 0 Qc t Q 1.2 4 2091 e 63 cos 15t 116 sin 15t l 0 as t l So, for large values of t , Qt _0.2 20 t Qp t 4 697 21 cos 10 t 16 sin 10 t and, for this reason, Qp t 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. FIGURE 8 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 dt 2 d 2Q L dt 2 m dx dt dQ R dt c kx Ft 1 Q C Et Spring system x dx dt m c k Ft displacement velocity mass damping constant spring constant external force Electric circuit Q I dQ dt L R 1C Et 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 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 c2 e r t, where r1 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 m s, 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 inductor 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 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 m s. (b) Graph the position function of the mass. ; a voltage of E t 6. For the spring in Exercise 4, find the damping constant that 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 m s. 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 sk m. Suppose that the damping constant is so small that the damping force is negligible. If an external force , use the method Ft F0 cos 0 t is applied, where 0 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 sk m. F0 cos t is applied (the applied If an external force F t frequency equals the natural frequency), use the method of undetermined coefficients to show that the motion of the mass is given by x t c1 cos t c2 sin t F0 2m t sin t. 11. Show that if , but 0 0 is a rational number, then the motion described by Equation 6 is periodic. 12 sin 10 t. Find the charge at time t. 16. The battery in Exercise 14 is replaced by a generator producing critical damping. ; 7. A spring has a mass of 1 kg and its spring constant is k 24 , an inductor 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 E t 12 sin 10 t. (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 x t A cos t . 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 d2 dt 2 t sin L 0 where t is the acceleration due to gravity. For small values of we can use the linear approximation sin 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 rad s. (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 y 1 2 xy y 0 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 cn x n fx c0 c2 x 2 c1 x c3 x 3 n0 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 y 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 y 0. y SOLUTION We assume there is a solution of the form y 2 c0 c2 x 2 c1 x c3 x 3 cn x n n0 We can differentiate power series term by term, so y c1 3c3 x 2 2 c2 x n cn x n 1 n1 3 y 2 c2 2 3c3 x nn 1 cn x n 2 n2 In order to compare the expressions for y and y more easily, we rewrite y 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 2 and began the summation at 0 instead of 2. y 4 n 1 cn 2 x n 2n n0 Substituting the expressions in Equations 2 and 4 into the differential equation, we obtain n 2n 1 cn 2 x n n0 cn x n 0 n0 or n 5 n0 2n 1 cn 2 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 cn 6 1 cn n 2 2n cn 1n 2 cn 2 n 0 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 c0 12 Put n 1: c3 c1 23 Put n 2: c4 Put n 3: c5 Put n 4: Put n 5: c2 c0 c0 4! 34 1234 c3 c1 45 2345 c1 5! c6 c4 56 c0 4! 5 6 c0 6! c7 c5 67 c1 5! 6 7 c1 7! By now we see the pattern: For the even coefficients, c2 n For the odd coefficients, c2 n 1 1 1 c0 2n ! n c1 n 2n 1! Putting these values back into Equation 2, we write the solution as y c0 c2 x 2 c1 x c0 1 x2 2! c1 x c0 1 n0 n x4 4! x3 3! x 2n 2n ! c3 x 3 c4 x 4 c5 x 5 x6 6! 1 x5 5! c1 x 2n 2n ! x7 7! 1 n0 n 1 n x 2n 1 2n 1 ! Notice that there are two arbitrary constants, c0 and c1. n x 2n 1 2n 1 ! 5E-18(pp 1196-1204) 1/19/06 3:31 PM Page 1201 SECTION 18.4 SERIES SOLUTIONS ❙❙❙❙ 1201 N OTE 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 ■ yx c0 cos x 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 y 2 xy y 0. SOLUTION We assume there is a solution of the form cn x n y n0 n cn x n 1 nn y Then 1 cn x n n1 y and 2 n n2 1 cn 2 x n 2n n0 as in Example 1. Substituting in the differential equation, we get n 1 cn 2 x n 2n 1 cn 2 x n 2n n1 n 2 ncn x n n0 cn x n 0 cn x n 0 1 cn x n 0 n0 2 ncn x n n0 2 ncn x n 1 n1 n n1 n cn x n 2x n0 2n 1 cn n0 2n 2 n0 This equation is true if the coefficient of x n is 0: n cn 7 2 2n n 1 cn 2 2n 1 cn 1n 2 We solve this recursion relation by putting n Put n 0: c2 1: c3 2: c4 3 c2 34 Put n 3: c5 0 0, 1, 2, 3, . . . 0, 1, 2, 3, . . . successively in Equation 7: 1 c1 23 Put n n 1 cn 1 c0 12 Put n 2n 5 45 c3 3 c0 1234 15 c1 2345 3 c0 4! 15 c1 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 c4 56 37 c0 4! 5 6 37 c0 6! Put n 5: c7 9 c5 67 Put n 6: c8 11 c6 78 3 7 11 c0 8! Put n 7: c9 13 c7 89 1 5 9 13 c1 9! 159 c1 5! 6 7 159 c1 7! In general, the even coefficients are given by 3 7 11 c2 n 4n 5 c0 2n ! and the odd coefficients are given by 159 c2 n 1 4n 1! 2n 3 c1 The solution is y c0 c1 x c2 x 2 12 x 2! c0 1 c1 x c3 x 3 c4 x 4 34 x 4! 376 x 6! 13 x 3! 155 x 5! 3 7 11 8 x 8! 1597 x 7! 1 5 9 13 9 x 9! or y 8 12 x 2! c0 1 37 4n n2 159 c1 x 5 2n ! 2n n1 4n 1! x 2n 3 x 2n 1 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 ■ ■ y1 x and y2 x 1 12 x 2! 37 n1 5 x 2n 2n ! n2 159 x 4n 2n 4n 1! 3 x 2n 1 5E-18(pp 1196-1204) 1/19/06 3:32 PM Page 1203 ❙❙❙❙ C HAPTER 18 REVIEW 2 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 y1 x , and we see how they converge to y1 . In this way we can graph both y1 and y2 in Figure 2. T¸ 2 _2 1203 T¡¸ NOTE 4 ■ If we were asked to solve the initial-value problem _8 y FIGURE 1 2 xy y 0 fi c0 _2.5 y0 y0 1 y0 0 c1 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 yx FIGURE 2 |||| y 3. y x2y 5. y xy 4n 1! 2n 3 y0 0, x 2n 1 Exercises 2. y y 6. y 1y 0 xy y 8. y xy y 0, 10. y x2y 0, y0 ■ y 2y x2y 18 Review 1, 1, y0 y0 ; 0 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. xy ■ 0, ■ ■ ■ y0 1 ■ ■ ■ ■ xy x2y 0 y0 1 y0 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 5, 5 . 0 y0 ■ 12. The solution of the initial-value problem 0 xy 9. y 3y ■ xy 4. x 0 x2y 11. y Use power series to solve the differential equation. 1. y 7. x 2 159 x n1 |||| 18.4 |||| 0 we would observe from Theorem 11.10.5 that 15 1–11 y0 ■ (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 y 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. 1. If y1 and y2 are solutions of y y 0, then y1 ■ y y2 is also a solution of the equation. 4. The equation y y c1 cosh x y 0 can be written as c2 sinh x e x has a particular solution of the form 2. If y1 and y2 are solutions of y c1 y1 6y 5 y x, then c2 y2 is also a solution of the equation. yp ■ 1–10 |||| 2y 15 y 0 2. y 4y 13 y 0 3. y 3y 4. 4 y 5. 4y dy dx 2 4 d 2y 6. dx 2 y 7. d 2y dx 2 2 8. d 2y dx 2 4y d 2y 9. dx 2 ■ ■ 11–14 dy dx |||| 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 m s. x2 y x cos x 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 6y 1 csc x, e 0 ■ 2x x ■ ■ 2 ■ ■ ■ ■ ■ ■ Solve the initial-value problem. 11. y 6y 0, 12. y 6y 25 y 13. y 5y 4y 14. 9 y 18. A spring with a mass of 2 kg has damping constant 16, and a e 2x sin 2 x ■ y ■ y1 3, 0, 0, e x, 3x ■ ■ y0 y0 2, 0, y0 ■ y1 ■ 1, ■ 12 y0 y0 y0 ■ 1 2 ■ ■ ■ xy y 0 y0 0 y0 1 ■ G Mr m r2 where G is the gravitational constant and Mr is the mass of the earth within the sphere of radius r . G Mm (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 y t of the particle from the center of the earth at time t is given by yt 1 15. Use power series to solve the initial-value problem y 0 19. Assume that the earth is a solid sphere of uniform density with dy dx y 2y 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. 5y 2y xy 17. A series circuit contains a resistor with R 0 dy dx dy dx d 2y 10. dx 2 y 0 2 ■ 16. Use power series to solve the equation Solve the differential equation. 1. y ■ EXERCISES Ae x k2y t where k 2 G M R 3 t R. (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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