Calculus > QUESTIONS & ANSWERS > Chapter 17: SECOND-ORDER DIFFERENTIAL EQUATIONS. Work and Answers (All)
Chapter 17: SECOND-ORDER DIFFERENTIAL EQUATIONS. Work and Answers
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17.1 Second-Order Linear Equations 1. The auxiliary equation is 2 − − 6 = 0 ⇒ ( − 3)( + 2) = 0 ⇒ = 3, = −2. Then by (8) the general solution is = 13 + � ... ��2−2. 2. The auxiliary equation is 2 − 6 + 9 = 0 ⇒ ( − 3)2 = 0 ⇒ = 3. Then by (10), the general solution is = 13 + 23. 3. The auxiliary equation is 2 + 2 = 0 ⇒ = ±√2. Then by (11) the general solution is = 0 1 cos√2 + 2 sin√2 = 1 cos√2 + 2 sin√2. 4. The auxiliary equation is 2 + − 12 = 0 ⇒ ( − 3)( + 4) = 0 ⇒ = 3, = −4. Then by (8) the general solution is = 13 + 2−4. 5. The auxiliary equation is 42 + 4 + 1 = 0 ⇒ (2 + 1)2 = 0 ⇒ = − 1 2. Then by (10), the general solution is = 1−2 + 2−2. 6. The auxiliary equation is 92 + 4 = 0 ⇒ 2 = − 4 9 ⇒ = ± 2 3 , so the general solution is = 0 1 cos 2 3 + 2 sin 2 3 = 1 cos 2 3 + 2 sin 2 3 . 7. The auxiliary equation is 32 − 4 = (3 − 4) = 0 ⇒ = 0, = 4 3, so = 10 + 243 = 1 + 243. 8. The auxiliary equation is 2 − 1 = ( − 1)( + 1) = 0 ⇒ = 1, = −1. Then the general solution is = 1 + 2−. 9. The auxiliary equation is 2 − 4 + 13 = 0 ⇒ = 4 ± √−36 2 = 2 ± 3, so = 2(1 cos 3 + 2 sin 3). 10. The auxiliary equation is 32 + 4 − 3 = 0 ⇒ = −4 ± √52 6 = −2 ± √13 3 , so = 1(−2+√13 )3 + 2(−2−√13 )3. 11. The auxiliary equation is 22 + 2 − 1 = 0 ⇒ = −2 ± √12 4 = −1 ± √3 2 , so = 1(−1+√3)2 + 2(−1−√3)2. 12. The auxiliary equation is 2 + 6 + 34 = 0 ⇒ = −6 ± √−100 2 = −3 ± 5, so = −3(1 cos 5 + 2 sin 5). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. 719 NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.720 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 13. The auxiliary equation is 32 + 4 + 3 = 0 ⇒ = −4 ± √−20 6 = − 2 3 ± √5 3 , so = −231 cos √35 + 2 sin √35 . 14. The auxiliary equation is 42 − 4 + 1 = (2 − 1)2 = 0 ⇒ = 1 2, so the general solution is = 12 + 22. We graph the basic solutions () = 2, () = 2 as well as = 22 + 32, = −2 − 32, and = 42 − 22. The graphs are all asymptotic to the -axis as → −∞, and as → ∞ the solutions approach ±∞. 15. The auxiliary equation is 2 + 2 + 2 = 0 ⇒ = −2 ± √−4 2 = −1 ± , so the general solution is = − (1 cos + 2 sin). We graph the basic solutions () = − cos, () = − sin as well as = − (−cos − 2sin) and = − (2 cos + 3 sin). All the solutions oscillate with amplitudes that become arbitrarily large as → −∞ and the solutions are asymptotic to the -axis as → ∞. 16. The auxiliary equation is 22 + − 1 = (2 − 1)( + 1) = 0 ⇒ = 1 2, = −1, so the general solution is = 12 + 2−. We graph the basic solutions () = 2, () = − as well as = 22 + −, = −2 − 2−, and = 2 − −. Each solution consists of a single continuous curve that approaches either 0 or ±∞ as → ±∞. 17. 2 + 3 = 0 ⇒ = ±√3 and the general solution is = 01 cos√3 + 2 sin√3 = 1 cos√3 + 2 sin√3. Then (0) = 1 ⇒ 1 = 1 and, since 0 = −√31 sin√3 + √32 cos √3, 0(0) = 3 ⇒ √32 = 3 ⇒ 2 = √33 = √3, so the solution to the initial-value problem is = cos√3 + √3 sin√3. 18. 2 − 2 − 3 = ( − 3)( + 1) = 0, so = 3, = −1 and the general solution is = 13 + 2−. Then 0 = 313 − 2−, so (0) = 2 ⇒ 1 + 2 = 2 and 0(0) = 2 ⇒ 31 − 2 = 2, giving 1 = 1 and 2 = 1. Thus the solution to the initial-value problem is = 3 + −. 19. 92 + 12 + 4 = (3 + 2)2 = 0 ⇒ = − 2 3 and the general solution is = 1−23 + 2−23. Then (0) = 1 ⇒ 1 = 1 and, since 0 = − 2 3 1−23 + 2 1 − 2 3 −23, 0(0) = 0 ⇒ − 2 3 1 + 2 = 0, so 2 = 2 3 and the solution to the initial-value problem is = −23 + 2 3 −23. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 17.1 SECOND-ORDER LINEAR EQUATIONS ¤ 721 20. 32 − 2 − 1 = (3 + 1)( − 1) = 0 ⇒ = − 1 3 , = 1 and the general solution is = 1−3 + 2. Then 0 = − 1 3 1−3 + 2, so (0) = 0 ⇒ 1 + 2 = 0 and 0(0) = −4 ⇒ − 1 3 1 + 2 = −4, giving 1 = 3 and 2 = −3. Thus the solution to the initial-value problem is = 3−3 − 3. 21. 2 − 6 + 10 = 0 ⇒ = 3 ± and the general solution is = 3(1 cos + 2 sin). Then 2 = (0) = 1 and 3 = 0(0) = 2 + 31 ⇒ 2 = −3 and the solution to the initial-value problem is = 3(2 cos − 3sin). 22. 42 − 20 + 25 = (2 − 5)2 = 0 ⇒ = 5 2 and the general solution is = 152 + 252. Then 2 = (0) = 1 and −3 = 0(0) = 5 2 1 + 2 ⇒ 2 = −8. The solution to the initial-value problem is = 252 − 852. 23. 2 − − 12 = ( − 4)( + 3) = 0 ⇒ = 4, = −3 and the general solution is = 14 + 2−3. Then 0 = (1) = 14 + 2−3 and 1 = 0(1) = 414 − 32−3 so 1 = 1 7 −4, 2 = − 1 7 3 and the solution to the initial-value problem is = 1 7 −44 − 1 7 3−3 = 1 7 4−4 − 1 7 3−3. 24. 42 + 4 + 3 = 0 ⇒ = − 1 2 ± √22 and the general solution is = −2 1 cos √22 + 2 sin √22 . Then 0 = (0) = 1 and 1 = 0(0) = √22 2 − 1 2 1 ⇒ 2 = √2 and the solution to the initial-value problem is = −2 0 + √2 sin √22 = √2−2 sin √22 . 25. 2 + 16 = 0 ⇒ = ±4 and the general solution is = 1 cos 4 + 2 sin 4. Then −3 = (0) = 1 and 2 = (8) = 2, so the solution of the boundary-value problem is = −3cos 4 + 2 sin 4. 26. 2 + 6 = ( + 6) = 0 ⇒ = 0, = −6 and the general solution is = 1 + 2−6. Then 1 = (0) = 1 + 2 and 0 = (1) = 1 + 2−6 so 1 = 1 1 − 6 , 2 = − 6 1 − 6 . The solution of the boundary-value problem is = 1 1 − 6 − 6 1 − 6 · −6 = 1 −16 − 16−−66 . 27. 2 + 4 + 4 = ( + 2)2 = 0 ⇒ = −2 and the general solution is = 1−2 + 2−2. Then 2 = (0) = 1 and 0 = (1) = 1−2 + 2−2 so 2 = −2, and the solution of the boundary-value problem is = 2−2 − 2−2. 28. 2 − 8 + 17 = 0 ⇒ = 4 ± and the general solution is = 4(1 cos + 2 sin). But 3 = (0) = 1 and 2 = () = −14 ⇒ 1 = −24, so there is no solution. 29. 2 − = ( − 1) = 0 ⇒ = 0, = 1 and the general solution is = 1 + 2. Then 1 = (0) = 1 + 2 and 2 = (1) = 1 + 2 so 1 = − 2 − 1, 2 = 1 − 1 . The solution of the boundary-value problem is = − 2 − 1 + − 1 . 30. 42 − 4 + 1 = (2 − 1)2 = 0 ⇒ = 1 2 and the general solution is = 12 + 22. Then 4 = (0) = 1 and 0 = (2) = 1 + 22 ⇒ 2 = −2. The solution of the boundary-value problem is = 42 − 22. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.722 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 31. 2 + 4 + 20 = 0 ⇒ = −2 ± 4 and the general solution is = −2(1 cos 4 + 2 sin 4). But 1 = (0) = 1 and 2 = () = 1−2 ⇒ 1 = 22, so there is no solution. 32. 2 + 4 + 20 = 0 ⇒ = −2 ± 4 and the general solution is = −2(1 cos 4 + 2 sin 4). But 1 = (0) = 1 and −2 = () = 1−2 ⇒ 1 = 1, so 2 can vary and the solution of the boundary-value problem is = −2(cos 4 + sin 4), where is any constant. 33. (a) Case 1 ( = 0): 00 + = 0 ⇒ 00 = 0 which has an auxiliary equation 2 = 0 ⇒ = 0 ⇒ = 1 + 2 where (0) = 0 and () = 0. Thus, 0 = (0) = 1 and 0 = () = 2 ⇒ 1 = 2 = 0. Thus = 0. Case 2 ( 0): 00 + = 0 has auxiliary equation 2 = − ⇒ = ±√− [distinct and real since 0] ⇒ = 1 √− + 2− √− where (0) = 0 and () = 0. Thus 0 = (0) = 1 + 2 (∗) and 0 = () = 1√− + 2−√− (†). Multiplying (∗) by √− and subtracting (†) gives 2√− − −√− = 0 ⇒ 2 = 0 and thus 1 = 0 from (∗). Thus = 0 for the cases = 0 and 0. (b) 00 + = 0 has an auxiliary equation 2 + = 0 ⇒ = ± √ ⇒ = 1 cos√ + 2 sin√ where (0) = 0 and () = 0. Thus, 0 = (0) = 1 and 0 = () = 2 sin√ since 1 = 0. Since we cannot have a trivial solution, 2 6= 0 and thus sin√ = 0 ⇒ √ = where is an integer ⇒ = 222 and = 2 sin() where is an integer. 34. The auxiliary equation is 2 + + = 0. If 2 − 4 0, then any solution is of the form () = 11 + 22 where 1 = − + √2 − 4 2 and 2 = − − √2 − 4 2 . But , , and are all positive so both 1 and 2 are negative and lim→∞ () = 0. If 2 − 4 = 0, then any solution is of the form () = 1 + 2 where = −(2) 0 since , are positive. Hence lim→∞ () = 0. Finally if 2 − 4 0, then any solution is of the form () = (1 cos + 2 sin) where = −(2) 0 since and are positive. Thus lim→∞ () = 0. 35. (a) 2 − 2 + 2 = 0 ⇒ = 1 ± and the general solution is = (1 cos + 2 sin). If () = and () = then (1 cos + 2 sin) = ⇒ 1 cos + 2 sin = − and (1 cos + 2 sin) = ⇒ 1 cos + 2 sin = −. This gives a linear system in 1 and 2 which has a unique solution if the lines are not parallel. If the lines are not vertical or horizontal, we have parallel lines if cos = cos and sin = sin for some nonzero constant or cos cos = = sin sin ⇒ cos sin = cos sin ⇒ tan = tan ⇒ − = , any integer. (Note that none of cos , cos, sin, sin are zero.) If the lines are both horizontal then cos = cos = 0 ⇒ − = , and similarly vertical lines means sin = sin = 0 ⇒ − = . Thus the system has a unique solution if − 6= . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS ¤ 723 (b) The linear system has no solution if the lines are parallel but not identical. From part (a) the lines are parallel if − = . If the lines are not horizontal, they are identical if − = − ⇒ − − = = cos cos ⇒ = − cos cos . (If = 0 then = 0 also.) If they are horizontal then cos = 0, but = sin sin also (and sin 6= 0) so we require = − sin sin . Thus the system has no solution if − = and 6= − cos cos unless cos = 0, in which case 6= − sin sin . (c) The linear system has infinitely many solution if the lines are identical (and necessarily parallel). From part (b) this occurs when − = and = − cos cos unless cos = 0, in which case = − sin sin . 17.2 Nonhomogeneous Linear Equations 1. The auxiliary equation is 2 + 2 − 8 = ( − 2)( + 4) = 0 ⇒ = 2, = −4, so the complementary solution is () = 12 + 2−4. We try the particular solution () = 2 + + , so 0 = 2 + and 00 = 2. Substituting into the differential equation, we have (2) + 2(2 + ) − 8(2 + + ) = 1 − 22 or −82 + (4 − 8) + (2 + 2 − 8) = −22 + 1. Comparing coefficients gives −8 = −2 ⇒ = 1 4, 4 − 8 = 0 ⇒ = 1 8, and 2 + 2 − 8 = 1 ⇒ = − 32 1 , so the general solution is () = () + () = 12 + 2−4 + 1 4 2 + 1 8 − 32 1 . 2. The auxiliary equation is 2 − 3 = ( − 3) = 0 ⇒ = 0, = 3, so the complementary solution is () = 1 + 23. We try the particular solution () = cos 2 + sin 2, so 0 = −2sin 2 + 2 cos 2 and 00 = −4cos 2 − 4 sin 2. Substitution into the differential equation gives (−4cos 2 − 4 sin 2) − 3(−2sin 2 + 2 cos 2) = sin 2 ⇒ (−4 − 6)cos 2 + (6 − 4)sin2 = sin 2. Then −4 − 6 = 0 and 6 − 4 = 1 ⇒ = 26 3 and = − 13 1 . Thus the general solution is () = () + () = 1 + 23 + 26 3 cos 2 − 13 1 sin 2. 3. The auxiliary equation is 92 + 1 = 0 with roots = ± 1 3 , so the complementary solution is () = 1 cos(3) + 2 sin(3). Try the particular solution () = 2, so 0 = 22 and 00 = 42. Substitution into the differential equation gives 942 + 2 = 2 or 372 = 2. Thus 37 = 1 ⇒ = 37 1 and the general solution is () = () + () = 1 cos(3) + 2 sin(3) + 37 1 2. 4. The auxiliary equation is 2 − 2 + 2 = 0 with roots = 1 ± , so the complementary solution is () = (1 cos + 2 sin). Try the particular solution () = + + , so 0 = + and 00 = . Substitution into the differential equation gives () − 2( + ) + 2( + + ) = + ⇒ °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.724 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 2 + (−2 + 2) + = + . Comparing coefficients, we have 2 = 1 ⇒ = 1 2 , −2 + 2 = 0 ⇒ = 1 2 , and = 1, so the general solution is () = () + () = (1 cos + 2 sin) + 1 2 + 1 2 + . 5. The auxiliary equation is 2 − 4 + 5 = 0 with roots = 2 ± , so the complementary solution is () = 2(1 cos + 2 sin). Try () = −, so 0 = −− and 00 = −. Substitution gives − − 4(−−) + 5(−) = − ⇒ 10− = − ⇒ = 10 1 . Thus the general solution is () = 2(1 cos + 2 sin) + 10 1 −. 6. The auxiliary equation is 2 − 4 + 4 = ( − 2)2 = 0 ⇒ = 2, so the complementary solution is () = 12 + 22. For 00 − 40 + 4 = try 1() = + . Then 0 1 = and 001 = 0, and substitution into the differential equation gives 0 − 4 + 4( + ) = or 4 + (4 − 4) = , so 4 = 1 ⇒ = 1 4 and 4 − 4 = 0 ⇒ = 1 4 . Thus 1() = 1 4 + 1 4 . For 00 − 40 + 4 = −sin try 2() = cos + sin. Then 0 2 = −sin + cos and 002 = −cos − sin. Substituting, we have (−cos − sin) − 4(−sin + cos ) + 4(cos + sin) = −sin ⇒ (3 − 4)cos + (4 + 3)sin = −sin. Thus 3 − 4 = 0 and 4 + 3 = −1, giving = − 25 4 and = − 25 3 , so 2() = − 25 4 cos − 25 3 sin. The general solution is () = () + 1() + 2() = 12 + 22 + 1 4 + 1 4 − 25 4 cos − 25 3 sin. 7. The auxiliary equation is 2 − 2 + 5 = 0 with roots = 1 ± 2, so the complementary solution is () = (1 cos 2 + 2 sin 2). Try the particular solution () = cos + sin, so 0 = −sin + cos and 00 = −cos − sin. Substituting, we have (−cos − sin) − 2(−sin + cos) + 5(cos + sin) = sin ⇒ (4 − 2)cos + (2 + 4)sin = sin. Then 4 − 2 = 0, 2 + 4 = 1 ⇒ = 10 1 , = 1 5 and the general solution is () = () + () = (1 cos 2 + 2 sin 2) + 10 1 cos + 1 5 sin. But 1 = (0) = 1 + 10 1 ⇒ 1 = 10 9 and 1 = 0(0) = 22 + 1 + 1 5 ⇒ 2 = − 20 1 . Thus the solution to the initial-value problem is () = 10 9 cos 2 − 20 1 sin 2 + 10 1 cos + 1 5 sin. 8. The auxiliary equation is 2 − 1 = 0 with roots = ±1, so the complementary solution is () = 1 + 2−. Try the particular solution () = ( + )2, so 0 = (2 + + 2)2 and 00 = (4 + 4 + 4)2. Substituting, we have (4 + 4 + 4)2 − ( + )2 = 2 ⇒ (3 + 4 + 3)2 = 2. Then 3 = 1 ⇒ = 1 3 and 4 + 3 = 0 ⇒ = − 4 9 , and the general solution is () = () + () = 1 + 2− + 1 3 − 4 9 2. But 0 = (0) = 1 + 2 − 4 9 and 1 = 0(0) = 1 − 2 − 5 9 ⇒ 1 = 1, 2 = − 5 9 . Thus the solution to the initial-value problem is () = − 5 9− + 1 3 − 4 9 2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS ¤ 725 9. The auxiliary equation is 2 − = 0 with roots = 0, = 1 so the complementary solution is () = 1 + 2. Try () = ( + ) so that no term in is a solution of the complementary equation. Then 0 = (2 + (2 + ) + ) and 00 = (2 + (4 + ) + (2 + 2)). Substitution into the differential equation gives (2 + (4 + ) + (2 + 2)) − (2 + (2 + ) + ) = ⇒ (2 + (2 + )) = ⇒ = 1 2, = −1. Thus () = 1 2 2 − and the general solution is () = 1 + 2 + 1 2 2 − . But 2 = (0) = 1 + 2 and 1 = 0(0) = 2 − 1, so 2 = 2 and 1 = 0. The solution to the initial-value problem is () = 2 + 1 2 2 − = 1 2 2 − + 2. 10. () = 1 + 2−2. For 00 + 0 − 2 = try 1() = + . Then 0 1 = , 001 = 0, and substitution gives 0 + − 2( + ) = ⇒ = − 1 2, = − 1 4, so 1() = − 1 2 − 1 4. For 00 + 0 − 2 = sin 2 try 2() = cos 2 + sin 2. Then 0 2 = −2sin 2 + 2 cos 2 002 = −4cos 2 − 4 sin 2, and substitution gives (−4cos 2 − 4 sin 2) + (−2sin 2 + 2 cos 2) − 2(cos 2 + sin 2) = sin 2 ⇒ = − 20 1 , = − 3 20. Thus 2() = − 20 1 cos 2 + − 20 3 sin 2 and the general solution is () = 1 + 2−2 − 1 2 − 1 4 − 20 1 cos 2 − 20 3 sin 2. But 1 = (0) = 1 + 2 − 1 4 − 20 1 and 0 = 0(0) = 1 − 22 − 1 2 − 10 3 ⇒ 1 = 17 15 and 2 = 1 6. Thus the solution to the initial-value problem is () = 17 15 + 1 6 −2 − 1 2 − 1 4 − 20 1 cos 2 − 20 3 sin 2. 11. The auxiliary equation is 2 + 3 + 2 = ( + 1)( + 2) = 0, so = −1, = −2 and () = 1− + 2−2. Try = cos + sin ⇒ 0 = −sin + cos, 00 = −cos − sin. Substituting into the differential equation gives (−cos − sin) + 3(−sin + cos) + 2(cos + sin) = cos or ( + 3)cos + (−3 + )sin = cos . Then solving the equations + 3 = 1, −3 + = 0 gives = 10 1 , = 10 3 and the general solution is () = 1− + 2−2 + 10 1 cos + 10 3 sin. The graph shows and several other solutions. Notice that all solutions are asymptotic to as → ∞. Except for , all solutions approach either ∞ or −∞ as → −∞. 12. The auxiliary equation is 2 + 4 = 0 ⇒ = ±2, so () = 1 cos 2 + 2 sin 2. Try = − ⇒ 0 = −−, 00 = −. Substituting into the differential equation gives − + 4− = − ⇒ 5 = 1 ⇒ = 1 5, so = 1 5 − and the general solution is () = 1 cos 2 + 2 sin 2 + 1 5 −. We graph along with several other solutions. All of the solutions except oscillate around = 1 5 −, and all solutions approach ∞ as → −∞. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.726 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 13. Here () = 12 + 2−, and a trial solution is () = ( + ) cos + ( + ) sin. 14. Here () = 1 cos 2 + 2 sin 2. For 00 + 4 = cos 4 try 1() = cos 4 + sin 4 and for 00 + 4 = cos 2 try 2() = ( cos 2 + sin 2) (so that no term of 2 is a solution of the complementary equation). Thus a trial solution is () = 1() + 2() = cos 4 + sin 4 + cos 2 + sin 2. 15. Here () = 12 + 2. For 00 − 30 + 2 = try 1() = (since = is a solution of the complementary equation) and for 00 − 30 + 2 = sin try 2() = cos + sin. Thus a trial solution is () = 1() + 2() = + cos + sin. 16. Since () = 1 + 2−4 try () = (3 + 2 + + ) so that no term of () satisfies the complementary equation. 17. Since () = −(1 cos 3 + 2 sin 3) we try () = (2 + + )− cos 3 + (2 + + )− sin 3 (so that no term of is a solution of the complementary equation). 18. Here () = 1 cos 2 + 2 sin 2. For 00 + 4 = 3 try 1() = 3 and for 00 + 4 = sin 2 try 2() = ( + )cos 2 + ( + )sin2 (so that no term of 2 is a solution of the complementary equation). Note: Solving Equations (7) and (9) in The Method of Variation of Parameters gives 0 1 = − 2 (120 − 210 ) and 0 2 = (1 20 −1 210 ) We will use these equations rather than resolving the system in each of the remaining exercises in this section. 19. (a) Here 42 + 1 = 0 ⇒ = ± 1 2 and () = 1 cos 1 2 + 2 sin 1 2 . We try a particular solution of the form () = cos + sin ⇒ 0 = −sin + cos and 00 = −cos − sin. Then the equation 400 + = cos becomes 4(−cos − sin) + (cos + sin) = cos or −3cos − 3 sin = cos ⇒ = − 1 3, = 0. Thus, () = − 1 3 cos and the general solution is () = () + () = 1 cos 1 2 + 2 sin 1 2 − 1 3 cos . (b) From (a) we know that () = 1 cos 2 + 2 sin 2 . Setting 1 = cos 2 , 2 = sin 2 , we have 120 − 210 = 1 2 cos2 2 + 1 2 sin2 2 = 1 2. Thus 0 1 = −cos sin 2 4 · 1 2 = − 1 2 cos2 · 2 sin 2 = − 1 2 2cos2 2 − 1sin 2 and 0 2 = cos cos 2 4 · 1 2 = 1 2 cos2 · 2 cos 2 = 1 2 1 − 2sin2 2 cos 2 . Then 1() = 1 2 sin 2 − cos2 2 sin 2 = −cos 2 + 2 3 cos3 2 and 2() = 1 2 cos 2 − sin2 2 cos 2 = sin 2 − 2 3 sin3 2 . Thus () = −cos 2 + 2 3 cos3 2 cos 2 + sin 2 − 2 3 sin3 2 sin 2 = − cos2 2 − sin2 2 + 2 3 cos4 2 − sin4 2 = −cos 2 · 2 + 2 3 cos2 2 + sin2 2 cos2 2 − sin2 2 = −cos + 2 3 cos = − 1 3 cos and the general solution is () = () + () = 1 cos 2 + 2 sin 2 − 1 3 cos. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS ¤ 727 20. (a) Here 2 − 2 − 3 = ( − 3)( + 1) = 0 ⇒ = 3, = −1 and the complementary solution is () = 13 + 2−. A particular solution is of the form () = + ⇒ 0 = , 00 = 0, and substituting into the differential equation gives 0 − 2 − 3( + ) = + 2 or −3 + (−2 − 3) = + 2, so = − 1 3 and −2 − 3 = 2 ⇒ = − 4 9. Thus () = − 1 3 − 4 9 and the general solution is () = () + () = 13 + 2− − 1 3 − 4 9. (b) In (a), () = 13 + 2−, so set 1 = 3, 2 = −. Then 120 − 210 = −3− − 33− = −42 so 0 1 = −( + 2)− −42 = 1 4 ( + 2)−3 ⇒ 1() = 1 4 ( + 2)−3 = 1 4 − 1 3( + 2)−3 − 1 9 −3 [by parts] and 0 2 = ( + 2)3 −42 = − 1 4 ( + 2) ⇒ 2() = − 1 4 ( + 2) = − 1 4[( + 2) − ] [by parts]. Hence () = 1 4 − 1 3 − 7 9 −3 3 − 1 4[( + 1)]− = − 1 3 − 4 9 and () = () + () = 13 + 2− − 1 3 − 4 9. 21. (a) 2 − 2 + 1 = ( − 1)2 = 0 ⇒ = 1, so the complementary solution is () = 1 + 2. A particular solution is of the form () = 2. Thus 42 − 42 + 2 = 2 ⇒ 2 = 2 ⇒ = 1 ⇒ () = 2. So a general solution is () = () + () = 1 + 2 + 2. (b) From (a), () = 1 + 2, so set 1 = , 2 = . Then, 120 − 210 = 2(1 + ) − 2 = 2 and so 0 1 = − ⇒ 1 () = − = −( − 1) [by parts] and 0 2 = ⇒ 2() = = . Hence () = (1 − )2 + 2 = 2 and the general solution is () = () + () = 1 + 2 + 2. 22. (a) Here 2 − = ( − 1) = 0 ⇒ = 0, 1 and () = 1 + 2 and so we try a particular solution of the form () = . Thus, after calculating the necessary derivatives, we get 00 − 0 = ⇒ (2 + ) − (1 + ) = ⇒ = 1. Thus () = and the general solution is () = 1 + 2 + . (b) From (a) we know that () = 1 + 2, so setting 1 = 1, 2 = , then 120 − 210 = − 0 = . Thus 0 1 = −2 = − and 0 2 = = 1. Then 1() = − = − and 2() = . Thus () = − + and the general solution is () = 1 + 2 − + = 1 + 3 + . 23. As in Example 5, () = 1 sin + 2 cos , so set 1 = sin, 2 = cos. Then 120 − 210 = −sin2 − cos2 = −1, so 0 1 = −sec2 cos −1 = sec ⇒ 1() = sec = ln (sec + tan) for 0 2 , and 0 2 = sec2 sin −1 = −sec tan ⇒ 2() = −sec. Hence () = ln(sec + tan) · sin − sec · cos = sinln(sec + tan) − 1 and the general solution is () = 1 sin + 2 cos + sin ln(sec + tan) − 1. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.728 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 24. As in Exercise 23, () = 1 sin + 2 cos, 1 = sin, 2 = cos, and 120 − 210 = −1. Then 0 1 = −sec3 cos −1 = sec2 ⇒ 1() = tan and 0 2 = sec3 sin −1 = −sec2 tan ⇒ 2() = − tan sec2 = − 1 2 tan2 . Hence () = tan sin − 1 2 tan2 cos = tan sin − 1 2 tan sin = 1 2 tan sin and the general solution is () = 1 sin + 2 cos + 1 2 tan sin. 25. 1 = , 2 = 2 and 120 − 210 = 3. So 0 1 = −2 (1 + −)3 = − − 1 + − and 1() = −1 +−− = ln(1 + −). 0 2 = (1 + −)3 = 3 + 2 so 2() = 3 + 2 = ln+ 1 − − = ln(1 + −) − −. Hence () = ln(1 + −) + 2[ln(1 + −) − −] and the general solution is () = [1 + ln(1 + −)] + [2 − − + ln(1 + −)]2. 26. 1 = −, 2 = −2 and 120 − 210 = −−3. So 0 1 = −(sin)−2 −−3 = sin and 0 2 = (sin)− −−3 = −2 sin. Hence 1 () = sin = −cos and 2() = −2 sin = cos − sin. Then () = −− cos − −2[sin − cos] and the general solution is () = (1 − cos)− + [2 − sin + cos]−2. 27. 2 − 2 + 1 = ( − 1)2 = 0 ⇒ = 1 so () = 1 + 2. Thus 1 = , 2 = and 120 − 210 = ( + 1) − = 2. So 0 1 = − · (1 + 2) 2 = − 1 + 2 ⇒ 1 = − 1 +2 = −12 ln1 + 2, 0 2 = · 2(1 + 2) = 1 +12 ⇒ 2 = 1 +12 = tan−1 and () = − 1 2 ln(1 + 2) + tan−1 . Hence the general solution is () = 1 + 2 − 1 2 ln(1 + 2) + tan−1 . 28. 1 = −2, 2 = −2 and 120 − 210 = −4. Then 0 1 = −−2−2 3−4 = − 1 2 so 1() = −1 and 0 2 = −2−2 3−4 = 1 3 so 2() = − 1 22 . Thus () = −2 − 2−22 = − 22 and the general solution is () = −2[1 + 2 + 1(2)]. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 17.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS ¤ 729 17.3 Applications of Second-Order Differential Equations 1. By Hooke’s Law (025) = 25 so = 100 is the spring constant and the differential equation is 500 + 100 = 0. The auxiliary equation is 52 + 100 = 0 with roots = ±2√5, so the general solution to the differential equation is () = 1 cos2√5 + 2 sin2√5. We are given that (0) = 035 ⇒ 1 = 035 and 0(0) = 0 ⇒ 2√52 = 0 ⇒ 2 = 0, so the position of the mass after seconds is () = 035 cos2√5. 2. By Hooke’s Law (04) = 32 so = 032 4 = 80 is the spring constant and the differential equation is 800 + 80 = 0. The general solution is () = 1 cos√10 + 2 sin√10. But 0 = (0) = 1 and 1 = 0(0) = √102 ⇒ 2 = √110, so the position of the mass after seconds is () = √110 sin√10. 3. (05) = 6 or = 12 is the spring constant, so the initial-value problem is 200 + 140 + 12 = 0, (0) = 1, 0(0) = 0. The general solution is () = 1−6 + 2−. But 1 = (0) = 1 + 2 and 0 = 0(0) = −61 − 2. Thus the position is given by () = − 1 5−6 + 6 5−. 4. (a) (025) = 13 ⇒ = 52, so the differential equation is 200 + 80 + 52 = 0 with general solution () = −21 cos√22 + 2 sin√22. Then 0 = (0) = 1 and 05 = 0(0) = √222 ⇒ 2 = 2 √122, so the position is given by () = 2 √122−2 sin√22. (b) 5. For critical damping we need 2 − 4 = 0 or = 2(4) = 142(4 · 12) = 49 12 kg. 6. For critical damping we need 2 = 4 or = 2√ = 2√2 · 52 = 4√26. 7. We are given = 1, = 100, (0) = −01 and 0(0) = 0. From (3), the differential equation is 2 2 + + 100 = 0 with auxiliary equation 2 + + 100 = 0. If = 10, we have two complex roots = −5 ± 5√3, so the motion is underdamped and the solution is = −51 cos5√3 + 2 sin5√3. Then −01 = (0) = 1 and 0 = 0(0) = 5√32 − 51 ⇒ 2 = − 101√3, so = −5−01cos5√3 − 101√3 sin5√3. If = 15, we again have underdamping since the auxiliary equation has roots = − 15 2 ± 5 √2 7. The general solution is = −1521 cos 5 √2 7 + 2 sin 5 √2 7, so −01 = (0) = 1 and 0 = 0(0) = 5 √2 72 − 15 2 1 ⇒ 2 = − 103√7. Thus = −152−01cos 5 √2 7 − 103√7 sin 5 √2 7. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.730 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS For = 20, we have equal roots 1 = 2 = −10, so the oscillation is critically damped and the solution is = (1 + 2)−10. Then −01 = (0) = 1 and 0 = 0(0) = −101 + 2 ⇒ 2 = −1, so = (−01 − )−10. If = 25 the auxiliary equation has roots 1 = −5, 2 = −20, so we have overdamping and the solution is = 1−5 + 2−20. Then −01 = (0) = 1 + 2 and 0 = 0(0) = −51 − 202 ⇒ 1 = − 15 2 and 2 = 30 1 , so = − 2 15 −5 + 30 1 −20. If = 30 we have roots = −15 ± 5√5, so the motion is overdamped and the solution is = 1(−15 + 5 √5) + 2(−15 − 5 √5). Then −01 = (0) = 1 + 2 and 0 = 0(0) = −15 + 5√5 1 + −15 − 5√5 2 ⇒ 1 = −5 − 3 √5 100 and 2 = −5 + 3 100√5, so = −5 100 − 3 √5 (−15 + 5 √5) + −5 + 3 100√5 (−15 − 5 √5). 8. We are given = 1, = 10, (0) = 0 and 0(0) = 1. The differential equation is 2 2 + 10 + = 0 with auxiliary equation 2 + 10 + = 0. = 10: the auxiliary equation has roots = −5 ± √15 so we have overdamping and the solution is = 1(−5 + √15) + 2(−5 − √15). Entering the initial conditions gives 1 = 2 √115 and 2 = − 2 √115, so = 1 2 √15 (−5 + √15) − 2 √115 (−5 − √15). = 20: = −5 ± √5 and the solution is = 1(−5 + √5) + 2(−5 − √5) so again the motion is overdamped. The initial conditions give 1 = 2 √1 5 and 2 = − 2 √1 5, so = 2 √1 5 (−5 + √5) − 2 √1 5 (−5 − √5). = 25: we have equal roots 1 = 2 = −5, so the motion is critically damped and the solution is = (1 + 2)−5. The initial conditions give 1 = 0 and 2 = 1, so = −5. = 30: = −5 ± √5 so the motion is underdamped and the solution is = −51 cos√5 + 2 sin√5. The initial conditions give 1 = 0 and 2 = √15, so = √15 −5 sin√5. = 40: = −5 ± √15 so we again have underdamping. The solution is = −51 cos√15 + 2 sin√15, and the initial conditions give 1 = 0 and 2 = √115. Thus = √115 −5 sin√15. 9. The differential equation is 00 + = 0 cos0 and 0 6= = . Here the auxiliary equation is 2 + = 0 with roots ± = ± so () = 1 cos + 2 sin. Since 0 6= , try () = cos 0 + sin0. Then we need ()−2 0(cos0 + sin0) + (cos0 + sin0) = 0 cos0 or − 2 0 = 0 and °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 17.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS ¤ 731 − 2 0 = 0. Hence = 0 and = − 0 2 0 = 0 (2 − 2 0) since 2 = . Thus the motion of the mass is given by () = 1 cos + 2 sin + 0 (2 − 2 0) cos0. 10. As in Exercise 9, () = 1 cos + 2 sin. But the natural frequency of the system equals the frequency of the external force, so try () = (cos + sin). Then we need (2 − 2)cos − (2 + 2)sin + cos + sin = 0 cos or 2 = 0 and −2 = 0 [noting −2 + = 0 and −2 + = 0 since 2 = ]. Hence the general solution is () = 1 cos + 2 sin + [0(2)] sin. 11. From Equation 6, () = () + () where () = 1 cos + 2 sin and () = 0 (2 − 2 0) cos0. Then is periodic, with period 2 , and if 6= 0, is periodic with period 20 . If 0 is a rational number, then we can say 0 = ⇒ = 0 where and are non-zero integers. Then + · 2 = + · 2 + + · 2 = () + + 0 · 2 = () + + · 20 = () + () = () so () is periodic. 12. (a) The graph of = 1 + 2 has a -intercept when 1 + 2 = 0 ⇔ (1 + 2) = 0 ⇔ 1 = −2. Since 0, has a -intercept if and only if 1 and 2 have opposite signs. (b) For 0, the graph of crosses the -axis when 11 + 22 = 0 ⇔ 22 = −11 ⇔ 2 = −1 1 2 = −1(1−2). But 1 2 ⇒ 1 − 2 0 and since 0, (1−2) 1. Thus |2| = |1|(1−2) |1|, and the graph of can cross the -axis only if |2| |1|. 13. Here the initial-value problem for the charge is 00 + 200 + 500 = 12, (0) = 0(0) = 0. Then () = −10(1 cos 20 + 2 sin 20) and try () = ⇒ 500 = 12 or = 125 3 . The general solution is () = −10(1 cos 20 + 2 sin 20) + 125 3 . But 0 = (0) = 1 + 125 3 and 0() = () = −10[(−101 + 202) cos 20 + (−102 − 201)sin20] but 0 = 0(0) = −101 + 202. Thus the charge is () = − 250 1 −10(6 cos 20 + 3 sin 20) + 125 3 and the current is () = −10 3 5 sin 20. 14. (a) Here the initial-value problem for the charge is 200 + 240 + 200 = 12 with (0) = 0001 and 0(0) = 0. Then () = −6(1 cos 8 + 2 sin 8) and try () = ⇒ = 50 3 and the general solution is () = −6(1 cos 8 + 2 sin 8) + 50 3 . But 0001 = (0) = + 50 3 so 1 = −0059. Also 0() = () = −6[(−61 + 82)cos 8 + (−62 − 81)sin8] and 0 = 0(0) = −61 + 82 so 2 = −004425. Hence the charge is () = −−6(0059 cos 8 + 004425 sin 8) + 50 3 and the current is () = −6(07375) sin 8. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.732 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS (b) 15. As in Exercise 13, () = −10(1 cos20 + 2 sin20) but () = 12sin10 so try () = cos10 + sin10. Substituting into the differential equation gives (−100 + 200 + 500) cos 10 + (−100 − 200 + 500)sin 10 = 12 sin 10 ⇒ 400 + 200 = 0 and 400 − 200 = 12. Thus = − 3 250, = 125 3 and the general solution is () = −10(1 cos 20 + 2 sin 20) − 250 3 cos 10 + 125 3 sin 10. But 0 = (0) = 1 − 250 3 so 1 = 250 3 . Also 0() = 25 3 sin 10 + 25 6 cos 10 + −10[(−101 + 202)cos 20 + (−102 − 201)sin 20] and 0 = 0(0) = 25 6 − 101 + 202 so 2 = − 500 3 . Hence the charge is given by () = −10 250 3 cos 20 − 500 3 sin 20 − 250 3 cos 10 + 125 3 sin 10. 16. (a) As in Exercise 14, () = −6(1 cos 8 + 2 sin 8) but try () = cos 10 + sin 10. Substituting into the differential equation gives (−200 + 240 + 200)cos 10 + (−200 − 240 + 200)sin10 = 12 sin 10, so = 0 and = − 1 20. Hence, the general solution is () = −6(1 cos 8 + 2 sin 8) − 20 1 cos 10. But 0001 = (0) = 1 − 20 1 , 0() = −6[(−61 + 82)cos 8 + (−62 − 81)sin8] − 1 2 sin 10 and 0 = 0(0) = −61 + 82, so 1 = 0051 and 2 = 003825. Thus the charge is given by () = −6(0051 cos 8 + 003825 sin 8) − 20 1 cos 10. (b) 17. () = cos( + ) ⇔ () = [coscos − sinsin] ⇔ () = 1 cos + 2 sin where cos = 1 and sin = −2 ⇔ () = 1 cos + 2 sin. [Note that cos2 + sin2 = 1 ⇒ 2 1 + 2 2 = 2.] 18. (a) We approximate sin by and, with = 1 and = 98, the differential equation becomes 2 2 + 98 = 0. The auxiliary equation is 2 + 98 = 0 ⇒ = ±√98, so the general solution is () = 1 cos√98 + 2 sin√98 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 17.4 SERIES SOLUTIONS ¤ 733 Then 02 = (0) = 1 and 1 = 0(0) = √982 ⇒ 2 = √1 98, so the equation is () = 02cos√98 + √198 sin√98 . (b) 0() = −02√98sin√98 + cos√98 = 0 or tan√98 = √598, so the critical numbers are = √198 tan−1 √5 98 + √98 ( any integer). The maximum angle from the vertical is √198 tan−1 √5 98 ≈ 0377 radians (or about 217◦). (c) From part (b), the critical numbers of () are spaced √98 apart, and the time between successive maximum values is 2 √98 . Thus the period of the pendulum is √298 ≈ 2007 seconds. (d) () = 0 ⇒ 02cos√98 + √198 sin√98 = 0 ⇒ tan√98 = −02√98 ⇒ = √1 98 tan−1−02√98 + ≈ 0825 seconds. (e) 0(0825) ≈ −1180 rads. 17.4 Series Solutions 1. Let () = ∞ =0 . Then 0() = ∞ =1 −1 and the given equation, 0 − = 0, becomes ∞ =1 −1 − ∞ =0 = 0. Replacing by + 1 in the first sum gives ∞ =0 ( + 1)+1 − ∞ =0 = 0, so ∞ =0 [( + 1)+1 − ] = 0. Equating coefficients gives ( + 1)+1 − = 0, so the recursion relation is +1 = + 1, = 0 1 2 . Then 1 = 0, 2 = 121 = 20 , 3 = 132 = 13 · 120 = 3! 0 , 4 = 143 = 4! 0 , and in general, = 0 !. Thus, the solution is () = ∞ =0 = ∞ =0 0 ! = 0 ∞ =0 ! = 0. 2. Let () = ∞ = 0 . Then 0 = ⇒ 0 − = 0 ⇒ ∞ = 1 −1 − ∞ = 0 = 0 or ∞ = 1 −1 − ∞ = 0 +1 = 0. Replacing with + 1 in the first sum and with − 1 in the second gives ∞ = 0 ( + 1)+1 − ∞ = 1 −1 = 0 or 1 + ∞ = 1 ( + 1)+1 − ∞ = 1 −1 = 0. Thus, 1 + ∞ = 1 [( + 1)+1 − −1] = 0. Equating coefficients gives 1 = 0 and ( + 1) +1 − −1 = 0. Thus, the recursion relation is +1 = −1 + 1, = 12, . But 1 = 0, so 3 = 0 and 5 = 0 and in general 2+1 = 0. Also, °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.734 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 2 = 0 2 , 4 = 2 4 = 0 4 · 2 = 0 22 · 2!, 6 = 4 6 = 0 6 · 4 · 2 = 0 23 · 3! and in general 2 = 2·0!. Thus, the solution is () = ∞ = 0 = ∞ = 0 22 = ∞ = 0 0 2 · ! 2 = 0 ∞ = 0 22 ! = 022. 3. Assuming () = ∞ = 0 , we have 0() = ∞ = 1 −1 = ∞ = 0 ( + 1)+1 and −2 = − ∞ = 0 +2 = − ∞ = 2 −2. Hence, the equation 0 = 2 becomes ∞ = 0 ( + 1)+1 − ∞ = 2 −2 = 0 or 1 + 22 + ∞ = 2 [( + 1)+1 − −2] = 0. Equating coefficients gives 1 = 2 = 0 and +1 = −2 + 1 for = 23, . But 1 = 0, so 4 = 0 and 7 = 0 and in general 3+1 = 0. Similarly 2 = 0 so 3+2 = 0. Finally 3 = 0 3 , 6 = 3 6 = 0 6 · 3 = 0 32 · 2!, 9 = 6 9 = 0 9 · 6 · 3 = 0 33 · 3!, , and 3 = 3·0!. Thus, the solution is () = ∞ = 0 = ∞ = 0 33 = ∞ = 0 0 3 · !3 = 0 ∞ = 0 3 3! = 0 ∞ = 0 33 ! = 033. 4. Let () = ∞ =0 ⇒ 0 () = ∞ =1 −1 = ∞ =0 ( + 1)+1. Then the differential equation becomes ( − 3) ∞ =0 ( + 1)+1 + 2 ∞ =0 = 0 ⇒ ∞ =0 ( + 1)+1+1 − 3 ∞ =0 ( + 1)+1 + 2 ∞ =0 = 0 ⇒ ∞ =1 − ∞ =0 3( + 1)+1 + ∞ =0 2 = 0 ⇒ ∞ =0 [( + 2) − 3( + 1)+1] = 0 since ∞=1 = ∞=0 . Equating coefficients gives ( + 2) − 3( + 1)+1 = 0, thus the recursion relation is +1 = ( + 2) 3( + 1) , = 012 . Then 1 = 230 , 2 = 3(2) 31 = 3320 , 3 = 3(3) 42 = 4330 , 4 = 3(4) 53 = 5340 , and in general, = ( + 1)0 3 . Thus the solution is () = ∞ =0 = 0 ∞ =0 + 1 3 . Note that 0 ∞=0 3+ 1 = (3 9−0)2 for || 3. 5. Let () = ∞ =0 ⇒ 0 () = ∞ =1 −1 and 00 () = ∞ =0 ( + 2)( + 1)+2. The differential equation becomes ∞ =0 ( + 2)( + 1)+2 + ∞ =1 −1 + ∞ =0 = 0 or ∞ =0 [( + 2)( + 1)+2 + + ] = 0 since ∞=1 = ∞=0 . Equating coefficients gives ( + 2)( + 1)+2 + ( + 1) = 0, thus the recursion relation is +2 = −( + 1) ( + 2)( + 1) = − + 2, = 012 . Then the even coefficients are given by 2 = −0 2 , 4 = − 2 4 = 0 2 · 4, 6 = − 4 6 = − 0 2 · 4 · 6, and in general, °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 17.4 SERIES SOLUTIONS ¤ 735 2 = (−1) 0 2 · 4 · · · · · 2 = (−1)0 2 ! . The odd coefficients are 3 = −31 , 5 = −53 = 3·15, 7 = −75 = −3 ·51· 7, and in general, 2+1 = (−1) 1 3 · 5 · 7 · · · · · (2 + 1) = (−2) !1 (2 + 1)! . The solution is () = 0 ∞ =0 (−1) 2 ! 2 + 1 ∞ =0 (−2) ! (2 + 1)! 2+1. 6. Let () = ∞ = 0 . Then 00() = ∞ = 2 ( − 1)−2 = ∞ = 0 ( + 2)( + 1)+2. Hence, the equation 00 = becomes ∞ = 0 ( + 2)( + 1)+2 − ∞ = 0 = 0 or ∞ = 0 [( + 2)( + 1)+2 − ] = 0. So the recursion relation is +2 = ( + 2)( + 1), = 0 1 . Given 0 and 1, 2 = 2·01, 4 = 4·23 = 4! 0 , 6 = 6·45 = 6! 0 , , 2 = 0 (2)! and 3 = 3·12, 5 = 5·34 = 5 · 4·13 · 2 = 5! 1 , 7 = 7·56 = 7! 1 , , 2+1 = (2+ 1)! 1 . Thus, the solution is () = ∞ = 0 = ∞ = 0 22 + ∞ = 0 2+12+1 = 0 ∞ = 0 2 (2)! + 1 ∞ = 0 2+1 (2 + 1)!. The solution can be written as () = 0 cosh + 1 sinh or () = 0 +2− + 1 −2− = 0 +2 1 + 0 −2 1 −. 7. Let () = ∞ =0 ⇒ 0 () = ∞ =1 −1 = ∞ =0 ( + 1)+1 and 00 () = ∞ =0 ( + 2)( + 1)+2. Then (−1)00() = ∞ =0 (+2)(+1)+2+1− ∞ =0 (+2)(+1)+2 = ∞ =1 (+1)+1− ∞ =0 (+2)(+1)+2. Since ∞ =1 ( + 1)+1 = ∞ =0 ( + 1)+1, the differential equation becomes ∞ =0 ( + 1)+1 − ∞ =0 ( + 2)( + 1)+2 + ∞ =0 ( + 1)+1 = 0 ⇒ ∞ =0 [( + 1)+1 − ( + 2)( + 1)+2 + ( + 1)+1] = 0 or ∞ =0 [( + 1)2+1 − ( + 2)( + 1)+2] = 0. Equating coefficients gives ( + 1)2+1 − ( + 2)( + 1)+2 = 0 for = 012, . Then the recursion relation is +2 = ( + 1)2 ( + 2)( + 1)+1 = + 1 + 2 +1, so given 0 and 1, we have 2 = 1 2 1, 3 = 2 3 2 = 1 3 1, 4 = 3 4 3 = 1 4 1, and in general = 1 , = 123, . Thus the solution is () = 0 + 1 ∞ =1 . Note that the solution can be expressed as 0 − 1 ln(1 − ) for || 1. 8. Assuming () = ∞ = 0 , 00() = ∞ = 2 ( − 1)−2 = ∞ = 0 ( + 2)( + 1)+2 and −() = − ∞ = 0 +1 = − ∞ = 1 −1. The equation 00 = becomes °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.736 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS ∞ = 0 ( + 2)( + 1)+2 − ∞ = 1 −1 = 0 or 22 + ∞ = 1 [( + 2)( + 1)+2 − −1] = 0. Equating coefficients gives 2 = 0 and +2 = −1 ( + 2)( + 1) for = 1 2, . Since 2 = 0, 3+2 = 0 for = 0 1 2 . Given 0, 3 = 0 3 · 2, 6 = 3 6 · 5 = 0 6 · 5 · 3 · 2, , 3 = 0 3(3 − 1)(3 − 3)(3 − 4) · · · · · 6 · 5 · 3 · 2. Given 1, 4 = 4·13, 7 = 4 7 · 6 = 1 7 · 6 · 4 · 3, , 3+1 = 1 (3 + 1)3(3 − 2)(3 − 3) 7 · 6 · 4 · 3. The solution can be written as () = 0 ∞ = 0 (3 − 2)(3 − 5) · · · · · 7 · 4 · 1 (3)! 3 + 1 ∞ = 0 (3 − 1)(3 − 4) · · · · · 8 · 5 · 2 (3 + 1)! 3+1. 9. Let () = ∞ = 0 . Then −0() = − ∞ = 1 −1 = − ∞ = 1 = − ∞ = 0 , 00() = ∞ = 0 ( + 2)( + 1)+2, and the equation 00 − 0 − = 0 becomes ∞ = 0 [( + 2)( + 1)+2 − − ] = 0. Thus, the recursion relation is +2 = + ( + 2)( + 1) = ( + 1) ( + 2)( + 1) = + 2 for = 012, . One of the given conditions is (0) = 1. But (0) = ∞ =0 (0) = 0 + 0 + 0 + · · · = 0, so 0 = 1. Hence, 2 = 0 2 = 1 2 , 4 = 2 4 = 1 2 · 4, 6 = 4 6 = 1 2 · 4 · 6, , 2 = 1 2!. The other given condition is 0(0) = 0. But 0(0) = ∞ =1 (0)−1 = 1 + 0 + 0 + · · · = 1, so 1 = 0. By the recursion relation, 3 = 1 3 = 0, 5 = 0, , 2+1 = 0 for = 0, 1, 2, . Thus, the solution to the initial-value problem is () = ∞ = 0 = ∞ = 0 22 = ∞ = 0 2 2! = ∞ = 0 (22) ! = 22. 10. Assuming that () = ∞ = 0 , we have 2 = ∞ = 0 +2 and 00() = ∞ = 2 ( − 1)−2 = ∞ =−2 ( + 4)( + 3)+4+2 = 22 + 63 + ∞ = 0 ( + 4)( + 3)+4+2. Thus, the equation 00 + 2 = 0 becomes 22 + 63 + ∞ = 0 [( + 4)( + 3)+4 + ]+2 = 0. So 2 = 3 = 0 and the recursion relation is +4 = − ( + 4)( + 3), = 01 2, . But 1 = 0(0) = 0 = 2 = 3 and by the recursion relation, 4+1 = 4+2 = 4+3 = 0 for = 012, . Also, 0 = (0) = 1, so 4 = − 0 4 · 3 = − 1 4 · 3, 8 = − 4 8 · 7 = (−1)2 8 · 7 · 4 · 3, , 4 = (−1) 4(4 − 1)(4 − 4)(4 − 5) · · · · · 4 · 3. Thus, the solution to the initial-value problem is () = ∞ = 0 = 0 + ∞ = 0 44 = 1 + ∞ = 1 (−1) 4 4(4 − 1)(4 − 4)(4 − 5) · · · · · 4 · 3. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 17.4 SERIES SOLUTIONS ¤ 737 11. Assuming that () = ∞ = 0 , we have = ∞ = 0 = ∞ = 0 +1, 20 = 2 ∞ = 1 −1 = ∞ = 0 +1, 00() = ∞ = 2 ( − 1)−2 = ∞ =−1 ( + 3)( + 2)+3+1 [replace with + 3] = 22 + ∞ = 0 ( + 3)( + 2)+3+1, and the equation 00 + 20 + = 0 becomes 22 + ∞ = 0 [( + 3)( + 2)+3 + + ]+1 = 0. So 2 = 0 and the recursion relation is +3 = − − ( + 3)( + 2) = − ( + 1) ( + 3)( + 2), = 012, . But 0 = (0) = 0 = 2 and by the recursion relation, 3 = 3+2 = 0 for = 0, 1, 2, . Also, 1 = 0(0) = 1, so 4 = − 21 4 · 3 = − 2 4 · 3, 7 = − 54 7 · 6 = (−1)2 7 · 2 6 · · 5 4 · 3 = (−1)2 227! 52 , , 3+1 = (−1) 2252 · · · · · (3 + 1)! (3 − 1)2 . Thus, the solution is () = ∞ = 0 = + ∞ = 1 (−1) 2252 · · · · · (3(3+ 1)! − 1)23+1 . 12. (a) Let () = ∞ = 0 . Then 200() = ∞ = 2 ( − 1) = ∞ = 0 ( + 2)( + 1)+2+2, 0() = ∞ = 1 = ∞ =−1 ( + 2)+2+2 = 1 + ∞ = 0 ( + 2)+2+2, and the equation 200 + 0 + 2 = 0 becomes 1 + ∞ = 0 {[( + 2)( + 1) + ( + 2)]+2 + }+2 = 0. So 1 = 0 and the recursion relation is +2 = − ( + 2)2 , = 012, . But 1 = 0(0) = 0 so 2+1 = 0 for = 0 1 2 . Also, 0 = (0) = 1, so 2 = − 1 22 , 4 = − 2 42 = (−1)2 42122 = (−1)2 24 (2!) 1 2 , 6 = −642 = (−1)3 26 (3!) 1 2 , , 2 = (−1) 1 22 (!)2 . The solution is () = ∞ = 0 = ∞ = 0 (−1) 2 22 (!)2 . (b) The Taylor polynomials 0 to 12 are shown in the graph. Because 10 and 12 are close together throughout the interval [−5 5], it is reasonable to assume that 12 is a good approximation to the Bessel function on that interval. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.738 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 17 Review 1. True. See Theorem 17.1.3. 2. False. The differential equation is not homogeneous. 3. True. cosh and sinh are linearly independent solutions of this linear homogeneous equation. 4. False. = is a solution of the complementary equation, so we have to take () = . 1. The auxiliary equation is 42 − 1 = 0 ⇒ (2 + 1)(2 − 1) = 0 ⇒ = ± 1 2. Then the general solution is = 12 + 2−2. 2. The auxiliary equation is 2 − 2 + 10 = 0 ⇒ = 1 ± 3, so = (1 cos 3 + 2 sin 3). 3. The auxiliary equation is 2 + 3 = 0 ⇒ = ±√3. Then the general solution is = 1 cos√3 + 2 sin√3. 4. The auxiliary equation is 2 + 8 + 16 = 0 ⇒ ( + 4)2 = 0 ⇒ = −4, so the general solution is = 1−4 + 2−4. 5. 2 − 4 + 5 = 0 ⇒ = 2 ± , so () = 2(1 cos + 2 sin). Try () = 2 ⇒ 0 = 22 and 00 = 42. Substitution into the differential equation gives 42 − 82 + 52 = 2 ⇒ = 1 and the general solution is () = 2(1 cos + 2 sin) + 2. 6. 2 + − 2 = 0 ⇒ = 1, = −2 and () = 1 + 2−2. Try () = 2 + + ⇒ 0 = 2 + and 00 = 2. Substitution gives 2 + 2 + − 22 − 2 − 2 = 2 ⇒ = = − 1 2, = − 3 4 so the general solution is () = 1 + 2−2 − 1 22 − 1 2 − 3 4. 7. 2 − 2 + 1 = 0 ⇒ = 1 and () = 1 + 2. Try () = ( + )cos + ( + )sin ⇒ 0 = ( − − )sin + ( + + )cos and 00 = (2 − − )cos + (−2 − − )sin. Substitution gives (−2 + 2 − 2 − 2)cos + (2 − 2 + 2 − 2)sin = cos ⇒ = 0, = = = − 1 2. The general solution is () = 1 + 2 − 1 2 cos − 1 2( + 1) sin. 8. 2 + 4 = 0 ⇒ = ±2 and () = 1 cos 2 + 2 sin 2. Try () = cos 2 + sin 2 so that no term of is a solution of the complementary equation. Then 0 = ( + 2)cos 2 + ( − 2)sin2 and 00 = (4 − 4)cos 2 + (−4 − 4)sin 2. Substitution gives 4 cos 2 − 4sin 2 = sin 2 ⇒ = − 1 4 and = 0. The general solution is () = 1 cos 2 + 2 sin 2 − 1 4cos 2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.CHAPTER 17 REVIEW ¤ 739 9. 2 − − 6 = 0 ⇒ = −2, = 3 and () = 1−2 + 23. For 00 − 0 − 6 = 1, try 1() = . Then 0 1() = 001 () = 0 and substitution into the differential equation gives = − 1 6 . For 00 − 0 − 6 = −2 try 2() = −2 [since = −2 satisfies the complementary equation]. Then 0 2 = ( − 2)−2 and 002 = (4 − 4)−2, and substitution gives −5−2 = −2 ⇒ = − 1 5 . The general solution then is () = 1−2 + 23 + 1() + 2() = 1−2 + 23 − 1 6 − 1 5 −2. 10. Using variation of parameters, () = 1 cos + 2 sin, 0 1() = −cscsin = −1 ⇒ 1() = −, and 0 2() = csccos = cot ⇒ 2() = ln|sin| ⇒ = −cos + sinln|sin|. The solution is () = (1 − )cos + (2 + ln|sin|)sin. 11. The auxiliary equation is 2 + 6 = 0 and the general solution is () = 1 + 2−6 = 1 + 2−6(−1). But 3 = (1) = 1 + 2 and 12 = 0(1) = −62. Thus 2 = −2, 1 = 5 and the solution is () = 5 − 2−6(−1). 12. The auxiliary equation is 2 − 6 + 25 = 0 and the general solution is () = 3(1 cos 4 + 2 sin 4). But 2 = (0) = 1 and 1 = 0(0) = 31 + 42. Thus the solution is () = 32cos 4 − 5 4 sin 4. 13. The auxiliary equation is 2 − 5 + 4 = 0 and the general solution is () = 1 + 24. But 0 = (0) = 1 + 2 and 1 = 0(0) = 1 + 42, so the solution is () = 1 3(4 − ). 14. () = 1 cos(3) + 2 sin(3). For 900 + = 3, try 1() = + . Then 1() = 3. For 900 + = −, try 2() = −. Then 9− + − = − or 2() = 10 1 −. Thus the general solution is () = 1 cos(3) + 2 sin(3) + 3 + 10 1 −. But 1 = (0) = 1 + 10 1 and 2 = 0(0) = 1 3 2 + 3 − 10 1 , so 1 = 9 10 and 2 = − 27 10 . Hence the solution is () = 10 1 [9 cos(3) − 27 sin(3)] + 3 + 10 1 −. 15. 2 + 4 + 29 = 0 ⇒ = −2 ± 5 and the general solution is = −2(1 cos 5 + 2 sin 5). But 1 = (0) = 1 and −1 = () = −1−2 ⇒ 1 = 2, so there is no solution. 16. 2 + 4 + 29 = 0 ⇒ = −2 ± 5 and the general solution is = −2(1 cos 5 + 2 sin 5). But 1 = (0) = 1 and −−2 = () = −1−2 ⇒ 1 = 1, so 2 can vary and the solution of the boundary-value problem is = −2(cos 5 + sin 5), where is any constant. 17. Let () = ∞ =0 . Then 00 () = ∞ =0 ( − 1)−2 = ∞ =0 ( + 2)( + 1)+2 and the differential equation becomes ∞ =0 [( + 2)( + 1)+2 + ( + 1)] = 0. Thus the recursion relation is +2 = −( + 2) for = 012, . But 0 = (0) = 0, so 2 = 0 for = 012, . Also 1 = 0(0) = 1, so 3 = −1 3, 5 = (−1)2 3 · 5 , °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.740 ¤ CHAPTER 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 7 = (−1)3 3 · 5 · 7 = (−1)3233! 7! , , 2+1 = (−1) 2 ! (2 + 1)! for = 0 1 2 . Thus the solution to the initial-value problem is () = ∞ =0 = ∞ =0 (−1) 2 ! (2 + 1)! 2+1. 18. Let () = ∞ =0 . Then 00 () = ∞ =0 ( − 1)−2 = ∞ =0 ( + 2)( + 1)+2 and the differential equation becomes ∞ =0 [( + 2)( + 1)+2 − ( + 2)] = 0. Thus the recursion relation is +2 = + 1 for = 0, 1, 2, . Given 0 and 1, we have 2 = 0 1 , 4 = 2 3 = 0 1 · 3, 6 = 4 5 = 0 1 · 3 · 5, , 2 = 0 1 · 3 · 5 · · · · · (2 − 1) = 0 2−1( − 1)! (2 − 1)! . Similarly 3 = 21 , 5 = 43 = 2·14, 7 = 5 6 = 1 2 · 4 · 6, , 2+1 = 1 2 · 4 · 6 · · · · · 2 = 1 2 !. Thus the general solution is () = ∞ =0 = 0 + 0 ∞ =1 2−1( − 1)! 2 (2 − 1)! + ∞ =0 2+1 2 ! . But ∞ =0 2+1 2 ! = ∞ =0 1 2 2 ! = 22, so () = 122 + 0 + 0 ∞ =1 2−1( − 1)!2 (2 − 1)! . 19. Here the initial-value problem is 200 + 400 + 400 = 12, (0) = 001, 0(0) = 0. Then () = −10(1 cos 10 + 2 sin 10) and we try () = . Thus the general solution is () = −10(1 cos 10 + 2 sin 10) + 100 3 . But 001 = 0(0) = 1 + 003 and 0 = 00(0) = −101 + 102, so 1 = −002 = 2. Hence the charge is given by () = −002−10(cos 10 + sin 10) + 003. 20. By Hooke’s Law the spring constant is = 64 and the initial-value problem is 200 + 160 + 64 = 0, (0) = 0, 0(0) = 24. Thus the general solution is () = −4(1 cos 4 + 2 sin 4). But 0 = (0) = 1 and 24 = 0(0) = −41 + 42 ⇒ 1 = 0, 2 = 06. Thus the position of the mass is given by () = 06−4 sin 4. 21. (a) Since we are assuming that the earth is a solid sphere of uniform density, we can calculate the density as follows: = mass of earth volume of earth = 4 3 3 . If is the volume of the portion of the earth which lies within a distance of the center, then = 4 3 3 and = = 33 . Thus = −2 = − 3 . (b) The particle is acted upon by a varying gravitational force during its motion. By Newton’s Second Law of Motion, 2 2 = = − 3 , so 00() = −2 () where 2 = 3 . At the surface, − = = − 2 , so = 2 . Therefore 2 = . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.CHAPTER 17 REVIEW ¤ 741 (c) The differential equation 00 + 2 = 0 has auxiliary equation 2 + 2 = 0. (This is the of Section 17.1, not the measuring distance from the earth’s center.) The roots of the auxiliary equation are ±, so by (11) in Section 17.1, the general solution of our differential equation for is () = 1 cos + 2 sin. It follows that 0() = −1 sin + 2 cos. Now (0) = and 0(0) = 0, so 1 = and 2 = 0. Thus () = cos and 0() = −sin. This is simple harmonic motion (see Section 17.3) with amplitude , frequency , and phase angle 0. The period is = 2. ≈ 3960 mi = 3960 · 5280 ft and = 32 fts2, so = ≈ 124 × 10−3 s−1 and = 2 ≈ 5079 s ≈ 85 min. (d) () = 0 ⇔ cos = 0 ⇔ = 2 + for some integer ⇒ 0() = −sin 2 + = ±. Thus the particle passes through the center of the earth with speed ≈ 4899 mis ≈ 17,600 mih. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved. [Show More]
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