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Chapter 18 - ET 17

# Chapter 18 - ET 17 - 18 CI SECOND-ORDER DIFFERENTIAL...

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Unformatted text preview: 18 CI SECOND-ORDER DIFFERENTIAL EQUATIONS CI ET 17 18.1 Second-Order Linear Equations 10. 11. 12. 13. 14. 15. . The auxiliary equation is 1'2 ~ 61' + 8 2 0 -> ET 17.1 (1' 4)(1' 2) 0 > 1' 4,1" 2. Thenby(8)thegeneral solution is y 2 cle43E + C2621”. The auxiliary equation is 1'2 — 415+ 8 2 0 => 1' 2 2 :: 21'. Then by (11) the general solution is y 2 e2z(c1 cos 21: + C2 sin 2w). . The auxiliary equation is 1'2 + 81' + 41 2 0 => 1' 2 A4 :: 5i. Then by (l 1) the general solution is y 2 6’” (C1 cos 556 + C2 sin 5w). The auxiliary equation is 21'2 — 1' — ‘1 2 (21' + 1)(1' — 1) 2 0 => 1' 1, 1' 2 —%. Then the general solution is _ 2 y:Clem+C2e W. The auxiliary equation is 1'2 — 21' + 1 2 (1' — 1)2 2 0 => 1' 2 1. Then by (10). the general solution is y 2 Clem + ngez. ,SOy 2 c1 +02e5I/3. “It” The auxiliary equation is 31'2 — 51' 2 1(31' — 5) 2 0 :> 1' 1 The auxiliary equation is 41'2 + 1 2 0 => 1' 2 :51} so y 2 c1 cos + C2 sin The auxiliary equation is 161'2 + 241' + 9 2 (41' + 3)2 2 0 => 1' 2 —%, s0 y 2 ale—3W4 + nge’g’z/‘l. The auxiliary equation is 41'2 + 1' 2 1'(41' +2 0 2> 1' 2 0, 1' 2 —%, so y 2 cl + Cge‘m/‘l. The auxiliary equation is 91'2 + 4 2 0 => 1' 2 ::%i, so y 2 c1 c0s(§w) + C2 sin(§x). The auxiliary equation is 1'2 — 21' — 1 2 0 => 1' 2 1 :: x/i, so y 2 cle(1+‘/§)t + Czeoﬂﬂ)? The auxiliary equation is 1'2 2 61' + 4 2 0 => 1' 2 3 :: x/g, so y 2 cle(3+‘/5)t + wee—ﬂ)? ﬁi, so y — (ft/2 [c1 c0s(3§t) + Cg sin(§t)]. “2 The auxiliary equation is 1'2 + 1' + 1 _ 0 _> 1- _ _ [oh—t 61'2~1'—22(21'+1)(31'—2)20so ’ y 2 Cle-I/2 + €262I/3. The solutions (01,02) 2 (0,1), (1, 0). (1, 2), (—27 1) are shown. Each solution consists of a single continuous curve that approaches either 0 0r :00 as :L‘ —> :00. 4x T2—8T+16=(T—4)2=080y=C164I+C2\$C The graphs are all asymptotic to the av-axis as w —> —oo, and as w —> co the solutions tend to :00. 640 16. 17. 18. 19. 20. 21. 23. 24. 25. 26. CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ET CHAPTER 17 7‘2 ~2r+5 :0 2 r = 1i2z’andthesolutionis y 2 51(c1 cos 23: + 62 sin 23:). Graphs for (01,62) 2 (1,0), (0,1), (1,—1), (—1,2) are shown. The solutions are all asymptotic to the :c—axis as a: —> —oo and they all oscillate. The amplitudes of the oscillations become arbitrarily large as a: —> 00 and arbitrarily small asa: —’ —OO. 2r2 + ST + 3 2 (2r + 3)(r + 1) 2 0, so r2 —%, r 2 —1 and the general solution is y 2 c1637”2 + me”. Then y(0) 2 3 => 01 + C2 2 3 and y'(0) 2 —4 => —%c1 2 02 2 —4, so 01 2 2 and 62 2 1. Thus the solution to the initial—value problem is y 2 26—3m/2 + 6—1. r2 + 3 2 0 => r 2 ix/3i and the general solution is y 2 60:” (c1 cos(\/3:c) + 02 sin(\/3m)) 2 c1 cos(\/3:c) + 62 sin(‘/3:c). Then 11(0) 2 1 => cl 2 1 and y’(0) 2 3 => 02 2 \/3, so the solution to the initial—value problem is y 2 c0s(\/3:c) + x/35in(\/3 4T2 4T + 1 _ (2r 1)2 — 0 —> r — % and the general solution is y 2 gem/2 + 02:661/2. Then y(0) 2 1 => C12 1 and y’(0) y 2 61/2 — 211361/2. —1.5 => %C1 + C2 2 —1.5, so 02 2 —2 and the solution to the initial-value problem is 27/2 33: 2T2 + 5r — 3 2 (2r — 1)(r + 3) 2 0 => r 2 %, r 2 —3 and the general solution is y 2 616 + 6267 Then 1 2 34(0) 2 C1 + C2 and 4 2 y’(0) 2 %C1 — 3C2 so C1 2 2.02 1/2 _ —1 and the solution to the initial-value 73cc problem is y 2 26 e r2+1620 2 0% ThenyG) 3 -> —C1——3 —> C1-3andy'(%)—4 —> solution to the initial—value problem is y 2 3 cos 4:6 — sin 43:. 01 cos 4:6 + Cg sin 43:) 2 C1 cos 4:6 + 02 sin 43:. 1, so the r 2 i4z' and the general solution is y 2 e 4C2—‘4 —> C22 r 2 1 :t 21' and the general solution is y 2 6:”(01 cos 2:6 + 62 sin 22:). Then => C1 2 0 and 2 2 y'(7r) 2 (c1 + 202)e7r (E , . . . e . _ . initial—value problem IS 3; 2 7 Sin 22: 2 (2m 7' sm 2m. 6 r2—2r+520 => 0 2 y(7r) 2 e"(C1 + 0) :> 02 2 1/e7r and the solution to the r2 + 2r + 2 2 0 => r 2 —1 :t 2' and the general solution is y 2 e_1(c1cosa: + 62 sin Then 2 2 y(0) 2 c1 and 1 2 y’(0) 2 C2 — C1 :> C2 2 3 and the solution to the initial—value problem is y 2 e_x(2 cos cc + 3 sin r2 + 12r + 36 2 (r + 6)2 2 0 => r 2 —6 and the general solution is y 2 Ole—GI + 02336431. Then 0 2 y(1) 2 016—6 + C26_6 => c1 + C2 — 0 and 1 — y'(1) — 6 => 6C1 + 5C2 C1 2 266 and 62 2 66. The solution to the initial—value problem is y 2 “666—61 + 661136—61 2 (a: — 1)66—6m. 2 —ee, so 6C1€_6 5C2€_ 4r2 + 1 2 0 => r 2 iéi and the general solution is y 2 c1 cos(%cc) + C2 sin(%m). Then 3 2 y(0) 2 C1 and —4 2 y(7r) 2 C2, so the solution of the boundary—value problem is y 2 3 cos(%cc) — 4 sin(%a:). r2 + 2r 2 r(2 + r) 2 0 2 r 2 0, r 2 —2 and the general solution is y 2 C1 + 026—2x. Then 62 1 — 2e2 , C1 2 . The solution of the 1 — 62 1 2 y(0) 2 C1 +C2 and2 211(1) 2 C1 +C26_2 SOC2 2 1 — 262 e2 —2:c 1—.er"—I—1—~e2 . boundary—value problem is y 2 1 e 27. 30. 31. 33. .r2—6r+25=0 :> .9r2—18r+10=0 :> SECTION 18.1 SECOND-ORDER LINEAR EQUATIONS ET SECTION 17.1 641 r2 — 3r + 2 : (r — 2)(r — 1) : 0 :> r = 1, r = 2 and the general solution is y = C161 + Cgeh. Then 1: 71(0) 2 C1 + C2 and 0 2 74(3) = C163 + C266 so C2 =1/(1— 63) and C1 = 63/(63 — 1). The solution ofthe m+3 21 e e 63—]. + boundary—value problem is y : 1 3 . ~ 6 r2 + 100 : 0 :> 5 = y(7r) = C1, so there is no solution. 7" 2 i107 and the general solution is y = (:1 cos 1092 + C2 sin 10w. But 2 : 71(0) 2 C1 and r = 3 j: 4i and the general solution is y = 63\$(C1 cos 4m + C2 sin 43:). But 1 : 74(0) 2 cl and 2 : 74(7r) : C1637r => C1 ': 2/63", so there is no solution. r2 — 6? + 9 = (r W 3)2 = 0 :> r = 3 and the general solution is 74 2 C1631 + ngeh. Then 1 2 74(0) : 01 33: 33: and 0 2 74(1) 2 0163 + C263 :> C2 : '1. The solution of the boundary-value problem is 74 = e — ave r2+4r+13=0 => _21( 2 = 74(0) 2 C1 and: 1 = : e“"(—C2), so the solution to the boundary—value problem is r : —2 j: 3i and the general solution is 74 = 6 C1 cos 3m + C2 sin 3m). But 74 2 eT2m(2 cos 33: ~ e7r sin 33:). ii and the general solution is y : ex (C1 cos g + C2 sin Then 2 \/§ 6" 7‘21 :l: 0 2 74(0) : C1 and 1 : 74(7r) 2 e7r Gal + g . The solution of the boundary-value Cg) => C2 2 26¢ ﬁe” sin(§) = 1363"" sin(§). (a) Case I (A 2 0): y" + A74 2 0 => 74” = 0 which has an auxiliary equation r2 = 0 => r = 0 => problem is y = 74 = C1 + 0213 where 74(0) : 0 and 74(L) : 0. Thus, 0 2 74(0) 2 C1 and 0 : 74(L) : C2L => C1 = C2 : 0. Thus, 74 = 0. 1 Case 2 (A < 0): 74" + A74 2 0 has auxiliary equation r2 = —A => r = i\/——A (distinct and real since A < 0) => 74 : Cleﬂi + C2€_ "” where 74(0) 2 0 and y(L) = 0. Thus, 0 2 74(0) = C1 + C2 (*) and 0 = 74(L) = Cle‘/_—’\L + 626_‘/_—/\L (T). JTAL (BEL _ eiJw—AL> : 0 :> Multiplying (*) by e and subtracting (’7) gives 02 C2 ; 0 and thus C1 = 0 from (*). Thus, 74 = 0 for the cases A = 0 and A < 0. (b) y" + Ay : 0 has an auxiliary equation r2 + A = 0 :> r = iix/A :> 74 = C1 cosx/sz + 02 sin x/Am where 74(0) : 0 and 74(L) = 0. Thus, 0 : 74(0) : C1 and 0 = 74(L) = C2 sin x/AL since 01 = 0. Since we cannot have a trivial solution, C2 # 0 and thus sin \/A L = 0 :> L : mr where n is an integer :> A = 7227r2/L2 and y = C2 sin(n7rav/L) where n is an integer. 34. The auxiliary equation is ar2 + br + C = 0. Vlf b2 — 4ac > 0, then any solution is of the form —b — Vb2 — 4aC d - T2 2a so both T1 and T2 are negative and limmnoo = 0. If b2 — 4ac : 0, then any solution is of the form 7‘11 —b + Vb? — 4aC an 2a 74(3)) 2 C16 + CgeT21 where n — . But a, b, and C are all positive 74 2 C16” + Cgme” where r = —b/ (2a) < 0 since a, b are positive. Hence limp+00 74(av) = 0. Finally if b2 A 4ac < 0, then any solution is of the form y(a:) = eaI(C1 cos ﬂan + C2 sin ﬁm) where a = —b/(2a) < 0 since a and b are positive. Thus limmaw y(:v) : 0. 642 CHAPTER 18 SECONDvOFIDEFI DIFFERENTIAL EQUATIONS ET CHAPTER 17 18.2 Nonhomogeneous Linear Equations ET 17.2 1. The auxiliary equation is r2 + 37" —I— 2 2 ('r —I— 2)(r —I— 1) 2 0, so the complementary solution is yc(\$) 2 616—2z —I— C2€_m. We try the particular solution yp(\$) 2 A91:2 + Ba: —I— C, so 3;; 2 2A\$ —I— B and 111’; 2 2A. Substituting into the differential equation, we have (2A) + 3(2Aa: + B) + 2(A292 + B2: + C) 2 :32 or 2A9:2 —I— (6A + 2B)a: + (2A + 33 + 20) 2 m2. Comparing coefﬁcients gives 2A 2 1, 6A + 2B 2 0, and 2A -I— 33 —I— 2C 2 0, so A 2 B 2 —g, and C 2 . Thus the general solution is Ma 11(20): yc(m) —I— yp(m) 2 C1641 —I— Cge‘It—I— éaE? — gm +71. 2. The auxiliary equation is r2 —I— 9 2 0 with roots r 2 i3z', so the complementary solution is yc(a:) 2 c1 cos(3a:) + 02 sin(3m). Try the particular solution yp(a:) 2 A1337”, So 3/; 2 314631 and y; 2 9A6“. Substitution into the differential equation gives 9A8” —I— 9(A33") 2, 63x or 18Ae3m 2 63". Thus A 2 1—18 and the general solution is y(:c) 2 yc(m) —I— yp(:c) 2 cl cos(3:c) —I— C2 sin(3m) —I— ﬁew. 3. The auxiliary equation is r2 — 2r 2 r(r —- 2) 2 0, so the complementary solution is yC 2 c1 —I— age”. Try the particular solution yp(:c) 2 A cos 49: —I— B sin 49:, so :11; 2 ~4A sin 4m + 4B cos 4:5 and y; 2 —16A cos 4:5 — 163 sin 4:5. Substitution into the differential equation gives (—16Ac0s4.r — 16B sin 4:5) — 2(~4A sin 4:5 —I— 4B cos 4m) 2 sin 4:5 2 (—16A — 8B) 0054.1 —I— (8A — 16B) sin4\$ 2 sin4x. Then —16A — 8B 2 0 and 8A — 16B 21 :> = \$7 2m and B 2 —§15. Thus the general solution is y(:c) 2 yc(:c) —I— yp(:c) 2 c1 + C26 + % cos 4:5 — 21—0 sin 4i. 4. The auxiliary equation is r2 + 6r —I— 9 2 (r —I— 3)2 2 0, so the complementary solution is yc(:c) 2 cle_3“E —I— (229313—31. Try the particular solution yp(:c) 2 Arc —I— B, so y’p 2 A and 1;; 2 0. Substitution into the differential equation gives 0 —I— 6A —I— 9(A\$ —I— B) 2 1 —I— x or (9A)x —I— (6A —I— 9B) 2 1 + x. Comparing coefﬁcients, we have 9A 2 1 and 6A —I— 9B 2 1, so A 2 g and B 2 Thus the general solution is 2 ale—3“E —I— ewe—3m —I— gay —I— 5. The auxiliary equation is r2 — 4r + 5 2 0 with roots 7" 2 2 i i, so the complementary solution is yC 2 e21 (c1L cosx + 62 sin Try 31,, (It) 2 Ae’x, so y,’, 2 —Ae'\$ and y; 2 Ae_1. Substitution gives Ae’z — 4(—Ae‘z) + 5(Ae’x) 2 671. 2> 10Ae‘“E 2 6"” 2> A 2 1—10. Thus the general solution is 2 621(01 cosx + 02 sin m) —I— ﬁe”. ’ 6. yc(m) 2 e‘z(cla: + Cg). Try yp(m) 2 9:2 (Am + B) 6"” so that no term in yp is a solution of the complementary equation. Then y; 2 [—AJ:3 + (3A — B)ar2 + 2B\$Ie"‘, yg 2 [Am3 —I— (B — 6A)a:2 -I- (6A — 4B):c —I— ZB]e_“E and substitution gives [142:3 + (B 2 6A)z2 + (6A — 41m + 23] + 2[—A\$3 + (3A — BM + 23\$] + (A353 + Bx?) = \$ 2 6A9: —I— 2B 2 9:. So yp (3:) 2 2102(éa:)e_z and the general solution is y(:c) 2 e_I(C1IE —I— Cg)+(-15\$3€_x. 7. 10. SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ET SECTION 172 3 The auxiliary equation is 7‘2 + 1 = 0 with roots 7‘ = :l:z', so the complementary solution is yc(.z') = c1 cosav + Cg sin x. For y" + y = 6" try yp1 2 A61. Then ygl = ygl :- AeI and substitution gives A61 + Aem 2 6" => A = %, so yp1 — éer. For y" + y 2 :53 try yp2(a:) 2 AI3 + Bx2 + Cm + D. Then yz’,2 = 3Aav2 + 2Bx + C and 71;; = 6Ax + 2B. Substituting, we have 6Aas+2B+Ax3+Bx2+Car+D:x3,soA:1,B=0,6A+C=0 ~=> C:—6,and2B+D=0 => D = 0. Thus yp2 = I3 ~ 630 and the general solution is tolcn y(m) = yc(m) + yp1 + yp2 C1 cosz + Cg sinx + ﬁe" +m3 ‘~ 62:. But 2 = y(0) = C1 +% => C1 and O — y'(0) — Cg + % 6 _> C2 ~ Thus the solution to the initial-value problem is 3 11 - 1 z 3 y(m) 5cosm+7smx+§e +2: —6x. The auxiliary equation is 7‘2 — 4 = 0 with roots 7‘ = 12, so the complementary solution is yc(ar) = 61621 + C26_2.I. Try yp(av) = 61(ACOS.Z' + B sin at), so y; 2 61(14 cos x + B sins: + B cosz — Asin x) and y; 61(2B cos x — 2A sin Substitution gives 61(2BCOSl' — 2Asinav) — 461(ACOSl' + Bsinar) = 6”” cosm :> (2B — 4A)eI cosx + (—2A — 4B)eI sinm 2 6I cosx => A = —%, B = 1—104 Thus the general solution is = 61621 + 626—21 + eI(—% cosm + lio sinx). But 1 = y(0) = C1 + Cg — % and A 40’ 2 :.y’(0) = 2C1 — 2Cg — liol Then C1 3, Cg : and the solution to the initial-value problem is : 362’” + 4—306—21 + e"‘(—% cosa: + 1—10 sinm). The auxiliary equation is 7‘2 — 7‘ = 0 with roots 7‘ z 0, 7‘ = 1 so the complementary solution is yc(m) : C1 + Cge". Try yp(m) = 22(Ax + B)e" so that no term in 3,1,, is a solution of the complementary equation. Then y; = (A2:2 + (2A + B)m + B)e”” and y; : (Ax2 + (4A + B)m + (2A + 2B))e””. Substitution into the differential equation gives (Ax2 + (4A + B)m + (2A + 2B))e" — (Ax2 + (2A + B)m + B)e" = \$61 => (2Ax + (2A + B))eI = \$61 => A = 32*, B = —1. Thus yp(z) 2 (ﬁx2 — ﬂex and the general solution is = C1 +626I + (ﬁxQ — ﬂex. But 2 = y(0) = 61 +62 and 1 = y’(0) = Cg A 1, s0 C2 = 2 and C1 = 0. The solution to the initial—value problem is y(x) = 26" +. (%m2 — an) CI = €I(%l'2 — a: + 2), yc(:1;) : c133” + Cge‘2z. For y" + y’ — 2y 2 z try ym = Ax +KB. Then ygl = A, ygl : 0, and substitution givesO+A—2(AI+B) =z :> A: —%,B= —i,soyp1(\$) — —%x— iFOryH—l—y’ —2y=sin2xtry yp2 = A cos 2m + B sin 2m. Then y;,2 :7 —2A sin 2m + 2B cos 2m, yg2 = —4A cos 2m — 4B sin 2m, and substitution gives (—4A cos 2m — 4B sin 2m) + (—2A sin 22: + 2B cos 2x) — 2(A cos 22: + B sin 2m) : sin 2m _i => A = —%, B = —2—30. Thus yp2 = cos 2m + —2—30 sin 2m and the general solution is 2o y(m) : C16I +C2e_21 — %m w i — 2—10 c0522: — % sin2ar. But 1 = y(0) = C1 +Cg — i — i0 and 0 — y'(0) — C1 2C2 % 130 _> C1 — % and Cg = %. Thus the solution to the initial-value problem is = 1—261 + %e_2”” — %m — i — 2—10 cos2m — 2—30 sin2m. 644 11. 12. 13. 14. 15. 16. 17. CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ET CHAPTER 17 ‘1/4 + C25”. Try yp(m) : Ael. Then yc(;c) : cle 10Aez : em, so A : T15 and the general solution is ﬁe? The solutions are all composed y(;c) : ole—W4 —I— Cg€_\$ —I— of exponential curves and with the exception of the particular solution (which approaches 0 as m —+ —oo), they all approach either 00 or —00 as x —+ —00. As m —+ 00, all solutions are i69:. asymptotic to yp : 10M —m/2 The auxiliary equation is (2r —I— 1)(r + 1) = 0, so 7” : —1,—% and yc(m) = 01871 —I— C28 . For 2g" + 33/ —I— y = 1, try ypl = A; substituting gives yp1i(\$) = 1. For 23/" + 39" + y = cos 235 try gm : A cos 2x —I— B sin 2m => y];2 = —2A sin 2x + 2B cos 2m, 11;; = —4A cos 2x — 4B sin 2x. Substituting into the differential equation gives cos 2x 2 (6B — 7A)'c0s 2x —I— (—7B — 6A) sin 2x. Then solving the equations 6B — 7A 2 1 and —7B — 6A 2 0 gives A = —8—75, B : %. Thus, gm 2 —% cos 2:3 —I— 8—65 sin 2m and the general solution is y = ale—I + 626_I/2 +1 — 8—75 cos 2m + 8—65 sin 2m. The graph shows 11;; = ypi + 11192 and several other solutions. Notice that all solutions are asymptotic to yp as \$400. Here yc(m) 2 c1 cos 3m + (:2 sin 3x. For y" + 9y : e27” = A621 and for y” —I— 9y = x2 sin x try yp2 = (Bx2 + Cm + D) coscc —I— (Em2 + Fm —I— G) sin x. Thus a trial solution is A621-+(Bi2+Cm+D)COS\$+(E12+Fm-I—G)sin;c. yp(\$) : yP1+ 9P2 Since yc(;c) = (:1 + 026’”, try yp(m) : (Am + B)e‘r cos 7m —I— (Cm + D)e_\$ sin 7m. Here 746(1) 2 cl + cze‘gr. For y" —I— 9y.’ = 1 try yp1 2 Am (since y = A is a solution to the complementary equation) and for y" —I— 93/ 2 mega” try 111.,2 : (Bm —I— C)eQI. Since yea) 2 C161 —I— C28_4I try yp(m) = 1(Ax3 —I— Bx2 —I— Cm —I— D)e”” so that no term of yp(m) satisﬁes the complementary equation. Since yc(m) 2 671(C1 cos 3x —I— C2 sin 3w) we try yp(w) = m(Aac2 —I— Bm —I— C)e_r cos 3m -I— :4sz -I— Em —I— F)e‘r sin 3w (so that no term of yp is a solution of the complementary equation). SECTION 18.2 NONHOMOGENEOUS LINEAR EQUATIONS ET SECTION 17.2 645 18. Here yc(:c) 2 cl cos 2:5 + C2 sin 2:5. For y" + 4y 2 es“E try yp1 2 A631 and for y" + 4y 2 1: sin 2:5 try ypz, 2 :c(Bm + C) cos 21: —l- :c(D1: + E) sin 21: (so that no term of yp2 is a solution of the complementary equation). Note: Solving Equations (7) and (9) in The Method of Variation of Parameters gives _§y2_ _Gyl._ a(y1yé 2 1121/1) a(y1y’2 — 112211) We will use these equations rather than resolving the system in each of the remaining exercises in this section. I I U1 : ‘— and 11.2 = 19. (a) The complementary solution is yc 2 cl cos 2:5 + 02 sin 21:. A particular solution is of the form yp(:c) 2 Arc + B. Thus, 4A1: + 4B 2 1: => A 2 i and B 2 O 2> yp(:c) 2 ix. Thus, the general 501mm" is y = ya + yp = C1 c0821: + C2 sin 2m + ix. (b) In (a), yc(x) 2 cl cos 2:5 + 02 sin 220, so set yl 2 cos 21:, 3/2 2 sin 21:. Then ylyé — 3123;12:2c0s2 2m + 2sin2 217 2 2 so 11/12 —%:csin2:c => u1(:c) 2 —% facsian dcc 2 —%(—:c c0s2cc + % sin2cc) [by parts] and u'2 2 éxcos 21: => uz(x) 2 §fcccos 2mdcc 2 %(ac sin2x + écos 21:) [by parts]. Hence yp(x) 2 2% (21: cos 21: + % sin 2:5) cos 23c + ﬂat sin 21: + % cos 23c) sin 21: 2 ix. Thus 2 yc(x) + yp(:c) 2 cl cos 21: + 02 sin 21: + ix. 20. (a) Here r2 — 3r + 2 2 O 2> r 2 1 or 2 and yc(x) 2 ole2m + 0261. We try a particular solution of the form yp(x) 2 Acoscc + B sinx => 3/122 —Asinx + Bcos x and y; 2 —Ac0sm — Bsin cc. Then the equation yy" — 33/ + 2y 2 sinx becomes (A — 3B) coscc + (B + BA) Since 2 sins: => A w 3B 2 O and B + 3A 2 1 => A 2 1% and B 2 1—10. Thus, yp(:c) 2 % coscc + lie sin ac. Therefore, the general solution is 31(55): yew) + yp(m) = c162“E + C26“E + 1—30 cosm + % sinm. 2w (b) From (a) we know that yc(\$) 2 0162’” + C261. Setting yl 2 e , 3/2 2 ex, we have I s1n we . _ yly'z ~ 31ng 2 es“E — 263\$ 2 —e3x .Thus u’l 2 — 3x 2 sm we 2" and —e sin .1762“E u'2 2 W 2 —sin me“. Then u1(1:)2 fee—2’” sinxdx 2 %e_21(—2sinx — cos x) [by —e _ —a: - _ 1 —a: ~ parts] and 1L2(1‘) _ —fe SlIlIEdIE 2 —§e (—s1nrc 2 cosx). Thus yp(:c) — §(—2 s1nx — cos x) + 2 (smx + cos x) — 10 s1ncc + 10 cosac and the general solutlon 1s 2 yc(x) + yp(:c) 2 016% + 626\$ + 1—10 sinac + 1—30 cos 1:. 21. (a) r2 — r 2 r(r — 1) 2 O 2> r 2 0, 1, so the complementary solution is yc(x) 2 cle“E + Cgmex. A particular solution is of the form yp(:c) 2 A621. Thus 414621 2 4A62“E + Ae2“E 2 62“E 2> Aezm 2 62’” 2> A 2 1 => yp(\$) = 62". So a general solution is y(x) 2 yc(x) + yp(x) 2 Clem + agate“E +6”. 2:: 1 62:: (b) From (a), yc(:c) 2 c167” + C2176", so set yl 2 61,112 2 me“. Then, ylyé — ygyi 2 e2m(1 + 1:) — we and so 113 2 —:cez => 1L1 2 — [er d1: 2 —(:c — 1)e“E [by parts] and 11\$ 2 cm => 2 fez date 2 6". Hence yp 2 (1 — @621 + \$621 2 e21 and the general solution is Mm) = yc(\$) ‘l‘ WW) = 0161 + szez + 621. 22. (a) Here r2 — 2r + 1 2 (r + 1)2 2 0 => r‘2 —1 and yc(:1:) 2 cl + agedE and so we try a particular solution of the form yp(:c) 2 Annex. Thus, after calculating the necessary derivatives, we get y” — y' 2 em => Ae“E (2 + m) — Ae“(1 + m) 2 e“ 2> A 2 1. Thus yp(\$) 2 mew and the general solution is y(\$) 2 (:1 + 026‘” + \$61. 646 23. 24. 25. 26. 27. 28. CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ET CHAPTER 17 (b) From (a) we know that yam) 2 01 + 0263”, so setting y1 = 1, 3/2 2 6””, then yiylg — 3/2313 2 er — 0 2 6””. Thus u’l —622/61 2 —e”” and 1/2 2 eI/er 2 1. Then u1(\$) — fexdac 2 —ez and u2(m) 2 ac. Thus yp_(:1:) 2 —e\$ + 11361 and the general solution is 2 c1 + C261 — 6”” + mew 2 c1 + 036\$ + \$61. As in Example 6, yc(\$) 2 cl sin x + 02 cos x, so set yl 2 sin 3/2 2 cos x. Then SEC T COS E 1 :1 : u1(m)2mand yly/g — 3/23/12 —sin2\$ — cosgm 2 —1, so 1/1 2 - sec m sin as —1 yp(\$) 2 ac sinm + cos ac ln(cos ac) and the general solution is 2 (c1 + 1:) sinm + [02 + ln(cos cos m. 1/2.: tanm 2 U2(\$) — ftanmdm 2 ln |cosm| 2 ln(c0sm) on 0 < m < Hence , , , cot 2: cos ac cosgm Sett1ng yi 2 s1n ac, 3/2 2 cos m, then M; — yzy’l 2 2 s1n2m — cosgac 2 —1. Thus u’l 2 ———f— 2 sinm cot s‘nm coszm .\ and 1/2 2 ~%—— 2 — cos :L‘. Then u12/ _ dm 2 f(csc;r —— s1n m) dm 21n(csc;v — cot m) + cosz _ s1n m and U2(\$) 2 — sin ac. Thus yp(ac) 2 [cos x + ln(csc ac — cot sin’m + (— sin :L‘)(cos ac) and the general solution is y(;L‘) 2 01 sinm + C2 cosm + sin a; ln(csc :L‘ — cot 2:: —:c I z I “e e y1 = 6 7212 = 62 and lily/2 22/21/1 = 63 ~50Ui=(1+6_1)63z _ "1—4:; and 87:: _z I ex ex u1(\$) / 6—: d1: n( + e ) “’2 + e—z)63:c 63:0 + 622 SO u2(m) 2 / 63xieh dac 2 ln<e 6:1) 6” — ln(1 + e”) — 6”. Hence yp(;L‘) 2 ex ln(1 + e”) + 62\$[1H(1 + e71) — e‘z] and the general solution is y(;L‘) 2 [01 +ln(1+ e_\$)]ez + [02 —— e“ +ln(1+ e_\$)]621. - a: —2:: _ _ n s1ne e , y12e \$44226 2Iandy1yé—y2yi2— 3" Sou'1——-( )sgc 2exs1nezand _ei ' a: —a: Sine e . - UIQ 2 I )SE 2 —e” s1n 6’. Hence ul 2 fez s1n ezdm 2 —cos em and _e— u2(\$) 2 f —e2\$ sin ezdac 2 6’ cos em — sin 6’. Then yp(;L‘) = —e_\$ cos ex — 6—2I[sin ex — em cos 6’] and the general solution is y(;L‘) 2 (01 — cos cane—I + [C2 — sin ex + (2" cos axle—2:”. e” e" yl 2e_\$,y2 2ezandy1y’2—y2y’122.50u’1 =——,u’2= and 2x 2x em 6” yp(;r) 2 —e_1 — (130 + em dm. Hence the general solution is 2x 2x em _I 6—1 I y(;L‘)2 01— —d:z:e + 02+ dme. 2m 7 2x 721 —2.:c —21 A2: / I _ —4z I __ _6 me _ 1 . 21 y1 = e ,y2 2 me and mm — yzyl — e . Then ul .— \$36_4I —; so u1(\$) = as and —2z 72:: ~21 A21 '22: e e 1 1 6 me e , . . u 2 —— 2 —— so U2 as 2 ——. Thus as 2 — 2 and the eneral solutlon 1s 2 x3e‘41 x3 I I 2&3 ypI ) :L‘ 2:32 2x g y(ac) 2 eT2I[cl + mm + SECTION 18.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS ET SECTION 17.3 647 18.3 Applications of Second-Order Differential Equations ET 17.3 1. By Hooke’s Law k(0.6) 2 20 so k 2 1 CASIO 0 is the spring constant and the differential equation is 310” + %0m 2 0‘ mIS The general solution is m(t) 2 cl cos( t) + 02 sin(%t). But 0 2 :c(0) 2 cl and 1.2 2 :c’(0) 2 \$02, so the position of the mass after t seconds is :c(t) 2 0.36 sin(1—30t). . k(0.3) 2 24.3 or k 2 81 is the spring constant and the resulting initial-value problem is 4m” + 81x 2 0, z(0) 2 —0.5 (since compressed), z’(0’) 2 0. The general solution is 2 cl c0s(%t) + 02 sin(%t). But —0.2 2 93(0) 2 c1 and 0 2 arc/(0) 2 gag. Thus the position is given by \$(t) 2 —0.2 cos(4.5t). . k(0.5) 2 6 or k 2 12 is the spring constant, so the initial-value problem is 2w" —— 149vI + 12\$ 2 0, 10(0) 2 1, \$'(0) 2 O. The general solution is 2 ole—6t + age—t. But 1 2 MO) 2 01 —— 02 and 0 2 w’(0) 2 —6cl — 02. Thus the position is given by :c(t) 2 2%?“ + £6”. . (a) The differential equation is 3x” + 3010' + 12310 2 0 with (b) 0.15 general solution :c(t) 2 e_5t(cl cos 4t + 02 sin 4t). Then 0 2 z(0) 2 cl and 2 2 ac'(0) 2 402, so the position is 0 ~ 2 —0.05 given by m(t) 2 %e_5t sin 4t. . For critical damping we need 02 — 4mk 2 0 orm 2 02/(4k) = 142/(4‘ 12) = % kg- . For critical damping we need 02 2 4mk or c 2 2 mG 2 2 V3 ~ 123 2 6 V41. . We are given m 2 1, k 2 100, z(0) 2 —0.1 and \$,(0) 2 0, From (3), the differential equation is 2 ((117: + c (‘11—: + 100x 2 0 with auxiliary equation r2 + or + 100 2 0. If c 2 10, we have two complex roots r 2 —5 i 5 x/3i, so the motion is underdamped and the solution is x 2 e_5t [cl cos(5 x/3t) + 02 sin(5 Then —0.1 2 30(0) 2 01 and o : 5(0) = 5\/§c2 — 5C1 => 02 2 — 101/3, so I 2 65” [—0.1cos(5 x/3t) — ﬁg sin(5 . Ifc 2 15, we again have underdamping since the auxiliary equation has roots 7‘ 2 —1—25 i The general solution is an 2 e45”2 [cl cos(542ﬂt)n+ C2 sin(542ﬁt)] , SO —0.1 2 10(0) 2 c1 andO 2 Il(0) 2 \$02 —- 1—2501 => 02 2 —-103ﬁ. Thus an 2 6—15t/2[—0.1COS(542ﬂt) — losﬁ sin(542ﬂt)]. For c 2 20, we have equal roots 7‘1 2 r2 2 —10, so the oscillation is critically damped and the solution is an 2 (cl + 02t)e_10t. Then —0.1 2 10(0) 2 01 and 0 2 :c’(0) 2 1001 I 02 02 — 1, so an — ( 0.1 t)e_10t. Ifc 2 25 the auxiliary equation has roots 7‘1 2 —5, r2 2 —20, so we have overdamping and the solution is x 2 ole—5t + age—20”. Then —0.1 2 \$(0) 2 cl +02 andO 2 z’(0) 2 —501 — 2002 2 cl 2 —1—25 and 02 2 %, 548 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ETCHAPTER 17 so a: : —%e“5t + 3i0€~20t. Ifc = 30 we have roots 7‘ = 715 :E 5 J5, so the motion is overdamped and the solution is a: : cle(_15+5‘/g)t + €2€(_15_ 5‘5)‘. Then —0.1 : \$(0) = cl —I— 02 and 0 = 5(0) = (—15+5\/§) at + (—15 — 5V5) 02 :> Cl 2 75—3\/5 andc2 : —5i315’so 100 100 1: : (-5I03J5) e(—15+5\/5)t+ e(—15—5\/5)t’ _0~11 . , . . ._ . d22: dd: , 8. We are g1ven m = 1, c = 10, \$(0) = 0 and an (0) = 1. The differential equation ls —dt~2 —I— 10 a; —I— km 2 0 w1th auxilia e uation r2 —I— 107" +' k = 0. k = 10: the auxiliary e uation has roots 7" = —5 :I: x/ 15 so we have ry q ‘1 overdamping and the solution is z = (316(‘5 + m I’ + Cg€(_5 ‘ m )t. Entering the initial conditions gives cl — 2‘25 and02 — —2\}15, son: — Til—5 €(_5+\/ﬁ)t — ﬁeI—“ml‘. k = 20: r = —5:I: x/5and the solution is a: = 016(55 + ‘5 )t —I— €26(—5 _ ‘5 )t so again the motion is overdamped. The initial conditions give Cl 2 2—3/5: and C2 = —ﬁ, soar = ﬁe(_5+ﬁ)t — \$44” J5» k = 25: we have equal roots r1 : r2 = #5, so the motion is critically damped and the solution is a: : (c1 + Cgt)e“5t. The initial conditions give 01 = 0 and 02 = 1, so a: = the—5t. k = 30: r = —5 :I: so the motion is underdamped and the solution is at = 6"“ [cl cos(\/5t) + 02 sin(\/5t)]. The initial conditions give c1 = 0 and C2 = so a: = f 6—5‘ sin(\/5t). k = 40: r = —5 j: so we again have underdamping. The solution is m 2 6‘5‘ [01 cosh/1515) —I— 62 sin(\/ 1515)], and the initial conditions give cl = 0 and 62 = % Thus 15' a: : ﬁe’ﬁ sin(\/1_5t). —0.01 9. The differential equation is mat” —I— k1: 2 F0 cos wot and (do 75 w = w/ k/ m. Here the auxiliary equation is mr2 —I— k = 0 with roots :I: k/mz' = iwi so \$C(t) = 01 cos wt + 02 sinwt. Since wo 75 w, try 10. 11. 12. 13. 14. SECTION 18.3 APPLICATIONS OF SECOND—ORDER DIFFERENTIAL EQUATIONS ET SECTION 17.3 649 xp(t) = Acoswot —I— B sinwot. Then we need (m) (—wg) (A cos wot —I— Bsinwot) —I— k(Acosw0t -I- Bsinwot) : F0 coswot or A(k ~ mwg) 2 F0 and F . k . B(k — mwg) = 0. Hence B 2 0 and A : —FO———3 = 0 2 Slnce w2 = —. Thus the motion of the . k —— mwo m(o.22 — wo) m mass is iven b (t) — o t —I— ‘ t —I— F0 cos t g ya: —c1c so.) C2s1nw m(w2_wg) we As in Exercise 9, xc(t) : c1 cos wt —I— C2 sin wt. But the natural frequency of the system equals the frequency of the external force, so try 2:1,(15) : t(A cos out —I— B sin wt). Then we need m(2wB — w2At) coswt — m(2wA —I— szt) sinwt —I— kAt cos out —I— kBt sin out 2 F0 cos wt or 2W3 : F0 and —2mwA = 0 (noting —mw2A —I— kA : 0 and —mw2B —I— kB 2 0 since w2 2 19/771). Hence the general solution is \$(t) = C1 cos wt +:C2 sin cut —I— [Fat/(2mw)] sin wt. F0 —— cos t. Then m(o.22 — tug) MO From Equation 6, x(t) = f(t) —I— g(t) where f(t) = C1 cos out —I— C2 sin wt and g(t) = :4. If f0 is a rational number, then we can f is periodic, with period 2‘7", and if to ¢ we, g is periodic with period 0 say Mic = % => a = Z—“C: where a and bare non—zero integers. Then \$(t+a-27")=f(t+a~27") +g(t+a‘27") =f(t)+g( = N) +g(t+b- 3—3) = W) +g<t> = we) so \$(t) is periodic. (a) The graph ofa: 2 C16" —I— C2te" has a t-intercept when 016" —I— C2te" = 0 <=> 6” (C1 —I— C2t) = 0 <=> C1 = —C2t; Since t > 0, a: has a t-intercept if and only if C1 and C2 have opposite signs. ‘ <=> (b) Fort > 0, the graph of a: crosses the t—axis when 016’” —I— C26T2t : 0 <=> C26T2t — —C16T1 rlt e . _ C2 = —C1 r2! — —C1€(r1_r2)t. But 7‘1 > 7‘2 => 7‘1 w 7‘2 > 0 and SlnCCt > 0, €(T1 r2)tl> Thus 6 , |C2| = |C1| €(rlir2ﬂ > |C1I, and the graph ofa: can cross the t-axis only if |C2| > |c1|. Here the initial-value problem for the charge is Q” + 20Q’ —I— 500Q = 12, 62(0) 2 Q'(0) = 0. Then Qc(t) = e‘lot(c1 cos 20t —I— C2 sin 20t) and try Q1, (t) = A => 500A =12 or A = (2(0) Q’(t) = I(t) = €710t[(—1061 —I— 20C2) cos 20t —I— (—10C2 — 20C1) sin 20t] but 0 : Q’(0) : C1"—I— i and 3 _ — But 0 —- 125 The general solution is Q(t) : e’10‘(cl cos 20t + C2 sin 20t) —I— 125. = —10C1 —I— 20C2.‘Thus the charge is Q(t) z ~ﬁloe’10t(6 cos 20t + 3sin 20t) —I— % and the current is I(t) = eilmg) sin 20t. (a) Here the initialvvalue problem for the Charge is 2Q” —I— 24Q’ —I— 20062 : 12 with Q(0) : 0.001 and Q’(0) = 0. Then Qc(t) = 6—6t(61 cos 8t —I— C2 sin 8t) and try Qp(t) = A => A = 5—30 and the general solution is Q(t) = e‘6t(C1 cos 8t —I— C2 sin 8t) —I— 5—30. But 0.001 = Q(0) : C —I— 5—30 so C1 = —-0.059. Also Q'(t) = I(t) = 6‘6‘[(—6c1 —I— 8C2) cos 8t —I— (—GC2 — 8C1) sin 8t] and 0 = Q’(0) = —6c1 —I— 8C2 so 650 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS El' CHAPTER 17 C2 : —0.04425. Hence the charge is Q(t) = —e_6’(0.059 cos 8t + 0.04425 sin 8t) + % and the current is 1(t) = e'6t(0.7375) sin 8t. (b) 0.08 0.35 0 1.5 1.5 0 —0.05 charge, Q(t) ‘ current, [(2) = Q’(t) 15. As in Exercise 13, Qc(t) = 6—10t(61 cos 201: + C2 sin 20t) but E(t) : 12 sin 101: so try Qp(t) = A cos 1073 + B sin 10t. Substituting into the differential equation gives (7100A + 2003 + 500A) cos 1075 + (—1003 — 200A + 500B) siri 1075 = 12 Sin 1075 => 400A + 200B 2 0 and 400B — 200A 2 12. Thus A = —-zg—0, B = % and the general solution is Q(t) = 6’10t(C1 cos 2075 + C2 sin 2075) — % cos 1073 + %5 sin 1075. But 0 = 62(0) 2 C1 — % so C1 = 52—0. Also Q’(t) : 2—35 sin 102: + % cos 102: + e-1°t[(—10ci + 2062) cos 207: + (—10c2 — 20ci) sin 202:] and 0 = Q'(0) = 2—65 — 10C1 + 20C2 so C2 = —5—&). Hence the charge is given by Q(t) = 6—10tIZ—go cos 20s — 5—3“) sin 207:] * 2—go cos 1075 + % sin 10s. 16. (a) As in Exercise 14, Qc(t) = e_6t(C1 cos 823 + C2 sin 8t) but try Q1703) : A cos 1025 + Bsin 10t. Substituting into the differential equation gives (—200A + 2403 + 200A) cos 1075 + (—2003 — 240A + 200B) sin 10t : 12 sin 1075, so B = 0 and A : —%. Hence, the general solution is Q(t) = e_6t(cl cos 8t + C2 sin 8t) — 2—10 cos 1075. But 0.001 2 62(0) 2 01 — 2—10,Q'(t) = e‘6t[(—601 + 8C2)cos 8t + (—602 — 8C1) sin 8t] — % sin 10t and 0 : Q’(0) = —6C1 + 802, so C1 = 0.051 and C2 = 0.03825. Thus the charge is given by Q(t) = e*6*(0.051 cos 8t + 0.03825 sin 8t) — % cos 107:. (b) 0.06 —0.06 17. \$(t) : Acos(wt + 6) <=> \$(t) = AIcoswt cos6 - sinwt sin 6] <=> x(t) = A(CK1 coswt + 6Z2 sinwt) where c0s6 : 01/14 and sin6 = —C2/A <=> x(t) = C1 coswt + C2 sinwt. (Note that cos2 6 + sin2 6 = 1 => of + 0% = A2.) SECTION 18.4 SERIES SOLUTIONS ET SECTION 17.4 3 651 . . . . . , (120 18. (a) We approximate sm0 by 0 and, w1th L : 1 and g = 9.8, the dlfferentlal equatlon becomes w + 9.80 = 0. The auxiliary equation is 72 + 9.8 = 0 => 7 = :tx/9.81', so the general solution is 9(1) 2 cl cosh/9.81) + 62 sin(\/9.81). Then 0.2 = 0(0) 2 c1 and 1 = 0'(0) = V9.89 => 02 : ale (D so the equation is 0(1) 2 0.2 cosh/9.81) + sin(\/ 9.8 t (b) 0’(t) : —0.2 v9.8 sin(\/9.8t) + cosh/9.81.) = 0 or tan(\/9.8t) 2 i so the critical numbers are 9. ’ t = ﬁ tan—1 (¢%) + k 7r (71 any integer). The maximum angle from the vertical is 0(ﬁ tan’1 N 0.377 radians (or about 21.7°). (C) From part (b), the critical numbers of 0(1) are spaced ﬁ apart, and the time between successive maximum ‘rr values is 2 (m). Thus the period of the pendulum is 3% :e 2.007 seconds. (d)0(t):0 :> 0.2cos(Mt)+ﬁsin(x/9.—8t)=0 :> tan(¢9§t)=—0.2m :> t = ﬁ [tan—1(v02 V9.8) + 7r] m 0.825 seconds. (e) 0’(0.825) N —1.180 rad/s. 18.4 Series Solutions ET 17.4 00 co 1. Let y(\$) = E 07.33". Then y'(ar) = Z ncnm"'1 and the given equation, 3/ — y = 0. becomes TL:0 71:1 2 mourn—1 — Z cnm" = 0. Replacing 71 by n + 1 in the ﬁrst sum gives 2 (n + 1)cn+1a:" — 2 cum" 2 0. 71:0 71:0 71:1 71:0 1 00 so 2 [(71 + 1)cn+1 — cub?" : 0. Equating coefﬁcients gives (71 + 1)cn+1 ~ on : 0, so the recursion relation is ‘ 71:0 n 1 1 1 1 1 cu“:TL:1,7L:0,1,2,....Thenc1 c0,02 501 620,03 E02 §~§co:%,C4:ZC3:%,and in general, c" : :01. Thus, the solution is 71. oo 00 C 00 \$7, 71 0 31(17)::cnm 227\$"2cozgzcoel 71:0 71:0 71:0 00 00 co 2. Lem/(m) : 2 Cum". Then 3/ : 13y => 3/ — my 2 0 => 2 ncnarn‘l — x Z 0711:" = Oor 71:0 71:1 71:0 00 so ‘ Z nonmn‘l — Z cum"+1 : 0. Replacing 71 with n + 1 in the ﬁrst sum and n with n — 1 in the second' 71: 1 71:0 00 OO 00 00 gives 2 (n +1)cn+1\$" — Z 671:11‘" : Oor 01 + Z (n + 1) cn+1\$" — Z cn_1\$" : 0. Thus, 71:0 71:1 71:1 71:1 652 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ET CHAPTER 17 00 C1 + Z [(71 + 1)cn+1 —— cn_1] m" : 0. Equating coefﬁcients gives (:1 : 0 and (n + 1) cn+1 — c7121 : 0. Thus, 71 2 1 the recursion relation is cn+1 : (5:11, 71 : 17 2, . . . . But c1 = 0, so C3 = 0 and C5 : 0 and in general C2,,“ = 0. 71 co CZ co co C4 co ‘_ c0 . ~ co AISO,C2 — 3,C4 — 1 — — 22.2!,c6 — 6 — 6.4.2 — 23 .3! and1ngenera1c2n — 2n _n!.Thus,the solution is 00 co co co 2 71 C 1‘ 2 2 y(x)zzcnx :Zc2nx2n: 2nonlm2nZCOZI :coez/z 71 = O n : O 71 : 71 : 0 oo oo ‘ 00 Assuming y(x) : 2 Cum", we have y’(m) : Z ncn\$"_1 = Z (n + 1)cn+1\$" and 11:0 71: 71:0 00 co —\$2y : — cum"+2 : — 2 (27,417". Hence, the equation y”: \$27; becomes 71:0 71:2 (71 +1)cn+1\$" — Z cn_2x" : 0 or Cl + 2C2x + E +1)cn+1 — cn_2I Iv" : 0. Equating coefﬁcients 0 11:2 71:2 gives c1 : 02 : 0 and cn+1 : 61:21 for 71 = 27 3, . . .. But c1 : 0, so C4 : 0 and C7 : 0 and in general 71 can+1 = 0. Similarly c2 2 0 so can+2 = 0. Finally ca _ as F 6—; _ 60.03 _ 3203!, _ C6 _ Co _ C0 _ CO . . Cg — 3 — 9.6.3 — 33_3!,...,and03n — 3n.n!.Thus,thesolutionis 00 oo 00 oo 3" oo 3 n _ n _ 3n _ CO 371 _ \$ __ (x /3) _ 23/3 y<m>— 2w — 2w — z —co: —co: 71:0 71:0 71:0 71:0 71:0 Let y : 20:0 cum" => 3/ : 22:1 ncnx"’1 : 2020(71 + 1)cn+1x". Then the differential equation becomes (x — 3) 22:00; + 1)cn+1z" + 2 22°20 cum" 2 0 _=> 2:001 + 1)Cn+117"+1 — 3 2:001 + 1)Cn+1\$" + 2 226:0 CM" 2 0 => 22°21 “W” * 2210 3(n + 1)Cn+1\$" + 2210 2w" = 0 : \$310K” + 2)0n — 3(n + 1)cn+1l 96" = 0 (since 22:1 ncnm" : 22:0 nouns") .' Equating coefﬁcients gives (71 + 2)cn — 3(n + 1)cn+1 : 0, thus the . . . 2 n 2 3 3 recurs10nrelatlon1scn+1 : %T_:—T)1€T,n : 0,1,2,...‘ Then (:1 : %, 02 : r621) : 302—0, 4 4 5 . . . 63 = ﬁg) = 3%), C4 — % — g, and In general, cn : (—71%)2. Thus the solutlon 1s y(m):ic ac":coi "+117" Notethatcoi n+1m"= _960_f0r <3 71:0 n 71:0 3” I 71:0 3" —\$)2 ' SECTION 18.4 SERIES SOLUTIONS ET SECTION 17.4 653 5. Let y = 2‘” Cum" :> y’ 2 22:1 11(:nav"_1 and y” = 22:0(11 + 2)(11 + 1)cn+2w". The 11:0 differential equation becomes 2::001 + 2)(11 + l)cn+2x" + 21 22°21 ncnm"_1 + 22°20 cnm" = 0 or 2°C) [(11 + 2)(11 + 1)cn+2 + 11c" + cub" (since 22:1 ncnx" : 22:0 110nm"). Equating coefﬁcients gives 11:0 . “ 1 11 11 (11 + 2)(n + 1)cn+2 + (11 + 1)::n = 0, thus the recursion relation is cn+2 = mfg—Twig = — n12, n = 07 1,2, . . . . Then the even coefﬁcients are givenby C2 = —%0, C4 : —% = %, c6 = —%4 = —2 6, . i n '00 _ (‘1)"00 .1 . _ Ci _ C3 _ Cl and In general, 02,, — (*1) : The odd coeiﬁments are as — ~E, cs - —g — c = —C—5 — — Cl and in eneral c —( 1)" Cl — (_2)nn!c1 The solution is 7 7‘ 3-5-7’ g ’2"+1 3-5-7-~--(211+1) (2n+1)!‘ _ m (—1.)n 211 00 »(_2)n n! 211+1 we) —coZ M 96 “127217117195 - 11:0 11:0 00 00 DO . Let = 2 Cum". Then y"(m) : 2 11(11 — 1)cnz"‘2 = Z (11 + 2)(11 +1)cn+2m". Hence, the equation 11:0 11:2 2 11 0 CO 00 00 y” = ybecomes Z (11 + 2)(11 +1)cn+2x" — Z cum" 2 0 or Z [(11 + 2)(11 +1)cn+2 ~ cn]z" = 0. So the 11:0 11:0 n:0 recursion relation is cn+2 = m, 11 : 0,1,.... Given c0 and C1, 02 = 20%, C4 2 : %, c-‘J—Ic‘1 —C—0 c A CO andc Cl C —cs ——cl —C—lc——c5 :c—l 6— 6-5 _ 6!’"" QT“ (211)! 3 _ 3.2’ 5“ 5-4 _ 5.4.3.2 _ 5!’ 7— 7-6 _ 62n+1 = Thus, the solution is 00 oo 2 oo 2 +1 00 \$211 0° \$211+1 11(1) : ":0 cum : §0C2n\$ + "gownﬂw = 60 "go (2”)! + Cl "20 —‘—(2n + 1)! The solution can be written as : c0 cosh x + c1 sinhm 61 + CAaE 6z — 67\$ ,Co + C1 m Co —- C1 ,I lory(m) —c0 2 +C1 2 — 2 e + 2 e . . Let y = 2:0 cum". Then y” : 22:0 11(11 ~ 1) cum"_2, my’ : 22°20 ncnm" and (m2 + 1) y” 2 22:0 11 (11 — 1) cnx" + 22°20 (11 + 2) (11 + 1) cn+2m". The differential equation becomes . . —1 n Zfzo[(11+2)(11+1)cn+2+[11(11—1)+n—1]cn]m"=0.Therecursionrelat10n1scn+2=—%, 11 11—0 12 Givenc andc c 6—0 c 6—2 C“ c 3‘34 _( 1)2 3c” — 7 7 a'H- 0 1: 2 27 V4 4 22.2!) 6— 23.3!au'a , 1-3~~~~~(211—3)co _1 (211—3)!c0 _1 (211—3)!c0 _ n: 1" 1 v —1 n ————: —1n —f c2 ( ) 2"11! ( ) 2"2”—211!(11—2)! ( ) 22n—211!(11—2)l or 11:2,3,....C3= (Jim :0 :> C2n+1 :Ofor11:1,2,....Thusthesolutionis 3 2 _ m °° (—1)"*1 (211—3)! 2" y<\$>-CO+CI\$+C°7+COZW\$ ' 11:2 654 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ET CHAPTER 17 DO 00 00 8. Assuming = Z cum", y”(m) = Z n(n — 1)cn:c"‘2 = Z (n + 2)(n + 1)cn+2m" and n:0 n=2 11:0 00 00 —my(m) : — Z cnmnll = — Z cnilm". The equation y” = my becomes n:0 n:1 oo oo 00 Z (n + 2)(n + 1)cn+2m" — 2 anlm" = 0 or 202 + Z [(71 + 2)(n + 1)cn+2 — cnv1]m" = 0. Equating 'n.=0 11:1 11:1 Cn—l —————forn=1,2,....Sincec =0, (n+2)(n+1) 2 coefﬁcients gives C2 = 0 and cn+2 = 63””=0forn=0,.1,2,....Givenco,C3= ﬁwezﬁ 2%,..q CO Givenc c Cl 04 C1 3n(3n—1)(3n—3)(3n—4)-----6-5-3-2' . 1"I_4-:3’ C1 The solution can be written as (3n+1)3n(3n—2)(3n—3)...7-6-4-3' g C311 — C3n+1 — oo °° 3n—2 3n—5 -----7-4.1 n 371—1 371—4 -----8-5-2 n 1 y(m)ZCOZ( CB3 +CIZ( CB3 + n20 00 w 00 DO 9. Lety(m) : Z cum". Then —my'(m) = —m E ncnm"’1 — Z ncnm" = * Z ncnm", n20 'n.=1 n21 n:0 II 00 y” = Z (n + 2)(n + 1)cn+2:c", and the equation y" — my’ - y : 0 becomes n :0 w E [(71 + 2)(n + 1)cn+2 —~ ncn — cn]m" : 0. Thus, the recursion relation is 'n. :0 71cn + CH _ 0n (n + 1) CH _ c for n = 0, 1, 2, . . . . One of the iven conditions is n+2)(n+1) (n+2)(n+1) n+2 g Cn+2 _ ( °° 1 y(0) = 1. Buty(0) =ch(0)" :c0+0+0+~-- =co,soco = 1. Hence,62 = 229 = 5,64 n=0 II | C4 1 1 6 2 ' 4 _ 6, ..., 62” = 2H7”. The other given condition is y’(0) = 0. But y’(0) = Z ncn(0)"_1 2 c1 + 0 + 0 + --- = 61, so cl = 0. By the recursion relation, C3 :. 0—31 = 0, C5 = 0, . . . , 'n.=1 02n+1 = 0 for n = 0, 1, 2, . . . A Thus, the solution to the initial—value problem is °° (ff/2)" oo oo 00 \$2.” 2 2 y(\$)=ZCn\$"=ZC2nm"=Z :2 :e“”/2 2"n! n! n20 I n=O ' n=0 n=0 +2 00 00 10. Assuming thaty(ac) = Z cum",we have \$21; = Z cum" and n: n=0 0 was): i n<n—1>cnm"—2 = i (n+4><n+3>cn+4w"+2 'n.=2 112—2 = 202 + 603m + Z (n + 4) (n + 3)cn+4m"+2 n :0 SECTION 18.4 SERIES SOLUTIONS ET SECTION 17.4 Thus, the equation y” + 3327; = 0 becomes 2C2 + 663\$ + Z + 4) (71 + 3)cn+4 + en] 93"” = 0, So 71=0 Cn —-————-——-—-—-, :0,1,2,.... (n+4)(n+3) n 02 = C3 = 0 and the recursion relation is cn+4 = — But c1 : y’(0) = 0 = C2 = C3 and by the recursion relation, C4n+1 = C4n+2 : C4n+3 : 0 for n = 07 17 2, . . .. Also, c0 = y(0) = 1, so C_4_ (-1)2 8;7‘8-7~4-3"“’ (—1)" 471(471 — 1)(471 — 4)(4n — Co 1 T3 ‘ = C4: ———Cg:— 4-3, Thus, the solution to the initial—value problem is 5).....4.3‘ 00 71”: DO 714n_—'1 I _1n 20\$ CO+n§OC4\$ +Z( )4n(4n—1)(471— 71:0 71:1 0° 471 4)(4n_5).....4.3 WE) Assuming that y(a:) = Z c711", we have 93y = :1 Z cnwn 71:0 11. co 1 : E Cn\$n+ ‘ 71:0 71:0 00 m 1 — 1 \$23; 2:132 5 ncnatn 1: é ncnmn+ , 71:1 71:0 00 2 71(71 — 1)cn:v"_2 Z (n + 3)(n + 2)cn+3\$"+1 71:2 71:71 [replace n with n + 3] ry/l(\$) 00 202 + Z (n + 3) (n + 2)cn+3\$"+1, 71:0 [(n + + 2)cn+3 + ncn + C77,] \$n+1 : m and the equation y" + 3327/ + 93y = 0 becomes 262 + Z 71 = 0 —ncn — an (n + 1)cn (n+3)(n+2) (n+3)(n+2)’ But c0 = y(0) = 0 = 02 and by the recursion relation, 63,; = C3n+2 = 0 for n = 0, 1, 2, . .. . Also, c1 : y’(0) = 1, so n:0,1,2,.... So 62 : 0 and the recursion relation is cn+3 = 22252 7! 201 2 c 504 4-3‘ 43’7 7-6w ' 2252~---(3n—1)2 n : —1 n—._— 03 +1 ( ) (371+1)! WU) = 2 Cum" 71:0 2-5 C4— (,1)27'6~4.3:(1) . Thus, the solution is n 2252 - - -- - (3n —- D2373"+1 (371+ 1)! 00 DO 12. (a) Let y(a:) = Z cnarn. Then \$2y”(a:) = 2 71(71 — Dena?" = Z (n + 2)(n + 1)cn+2\$"+2, 71:0 71:2 71:0 00 00 n + 2 cn+233n+2 : clay + n + 2 cn+2\$"+2, and the e uation ‘1 71:0 00 (By/(:17) : Z ncnatn : 71:1 71=—1 21/ a: y + aty' + 12y = 0 becomes C13: + Z + 2)(n + 1) + (n + 2)]cn+2 + cn}a:"+2 = 0. So c1: 0 71:0 C71 —,n=0,1,2,....Butc = ’0 =OSOCn :Ofor (n+2)2 1 y() 2+1 and the recursion relation is cn+2 = — 655 656 CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ET CHAPTER 17 _ _ _ __ 1 Cg 2 1 2 1 n-0,1,2,....A1so,c0_y(0)_1,sOC2_ 22,04— 54 1) 4222—( 1) 24(2!)2. _ C4 _ 3 1 _ n 1 . . 06,— —6—2 _ (—1) 26602, . . . , 02" — (~1) The solution is oo 00 I211 W) ZED” 2.20M 22%)? (b)-The Taylor polynomials To to T12 are shown in the graph. Because T10 and T12 are close together throughout the interval [—57 5], it is reasonable to assume that T12 is a good approximation to the Bessel function on that interval. 18 Review ET17 ———————————— CONCEPT CHECK —————————-————— 1. (a) ay" + by’ + cy = 0 where a, b, and c are constants. (b)ar2+br+c=0 (c) If the auxiliary equation has two distinct real roots T1 and T2, the solution is y = 01 ele + 026T”. If the roots are real and equal, the solution is y 2 C16” + 02176” where r is the common root. If the roots are complex, we can write r1 2 (1 +23 and r2 = a — i5, and the solution is y = e”(cl cos Bx + C2 sin 51;). 2. (a) An initial—value problem consists of ﬁnding a solution y of a second-order differential equation that also satisﬁes given conditions y(zo) 2 yo and y’(mo) = y1, where yo and y1 are constants. (b) A boundary-value problem consists of ﬁnding a solution y of a second-order differential equation that also satisﬁes given boundary conditions y(:no) : yo and y(x1) = y1. 7 I 3. (a) ay” + by' + cy = G where a, b, and c are constants and G is a continuous function. (b) The complementary equation is the related homogeneous equation ay” + by’ —I— cy = 0. If we ﬁnd the general solution yC of the complementary equation and yp is any particular solution of the original differential equation, then the general solution of the original differential equation is y(:t) = yp(x) + (c) See Examples 1—5 and the associated discussion in Section 18.2 [ET 17.2]. (d) See the discussion on pages 1188—1190 [ET 1152-1154]. 4. Second-order linear differential equations can be used to describe the motion of a vibrating spring or to analyze an electric circuit; see the discussion in Section 18.3 [ET 17.3]. 5. See Example 1 and the preceding discussion in Section 18.4 [ET 17.4]. 1. 2. 3. 4. 4 10. . The auxiliary equation is r2 — 2r —— 15 2 0 => (7“ — 5)(r + 3) 2 0 => 1“ CHAPTER 18 REVIEW ETCHAPTER 17 657 TRUE-FALSE OUIZ True. See Theorem 18.1.3 [ET 17.1.3]. False. The differential equation is not homogeneous. True. cosh m and sinh m are linearly independent solutions of this linear homogeneous equation. False. y = Aer is a solution of the complementary equation, so we have to take yp(m) : Axel. EXERCISES 5, 1“ — —3. Then the general solution is y 2 C1651 + C2e’3m. The auxiliary equation is r2 —— 4r + 13 = 0 => 1“ = —2 :l: 32', so 3/ = 6—21(C1 cos 31 + C2 sin 3m). The auxiliary equation is r2 —— 3 = 0 => 1“ = 213/321 Then the general solution is y = C1 cos(\/3\$) + C2 sin(\/3 The auxiliary equation is 41“2 + 4r + 1 = 0 => (21“ + 1)2 = 0 => 1“ = —%, so the general solution is y = Clei‘m/2 + C2\$€_I/2. r2 — 4r + 5 = 0 => 1“ = 2 :l: i, so yC 2 621(C1 cosm + C2 sinm). Try yp = Aeh => 31;: 2A62\$ 2a: and y; = 414621. Substitution into the differential equation gives 4Ae2m — 8A6“ + 514621 = 6 => A = 1 andthe general solution is = €2I(01 cos x + C2 sin an) + e21. r2+r—2:0 => r: 1,1“: —2andyc(m) =01€I+C26721.Tryyp(1') =Am2+Bm+C => y; : 2A1 + B and y; = 2A. Substitution gives 2A + 2A\$ + B — 2A12 — 2B\$ — 2C 2 \$2 => A = B = —%, C : —% so the general solution is 2 C161 + (126—22 — £162 — %:c — %. r2—2r+1='0 => r=1andyc(r):cleI+C2reI.Tryyp(r)=(A\$+B)cosm+(C\$+D)sinr => yg:(C—Am—B)sinm+(A+C\$+D)cosmandyg=(2C—B—A\$)cosr+(~2AvD—C\$)sin\$. Substitution gives (—201 + 20 — 2A — 2D) cos 1' + (2Am — 2A + 2B — 2C) sinm = mcosm => A = 0, 1 1 —c0sr §(\$I 2 1)sinm.v B = C = D = —l. The general solution is — C161 : C2mez 2 z r2 + 4 = 0 => 1“ = 21:22" and yc(m) = C1 cos 2m + C2 sin 2m. Try yp(r) = Am cos 2m + Br sin 2m so that no term of yp is a solution of the complementary equation. Then y; = (A + 2B1) cos 2m + (B — 2A1) sin 2m and y; : (4B — 4Am) cos 2m + (—4A — 43m) sin 21. Substitution gives 43 cos 2m v 4A sin 2m = sin 2m => A = —i and B : 0. The general solution is = Cl cos 2m + C2 sin 2m — 711m cos 2m. 2 —2m r — r — 6 = 0 => +C2631. Fory” —y’ — 6y 2 1, try ypl : A. Then W T = —2, r = 3 and yc(m) 2 C16 yz',1 = 3/51 = 0 and substitution into the differential equation gives A = —%. For 3/” A 3/ ~ 63/ = e‘” t (B — 2Bm)e'2m and => B = —%. The general solution then is yp2(m) 2 Bare—21 (since 3/ 2 Be'21 satisﬁes the complementary equation). Then 34,2 y; 2 (43m — 4B)e_21, and substitution gives —SBe_21 = 6‘21 v21 \$6 y<w> = + + 21.. (r) + yam = + —é % '«m, and Using variation of parameters. yc(a:) 2 c1 cos x + C2 sinm. u’l — csc m sinm = —1 => ul CSC 1' COS \$ 2 cotr :> 112(1) 2 1n |sinr| => yp = —rcosr —+— sinrln|sinr|. The solution is 1' 11W) = (Cl — I) COSI-l- (C2 -+- ln|sinr|)sinr. 11. 12. 13. 14. 15. 16. 17. CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ET CHAPTER 17 The auxiliary equation is r2 + 6r 2 0 and the general solution is = C1 + 626—63: 2 k1 + [we—NIT“. But 3 2 y(1) = k1 + kg and 12 = y’(1) = —6k2. Thus k2 = —2, k1 = 5 and the solution is (41(16): 5 — 2e_6(1‘1). The auxiliary equation is r2 — 6r + 25 = 0 and the general solution is y(\$) : 631(61 cos 4:5 + 62 sin 4:8). But 2 2 y(0) = C1 and 1 = y'(0) = 3C1 + 4C2. Thus the solution is 2 (231(2 cos 4x — gsin 42:). The auxiliary equation is r2 — ST + 4 = 0 and the general solution is = Clea” + C2641. But : (64:0 _ er). 0 y(0) : c1 + 62 and 1 = y’(0) 2 c1 + 4C2, so the solution is y(a:) % yc(a:) = C1 cos(a:/3) + C2 sin(:c/3). For 9y" + y = 32:, try yp1 2 Ax + B. Then yp1 = 3:5. For Ae-I. Then 9A6“I + Ate—7” ﬁe”. Thus the general solution 9y” + y = 6", try ypz (a?) e" or 311720;) is = C1 cos(J:/3) + C2 sin(1:/3)+ 32: + l—loe’x. But 1 = y(0) = C1 + 1—16 and 2 : y'(0) : %C2 + 3 — %, so c1 = 1% and 62 ; —f—(7). Hence the solution is y(ac) = 1i0[9 cos(a:/3) — 27sin(a:/3)] + 3x + ﬁe”. Let = 2:10 cum". Then y" 2 22:0 n(n —1)cnz'"_2 : 2:1001 + 2)(n +1)cn+2a:" and the differential equation becomes 22°20 [(n + 2)(n + 1)cn+2 + (n + 1)cn]\$" = 0. Thus the recursion relation is Cn+2 : —cn/(n+2)forn=0,1,2,....ButCO =y(0) =0,socr_;n =0forn=0,1,2,.... Also 1 (—1)'-’ (—1)3 (—1)3233! (—1)"2"n! 2,021, :——7 = ’ : :————, ’ n :—-———f Cl “I 5063 365 3.5 67 35-7 7! 0" +1 (2n+1)! or n = 0 1 2 Thus the solution to theinitial-value problemis = E00 c 2:" = E00 *(_1)n 2n n! 2’”? ” ":0 " ":0 (2n-I—1)! " Let : 22°20 cum". Then y" 2 22°20 n (n — 1) C712?”2 2:001 + 2)(n +1)Cn+22:n and the differential equation becomes 23:0 [(n + 2)(n + 1)cn+2 — (n + 2)cn]a:n : 0. Thus the recursion relation is Cn+2 = “:1 form: 0,1,2,....GivencoandC1,wehaveC2 : C—10,C4 : 632 = 1C—03, C _ C_4 _ CO C CO —c THITFIII Similar] c — ﬂ 6— ‘1-3-5""’ 2n—1-3-5~---(2n—1)_ 0 (2n—1)!' y 3— 2’ C3 C1 65 Cl C1 CI . . C5 4 2_4,C7 6 2.4.6, 7C2 +1 2.4.6.Hu2n 2n”! usthegeneralsolution1s 0° 0° 2n—1(n _1)!\$2n 0° - \$2n+1 0° \$2n+1 co (;\$2)n 2 2 n n Z —‘—" —— . B t = 2 I (c /2 2C a: C0+Cog (2n—1)! +6; 2"n! u 1;] 2"nI 2:; n! \$6 ’ 00. n-l 2n ' — M w soy(:c) —C1:ce +c0+Conz=:1 (2n_1)! Here the initial-value problem is 2Q" + 40QI + 400Q = 12, Q (0) = 0.01, Q’(0) : 0!. Then Qc(t) = e‘mt(c1 cos 10t + C2 sin 10t) and we try Qp(t) 2 AI. Thus the general solution is Q(t) : e'l‘”(o1 c0810t + Cg Sin 1025) + ﬁL But 0.01 = Q’(0) = C1 + 0.03 and 0 = Q”(0) = —IOC1 +10o2, so C1 = —0.02 = C2!. Hence the charge is given by Q(t) = —0.02e’10t(cos 10t + sin 1025) + 0.03. CHAPTER 18 REVIEW ETCHAPTER 17 659 18. By Hooke’s Law the spring constant is k = 64 and the initial-value problem is 22:” + 1621’ + 642: = 0, 27(0) 2 0, at’(0) = 2.4. Thus the general solution is x(t) = e_4t(c1 cos 4t + C2 sin 4t). But 0 2 27(0) 2 c1 and 2.4 = :L"(0) = —4c1 + 4C2 => c1 = 0, C2 = 0.6. Thus the position of the mass is given by :L'(t) : 0.687“ sin 4t. 19. (a) Since we are assuming that the earth is a solid sphere of uniform density, we can calculate the density p as mass of earth M follows: p = m ,: gsz . If V, is the volume of the. portion of the earth which lies within a . M 3 GM. M dlstance r of the center, then VT : gwrs and M, = [M = R: . Thus Fr 2 — T2 m = — GRsmr. (b) The particle is acted upon by a varying gravitational force during its motion. By Newton’s Second Law of d2y '_ _ am Motion, m W — Fy = R3 y, so y”(t) = —k2y (t) where k2 = At the surface, —mg 2 FR = v-Ggim, sog = Therefore k2 : (c) The differential equation y” + k2y = 0 has auxiliary equation 7‘2 + k2 = 0. (This is the r of Section 18.1 [ET 17.1], not the r measuring distance from the earth’s center.) The roots of the auxiliary equation are iik, so by (l l) in Section 18.1 [ET 17.1], the general solution of our differential equation for t is = C1 cos kt + C2 sin kt. It follows that y'(t) = —c1k sin kt + Cgk cos kt. Now y(0) = R and y’(0) = 0, so c1 = R and 02k = 0. Thus y(t) : Rcos kt and y’(t) : —kR sin kt. This is simple harmonic motion (see Section [8.3 [ET 173]) with amplitude R, frequency k, and phase angle 0. The period is T : 27r / k. R z 39601111 2 3960- 5280 ft and g = 32 ft/s2, so k = \/g/_R % 1.24 X 1043*1 and T : 27r/k % 5079 s w 85 min. (d) y(t) = 0 <=> cos kt = 0 <=> kt = g + 7th for some integer 71 => I y’(t) = —kR sin(12'- + 7m) 2 ikR. Thus the particle passes through the center of the earth with speed kR \$3 4.899 mi/s % 17,600 mi/h. Blank Page ...
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Chapter 18 - ET 17 - 18 CI SECOND-ORDER DIFFERENTIAL...

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