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# 4.2 - In Exercises 4.2 Reduction of Order Exercises 4.2...

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Unformatted text preview: In, Exercises 4.2 Reduction of Order Exercises 4.2 Problems 1-8 we use reduction of order to ﬁnd a second solution. In Problems 9-16 we use formula {5) from the text. 1. 4. Deﬁne y = u(:r)r32‘” so 3;” = Quegx + u’eQx, y” 2 6231.5” + 4e2xu’ + 4623155, and y” — 43;! + 43; = (32%” = 0. Therefore a” = 0 and a 2 C133 + 62. Taking c1 = 1 and c2 2 0 we see that a second solution is _ 23': yz — 338 . Deﬁne y = retype—x so .l' H 29‘ = (1 — rye—ma + rte—“’u’, y = are—mu" + 2(1— tie—fa’ — (2 u riﬂe—mu: and 2 yr; + 2y! + y : e—\$(xuﬂ + 2n!) : 0 Or “I” + E u: = If is = a” we obtain the linear ﬁrst-order equation to!" + — w = 0 which has the integrating factor 117 e2fdx/‘r = 5.522. Now at (£3: 2 [22210] = 0 gives :1: w = c. . . 1 Therefore to : u’ = c 3:2 and u z 01 :r. A second solution is 2 = — 9' Deﬁne y 2 am) cos 4:5 so I I! = n” cos 43: — 815’ sin 4:1: — lﬁu cos 4.9: y = —4u sin 4x + u' cos 41:, y and g y” + 163; = (cos 4m)a” — 8(sin dime" : 0 or a” — 8(tan 41:)a’ : 0. If ’LU = u’ we obtain the linear ﬁrst-order equation w’ — 8(tan 42:)w : 0 which has the integrating factor e‘gl ta” 433‘” = cos2 458. Now d d— (cos2 Aisha] = 0 gives (cos2 4:12)w = c. 11," Therefore in = a” = csec2 4:8 and a 2 c1 tan 43:. A second solution is y; = tan 4:1: cos 4:: = sin 4x. Deﬁne y = a(:r) sin 31: so y’ = 3a cos 33c + u' sin 311:, y” = a" sin 33: + 61:,” cos 3:1: — 9a sin 3x} 133 Exercises 4.2 Reduction of Order and y” + 9y : (sin 3:5)a” + 6(cos 3:131!I : U or a” + 6(cot 3:r:)a‘r = 0. If a; = u’ we obtain the linear ﬁrst-order equation a)" + 6(cot 31101:; = 0 which has the integrating 6f mt 33d": = sin2 Now (1" d—[(sin2 3m)w] = 0 gives (sin2 = c. :1: Therefore in = a’ = ccse2 3:1: and a = (:1 cot 3x. A second solution is y2 = cot 3:1: sin 3:1: : cos factor (3 5. Deﬁne y = u(:z:) coshx so 3,!" = a sinh :I; + “u! cosh 3:, y” = a” cosh I); + 2n" sinh m + a cosh :r: 5 and y” — y : (cosh 3:)u” + 2(sinh ﬂu" : 0 or a" + 2(tanh .13)-u,' = 0. If is = at" we obtain the linear ﬁrst—order equation it?" + 2(tanh33)’w = 0 which has the integrating 2] tanhxdx = cosh2 51:. Now of d— (cosh2 ache] = 0 gives (cosh2 :s)w = c. .1: Therefore to = u’ = csechgx and u = ctanhx. A second solution is yg : tanhrrcoshm = sinhzr. factor 8 6. Deﬁne y : u(:c)e5‘f so y’ = 5853’s + (35\$?LI, “y” = (355811,” + 10e5xuF + 25653711, and y” — 253; = e5m(u”+10u’) = 0 or a” + 10's" 2 0. If is = “u' we obtain the linear ﬁrstborder equation “to! + 1010 = 0 which has the integrating factor 610 f d9: 10:): :6 .Now of da: and a = ole—103”. A second solution is y2 = 571050953? : e 10 [ema’w] = 0 gives 6 33w : c. —lU.:I: —5-T_ Therefore to = u’ 2 cs 7. Deﬁne y = u(x)e2x/3 so 2 , 4 4 ., r: _P2\$/3u + can/3“: yr; : 822:3?er + _B2x/3ur + _€2x/3u 3" 3" = 3 9' and 93;" — 12y’ + 4y : gels/3a” : 0. Therefore a” = 0 and u 2 (31:1: + c2. Taking; c1 = 1 and c2 = 0 we see that a. second solution is yz __, main/3 8. Deﬁne y = u(:c)e\$f3 so 2 yr ___ gear/3a + era/3m!1 yr: 2 623/3“!!! + gem/Bur + éexf3u 134 Exercises 4.2 Reduction of Order and 63;” + y’ - y : (BI/3(6u” + 5n’) = 0 or u” + 72-13 = 0. If m = u” we obtain the linear ﬁrst-order equation 10" + gm = 0 which has the integrating factor 45:6) fer-r = 85:0{6' NOW d % [853/610] = 0 gives epic/aw 2 c. and u = C18_5\$/6. A second solution is yg = 6—5x/6e‘f/3 = eff/2. . Identifying P(:1:) = ~7/3: we have -—f(—7f1:)da: 92 =x4/8Tdm=\$4/%dx=x4lnlx|. Therefore '10 I u' = ce"5x/5 A second solution is y2 = 234111 . Identifying P(\$) = 2/1: we have —f(2/:c)d:1: 1 952 = 5132/6— da: 2 :32/3:_6d:1: = —g:c_3. \$41 A second solution is y2 2 53‘3. . Identifying Pm} : 1/3: we have 6—fdw/I d2: 1 : 1 ——v——‘ I‘ 2 : ——-—- = — I m nm/ (Inst)2 dip him/22(1an In\$( Inst) 1 A second solution is W = 1. . Identifying P(:I:) : 0 we have yg = xlﬂlnx/ A second solution is yg : \$132. 6— f0dx 33(11133)2 dz = x1/2 lnar: = —m’1/2. Ins: . Identifying P(x) = —1/ac we have B—I—dx/cc a : g — = - —3 m xqmun sin2(1n\$) d1: msmanﬂfﬂ sin2(1nx) d3: 2 l = :rsin(ln :12) / weft—Hm) da: 2 [11: sin(lna:)] [— cot(ln:c)] = —:1:cos(lnx). A second solution is yg = :1: cos(1nm). ' . Identifying P(x) = —3/ac we have 2 G—I—de/x 2 x3 = .J . i I , j—d = I ]— yg T cos(n3’) \$4 00820113) 3: x cos(nx) 134008201127) dz: 2 l = x2 cos(ln f w (£3: = \$2 cos(ln :L’) tanﬂn 1:) = :52 sin(ln 135 Exercises 4.2 Reduction of Order A second solution is yg = 2:2 15. Identifying P(3:) : 2(1 + e)/ (1 — 2:1: — 3:?) we have sin(ln €- 2(1+z)d\$/(1—2:E—\$2) eln(1—2\$—\$2) = 1/4“; [the 3’2 (27+ ) (:i:+1)2 dr (“—1) (:r:+1)2 T 1 — 2:1: — m2 2 =.,1—~d: 1/ —1dt, (r+) (x+1)2 a: (32+) [(\$+1)2 ]’r 2 2 —(\$+1)[ \$+1 x]— 2 cs 11:. A second solution is 93 = .122 + :r. + 2. 5- 16. Identifying 13(3) 2 —2:r:/ (1 — :32) we have i - 1 1 1 ' 92 = /8_f_2\$d\$/(1_Iz)d\$ =fe_l“(1_fld:c = 1— m2 d3: = £111 1 A second solution is ’92 = 1n |(1 + :r)/(1— 17. Deﬁne y z Merle—2‘" so yr : _2ue—2:r: + 2Life—2:33: yr; : “Ire—2x _ 4ure—2zr. + glue—~23 and y” - 4y = 64%” — die—2315’: 0 or u” — 4a’ = 0. If 11! : u’ we obtain the linear ﬁrst-order equation 10" — 41:} = 0 which has the integrating factor 6—4Id5‘" = 6—41”. Now at —[e_4xw] = 0 gives eﬂéxw = 0. dz Therefore in = u," z (3841: and u : c1843. A second solution is yg : 63—23849: : 62‘”. We see by observation that a particular solution is yp = —1/2. The general solution is 1 y = cue—2”" + age2m — E. 18. Deﬁne y : u(:r) - 1 so y! = u}, y” = u?! and y” +y.’ : u” + u! : If to = of we obtain the linear ﬁrstnorder equation 10' + w = 1 which has the integrating factor e-ldx = (29‘. Now d %[emw] = ex gives 6qu = ex + 0. Therefore '11: = u” = 1 + 08“” and u = :1: + ole—x + 02. The general solution is y=u=x+cle_I+02. 136 19. Deﬁne y = -u(x)e\$ so 21. Exercises 4.2 Reduction of Order I! y’ = neg" + u’ex, y = u’lew + 2a'e3‘ + ueI and y” — 33," + 2y 2 exit" — emu! = 5639’". If a: = ’a’ we obtain the linear first-order equation w’ — m : 582\$ which has the integrating factor {false = e‘“? Now d _ . _. a— [e 27w] 2 56¢ glves 6 “I’m : 561: + (:1. :1: Therefore to = u’ = 562\$ + clef and u z éezx + clef + c2. The eneral solution is 2 g x 5 3x 2x a: yzue =—e +c1e +626. 2 . Deﬁne y = M10625“ so If y' = new + item, 3; = aﬂe‘l’ + 2u'ex + new and y” — 43," + 33; = emu” — emu' = :3. If to = u” we obtain the linear ﬁrst-order equation w" -— 2w : me” which has the integrating factor 8—f2dw : e—Qn NOW (3 _ _ . _ 1 _ 1 23w] : are 3‘” gives e 2%: = —— me 3:” — —e_3'17 + cl. d2: 3 9 Therefore in 2 u" = —1 zine—m — 16—3 + 01621: and u 2 l are”: + 3e_"3 + (326323 + (33. The ‘eneral :- 3 9 3 9 3 solution is :r 1 4 31: x y=ue’=§\$+§+e2e +036. (a) For m1 constant, let y1 = emlm. Then yi : mlemlx and y’i’ = mgemli". Substituting into the differgntial equation we obtain eyil + + cyl = om?emlx + bml emlm + Gem"m 2 em]\$(am§ + bml + c) : 0. m1 .1: Thus.F y1 = e will be a solution of the differential equation whenever (on? + mm + e = 0. Since a quadratic equation always has at least one real or complex root, the differential equation must have a solution of the form y1 = emlx. (b) “rice the differential equation in the form 5 c H J' -— — :0 y+ay+ay ; 137 Exercises 4.2 Reduction of Order and let yl : e'mlI be a solution. Then a second solution is given by :92 = 6’”le e—bar/c d1: 6277113 = emlmfe—(bfa+2m1)a¢dx —b/a + 2m1 mb/a, + 21011 emlme—(bfa+2m1)x (m1 75 43/200 B—(b/a+m1)x. it Thus, when m1 yé —b/2e, a second solution is given by yg = em” where m2 é —b/a — m1. When m1 = —b/2a a second solution is given by (c) The functions sinx=— sinhx = - (ex — e-o y2 = em” [dz = stem”. _ . 1 . . (em — 6—”) oosx : 5(6“ + 6—“) 1 :1: —SC cosha: = 503 + e ) are all expressible in terms of exponential functions. 22. We have ya 2 1 and 3;? = 0, so my? — my’i + y1 = 0 — :15 + :5 : 0 and y1(a:) = a: is a solution of the differentiai equation. Letting y = u(x)y1[\$) = mu(x) we get I "y = SEEKS?) + and y” = mu”(3:) + 2n'(\$). Then IL‘ "—3: ’+y = xQu”+2xu’—x2u’wxu+xu = x2u”— :52 —2:.c u’ = 0. Ifwe make the y y substitution w = u’, the linear ﬁrst-order differential equation becomes 11:210’ — (m2 — mu) 2 0, which is separable: d 1 d—1:=(l—E)w d§=(l—%)drr Then u” = G em a: and n = cl exists: 3:. To inte rate em a: we use the series re resentation for ex. 1 g 138 Exercises 4.2 Reduction of Order Thus, a second solution is BIC y; = 1785(3) : c1x/— oh: 3': l 12 13 =clx/E(l+x+ix +§i\$ +~-)dx —c /(1+1+1+i2+ )d ‘13” :1: 213: 3!”: 93 —cx(lnx+r+;x2+—1—x3+ ‘ 1 ‘ 2(2!) 3(3!) 1 1 _ i 2 3 4 —cl(m]n£+:r +2(2!)\$ +3(3!)m + An interval of deﬁnition is probably (0, 00) because of the inn: term. . (a) We have 19’ = y” = 83’, so my” — +10)y'+ lﬂy = rem — (:3 +10)?” +10e\$ = 0, and y = 8‘” is a solution of the differential equation. (b) By (5) a second solution is €—fP(x) (in: er 33:10 (in: ef(l+10/x)da: {J I T 63-5—1119? Z 8.. E52:0 d3: = €\$/\$106_\$ d2: = ex(—3=628,800 — 3,628,800.12 — 1814,4003:2 — 60480083 — 151,2003;4 — 30,240.85 — 50:40:86 — 72087 — 90x8 — 102:9 — x1°)e*\$ = —3,628,800 — 3628,8009; — 1,814,400372 — 604,800323 — 151200394 6 — 30240325 — 5,0409;6 — 720.127 — 90:1:8 — 103:9 — 3:10. 1 10 1 (c) By Corollary (A) of Theorem 4.2, — 10' 102 = Z —l 3:“ is a solution. . n20 n. 139 ...
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