# MATH 235 - #13, Section 2.1, page 39 y'-y = 2m” 273—}...

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Unformatted text preview: #13, Section 2.1, page 39 y'-y = 2m” 273—} —y z 2162’ dt i): + (—l)y 2 2162’ d] W W) Multiply the whole equation byy(l) : e— l —1 21 ‘1 e —~~ye =2te e %(e_’y) = 2te’ e"'y : Ple'dt ei'y : 2 Ire’dt e"y = 2(te’ ~e’ + C) (fry 2 2te' — 26' + 2C y : 2te’ — 25’ + 2C 6 y = (216’ - 26’ + 2C)e’ y : 2tez’ — 262’ + ZCe' Find the integration constant C from the initial condition: I : 0 y(0) :1<—-+ y:l Substitute this information into the above expression: y = 2m” — 262’ + 2Ce' l22-0-62“—Zez'0+2Ce0 if T 3? 1:0—2+2( (:22 2 Therefore the solution is: y 2 2182, — 262' + 2Ce' 3 y=2162’ —282’ +2426l y = 2162’ m 2621+ 36' Integration by parts [fvggi‘dmfgv Jgg-dr it e’ dr:z‘e’— [61011 J L-v-’ L-w‘“ l f=l WK“: l i 37"“ 511:} i all gre’ : te’ — J‘e’ a’t l = 16’ —e' + (' #15, Section 2.], page 39 ty'+2y=t2~«t+i tﬂ+2y:I2—l+l d1 2 ﬂ+2 :z —1+1 dt 1‘ t dt L t 17(1) t2ﬂ+IZEy—12(t—l+1] t t \ t yt2 2 [03 —r2 +t)dt 14 13 {2 t2:———+~—+C y 4 3 2 , _4 3 2 . I JRJL_L+L+C+ﬁ \4 3 2 t“ I l‘C J/:—.~——.__+_y.7 4 3 2 I“ 1 12 i 1 C __-;___2_.H__+_7_ 2 4 3 2 I“ l:l_l+l_+c 2 4 3 2 gel 12 Therefore the solution is: ﬂﬁmz+i+£ y 4 3 2 t2 1 t2 t 1 E =H~~+«+w y 4 3 2 :2 t2 t 1 1 y=~«~+»—+ q 7 Note that: e‘ln® = Q9 2 =1 Initial condition: MD=%e—4;: é: MEX/"=0 Vi‘ ~ ULUMJ “Emqu 7:22 vatUi 2.0 Q: O'Ez‘°+C€ 2=O+C C21 2:; (1‘ ~ M. i=5? Vi ji m at NP- ,7 “F21 *gg‘CSML '33:! 3:”055‘2. “ttmkir [wilt “J: ant+ sint+ C ‘ I t 84: OH: w as; 1 .in :3” OH‘ a“: \$2; 3:.»6" ﬁg: ~‘te‘hj 'Ed‘o‘t —-‘t€"&+ : 433%" 6-1;+ C 'L ’+ i " 7 (+Hymi) )qﬂmg): (mus ’c-alnz I {H 2 7+ f y:% y I g! aﬁU‘i’)“ LHWJ pd) ML“: ejpwax : ejmﬂdk x 65%” He»: 2: awn : 61:. C‘nt t n he?“ =9 til te at + U+“’%))’t€tz itei oi a}: { y- te'“) ~'~“ “tat y‘tct ‘1 jte‘cit yte‘t 2 ’ce’HeWC 7: te‘w‘m PH“ tat mm {:3 ‘ j In) amok yzl {VJ}: : 1 z.— Jeuze‘m €‘“‘+C Jaw-eh“ I: 2M2~2+C 2AA 2&2 2 ‘= Zﬂhz—ug c:- 2 PM y: te‘» 6+2 tat : t~l+>2ejt we start W a! :36 3W" ilk-1- i) I ‘f 152?. alt ' 74‘ tat-bf at“. z t i6 - €t+Cx #20, Section 2.1, page 40 W40+Dy:t , 1+1 WTFI . v 1 y+(l+;]y=1 L._\r__/ 17(1) 1! I 1 l Zey+te(l+;jy=te d s 1 ~— 1 :t (#06) e yte' : J'te’dt W we have done this in #13 yte’ = te’ —e’ + C '1 C y:1——+~ t rte 121m; 67 1112 1n2-eh“ “‘5‘” 1:1"““L‘+ C 1n2 21n2 _I___ C m2 2m2 C22 Thcrcforc the solution is: I C y:1——+ , 1 le —1_l+_2_ y t te’ % W) = eI "W" w : te’ 1 Initial condition: Lu #21, Section 2.1, page 40 1 2'—— :ZCost J 2)” d 1 ~3+———y:2cosl dt 2 ‘—v—’ W) 1 I I __, d _, 2,, e 3 lmlye 3 =26 2[cost dt 2 d i 7-1 [ye ZI]:2e 2’ cost (I! J, #1, ye 2 : 2e 2 COSZJJZ 1 1 I ‘ ye 2 :2 e 2 (ghost—35in! +C 3 3 l l ‘ ye 2 =26 2 (%Cost~§sintj+2C .3 fl 1 y : 2[3(;<nst~££sintj+2Ce2 ‘3 3 I yuﬁcost—gsinHZCez .) 1 a:£c080-§sin0+2Ce20 3 Y 3 “o” “T” a=i~0+2C 3 Therefore the solution is: 2 4 . , ‘r y = 2 30051—351111 + 2C62 7 ~14 )x:2{;cosl :sintj : 2- 1[a 4je2 4 8 . 4) :—cost~—smt+ a——- e 3 3 3,, } —~I 1 1‘2’ e 2 cost dtze 2 -sintszint-~—e W h-v—d 2 "'1‘ \$7005] _/=L‘ 1' dr i ¢ 11! 2 . 1 . =82 sml+—J e ~ Slnldl 2 ,1, db . “a: 2 1;]?an dd 1 _l, (7—3005! _.:.“c 2 dz 2 . 1 a —:r \ 7 :ezsmt+ e- cost e- cost I 2 2 -%I . 1 it 1 4 =9 ~ smt+~ e 2 -~cost—~ e 2 costdt 2 2 I l 1 I ~t 1 1 m, _ [e 2 costdt=e 2 ‘, v1 —1 sint—~e 2 cost~— e 2 costdt 2 4 5 A1 —71 . —~I Zje 3 costdtze 2 smt~—e 2 cost 1 *1 je 2 cosle 2 5K 6 1 w—/ 7 2 _Ir 1 [l «I 4 . “’ ~w JG 2 costdIZ—{C 3 smt—me 2 cost +6 2 { A, \ e 2 sinr—Ee COSJ+C 1 A: ) U1 ...
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## This note was uploaded on 04/05/2009 for the course MATH 235 taught by Professor Chen during the Spring '08 term at Michigan State University.

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MATH 235 - #13, Section 2.1, page 39 y'-y = 2m” 273—}...

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