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e2a - 5 I A 145va EXAM 2 MATH 2233 SECTION 002 SPRING 2011...

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Unformatted text preview: 5' I; A 145va EXAM 2 MATH 2233 SECTION 002, SPRING 2011 INSTRUCTOR: WEIPING LI Print Name and Student # SHOW WORK FOR CREDIT !!! SHOW WORK FOR CREDIT H! (1) (15pts) (21) Solve the initial value problem: 7 yn—y=0, y(0)=2,y'(0)=—%. Yli =0 'Y -= ii . 3: :8f ‘r Czét y’ficief—CZEJC —5—‘F=Ew) = Ca 1“ Ca. “5: -310): La - CL (1.: 2’; (Z :1 v f ”f (b) )Determine the minimum value of the solution. (You need to verify it is minimum) wt ,_ 1,— “it "E Hey-1‘8 —e =0 gaze, {up 6 "fr-i112 't '1: PC) _|_ Z 5” few, HUFi? ' 1" i = l ‘5 M ”W “4‘16. WEIPING LI (2) (lOpts) Verify that two solutions 1/105) = at and y2(t) = te1t of the difierential equation y" — 23/1 + y = 0 form a fundamental set of solutions. EV( 7| 7L / a ' €1- {Tet 0/ fig) 72, )6f 5+th : €T’(€T+t€f> ,_ at (tat) 'l: 13‘ ‘65:? =5 0. Tim; 113/], fll) fawn a fw’wiawavdfii Skiff“ 57f S‘dfldfiZ-WS. (3) (10pts) Find the general solution of the nonhomogeneous equation y — 2y + y— _ 23‘. (1‘) H r:— C‘ (it ‘t Cat-€1— is “flag WZLJ Salulm‘D/l" LIOVHOWVLS 4:;er _ f f (B) Ft’v‘cl I1 (W170 :guficluifiw JP wage + Claw-t6 ,I‘ , :33}. ._—_ _ __E_iE_ : ,th cimp‘t‘L C‘ ‘1’)" WWZ) e.” 6‘ Cf C / .. :1 _____3 j.‘ .: i4?— =_ 2 0.106): 2f 1“?) W07. YZ) 621: 3P:*Jc2fe +2f£te) =tzef, it”, W “(@7544 of 14014 «Mkjmeus armada j 1 Ci elk“ Catéf + (4:161?) DIFFERENTIAL EQUATIONS 3 (4) (lOpts) If yl and y2 are a fundamental set of solutions of t2y" — 23/ + (3 + t)y : 0 and W(y1,y2)(2)=e_1,then find the value W(y1,y2)(4 5 By An mm my, W e: “W 3)” — $211+ 3°": 3: 0 her ‘39, fut—1: ff Wl7.,7a)~=Ce AT» Ceidicabi‘ W1x,yz)(z) ”Z": :6‘ (1:1 ‘5: ._L (5) (15pts) Consider the initial value problem 1/” +2y' +224 = 0 74(0) = 2 y'(0) = a- 1) Find the solution y(t )of this problem;( ii) )Find oz so that y— — 0 when ( (i) Chmmi‘Le/nslmiiai'hm Y +2Y +2 :1 O 7-1 z - N: :——z+22[ : ~iif __.t 3:C 8 ”Cost + C26 SM'LL —"l‘ _ ‘_ ..t 213(0) = C1 3/: Li ('3 Caifiécéim’f)‘+€z(‘€$thf #661an $2320): C.(—-I +o)~r Cz(0+ i) CL: OCT ci :— o(+2 -‘t‘ ”1,— 5: 26 Cast +(o(+2)€_ Smi‘ IL 7 \l 4 WEIPING LI (6) (10pts) Use the method of reduction of order to find a second solution of the differential equation t2y1I+3tyl+y = 0, t > 0, y1 = r1. ' ' —; 32: C(65)! :C(+)'t Hz]: Cleft ' ‘- CWT L - I/ ~‘ . a _5 fizzmflil~zciota+2C(f)t ; , - ,_. f ’— *- "I 'f ‘1 - 71:21 Crit) t i'ZCff-jt 2+ ZCCfit 3)+5{'(C(H‘t .. lejt )1L CPU-t :0 'tCltM“ Ch) = C) Hit)"‘Ci(f) £35: (:15) 7,. +3’H‘E +11 :0 3% :-j_i%7_ Ital/l 1—im—t 11:1,: 1 I Cec)=fu(t;d+ =[;dt :[ht L - I (7) (lOpts) Given yhyg satisfying the corresponding homogeneous equation. Find a particular solution of the nonhomogeneous equation, by Variation of Parameters, t2y” — 2y = 3152 — 1, t > 0; y1(t) = t2, y2(t) = 25—1. . 7. Ya ‘61 t" “2. if, : _ Wl‘lh Y2) : Vll y; : lZ't "t'l :- " f "t 2t t 5 Win = "2%; =9m_ 3?: GM + MM.»— i . ,- ~i '3 -— / 1 3'4“: t. 5t "'1 "-~—L 5 Ci(t):‘%y) ”’L :2 : 5 ’1 5t , —3 ,,.i '1 Cue: +5351 w W fl 3 raw—Lay; ail—v 2‘ ' Cr -1; _ t 2 dc ++ zit) WW7?) ,5 3 '5 3 .2 J. Chapman-1e “new DIFFERENTIAL EQUATIONS 5 (8) (lOpts) Identify p(t), q(t), g(t) of the following second order differential equation, determine the largest interval so that the equation is certain to have a unique twice differentiable solution (It - 3)y" + 252/ + (111 ltl)?! = 0, y(—1)= 2, y'(-1) = 4- ” “t I“ .~ M [U + _ O 3 + {3 y «we 5 T w in ’— 14 m 3 We, at w , m ,. m m WWW «excelai’ 1’: 3 ) f=0 t0;.., G(«z><> a) flaw-M— (9) (10pts) If the Wronskian W of f and g is tZet, and f = t. Find g(t)- WW3) 5!? J5] = M”? fig 21:26" t V6353 = {get 3’~-—é3 : tot (“tam/35 7%ch /(: 317”)”: if?” :65 SQ-M: (/wjy: ij"—i~23=%tct :6 7, ,uj : fetch :61: + C. %j({.)={€+c ...
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