MTH 205-Soultions of Test-02-B-SU09

MTH 205-Soultions of Test-02-B-SU09 - MTH-ZOS, July 13-2009...

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Unformatted text preview: MTH-ZOS, July 13-2009 Name: E e a Section: - ow all your work. There Wlll be a partlal credlt for each right step. 1. [20 pts] Use the method of undetermined coefficients to find the general solution of the non-homogenous differential equation dyz dy 3 ' ———3— = 3e ‘ +1051nx air: air 1 5 @mlm"3>:a Meg! Fm! mm => m -3"‘ ° Ax€}&—'ffiml +CSML wgclifieé l1) 2: 3k 1‘ £5”; __]__ ecO-‘Q 'm- A5 +3Axe .- 1? 31 1Axe3x—v5‘9i‘” 05",“ 71%“ :gAe + P 5 3:4: gégjk ... c.5244: 'Sx GA 6 ’C r ml“l fl 13A:&J _s_36 :19 H“?g t3C : I'Igw fizz-1 --/._13 :‘70 :75 _ J (2?:- 3C=q 3pl£>:xcgk+5051: +35».&_ )7 C13 m tall}: lédLé) (’ 2.[20 pts] Use the method of Variation of P arameters to find the general solution of the non-homogenous differential equation I 2 2- n : I .1: ‘6“) "' 61((61‘1’3’L 1' fits-“’1 - - {£59K - SKI/an; C d .2 1] 'F ’1‘. — \ ex‘o‘ €23va 3.(a) [1 Opts] Find the general solution of the differentiai equation: d5 d4 d3 d2 flaw w __ 2" _. m>a§mq+|§m‘—?6m ‘0 m’(m3—§‘m‘+b/m-163 m La 7. _ _._ 2—3 ‘ , m [M 60‘1'17U“ “L, H.172; [M-QDKM1_+L\\ :0 “jam? / \fikk) S e! {—CLX +EJeq‘td—Cg‘ 092): +C5-$ma,( (b) [1 Opts] Solve the initial-value problem (2’);2 dy . xZE—4xa+6y = 0, y(1) = I, y (0:0 Mkm~\)__"{m1'én =0 =>hl_gm+6=3 % (W ‘5)Q"*L ):-'_a fimzlls “‘1'? dex):C),rL—5—CLXJ ‘2. ‘a'f-z) : aqx +3 (bx 7.3) Ci +Cl 7:, @ZC,+ZC‘_:2 ‘60}:3 20' +3(L :a :9_Zc’,_,5£’,_:a —— Cl: 2. CL:-—L 4. (a) [10 pts] Let yl(x)=2x be a solution of the following differential equation 2 3x2 £7y+xfl~ y = 0 dx dx Find a second linearly independent solution yl (x) for the differential equation using the reduction of order. ‘ I ' -:: :L‘ that —-,;.u 9 no“) 3-,: PWJOLK 1 it.“ 1 “J _' I): ‘Jjéfik Li"); 6 Z 6 :5 C3 :3 e -JPGMA (b) [iopts] Find the general solution of the following differential equation: 2 div d3)» w 3 W (12:4 dx3 x / 5. [20pts] When a 2pounds weight is attached to a spring . coming to rest. A damping force equal to the instantane the system. The weight is set in motion from the equ ' ' velocity of 10fl.isec. throughut w 2 I )0! —- ___...- ——-----'-'--‘1 w = m :24 - 3- ‘—"—" 3 y? k" s l/ / :3 3L 19 Z I ‘— DIC(0):()/ )L (0)::I0 )fl #4 /r / _..- Iax+£w><w XII _t x/ + Lixs-a _..... X "f ’6 Viol-{'le 7—15? ":2 "’ 3’ (m “:10 f”? M:fllfl_g ‘3} ~84" c Jae. 7mm- Cac -+ L F“ , __ z c ta 'XKO),O ::)C, *0 J K“) '2 ’84 _3f 76(6):”; 2:) 10;»1L it 7f”):/0'£€3 ;/06 1 '0 e” rile" Xmas" 8’ ‘f’ ‘t — L— C) 0.... MHBX :0 1 {Hal} "°( {-900 ...
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MTH 205-Soultions of Test-02-B-SU09 - MTH-ZOS, July 13-2009...

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