HW4Key - Due: 1. Assignment #4 At the START of tutorial on...

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Unformatted text preview: Due: 1. Assignment #4 At the START of tutorial on June 29, 2011. No late assignments will be accepted. Consider the differential equation y’” + 5y” + Sy’ + 6y : ac. Given that e—m cosm is a solution to the homogeneous version of the equation, find the general solution. Give the differential equation which has the following solution. For each, you must show your work in order to receive any credit. i (a) y : 01 + 028—“? + 03629” — xe— (b) y = m[c1 cos(2 ln 3:) + c2 sin(2lnm)] + ln at + 3 Solve the following differential equations a) tzy” — 2y : t21nt Consider a nuclear decay sequence in which nucleus A decays to B with decay constant IQ, and B further decays to nucleus 0 with decay constant k2: A —> B and B ——> C (a) Describe this process as three first—order linear ODEs. and d—E we can obtain the 2nd ‘ ' - - dA (b) Show that by combining the two d.e. 1nvolv1ng E dt order linear ODE in B alone: d2B Wt dB (k1 + + [£1sz = O (c) If k1 # k2, solve this differential equation by knowing the initial condition A(0) 2 A0 and B (0) = 0. Mb lgkw Mz-"p ()0 {LUO fLGE aéM+‘-L)("‘f1‘f1~) s MZ+ZM1L . l f besmldégwd g (“WNW ezmw) ~57 M s—3 5 3° -SK ’X,‘ ) “a fig +2, (a ng—tCfiSkx 0 l 2’ For! [D‘Ve'Cu/(n/ 59U79v‘xi‘fioaa/ vac/bf‘V‘M/ D MWVZ\ SIMPLY ‘fi'cxcj {Lu Aim/5’1"“ 6% Pawwg 59 ‘ '2; 46 {‘3 fl; (4)36 («5? 7L (Up A3 «3 /// I' i ’ 12-" -— 9» f :r 19C“) :> {JL (0(3)) — (1%) ’ 10’4"»: a ("5) slim Marx [CI Cr: (flux) --t C; S-(L/41)k¢ [wk 93 D m Cavia, EKCL/ ((42. n12 M: (—1., 1L (VJ/(x aux.'(;.\¢3 911A «Kym/Ma =‘> (M~i +LL)(M-t—1L)s (MZ-2M+‘+4) “:3 £04} 5 5‘ [A 1 _t. Z A, c; ‘ [In x a la T 53 5 1L f =3) MZ'M,1’ :0 :7 (M-1)[Mf()!‘o ' z :3 MS*';V"‘=2- flvéa£d+CL€E c 2t 1 ‘7 f -3 HI ‘4’ UL}; [A] S —Ji/ if “ 0 £1 :t” WIN/«L Mi {fl/“1" 1+ l N "'r’ MIA “'3‘ 5 K S 5‘ 5 =3M-fi/ln£——Jiej’ ’JLln7l-i ‘ I5 15' .76 l5 +45 v; o l 'u’ K I". if“! S *anc z 5 3) ’> m _s 71‘ (A -j {l}; H 1 €( /2 ‘7 1V0) sLuzlnLJtL) ~ 4 ‘r v 1: (“kw , Z I 4 1‘ .'( g ‘3’3:_C_L+CL{ ‘S'(£[“"3’)+/é 2. -LM \ .f 0,- (71 1~L A. .‘v’\ e 2. f =7 " ’ r ‘1 1 ——:> . ., , 7 ‘E. C C 7L C c a 1‘ £860 11 , a 1 Law.) FM [Nu/4,: W V “jivfl/ s u‘ NA U\ N3 WWI‘U/é\ P a z 2. _ Cu, 1. “l.- x UJ = (3 Jam» a lg 6m 7L»: CH) \ \__~__’ ’k FR“ \8 Cnfi Q, 57.} W Cal—$64k) Ql($~'v‘¥\f (‘51) 2.1: ,1 _ s 57$ g” fi/C’V" : - éxzw ( 2.1,. cm“ 6 . 7 Z \ ‘ ‘7 R‘s-«j Lo 7‘" Ck : 5Cu3‘h o/w —J§€.C?L (id-3W :sfii— LwdsgoL+fann\ el\ 0 I (/6 IL 1 “L S 61L (0’31 —3-A1L) e {an L a 5"“W W t 2A» e1?" 9/ “7,5553%ka s «(63% L :QC’L “3P “ (5.”. =1. —— LA {Scram’efomx' “353°”? 5-]; m-Z IV .— M(M“)7\’ m ' “b “3”” fflw—wm-zh swam-x) + w +2) :9 ’L ’5 z.) *AMQ%$ ‘5 =5 M’*(Mtz)+q(m+z), (maqfimflpo L”) VvtsiiL‘fl W‘s’l In (L Slkfi) CL /n(Z.CD7L) 1” was M-i)(m-Z))L 2/) ’j | I, ._ «3 7, “3 , 59MB , P , / J U\ z «6/ UK = 43/ 7 .4; g A 5 WASH!» . , 7V. , M79 , , , , ' 7' \-I' '_'” 7 " ' '7 670, V? 1.x = ,.\A j =7), = '_L.,, 7 .fli” "(‘9 0’?” . J9. MW, , 9', ,, $14., o/Lm) w‘i: _ L w s a. a; Jude H .u1 _,é/w ,, 1', ,,,, ,4“, ,, ,, V V. . ll,‘ 7 3 7 Wild-Avg“, IV A 49—; TQM ' * ' '7 ' *H “i r r 7 r ‘P Lw’ I‘r/t fik—r 7 ‘fim mm m In x 9/ " g a v4 9/ '3' 7" ’ j 1 ,V. W 1 4 c #9 a [k/V‘ s ‘ *5 7k 3 1* ‘r C ,L( ,LVTS X Jw x. aaiiwzm 1 i ‘3 1—! 4‘ 75/)» r L - C. J...) = ) __L_— = 3W4 1 7 I in/W ‘ l“ C 79, 7? C1, ...
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HW4Key - Due: 1. Assignment #4 At the START of tutorial on...

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