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HW4Key_revised

# HW4Key_revised - Assignment#4 Due At the START of tutorial...

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Unformatted text preview: Assignment #4 Due: At the START of tutorial on June 29, 2011. No late assignments will be accepted. 1. Consider the differential equation 3/” + 5y” + 83/ + 6y 2 :13. Given that 6'93 cosm is a solution to the homogeneous version of the equation, find the general solution. ,2, Give the differential equation which has the following solution. For each, you must show your work in order to receive any credit. —:1: (a) y = 01 + 626—32 + cgez’” — me (b) y = m[cl cos(2 1n 3:) + c2 sin(2lm:)] + 11130 + 3 3. Solve the following differential equations (a) t2 ” — 2y 2 t2 lntL (b) y” — 2y’ + 2y = 6”” tanm (c)\$2y”—my’+y=x2 ((1) x35? +5m2d—22my +7m% +8y=0 ( (.912 e)?!” "gyl=— m 4. Consider a nuclear decay sequence in which nucleus A decays to B with decay constant k1, 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. (b) Show that by combining the two d e. involving— and ”1—3 we can obtain the 2nd order linear ODE 1n B alone: d2B dB F + (k1+k2)?+k1k23 =0 (c) If k1 7é k2, solve this differential equation by knowing the initial condition A(0) 2 and 3(0) = 0. Jawshgq‘hgj‘ E» J ;;J 1%,. mt ’CWrMFMFﬂwJ' " ” W, i, , m, M (tilt/griqwﬂwwd ,i§9,,,(ei‘c’£'fjiv ,ififl‘”, ,A 055.102.0(15/ ”i , ,, , ”my/‘9M ,).4€,, 1),,W ,7. :h xFa’M , .‘ ‘s' 7,6”ij ;Mo/775.3K'CMJ, its .. mm-“ _ ~~3m/Oa/d~\$LL~~fwélatw¥dusj W 724”,“ , i rM/«f: .,,, ’”ﬁg?""‘fggdsm: PM w—ﬂalnf’é‘“ '* '* ~ W"’ #«M% ,,,_ h W M ‘ v ”‘30 "HfﬂO'W‘fG'G’g'W’ ' " W M iffmmi 8n; 5 3.0 \$*9_1/L‘<. CM/[v—N__ l _Aa__9,c _{46/’ 3. c 42, ,1 a ”(c.7240 it {CESAHJ Jré’ﬁfz/y. -1 ' 2‘ -1. AM __ )3? ' l3(; 5) s;£"\) : e: 77 (/le ,i M < 9» 'F.-~(—(/,< 5);“ (1 ,, :7 e 7, M A 2:) 16015 {a J 757 [Alf-t '1 7 045 7 i 513; 5/ 7““ ’ 5L ' x u’ = > (a {MN . 7 a [C guﬁii (at) l M e’%( can-\$l.hl). QKC'S'V‘W'f (ﬁnk). 2.1L 1 r V s -—6 ski/5"" : 4'5“.sz " 2;,” Cm‘h 6 2. \ . .. , W, . (56*- : SA‘L - Ln QScOL—t‘tanw\ 1k , 6' 00 L 0 7' 7’ 1£ 1, l . a 5 6 (001—3‘k) e “In 5 5MK L . , ,S 3' SM‘W L1» 21"" a Wpst’Mm—r/w s —Cd3*' N33 étcoié5n~1~LA‘SC—C’Yk’ffow'sl' P “5%9+3r _L) sick ”é g ~c x « C MW -* L ’1, ()w" “A“ 5‘3“” ,M, Mr] ‘fa 4!th— i041 6- o’u/«c/(L’ [lab NZ V‘ \31,——.1 V7 ngl'v, J 3 I 0 -1iv\‘\\ 1.) m\ _ \\ “1‘s -L‘ww .__———————-"-’ 2'” k S " I. m ‘Kl \ 7‘ \\ \+‘Y¥“ w.) my - xiii Mk ‘ “x 0‘ ( 7, ,7_, u. r l‘ l s ‘ =7 u 5 9" Z' ,, Mm- ,,, L CL, 1 (‘3 ,3, W" ‘8mmiqu- \13' : "4’“?C’7-"In' i" j? K" ,, P M , x A, , z,,,,.,..,.,,--< , _ .- , , [=3 “6 “6;? «a (19¢ my; +qf._' MHI‘M 1.4 I 1 - 1 A 5 04 ‘3/6 rkAi-kﬂ F" => kH/L: fig—skzﬁ 7; l 7, Jf O/C , k 6 111 \$1 '72,; L 1‘ / k|é=6ak8 w 0’75 +l<279€ +1413; = 0 (Hz, « ’ J1‘ =° 9’13 +k olé_+_lz(z_/_é+kzi3)=° +‘ 7’ Off 4 0H» -3 0/26 I l ‘r 7' I I - -k,t -Ag‘ 7 60143-9 e +- CLB Cf-c (KK’ﬂj 73(0): 0 7—0 C,‘fczs G ’3 '“ 1‘ / 8(a): kﬁ(a)——IQLB(0)= mg I“? Q / .kt ’ L / -4 I< ( 73(4); vCIA'&.-I “CL/(Le 5:, 6(0P‘C’, 2,2. K:‘**v”( a“) C135,” Cﬁ,kl ”C‘LkLz? ...
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HW4Key_revised - Assignment#4 Due At the START of tutorial...

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