fv1 - Name—KEL— Math 307F Final Exam December 13, 2007...

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Unformatted text preview: Name—KEL— Math 307F Final Exam December 13, 2007 Instructions: There are ten problems, with the value of each problem indicated, for a total of 160 points. You are allowed the use of one page of handwritten notes, front and back, on standard size paper. You are also allowed use of a scientific calculator (but graphing calculators and other calculational devices are not allowed). The next page of the exam contains a table of Laplace transforms, which you may use in your solutions. 0 Work the problems in the space provided. If you need more space, use the back of the page, and clearly indicate that you are doing so. 0 Neatness counts! A well-organized solution, even with mistakes, will get more partial credit than a haphazard collection of unrelated calculations. 0 Put the answer you want considered in the provided. 0 You MUST show all your work and reasoning to receive credit. If in doubt, ask for clarification. 0 Turn off all cell phones and pagers. Problem 1 15 points Problem 2 15 points Problem 3 15 points Problem 4 15 points Problem 5 20 points Problem 6 15 points Problem 7 15 points Problem 8 15 points Problem 9 15 points Problem 10 20 points 6.2 Solution of Initial Value Problems 319 TABLE 6.2.1 Elementary Laplace Transforms f(:) = L" {Fm} ' F(s) = am): Notes I l. l -. .s'>() Scc.6.l;Ex.4 S | 2. c‘” —, .s- > a Sec. 6.1; Ex. 5 ' S — (I i 3. I", n = positive integer s > 0 Sec. 6.l; Prob. 27 I. 4_ II" I, > -1 s > 0 Sec. 6.I; Prob. 27 .5" ' . a 5. sum! ‘ , , s > () Scc. 6.l; Ex. 6 s- + u- 6. cos“! , S , , s > () Sec. (Li; Prob. 6 A" + (l' 7. sinhut ‘, a r. .s' > |u| Sec. 6.1;Prob. 8 s— — l - 8. cosh at , "- ,—. .s' > |u| Sec. 6.l; Prob. 7 A" - (l' . I) 9. H" S") bt ———,—,. s > u Sec. 6.l; Prob. l3 (.s' — a)~ + I)- s — a . H). v'” cos ht —,-———,, s > u Soc. 6.l; Prob. I4 (3 —— u)- + I)- v I L rut...“ u = positive integer FEW—f, s > a Sec. 6.1;Prob. 18 e—l‘)‘ I2. "((1) S , s > 0 Sec. 6.3 l3. u..(t)_/'(t — r') e‘”F(s) Sec. 6.3 l4. ("’fU) . F(s - c) Sec. 6.3 _ 1 s i 15. l/(ct) —F (—) , c > 0 Sec. 6.3; Prob. l9 c c I 16. f [(1 — t)g(t)(lt F(s)G(s) Sec. 6.6 ll [7. 8(1 —— c) c“'-" Sec. 6.5 l8. j"’”(r) .s'"F(s) — s"“f(0) — --~ —f‘"“"(()) Sec. 6.2 [9. (—()"_f(() F “"(s) ' Sec. 6.2; Prob. 28 1. (15 points). Solve the initial value problem 11(0) = 1- Your solution should give y explicitly as a function of 2:. Answer: ‘3lo\= \ =3) gob/M? for 5 '. 2. (15 points) Find the general solution to sin(t.2)y’ + [2t cos(t2)]y = e‘ + t. Answer: Fwd W 7m+ecjmfiiu3 faxbnr 53:325.: 3TH H3- }4H1)= e, ‘ We M K: S’Wtftl}! A“: it Cod/i1.) 0H; 5 ‘ 1 A ‘ o lmfgcifgz A1? = J37}: [v1 UL: In BM [—8- lv» tweet» F‘Hflze = {M (’12?) Ge“? t \ 7. /___ E, +Ft‘ I I 2 61+ [gmlt ) -— ‘ gufifl-S‘uH) "I; ' 2 => LS‘iwl'TB—33tj : 61-4, :ta. +C ’2’ Cir—1L 1‘; + C. 3. (15 points) A tank initially contains 100 liters of pure water. A mixture containing salt in the concentration of 5 grams/ liter enters the tank at a rate of 2 liters/min, and the well-stirred mixture leaves the tank at the same rate. Let S(t) denote the amount (in grams) of salt in the tank at time t. (a) Find a formula for S(t). (b) What is the limiting amount of salt in the tank as t —> 00? .—t 0 Answers: (a) S']t]:goo[1-e “"3 €00 grams 2 Qfli‘er/m‘mlxlgfim MIR") "' (lakr/M‘M) gaggu/Q‘JW) _- IO ~ 5% 51-17). (gm/ml“). 45 s \ "‘13 jg E55 =lD CldewrQwear, #117): e ) N0 Soil: "\m‘x'HO-QQAj) S‘a Shflf—a % o= Sco)= ea'oszo)=5ooe‘° +d % d= ~S‘oo a. -29. gal “‘31 Slt)=5‘00 —sooe 5° => say) 2900 ~ 6' M] 0&0 tfioO. M Ts TQaSmaeyQo 51m H‘ e W WM’Y‘PN 66 We, M'flbto "fin/tea VWCK We 'l‘aMlc = “Saw/«95% ) "QOO «Uta/x). 4. (15 points) Solve the initial value problem 3/” + 6y’ + 10y = 0, y(0) = 1,y’(0) = 0- Answer: 5. (20 points) Find the general solution to y” e 43/ + 4y = sin(2t) + 62‘. 7;! U: 2 '2. Answer: a :. q 6 +C2f€ + igcosfl’t) + )7: t 6 Ulfi z—D—A ms‘m (14:) + 338 £06 {1—5) H" hum Cosfzt) ~ch a; (21:3 ‘32, “Wigfi “34>, : ["W’ '83 ‘5‘ C04 [149) I 4’ [’33 WW +528] 5m 1») So 850/ A45» rm 39‘2—33, cosllt) . S’mu 100W ewm teztavve SJMMS 727742 WW5 «6.4) We m 391 = 1:26,“, 50 'b _ 3gb: Cfltz€2t+ zte’z l 932: Cfillztzezt+liezt) 'L' rilu ez +6759]. => 0 o u 7-7 1 2.1: 7 21 Q, ’+\.‘ 2 L“, H6 ta + <8-~8c]te Spa 3532 3?: L g: 311: 1C8 = E 6. (15 points) A mass weighing 5 lb stretches a sping 6 inches. Suppose that the weight is pulled down 1 ft and given a downward velocity of 8 ft/sec. There is no damping, nor are there external forces. Determine the subsequent motion u(t), and also the amplitude R of this motion. Answers: “17]: cos + {l \4 { €47) J E: m3=5=’l’el.4= 2H: [6M=% 17001,) é ¥¢=m MAS MW vafimrfidmwwava W3 P33431709} ’5': "‘l‘ lDu=0J u!o)=l’ q’fojzg 32 Or u" + Mar—0. Omarafims rH—wzo—a \ W>i8\v) 50 W WSVQufim/L \3 mm =— q mké’th— Q1 swat) l=u103=cl u'rt]= —844:u(8$)+ ‘o’Clcoal‘cHfl <2: We 1: See cg} ‘9 Mt);— coa/8Jc)+ swat) R=W=VE 7. (15 points) Let uc be the standard unit step function defined by . 0 if t < c c t = . u ( ) {1 1ft 2 0 Let f“) = “105) " “205) + (t — 3)u3(t) - (t — 4)U4(t)- (a) On the axes below sketch the graph of f (b) Compute the Laplace transform F(s) of f 8. (15 points) Find the inverse Laplace transform of 32 + s + 3 Answer: M56 Parfiail (mg: grimy-3 _ J: 35.”; [s— \) I 51%) 5“ szvr H “"> S°3rS¢3= Pr{5°‘+‘f)+ (Bin-o) (3'1) 7' [PH—73] 32+ [C- 835 +— Dyan-c] N H \ | l 3 S0 Pr +33 =\ ~T3 +C‘:\ "IA ac, = 3 51A =3”? A21 =) 8:0 ’9 C21 _ __\_ \ Pl$)_ 8—! + $734N M" H” {M = et + ilsmzt) 9. (15 points). Use Laplace transforms to solve the initial value problem 1/” - y’ .- 2y = 0, y(0) = 2, y’(0) = 1- You can check your answer! Answer: AM: I [s1 YM—sglo) Aj’lofl - [SYN 11m] ~1YIS) :0 W3, \vflfiofl wai'fiws) (s1. 5’ 2) YZS) = 134 _ 25-) __ 2.8—) . ’31- 94. ’ (S—LMSM) W1; pm‘fiofl {Ira/ohms : ,ZS ) A V B (govern-mp“) " 2 ————- Jr — % =| :1 (S-‘LMS—H) 5'7, 3-H A ’ B 2.1.! .__L 1:9 ‘79) 5—2 + S'H it" w“ \ ~t Oneokt L3‘~3’~23= E—z-1]€Zt+[\H—?—]€ =2) \/ *3,lo)=i+)=2 v/ 11: —’IT {me —e , ’[o)=l*l‘='\ \/ 10. (20 points) Let f (t) be the forcing function defined by f (t) = 0 if 0 S t < 2, ‘ and f (t) = 2 if t Z 2. Solve the initial value problem 3/” + 331’ +' 2y = f (t), 11(0) = 0, y’(0) = 0- Answer: APM I 3 we): Zulit) SZYIS) + 35 Yls) + 2W3): MW} = 2 if S =3 _ ~15 1 \fls)” e stsI +3.92) W His) PM Campuhs w»): {Wm}, MWM 19mm 1 __ L _ A B C H15): 5 /s‘+35n) Stsmmu — T + 5:" + 3T2. ‘ W? “Covev-Hp“ W Wear ecbrvé, git Pr: 1, B: "2, (i=1- ~) 1 3 ) Nth at E‘ZTH’" 5,7} ...
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This note was uploaded on 04/08/2008 for the course MATH 307 taught by Professor Doran during the Winter '08 term at University of Washington.

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fv1 - Name—KEL— Math 307F Final Exam December 13, 2007...

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