FinalSolutions

# FinalSolutions - M427K Final Exam May 15th 2007 Question 1...

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Unformatted text preview: M427K Final Exam May 15th, 2007 Question 1 A tank with a capacity of 500 liters originally contains 300 liters of water with 1500 grams of potassium bromide salt in solution. A stronger potassium bromide solution With a concentration of 10 grams of salt per liter ﬂows in a rate of 2 liters per minute. The well mixed solution flows out at a rate of 1 litre per minute. 1. [7 Points] Find an expression for the quantity of potassium bromide salt in solution in the tank after t minutes. Lei GOG be the amounl o? bromide SOd’l' {ﬂ soluhon oi lime 4:. 61(0):le qr Let \fCt) be Hie volume 09 water In the tankdol time, )0 VCtl = Boo «He. £1.63. _. GOG _ N _ Q99) out ‘ 20 — VC’H ‘10 ’?>00+t 3% + ggéﬂg :20 lsl order linear: lnlegrohng tacior is (300 ft} 0% [(300 + @6160] = 20 (gem-t) 300%) @021 = 60001: + 106‘ +Q 10 J? -r 6000*: +Q QQ‘) = 300 +5; gram) 2 - , r .: (0t +6000} ~r 450000 9mm 2. [2 Points] Determine the concentrazttagnlolf potassium bromide in the tank at the moment the tank is full. ~ I0 (400%) + n ooooo~+ ssooma Tank {3 Hill 0+ -t=2oo. QCzoo): 5w : 20550—0 DIVI‘ClU/lg 8Q+ Hag C‘Onmﬂﬁal’on ., 4‘ l (a! WQ gel /5. / 3. [1 Point] Determine the theoretical limiting concentration if the tank had inﬁ- nite volume. 30wa a) {Momma Plow lie. 10 3/3. M427K Final Exam May 15th, 2007 Question 2 [10 Points] Find the solution of the initial value problem dy_ 332 dx_1+y’ y(0)=0 explicitly. Seyorobié (I’rgjidfj = Xildx- £59232 = 5b *Q 15 H; = 7338’ H3l L9+Dl~t =%x3+c‘ (w? = 7%? +0“ Lj-H =f(%xg+C“ V1 U) a“) 1' (1/3 xg’rC'OVZ M427K Final Exam May 15th, 2007 Question 3 Consider the autonomous ordinary differential equation j—:’ = f(y) where f(y) = 2M — my - 2m + 2m. = 2 (3+ 0 C3- 0 C3 4-)?) (3 +229 1. [4 Points] Sketch f (y) vs y. g and nghlaﬁ’l powu {3 rj (l hm C\ yosilwe coellic Mani . 2. [4 Points] Find the equilibrium solutions and classify each according to its stability. 5 pom’rg Bug—1,0, [,1 if «2 IS semisiolale {j =»—\ (q (1)3mpi'0hcollxj 93mth 3» O is um’rolole ‘ \j >l is dgmsel‘biicc‘llg Sicile B = 1 l8 unsi <1th 3. [2 Points] Sketch y(t) vs t. M427K Final Exam May 15th, 2007 Question 4 1. [6 Points] Find the general solution of the inhomogeneous second order ordinary differential equation y" — 5y’ + 6y = (2t + 5)e3t, Cowmant Rim an m”'~5m+6 =Cm~37Cm~7D min are mzlimts 9m: one +C2e Form 0? porhCuior sol”, ‘n‘l‘b' 9%“ (AW)? OVU‘WW 30 aﬂéuagﬂe“ Nb“ ‘5; = (2M + 1335” + 3(AP+BJC3Q3* ‘59‘ = (1A3. 63" + 6(1AJC+B‘>Q_3JC + CiCAtABJQeg'k' Look 0dr mama‘in o£ 83+. 2A-r6i5 5(8) =ZA+B =5. Look 0dr coeWanl 0? Joe?” nA+qe+ -5(2A+BB)-+e{133:2 QA=1 A ll ’- Tkm B=3 3?: (964439316. Generoi Soiuhon is (j: 0‘ 9%»: Clerc + (9+3?) a” 2. [4 Points] Find the complementary function and a suitable form for a particular solution of y” + 9y 2 tcos(3t) + 63‘ sin(3t) Do not ﬁnd the undetermined coeﬂicients. companiede Rmchon kg; q @0589 fczsm (3 t) (5% = COSB’B +<QJC+D33lYIBil OVQAQpS 30 9v.=(Ak1+ek\ COS (29:5 4: (CRUDE SMQE’O km =E€3£ismigﬂ + Fezkeosféﬂ 5 Varl’icuim l‘orm (AUHSH COSBt + (Qth D3 smCét) + Eezis‘mOt) T FQ'SJVCQ‘BQ M427K Final Exam May 15th, 2007 Question 5 [10 Points] Find the general solution of the ordinary differential equation my” + (3:1: —1)y'— 3y 2 0 given that —3z 311(90) = e is a solution. Reduclion 0? order 900 ’= V90 {BX W60 = V‘Cx) 53" ~ '3 VCDO 9%“ 3 {3“(0 = \J “(20 «2'3" - (3 v loo 0:3" + Olvbd e" x Subsiﬂulihg Jrlm‘g (remembering all the V term! Cancel) 7LCVHCL) 6:3" — 6‘v'Cx3e—2X) + (3x *\)< V‘CDQ e‘3"> I: 0. 3x Expanding 0nd ernoVMB the e' comrnon Poclor' IVHOO ﬁxm woo =0 . WOO ‘C34 l/><l VIC-70 =0 lVlnglallVlg PQCWLOI" €‘(3x-Hn r3 3 efgx- 0% [ iQ—ng‘hﬂ = O i 63" v‘CvO :Q M427K Final Exam May 15th, 2007 Question 6 Consider the second order linear ordinary differential equation y” -m2y’ -y = 0. 1. [6 Points] Find the recurrence rule for the series solution about the point :50 = 0. 00 ~ 2 a x“ B . 11:0 n co ‘ ’30 ‘ T ‘_ h” VH' . h g ' 2 an n X 383‘ = Z Onhx = L on,'t,h—\)>< n: n:\ h=1 3.? __Z ’50 h 3% Z QnHCn~D x" “= Z Cinﬂ(n+z)(h+\3x ‘ ' “=1 {j n=0 [)0 n B=OKO+CMX "+2an h=1 co Xiij‘: Z QMCH‘W Kn. 34:2. ‘3“[email protected]+ max ‘* Z cxmdmﬁmmxw- h=2. 106090 haqu‘éO‘ Qn+2(n+2)(n+\3 ~an. (n—\\ —o\,\ :0 010 _. q‘ , 01 = - Ct - /e ’ 2 7‘ Q (l: Ci“ + (n43 Om—u “*1 ’ n+2)LnH\. 2. [2 Points] Find the ﬁrst hree non—zero terms of the solution corresponding to y(0) = 1 and y’(0) = 0. l-+-51‘XQ+ tﬁxq+~~ Qo:\ q\:0 3. [2 Points] Find the ﬁrst three non—zero terms of the solution corresponding to y(0) = 0 and y’(0) = 1. go:- 0 o“ =\ M427K Final Exam May 15th, 2007 Question 7 [10 Points] Solve the initial value problem 1/” + 22/ + 2y = W)» y(0) = 0, y’(0) = 0 Where 10 cos(2t) 0 g t < 7r W) = 0 7r 3 t using the partial fractions expansion 103 3—4 3—2 (52+4)(s2+2s+2) "32+4 + 52+2s+2 Lapioce hmnsii‘orm 0? We LHS 1"; (31+ 7.3 +23 RH‘ﬁ {g Chunﬁw \O cosClJc) Laplace iYQmRnrm 09 RHS‘ {3- 3,23: “€qu Sch/Mg RV Ki? we ob‘imn :____ﬂ;___ _ 4Y3 \Os tiin @HiCSI’rZS’rﬁ Q (SIHXCSIQHH Now- 4 [0g ‘ ~\ 5 g ’2 >, 3+] _ i i i i \$1M) (514 23+2)‘g )5 i 31% +2 31+“ + LSHYH BQHYHE a: - QOSQJQ +1 gmlt + eftcosjc :5 {ksm JC Pu’Fi’ihg Hm‘s kgeiirvu omoi Yemembexmg “‘2 915% 3-, 403(2):) ’vlsmélﬂ ~+ e’tcost—Be’itsmt A _ — u“ 0:) [’C03 cm + 2 s m 030 + e‘ckﬂcost’c if) ‘3Qﬂtqﬂsmﬁ-ﬁﬂ =<1-uﬁc+>)(i 3mm» ~Coﬂﬂ) +Q+QOSJC Jggkgm‘f + Hum [ 5H“ COch) ~3 6”“ 3mg M427K Final Exam May 15th, 2007 Question 8 1. [5 Points] Solve the eigenvalue problem 11” + /\y = 0, y’(0) = (Li/(Tr) = 0- You may assume that A Z 0. l1(‘>\=O lhen 3:0‘9CJrC1. 3‘(.O)=<j‘(1ﬂ=0 means 9,20. Q1 Qcm be Cmglhx‘ng 30 there orenowlﬂviql soluliom. A30 ‘3 0“ Eigenvale wd’h QlS-QH'RAHC’lOH €o‘=\ WAX) liken lei Ag}. B:C,cos(M><31Czsma<> ‘5‘ = ‘Cs,M3\Yl(M><> 't'C'Z/U cosUwQ g‘CO’FO =a C1=O. g‘Lﬂ=O %. either Q,=O or 3410041020. For hon-lr‘chil golqhon we mwl’ have MHCMTO=O This meow. Min. Aha-n“ is om QABQIh/Oilbkﬂ Mlllx eigenlimclion QOSCH><> 2. [5 Points] Use the method of separation of variables to replace the partial differential equation mum + at = 0 with a pair of ordinary differential equations. Lei uCX ,Jc3 = XCXX T06 xX“C>O TCH + XCX) T‘CB =0 LXL)‘_M-_)\ ‘. _ YGO TLH Ciel your own equoih‘om. ~DONCXJ +RXCX) =0 T‘Gf) " KTCH =0 9 M427K Final Exam May 15th, 2007 Question 9 Compute the Fourier sine series for the function f deﬁned on [0, 2] by f(m)={x ng<1 2—33 13mg? bn -= E“ L Rx) sm CHI—xhﬁx Eda-Fourier 'ﬁbrmulq in our QOOQ L=Z 1. bn = JO \$300 3m( “91? >O\x - ‘ nnx 1 'hTTX "J‘ x smi’ 1 >Oi>< ‘ii‘ CX~DS\Y\C—f> (AX O — ><Z—7——-——QOSCHT[X) V + fl-ZLQOS (nix ) 0i>< WW ohﬁ z 7. - _ , _’lcosCﬂ) I g ’1 ———n“x 5 [6(2) (W ‘+ MW 203(1)o\.>< i m mw I .__Z_ 31)....igmﬂ 17‘ =—mr (103(1 +W3ln(__l >0 +m Q03 ; hlﬁm a. ‘ - 4 L 4 El ‘ W1 3M1 mini slni l) - ﬂI. * REP: M2 = {O m h even. 1-15) n=1mH Thu) We Fourier' Ema. serum \is z Mold is rmr nrrx hqﬁq 3m (3:) SIHCT 10 M427K Final Exam May 15th, 2007 Question 10 Suppose the temperature u(:c,t) of a metal bar of length 2m satisﬁes the partial differential equation €32 _ an 8362 — 8t. Suppose that both ends of the bar are held at 0°C, i.e. u(0, t) = u(2, t) = 0. 1. [4 Points] Give the basic solutions un(m, t) of the boundary value problem. ‘ Iln’lk n UhLX)t> : e 4 Sln( 1x) 2. [3 Points] Find the solution u(a:, t) Which satisﬁes the initial condition u(:c, O) = 3sin(%) — sin(37rT\$) ULch) = 3u.C><ﬁc\~ ugbqﬂ —T}t 1r>< ~C‘1T2/Jc 311x =3e, 4 stn(-5_-)-«e 4 3m 1 . 3. [3 Points] Find the solution u(x, t) which satisﬁes the initial condition as 0§m<1 u(m’0)_{2——m 13x32. Note: This is the same function that appears in Question 9. °° g nln‘l/Jc vmx v mr * 4 ULX‘H 1?; nlvl 3M: 3 SIM 1 Pulrth t4) N2 gel the Fourier 3mg gang) Pmm HA2 pmzvmm WoleWl. 11 ...
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