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UT M 408D - HW05-solutions

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tapia (jat4858) – HW05 – clark – (52990) 1This print-out should have 11 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 10.0 pointsIf y1is the particular solution of the differen-tial equationdydx−2yx= 4x2− 5which satisfies y( 1 ) = 5, determine the valueof y1(2).1. y1(2) = 242. y1(2) = 233. y1(2) = 26 correct4. y1(2) = 225. y1(2) = 25Explanation:The integrating factor needed for the firstorder differential equati on (‡) isµ = e−R2xdx= e−2 ln x=1x2.After multiplying both sides of (‡) by 1/x2wecan thus rewrite the equation asddxyx2= 4 −5x2.Consequently, the general solution of (‡) isgiven byyx2= 4x +5x+ Cwhere C is an arbitrary constant. For theparticular solution y1the value of C is deter-mined by the condition y(1) = 5 sincey(1) = 5 =⇒ 5 = 9 + C,and soy1(x) = x24x +5x− 4.At x = 2, therefore,y1(2) = 26.002 10.0 pointsDetermine whether the differential equati onxy′+ ln x − x5y = 0 is linear.1. The equation is not linear.2. The equation is linear. correct3. The equation is neit her l inear nor nonlin-ear.Explanation:003 10.0 pointsSolve the differential equation(6 + t)dudt+ u = 6 + t, t > 0 .1. u =t2+ 12t + 2C2(t + 6)correct2. u =t2+ 6tt + 6+ C3. u =t2+ 6t2(t + 6)+ C4. u =t2+ 6t + 2C2(t + 6)5. u =t2+ 6t + Ct + 6Explanation:004 10.0 pointsUse the Bernoulli’ s method to sol ve thedifferential equati on y′+8xy =y3x4.1. y =Cx16±219x31/2tapia (jat4858) – HW05 – clark – (52990) 22. y = ±Cx16−119x3−1/23. y =Cx16±219x3−1/24. y = ±Cx16+219x3−1/2correct5. y = ±Cx16+219x31/2Explanation:keywords:005 10.0 pointsThe performance level of someone learning askill P (t) is a function of the training time tand given by the differential equationdPdt=k(M − P (t)), where k is a positive constant.Two new workers were hired for an assemblyline. Jim processed 27 units during the firsthour and 43 units during the second hour.Mark processed 32 units during the first hourand 48 units the second hour.Estimate the maximum number of unit s perhour that each worker is capable of processing.Assume that P (0) = 0.1. 66 for Jim a nd 64 for Mark correct2. 75 for Jim a nd 68 for Mark3. 53 for Jim a nd 80 for Mark4. 62 for Jim a nd 66 for Mark5. 68 for Jim a nd 70 for MarkExplanation:006 10.0 pointsThe figure shows a circuit containing anelectromotive force, a capacitor with a capac-itance of C farads (F), and a resistor with aresistance of R ohms (Ω). The voltage dropacross the capacitor isQC, where Q i s thecharge (in coulombs), so in this case Kirch-hoff’s Law gives R I+QC= E(t). B ut I =dQdt,so we have RdQdt+1CQ = E(t). Suppose theresistance is 10 Ω, the capacitance is 0.05 F, abattery gives a constant voltage of 60 V, andthe initial charge is Q(0) = 0 C.0.05 F10 Ω60 VSwitchFind the charge and the current at time t.1. Q(t) = 3 e−2t, I(t) = 61 − e−2t2. Q(t) = 61 − e−2t, I(t) = 3 e−2t3. Q(t) = 31 − e−2t, I(t) = 6 e−2tcor-rect4. Q(t) = 6 e−2t, I(t) = 31 − e−2t5. Q(t) = 31 + e−2t, I(t) = 6 e−2tExplanation:007 10.0 pointsPopulations of aphids and ladybugs are mod-eled by the equationsdAdt= 2A − 0 .01ALdLdt= −0.7L + 0.0001ALFind an equilibrium soluti on.1. A = 7000, L = 200 correct2. A = 6000, L = 200tapia (jat4858) – HW05 – clark – (52990) 33. A = 4000, L = 1004. A = 7000, L = 05. A = 9000, L = 200Explanation:008 10.0 pointsPopulations of aphids and ladybugs are mod-eled by the equationsdAdt= 5A − 0 .01ALdLdt= −0.4L + 0.0001ALFind an expression fordLdA.1.dLdA=−0.04A + 0.00001AL(5A − 0.01AL)22.dLdA=−0.04L + 0.00001AL5A − 0.01AL3.dLdA=−0.4L + 0.0001AL5A − 0.01ALcorrect4.dLdA=5L − 0.01AL−0.4L + 0.0001AL5.dLdA=5A − 0.01AL−0.4L + 0.0001ALExplanation:009 10.0 pointsPopulations of aphids and ladybugs can bemodeled with a Lotka-Volterra system as fol-lows:dAdt= 3A(1 − 0.0001A) − 0.01ALdLdt= −0.5L + 0.0001ALFind the equilibrium solutions.1. A = 7000, L = 502. A = 7000, L = 1503. A = 5000, L = 150 correct4. A = 5000, L = 3505. A = 8000, L = 200Explanation:010 10.0 pointsPopulations of aphids and ladybugs can bemodeled with a Lotka-Volterra system as fol-lows:dAdt= 5A(1 − 0.0001A) − 0.01ALdLdt= −0.7L + 0.0001ALFind an expression fordLdA.1.dLdA=5A(1 − 0.0001A) − 0.01AL−0.07L + 0.00001AL2.dLdA=−0.7L + 0.0001AL5A(1 + 0.0001A) − 0.01AL3.dLdA=−0.07L + 0.00001AL5A(1 − 0.0001A) − 0.01AL4.dLdA=5A(1 − 0.0001A) − 0.01AL−0.7L + 0.0001AL5.dLdA=−0.7L + 0.0001AL5A(1 − 0.0001A) − 0.01ALcor-rectExplanation:011 10.0 pointsPopulations of aphids and ladybugs can bemodeled with a Lotka-Volterra system as fol-lows:dAdt= 2A(1 − 0.0001A) − 0.01ALdLdt= −0.5L + 0.0001ALSuppose that at t ime t = 0 there are 1 000aphids (A) and 200 ladybugs (L).tapia (jat4858) – HW05 – clark – (52990) 4Choose the corresponding phase trajectory.1.2000 4000 6000200400AL2.2000


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