Page 1Math 217Exam 2Name:ID:Section:This exam has 14 questions:• 12 multiple choice worth 6 points each.• 1 hand graded worth 28 points.Important:• No graphing calculators!• For the multiple choice questions, mark your answer on the answer card.• Show all your work for the written problems. You will be graded on the ease of reading yoursolution.• You are allowed a 3 × 5 note card for the exam.1. Multiply the matrices1 2 −2−3 2 05 −2 10 1 −2−1 2 14 4 0Add up all the entries in your answer and select the closest answer below.(a) −28(b) −21(c) −14(d) −7 −→ CORRECT(e) 0(f) 7(g) 14(h) 21(i) 28(j) 34(k) 43Solution:−10 −3 0−2 1 86 5 −12−→ −7Page 2Math 217Exam 22. Consider the systemX0=3 −2 0−1 3 −20 −1 3XThis system has solutions belowX1(t) = e2t221X2(t) = e4t20−1X3(t) = e6t2−21Let W (t) = W (X1, X2, X3) be the Wronskian. Find W (0). Choose the closest answer.(a) −20(b) −16 −→ CORRECT W = −16e12t(c) −12(d) −8(e) −4(f) 0(g) 4(h) 8(i) 12(j) 16(k) 20Solution:W =2e2t2e4t2e6t2e2t0 −2e6t1e2t−1e4t1e6t= −16e12tPage 3Math 217Exam 23. Consider the initial value problemx0= x2− 2y −t x(0) = 1y0= 5x − 4y + t y(0) = −2Using Runge-Kutta with step size h = 0.5 to approximate y(0.5). Choose the closest answer.(a) −2.0(b) −1.5(c) −1.0(d) −0.5(e) 0(f) 0.5(g) 1.0(h) 1.5(i) 2.0 −→ CORRECT 1.914683(j) 2.5(k) 3.0Solution:n tnXnK1K2K3K40 0.01−2 513 2.31256.5 2.99059.6406 0.0855941.69491 0.52.30761.91468Page 4Math 217Exam 24. Convert the initial value problem to a first order system and use Euler’s method with step sizeh = 0.5 to approximate x(1.0)x00− 2(x0)2+ x2t = −1 x(0) = 1, x0(0) = −1Choose the closest answer.(a) 4.40 −→ COR REC T(b) 4.45(c) 4.50(d) 4.55(e) 4.60(f) 4.65(g) 4.70(h) 4.75(i) 4.80(j) 4.85(k) 4.90Solution:Equation transforms tox01= x2x1(0) = 1x02= 2x22− x21t − 1 x2(0) = −1n tnXnK = F (t, Xn)0 0.01−1 −111 0.50.5−0.5−0.5−0.6252 1.00.25−0.8125So, x(0) ≈ 0.25.Page 5Math 217Exam 25. Which of the following triplets of functions are linearly independent?I. f(x) = ex, g(x) = e−x, h(x) = 1II. f(x) = ex, g(x) = cos x, h(x) = sin xIII. f(x) = x, g(x) = x2, h(x) = x2− xIV. f(x) = cos2x, g(x) = sin2x, h(x) = 1(a) II only(b) IV only(c) I and II only −→ CORRECT(d) I and IV only(e) II and III only(f) I, II and III only(g) I, II and IV only(h) II, III and IV only(i) I, II, III and IVSolution: We can compute Wronskians to answer this question (you can also see some of themare linearly independent be seeing that one of the functions is a combination of the others).I. W (f, g, h) = 2 6= 0II. W (f, g, h) = 2ex6= 0III. W (f, g, h) = 0, also note h = f − g.IV. W (f, g, h) = 0, also note h = f + g.Page 6Math 217Exam 26. Consider a mass-spring with mass m = 1, c = 0 (no damping) and k = 25. An input forcesatisfies F (t) = F0sin ωt. For what value of ω, if any, is there resonance? Choose the closestanswer.(a) ω = 0(b) ω = 1(c) ω = 4(d) ω = 5 −→ CORRECT(e) ω = 8(f) ω = 9(g) ω = 16(h) ω = 25(i) No resonance possible(j) None of the aboveSolution: Resonance occurs when ω = ω0=qkm= 5.7. Choose the correct form of a particular solution to the diffe rential equation below using themethod of undetermined coefficients.y(3)− y00− 4y0= sin x + x2− 5(a) yp= A sin x + B + Cx2(b) yp= A cos x + B + Cx2(c) yp= A sin x + B + Cx + Dx2(d) yp= A cos x + B + Cx + Dx2(e) yp= A cos x + B sin x + C + Dx + Ex2(f) yp= A cos x + B sin x + x(C + Dx + Ex2) −→ CORRECT(g) yp= x(A cos x + B sin x) + C + Dx + Ex2(h) yp= x(A cos x + B sin x) + x(C + Dx + Ex2)(i) yp= x(A cos x + B sin x) + C + Dx + Ex2+ F x3(j) yp= x(A cos x + B sin x) + x(C + Dx + Ex2+ F x3)(k) The method of undetermined coefficients can not be used with this differential equation orsome other answer.Solution: It is important to note that the solution to the homogeneous equation isyc= c1+ ex/2"c2cos t√172!+ c3sin t√172!#Actually, all you have notice is that y = c1is part of ycand y = sin x is not part of yc. Once youdo this, you should be able to see the correct form.Page 7Math 217Exam 28. Select the differential equation that has y = 4e2x− e−2xcos 2x as a solution.(a) y(4)− 6y000+ 15x00− 17y0+ 10 = 0(b) y000+ 2y00− 16y = 0 −→ CORRECT(c) y000+ 2y00− 3y0− 10y = 0(d) y000+ y00− 4y0− 4y = 0(e) y000− y00− 4y0+ 4y = 0(f) y000− 3y0+ 2y = 0(g) y000+ 2y00− y0− 2y = 0(h) y000+ 4y00+ 5y0+ 2 = 0(i) All of the above have the given solution(j) None of the above have the given solutionSolution: If this is a solution then the roots of the characteristic polynomial must be 2 and−2 ± 2i. Therefore the characteristic polynomial must have, as a factor:(r − 2)[r − (−2 + 2i)][r − (−2 − 2i)] = r3+ 2r2− 16Page 8Math 217Exam 29. To solve non-homogeneous equations we discussed the method of undetermined coefficients andvariation of parameters. Which of the following equations does the method of undeterminedcoefficients apply?I. 3y000− 2y00+ 1y0= e2x+ cos(x2)II. xy000− x3y00+ x2y0= cos x + sin xIII. 2y000+ 5y00− πy0= x cos x + sin xIV. 4y000+ 7y0− 13y = xex2(a) I only(b) II only(c) III only −→ CORRECT(d) IV only(e) I and II only(f) I and III only(g) I and IV only(h) II and III only(i) II and IV only(j) III and IV only(k) Some other combination or none of the aboveSolution: I. and IV. can not be solved using undetermined coefficient because the derivativesof the left hand side are not spanned by a finite set of linearly independent functions.II. is ruled out because of the x2in the differential equation—you need constant coefficients touse the method of undetermined coefficients.Page 9Math 217Exam 210. Solve the initial value problemy00− 4y0+ 4y = e2x; y(0) = 0, y0(0) = 0.What is y(2)?(a) 0(b) 1(c) 2(d)12e2(e) 2e2(f)14e4(g)12e4(h) e4(i) 2e4−→ CORRECT y =12x2e2x(j) 4e4Solution: Factor the characteristic equation and find complementary solution yc= c1e2x+c2xe2x. Use the method of undetermined coefficients to find yp. The correct form for ypshouldbe yp= x2(Ae2x). Plugging in and solving gives A =12which gives a general solutiony = c1e2x+ c2xe2x+x22e2xUsing the initial conditions to solve for c1and c2gives c1= c2= 0 and thereforey =x22e2xAnd therefore y(2) = 2e4.Page 10Math 217Exam 211. A
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