Page 1 of 7Student Name:Student Net ID:MATH 286 SECTION G1 – Introduction to Differential Equations PlusMIDTERM EXAMINATION 2October 18, 2012INSTRUCTOR: M. BRANNANINSTRUCTIONS• This exam 50 minutes long. No personal aids or calculators are permitted.• Answer all questions in the space provided. If you require more space to write youranswer, you may continue on the back of the page. There is a blank page at the endof the exam for rough work.• EXPLAIN YOUR WORK! Little or no points will be given for a correct answer withno explanation of how you got it. If you use a theorem to answer a question, indicatewhich theorem you are using, and explain why the hypotheses of the theorem are valid.• GOOD LUCK!PLEASE NOTE: “Proctors are unable to respond to queries about the interpretation ofexam questions. Do your best to answer exam questions as written.”Student Net ID: MATH 286 G1 Page 2 of 7Question: 1 2 3 4 TotalPoints: 15 15 12 12 54Score:1. Consider the ordinary differential equationLy = y(3)− 2y00+ 2y0= x + xex.(a) (5 points) Determine the complementary function yc(x) for the above ODE.Solution: The characteristic polynomial isP (r) = r3− 2r2+ 2r = r(r2− 2r + 2) = r(r − 1 − i)(r − 1 + i).Therefore the complementary function (which is the solution to Ly = 0) isyc(x) = c1+ c2excos x + c2exsin x.(b) (8 points) Find a particular solution yp(x) to the above ODE.Solution: Since F (x) = x + xex, a first guess at a trial solution would be of theform Ax+B+(Cx+D)ex. But there is overlap with the complementary functionyc(x) coming from the constant B. To eliminate this we multiply Ax + B by xand get a particular solution of the formyp(x) = x(Ax + B) + (Cx + D)ex=⇒ y0p(x) = 2Ax + B + Cex+ Cxex+ Dex= 2Ax + B + Cxex+ (C + D)ex=⇒ y00p(x) = 2A + (C + D)ex+ Cex+ Cxex= 2A + Cxex+ (2C + D)ex=⇒ y(3)p(x) = (2C + D)ex+ Cex+ Cxex= Cxex+ (3C + D)ex.Setting Lyp= F and comparing coefficients, we get(coefficient of xex) : 1 = C − 2C + 2C =⇒ C = 1(coefficient of ex) : 0 = 3C + D − 2(2C + D) + 2(C + D) =⇒ D = −1(coefficient of x) : 1 = 4A =⇒ A =14(coefficient of 1) : 0 = −4A + 2B =⇒ B =12.Student Net ID: MATH 286 G1 Page 3 of 7(c) (2 points) Write down the general solution to the above ODE.Solution: The general solution isy(x) = yc(x) + yp(x) = c1+ c2excos x + c2exsin x +14x2+12x + (x − 1)exStudent Net ID: MATH 286 G1 Page 4 of 72. Consider the ordinary differential equationLy = y00−1xy0+1x2y = x3.(a) (4 points) Verify that y1(x) = x and y2(x) = x ln |x| are solutions to the associatedhomogeneous ODELy = 0 (x 6= 0).Solution:y001−1xy01+1x2y1= 0 −1x+1x= 0.y002−1xy02+1x2y2=1x−1x(ln |x| + 1) +1xln |x| = 0.(b) (5 points) Compute the Wronskian W (y1, y2) for the pair of functions y1, y2above.Are y1and y2linearly independent on the interval I = (0, ∞)? Why or why not?Solution:W (x) = dety1(x) y2(x)y01(x) y02(x)= x(ln |x| + 1) − 1(x ln |x|) = x.Since y1, y2are solutions to the ODE y00+ p(x)y0+ q(x) = 0 and W (y1, y2) 6= 0on I, the theorem on linear independence and Wronskians implies that thesefunctions are linearly independent on I.(c) (6 points) Find a particular solution toy00−1xy0+1x2y = x3(x > 0).(Hint: One possible approach is to use the variation of parameters method.)Solution: Letu1(x) = −Zy2(x)x3W (x)dx = −Zx3ln |x|dx =−x44ln |x|+Zx34dx =−x44ln |x|+x416.u2(x) =Zy1(x)x3W (x)dx =Zx3dx =x44.Then the method of variation of parameters givesyp(x) = u1(x)y1(x) + u2(x)y2(x) =x5414− ln |x|+x54ln |x| =x516.Student Net ID: MATH 286 G1 Page 5 of 73. (12 points) For the following endpoint problem, determine all eigenvalues λ ∈ R andtheir associated eigenfunctions:y00− 4y0+ λy = 0, y(0) = y(1) = 0.Solution: The characteristic equation is0 = P (r) = r2− 4r + λ ⇐⇒ r = 2 ±√16 − 4λ2= 2 ±√4 − λ.Case 1: If λ < 4, the general solution isy(t) = c1e(2+√4−λ)t+ c2e(2−√4−λ)t.Plugging in the endpoint conditions, we find c1= c2= 0. So, there are no eigenvaluesλ < 4.Case 2: If λ = 4, the general solution isy(t) = c1e2t+ c2te2t.Plugging in the endpoint conditions, we find0 = y(0) = c10 = y(1) = c2e2=⇒ c1= c2= 0.So λ = 4 is not an eigenvalue.Case 3: If λ > 4, the general solution isy(t) = c1e2tcos√λ − 4t + c2e2tsin√λ − 4t.The endpoint conditions then give0 = y(0) = c10 = y(1) = c2e2sin√λ − 4 =⇒√λ − 4 = nπ (n = 1, 2, 3, . . .).So the eigenvalues areλn= 4 + n2π2(n = 1, 2, . . .),and the eigenfunctions areyn(t) = e2tsin(nπt)Student Net ID: MATH 286 G1 Page 6 of 74. Consider a mass-spring-dashpot system with mass m = 1kg, spring constant k = 4 N/mand dashpot damping constant β > 0 N s/m. Let x(t) denote the displacement (inmetres, at time t) of the mass from its equilibrium resting position.(a) (4 points) For what values of β is the system underdamped?Solution: The characteristic polynomial for this system isP (r) = mr2+ βr + k = r2+ βr + 4,which has roots r =−β2±√β2−162. The system is underdamped when the rootsare complex. I.e., when β2− 16 < 0 ⇐⇒ 0 < β < 4.For the remainder of the problem, assume that the dashpot is disconnected fromthe system (i.e., set β = 0) and that an external force F (t) = 2 sin ωt Newtons isapplied to the mass.(b) (2 points) At what forcing frequency ω will resonance occur in the forced system?Solution: Resonance will occur when ω = ω0=pk/m = 2, which is thenatural frequency of the system.(c) (6 points) Write down the general solution x(t) in this case.Solution: The complementary function for this equation isxc(t) = C cos(2t − α).A particular solution will be of the form xp(t) = At cos(2t) + Bt sin(2t), settingy00p+ 4yp= 2 sin(2t), we get2 sin(2t) = −4A sin(2t)−4At cos(2t)+4B cos(2t)−4Bt sin(2t)+4At cos(2t)+4Bt sin(2t),giving A = −1/2 and B = 0. Thus the general solution isx(t) = xc(t) + xp(t) = C cos(2t − α) −t2cos(2t).Student Net ID: MATH 286 G1 Page 7 of 7(Extra work