Page 1 of 16Student Name:Student Net ID:UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGNDEPARTMENT OF MATHEMATICSMATH 286 SECTION X1 – Introduction to Differential Equations PlusFINAL EXAMINATIONDECEMBER 17, 2013INSTRUCTOR: M. BRANNANINSTRUCTIONS• This exam is three (3) hours 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.”Question: 1 2 3 4 5 6 7 TotalPoints: 14 25 9 9 14 15 14 100Score:Student Net ID: MATH 286 X1 Page 2 of 16SOME USEFUL FORMULAS:eB=∞Xk=01k!Bk= I + B +12!B2+13!B3+ . . .x(t) = Φ(t)Φ(a)−1x(a) + Φ(t)ZtaΦ(s)−1f(s)dsa bc d−1=1ad − bcd −b−c aan=1LZL−Lf(t) cosnπtLdt, bn=1LZL−Lf(t) sinnπtLdt1. Let f(t) be the π-periodic function defined byf(t) =(1 −π2< t < 0,t 0 ≤ t ≤π2.(a) (2 points) Sketch the graph of f over a few periods.(b) (6 points) Leta02+P∞n=1(ancosnπtL+bnsinnπtL) be the Fourier series for f. Calculatethe Fourier coefficients a0, anand bn(n ≥ 1).Student Net ID: MATH 286 X1 Page 3 of 16Solution:a0=2πZπ/2−π/2f(t)dt =2ππ2+π28= 1 +π4,an=2πZπ/2−π/2f(t) cos(2nt)dt =2πZ0−π/2cos(2nt)dt +2πZπ/20t cos(2nt)dt=sin(2nt)nπ0−π/2+2πt sin(2nt)2n+cos(2nt)4n2π/20=cos(nπ) − 12πn2,bn=2πZπ/2−π/2f(t) sin(2nt)dt =2πZ0−π/2sin(2nt)dt +2πZπ/20t sin(2nt)dt=−cos(2nt)nπ0−π/2+2π−t cos(2nt)2n+sin(2nt)4n2π/20=cos(nπ) − 1nπ−cos(nπ)2n.Student Net ID: MATH 286 X1 Page 4 of 16(c) (2 points) Does the Fourier series for f converge to f(t) at every point t? Whatdoes the Fourier series converge to when t = 0?Solution: Since f is piecewise smooth and has jump discontinuities, the FS forf does not converge at every point t. For example, at t = 0 the FS convergestof(0+)+f(0−)2= 1/2 6= f(0).(d) (4 points) A 1 kg cart is connected to a wall by a spring with unknown springconstant k > 0 N/m, and is periodically forced by f(t) Newtons (where f is theperiodic function defined above). Assuming there is no friction in the system, theresulting equation of motion for the displacement x(t) of the cart from rest is givenbyx00+ kx = f(t).Find all values of k that will cause resonance in the forced mechanical system.Solution: When the Fourier Seriesa02+P∞n=1(ancos(2nt) + bnsin(2nt) for fcontains a non-zero sine or cosine component with frequency equal to the naturalfrequency ω0=√k of the system, there will be resonance. Since all the bn’s(and all the odd an’s) are non-zero, this happens precisely when√k = 2n ⇐⇒ k = 4n2(n = 1, 2, 3, . . .).Student Net ID: MATH 286 X1 Page 5 of 162. In this multi-part problem, we will derive the solution to a one-dimensional heat equationwith mixed boundary conditions (with one endpoint held at a fixed temperature, and theother endpoint insulated). For the remainder of this problem, let L > 0 be fixed.(a) (6 points) Consider the constant functionf(x) = 100 defined on the interval [0, 2L].Sketch the graph of the 4L-periodic odd extension of f and compute its Fouriersine series.Solution: The expansion we seek is f(x) =P∞n=1bnsinnπx2L, wherebn=22LZ2L0f(x) sinnπx2Ldx =100LZ2L0sinnπx2Ldx=−200nπcosnπx2L2L0=n0 n is even400nπn is odd(b) (5 points) Consider the following eigenvalue problem for the function X(x) on theinterval [0, L]:X00+ λX = 0; X(0) = X0(L) = 0.Show that the λ is an eigenvalue if and only if λ = λn=(2n−1)2π24L2, where n =1, 2, 3, . . .. For each λn, write down the corresponding eigenfunction Xn(x).(Note: You may assume without proof that all the eigenvalues are positive.)Solution: Since λ > 0, the general solution to this ODE isX(x) = c1cos(√λx) + c2sin(√λx).From the BC’s, we getc1= 0, &√λc2cos(√λ2L) = 0 =⇒√λ2L = (2n−1)π ⇐⇒ λ =(2n − 1)2π24L2,where n = 1, 2, . . . . The corresponding eigenfunction is Xn(x) = sin(2n−1)πx2L.Student Net ID: MATH 286 X1 Page 6 of 16(c) (3 points) A laterally insulated metal rod of length L (with thermal diffusivityk = 2) is heated to a uniform temperature of 100 degrees Celsius. At time t = 0,the left end of the rod (x = 0) is placed in an ice bath at 0 degrees Celsius, and theright end (x = L) is insulated so that no heat flows in or out at this end. If u(x, t)denotes the temperature (in degrees Celsius) of the rod at position 0 < x < L andtime t > 0, then u satisfies the one-dimensional heat equationut= 2uxx(0 < x < L, t > 0).Write down the Boundary Conditions and Initial Condition for this problem.Solution:u(0, t) = 0 = ux(L, t) = 0 & u(x, 0) = 100.(d) (4 points) Using the method of separation of variables, show that ifu(x, t) = X(x)T (t)is a solution to the above heat equation satisfying the boundary conditions frompart (c), then X(x) must be a solution to the eigenvalue problem in part (b).Solution: If u(x, t) = X(x)T (t), thenut= 2uxx⇐⇒ XT0= 2X00T ⇐⇒X00X=T02T= −λ (λ ∈ R).This together with the BCs above yields the two ODEsX00+ λX = 0; X(0) = X0(L) = 0andT0+ 2λT = 0.(e) (3 points) For each eigenfunction Xn(x) from part (b), find the corresponding so-lution Tn(t).Solution:T0n+ 2(2n − 1)2π24L2Tn= 0 =⇒ Tn(t) = exp− 2(2n − 1)2π2t4L2Student Net ID: MATH 286 X1 Page 7 of 16(f) (4 points) Letu(x, t) =∞Xn=1αnXn(x)Tn(t).Then u(x, t) satisfies the heat equation and boundary conditions from part (c).Find the constants αnso that u(x, t) also satisfies the initial condition u(x, 0).(Hint: Use part (a).)Solution: We haveu(x, t) =∞Xn=1αnsin(2n − 1)πx2Lexp− 2(2n − 1)2π2t4L2,so100 = u(x, 0) =∞Xn=1αnsin(2n − 1)πx2L.This tells us thatαn= b2n−1(n = 1, 2, . . .).Student Net ID: MATH 286 X1 Page 8 of 163. (9 points) Solve the initial value problemdydx=x + 3yx − y; y(1) = 0.An implicit equation for y(x) is fine.Solution: Let v =yx. Thendydx= xv0+ v and the equation becomesxv0+ v =1 + 3v1 − v⇐⇒ xv0=1 + 2v + v21 − v⇐⇒(1 − v)dv(v +