Page 1 of 12Student Name:Student Net ID:UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGNDEPARTMENT OF MATHEMATICSMATH 286 SECTION G1 – Introduction to Differential Equations PlusFINAL EXAMINATIONDECEMBER 20, 2012INSTRUCTOR: 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 8 TotalPoints: 8 8 10 12 12 10 18 22 100Score:SOME USEFUL FORMULAS:eA=∞Xk=01k!Ak= I + A +12!A2+13!A3+ . . .x(t) = Φ(t)Φ(a)−1x(a) + Φ(t)ZtaΦ(s)−1f (s)dscos(a) cos(b) =12(cos(a + b) + cos(a − b))sin(a) sin(b) =12(cos(a − b) − cos(a + b))sin(a) cos(b) =12(sin(a + b) + sin(a − b)).Student Net ID: MATH 286 G1 Page 2 of 121. (8 points) Find an explicit solution to the initial value problemdydx= xexy2+ xex; y(0) = 0.Solution:dyy2+ 1= xexdx ⇐⇒ arctan(y) = xex− ex+ C ⇐⇒ y(x) = tan(xex− ex+ C)& y(0) = 0 = tan(C − 1) =⇒ C = kπ + 1 (k ∈ Z)=⇒ y(x) = tan(xex− ex+ kπ + 1) (k ∈ Z).Student Net ID: MATH 286 G1 Page 3 of 122. (a) (3 points) Prove that the differential equation1 + y3+ y cos(xy) +3y2x + x cos(xy)dydx= 0,is exact.Solution: Let M = 1 + y3+ y cos(xy) and N = 3y2x + x cos(xy). ThenMy= 3y2+ cos(xy) − xy sin(xy) = Nx.(b) (5 points) Find an implicit solution to the ODE in part (a).Solution: Since the ODE is exact, an implicit solution will be of the formF (x, y) = 0, whereF (x, y) =ZM(x, y)dx = x + xy3+ sin(xy) + g(y),for some function g(y). To get g, we note thatFy(x, y) = 3xy2+ x cos(xy) + g0(y) = N = 3y2x + x cos(xy) =⇒ g0(y) = 0.So an implicit solution isF (x, y) = x + xy3+ sin(xy) = 0.Student Net ID: MATH 286 G1 Page 4 of 123. The town of Abnormal, Illinois is infected with a population of zombies. Suppose thatthe zombie population P (t) grows by infecting healthy humans, while at the same timeis partially “harvested” (i.e., destroyed) by the local health authority. Assume that P (t)is modeled by the “harvested logistic equation”dPdt= (15 − P )(P − 5) (thousands of zombies per day).(a) (6 points) Sketch the slope field for this differential equation.(b) (2 points) If at day 0, the zombie population is 10 thousand, what will the long-term population P∞= limt→∞P (t) of zombies be?Solution: From part (a), we see that P∞= 15 thousand.(c) (2 points) For what range of initial populations P (0) will the zombie populationP (t) be eventually reduced to zero by the health authority?Solution: 0 ≤ P(0) < 5 thousand.Student Net ID: MATH 286 G1 Page 5 of 124. (a) (6 points) Find the general solution to the differential equationy000+ y00+ 3y0− 5y = 0.(Hint: r0= 1 is one of the roots of the characteristic polynomial.)Solution: From the hint, the characteristic polynomial isP (r) = r3+ r2+ 3r − 5 = (r − 1)(Ar2+ Br + C)= Ar3+ (B − A)r2+ (C − B)r − C=⇒ P (r) = (r − 1)(r2+ 2r + 5)=⇒ P (r) = (r − 1)(r − 1 − 2i)(r − 1 + 2i).From these roots, we get the general solutiony(x) = c1ex+ c2excos(2x) + c3exsin(2x).(b) (5 points) Find a particular solution to the differential equationy000+ y00+ 3y0− 5y = ex+ 1.Solution: Since the forcing function contains a part exwhich is part of thecomplementary function yc(x) from part (a), we take trial solution: yp(x) =Axex+ B. Plugging this trial solution in, we find A, B:ex+ 1 = −5Axex− 5B + 3Aex+ 3Axex+ 2Aex+ Axex+ 3Aex+ Axex.Comparing coefficients, we haveB =−15, 0 = −5A + 3A + A + A 1 = 3A + 2A + 3A =⇒ A =18.(c) (1 point) Write down the general solution to the differential equation in part (b).Solution: y(x) = yc(x) + yp(x) = c1ex+ c2excos(2x) + c3exsin(2x) +18xex−15.Student Net ID: MATH 286 G1 Page 6 of 125. Let P (t) = [pij(t)] be an n × n matrix of continuous functions (on R) and consider thehomogeneous first order linear systemx0= P (t)x (x(t) ∈ Rn).(a) (3 points) Explain what a fundamental matrix is for this linear system.Solution: A fundamental matrix is a matrix-valued function t 7→ Φ(t) ∈ Mn(R)that is invertible for all t and Φ0(t) = P (t)Φ(t). Equivalently, Φ(t) is a matrixwhose columns {x1(t), . . . , xn(t)} form a linearly independent set of solutionsto the system x0= P (t)x.(b) (3 points) Prove that Φ(t) := expRt0P (s)dsis a fundamental matrix for thislinear system. (Here exp(B) denotes the matrix exponential of a matrix B.)Solution: From the properties of matrix exponentials,Φ0(t) =d expRt0P (s)dsdt= P (t) expZt0P (s)ds= P (t)Φ(t),andΦ(t)−1= exp−Zt0P (s)ds.(c) (6 points) Solve the initial value problemx0=0 t20 0x +t0; x(0) =01.Solution: Here P (t) =0 t20 0, so Φ(t) = exp0 t3/30 0=1 t3/30 1, andΦ(t)−1=1 −t3/30 1. Now, by variation of parameters:x(t) =1 t3/30 101+1 t3/30 1Zt01 −s3/30 1s0ds=t3/31+1 t3/30 1t2/20=t3/3 + t2/21Student Net ID: MATH 286 G1 Page 7 of 126. (10 points) Find the general solution to the linear systemx0=1 −4 30 1 60 0 2x.Solution: Let A be the coefficient matrix for the above system. The eigenvalues aregiven bydet(A − λI) = −(λ − 1)2(λ − 2) = 0 =⇒ λ = 1( with mult. 2 ), 2( with mult. 1).For λ = 2, we find an eigenvector v = [a, b, c]T:(A − 2I)v = 0 ⇐⇒−1 −4 30 −1 60 0 0abc=000⇐⇒ v =−21c6cc.So v = [−21, 6, 1]Tis an eigenvector. Next consider λ = 1, and an eigenvectorw = [a, b, c]T. Then we have(A − I)w = 0 ⇐⇒0 −4 30 0 60 0 1abc=000⇐⇒ w =a00.So (up to scaling) there is only one eigenvector w = [1, 0, 0]T, and the defect of λ = 1is one. We therefore seek a length 2 chain of generalized eigenvectors {w, u}, whereu = [a, b, c] satisfies(A − I)u = w ⇐⇒0 −4 30 0 60 0 1abc=100⇐⇒ u =ab0.Taking a = 0, b = 1, we get u = [0, 1, 0].