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UIUC MATH 286 - midterm03_solutions

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N e t I D : N a m e :Midterm#3MATH 286—Differential Equations PlusThursday, April 17• No notes, p ersonal aids or calculator s are permitted.• Answer all questions in the space provided. If you require more space to write your answer, you may continueon the back of the page. There is a blank page at the end of the exam for rough work.• Explain your work! Little or no points will be given for a correct answer with no explanation of h ow you got it.Good luck!Problem 1. (10 poi nts) Let A be a 5 × 5 matrix with eigenvalues ±3i, 1, 1, 1.(a) Suppose that the eigenvalue λ = 1 has defect 1. Does the equation x′= Ax have (nonzero) so lutions of one ofthe following forms?(v1t + v2) etv1t22+ v2t + v3etv1t36+ v2t22+ v3t + v4et(v1t + v2) sin (3t) v1etcos(3t)Circle those that are solutions (for appropriate choices of the coefficients v1, v2, v3, v4).(b) Now, consider the differential equation x′= Ax +3t2, 0, co s (t), 0, −1T. Write down a particular solutionxpwith undetermined coefficients.Solution.(a) (v1t + v2) etis the only form, among the ones listed, of which t he re exists a nonzero solution.(b) xp= a1t2+ a2t + a3+ a4cos(t) + a5sin(t). Problem 2. (10 points) Three brine tanks T1, T2, T3are connected as indicated in the sketch below.The mixtur es in each tank are kept uniform by stirring.Suppose that the mixture circulates between the tank s atthe rate of 10gal/min. T1and T3contain 100gal of brineand T2contains 50gal .Denote by xi(t) the amount (in pounds) o f salt in tank Tiat time t (in minutes). D erive a system of linear differentialequations for the xi.(Do not solve the system.)T1100galT250galT3100gal10gal/minSolution. In the time interval [t, t + ∆t], we have:∆x1≈ 10 ·x3100· ∆t − 10 ·x1100· ∆tx1′=110x3−110x1∆x2≈ 10 ·x1100· ∆t − 10 ·x250· ∆tx2′=110x1−15x2∆x3≈ 10 ·x250· ∆t − 10 ·x3100· ∆tx3′=15x2−110x3Optional: in matrix form, writing x = (x1, x2, x3), this isx′=−1100110110−150015−110x. Michael Brannan Armin [email protected] [email protected]/4Problem 3. (20 points) Let A =1 −44 −7.(a) Find two linearly independent s olutions to the linear system x′(t) = Ax(t).(b) Compute etA.Solution.(a) The characteristic equation is (1 − λ)(−7 − λ) + 16 = λ2+ 6 λ + 9 = 0. So λ = −3 is an eigenvalue of A withmultiplicity 2. If v = (a, b)Tis an eigenvector, then a = b, so the de fect of λ is 1 and we must build a length2 chain {v1, v2} of generalized eigenvectors. Taking v1= (1, 1)T, one then gets v1= (A + 3I) v2, which hasv2= (1/4, 0)Tas a solution. We therefore obtain the following two linearly independent solutions:x1(t) = e−3tv1=11e−3tand x2(t) = e−3t(t v1+ v2) = t +14t!e−3t.(b) From the first part, we get the fundamental matrixΦ(t) = (x1(t), x2(t)) = e−3t 1 t +141 t!.Hence,etA= Φ(t) Φ(0)−1= 1 t +141 t!e−3t 1141 0!−1= e−3t 1 t +141 t!0 14 −4= e−3t1 + 4t −4t4t 1 − 4t= e−3t1 + 4t −4t4t 1 − 4t. Michael Brannan Armin [email protected] [email protected]/4Problem 4. (20 points) Let A be a 3 × 3 mat rix such that etA=1 + t −t −t − t2t 1 − t t − t20 0 1.(a) What are the eigenvalues of A and what are their defects?(b) Solve the initial value problem x′(t) = Ax(t), x(0) =001.(c) Find a particular solution to the inhomogeneous linear system x′(t) = Ax(t) +02/t30.(d) Find the matrix A.Solution.(a) After exam ining the columns of etA(which a re linearly indep endent solutions to the given DE) and noting thatany solution to the DE is a linear combination of eigenvalue solutions, we see that the only eigenvalue of A isλ = 0 with multiplicity 3. The defect is 2 since terms of the form t eλt, t2eλtappear in the columns of etA.(b) The initial value problem is solved byx(t) = etAx(0) =1 + t −t −t − t2t 1 − t t − t20 0 1001=−t − t2t − t21.(c) By variation of constants,xp(t) = eAtZe−At02/t30dt = eAtZ1 − t t t − t2−t 1 + t −t + t20 0 102/t30dt= eAtZ2/t22/t3+ 2/t20dt=1 + t −t −t − t2t 1 − t t − t20 0 1−2/t−1/t2− 2/t0=−2/t − 2 + 1/t + 2−2 − 1/t2− 2/t + 1/t + 20=−1/t−1/t2− 1/t0is a particular solution.(d) We find A asA =ddtetAt=0=1 −1 −1 − 2 t1 −1 1 − 2 t0 0 0t=0=1 −1 −11 −1 10 0 0. Michael Brannan Armin [email protected] [email protected]/4Problem 5. (15 points) Find four independent real-valued solutions ofx′=3 −4 1 04 3 0 10 0 3 −40 0 4 3x.You may use that the characteristic poly nomial has the repeated roots 3 ± 4i. Moreover, you may use thatv2= (0 0 1 i)Tis a generalized eigenvector of rank 2 for λ = 3 − 4i.Solution. We first find the corresponding eigenvector v1asv1= (A − λI)v2=4i −4 1 04 4i 0 10 0 4i −40 0 4 4i001i=1i00.The chain induces the two so lutionsx1= v1e(3− 4i)t=1i00(cos (4t) − i sin (4t))e3t=cos (4t)sin (4t)00e3t+ i−sin (4t)cos (4t)00e3tx2= (v1t + v2) e(3− 4i)t=tit1i(cos (4t) − i sin (4t))e3t=cos (4t) tsin (4t) tcos (4t)sin (4t)e3t+ i−sin (4t) tcos (4t) t−sin (4t)cos (4t)e3t.Taking real and imaginary part, this gives the four real- valued solutions:cos (4t)sin (4t)00e3t,−sin (4t)cos (4t)00e3t,cos (4t) tsin (4t) tcos (4t)sin (4t)e3t,−sin (4t) tcos (4t) t−sin (4t)cos (4t)e3t Michael Brannan Armin [email protected]


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