Page 1 of 8Student Name:Student Net ID:MATH 286 SECTION X1 – Introduction to Differential Equations PlusMIDTERM EXAMINATION 3November 20, 2013INSTRUCTOR: M. BRANNANINSTRUCTIONS• This exam 60 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.”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Student Net ID: MATH 286 X1 Page 2 of 8Question: 1 2 3 TotalPoints: 12 14 24 50Score:1. Consider the following first order linear system of differential equations:x01= −3x1+ 2x3x02= x1− x2x03= −2x1− x2.(a) (4 points) Write this system in the vector-matrix form x0= Ax.Solution:x0=−3 0 21 −1 0−2 −1 0x, x =x1x2x3(b) (8 points) The eigenvalues of the matrix A in part (a) are −2 and −1 ±(√2)i. Aneigenvector associated to the eigenvalue −1 − (√2)i isw =−√2i1−1 −√2i..Find three linearly independent real-valued solutions to this system.Solution: Let v =abcbe an eigenvector for λ1= −2. Then(A + 2I)v1= 0 ⇐⇒−1 0 21 1 0−2 −1 2abc= 0.The first two equations imply that a = 2c = −b, while the third equation is thefirst minus the second. Taking a = 1, we get an eigenvector v =1−11/2. Thisgives one solution x1(t) = e−2tv.We are also given the eigenvector w associated to the eigenvalue −1 − (√2)i.Student Net ID: MATH 286 X1 Page 3 of 8This yields a complex solutionz(t) = e(−1−√2i)tw = e−t(cos√2t − i sin√2t)−√2i1−1 −√2i.= e−t−√2 sin√2tcos√2t−cos√2t −√2 sin√2t.| {z }x2(t)+i e−t−√2 cos√2t−sin√2t−√2 cos√2t + sin√2t.| {z }x3(t)Then x1(t), x2(t), x3(t) are three linearly independent solutions real valuedsolutions.Student Net ID: MATH 286 X1 Page 4 of 82. (a) (10 points) Let λ be a fixed real number, and letA =λ 1 00 λ 10 0 λ.Show that etA=eλtteλtt22eλt0 eλtteλt0 0 eλt.Solution: Write tA = λtI + tN, whereN =0 1 00 0 10 0 0=⇒ N2=0 0 10 0 00 0 0=⇒ N3= 0.From this, we getetA= eλtI+tN= eλtIetN= eλtIetN= eλt(I + tN +t22N2) =eλtteλtt22eλt0 eλtteλt0 0 eλt.An alternate solution to this problem would be to show that the matrix Φ(t) =eλtteλtt22eλt0 eλtteλt0 0 eλtis a fundamental matrix for the system x0= Ax, and thatΦ(0) = I. Then etA= Φ(t)Φ(0)−1= Φ(t).(b) (4 points) Let A be the matrix from part (a). Solve the initial value problemx0(t) = Ax(t); x(0) =123.Solution: The solution isx(t) = etAx(0) =eλtteλtt22eλt0 eλtteλt0 0 eλt123= eλt1 + 2t +3t222 + 3t3.Student Net ID: MATH 286 X1 Page 5 of 83. (a) (9 points) Find two linearly independent solutions to the systemx0(t) = Ax(t); where A =7 1−4 3Solution: First, we find eigenvalues:0 = det(A −λI) = (7 − λ)(3 − λ) + 4 = 21 − 10λ + λ2+ 4 = λ2− 10λ + 25.So λ = 5 is a multiplicity 2 eigenvalue. Next, we look for eigenvectors: Letv =abbe an eigenvector, then(A − 5I)v = 0 ⇐⇒2 1−4 −2ab=00⇐⇒ v =a−2a(a 6= 0).From this we see that λ = 5 has defect 1 and we want to find a length 2chain {v1, w2} of generalized eigenvectors. Taking v1=1−2, we then have forv2=ab,v1= (A − 5I)v2⇐⇒1−2=2 1−4 −2ab⇐⇒1 = 2a + b−2 = −4a − 2b.Taking a = 1/2 and b = 0, we get v2=1/20. This gives the following twolinearly independent solutions:x1(t) = e5tv1& x2(t) = e5t(tv1+ v2).(b) (3 points) Write down a fundamental matrix Φ(t) for the system in part (a).Solution: Let Φ(t) = [x1(t) x2(t)] =e5te5t(t + 1/2)−2e5te5t(−2t).Student Net ID: MATH 286 X1 Page 6 of 8(c) (5 points) Compute the matrix exponential etA, where A is the matrix from part(a).Solution:etA= Φ(t)Φ(0)−1=e5te5t(t + 1/2)−2e5te5t(−2t)1 1/2−2 0−1=e5te5t(t + 1/2)−2e5te5t(−2t)0 −1/22 1=e5t(2t + 1) te5te5t(−4t) e5t(−2t + 1)(d) (7 points) Solve the following initial value problem:x0(t) = Ax(t) +e−t0; x(0) =11,where A is the matrix from part (a).Solution: We will use the matrix exponential from part (c).x(t) = etAx(0) + etAZt0e−sAe−s0ds= etA(x(0) +Zt0e−sAe−s0ds)=e5t(2t + 1) te5te5t(−4t) e5t(−2t + 1) 11+Zt0e−5s(−2s + 1) −se−5se−5s(4s) e−5s(2s + 1)e−s0ds!=e5t(2t + 1) te5te5t(−4t) e5t(−2t + 1) 11+Zt0e−6s(−2s + 1)e−6s(−4s)ds!=e5t(2t + 1) te5te5t(−4t) e5t(−2t + 1) 11+19e−6t(3t − 1) +19−19e−6t(6t + 1) +19!=e5t(3t + 1)e5t(−6t + 1)+e5t(2t + 1) te5te5t(−4t) e5t(−2t + 1)19e−6t(3t − 1) +19−19e−6t(6t + 1) +19=109e5t(3t + 1)109e5t(−6t + 1)+e−t9−1−1Student Net ID: MATH 286 X1 Page 7 of 8(BONUS PROBLEM (5 Points)).Let λ be an eigenvalue of an n × n matrix A and let {v1, v2, . . . , vr} be a length rchain of generalized eigenvectors associated to the eigenvalue λ.(a). Explain what it means to be a length r chain of generalized eigenvectors.For {v1, v2, . . . , vr} to be a length r chain of generalized eigenvectors, the vector vrmust satisfy (A −λI)rvr= 0, (A −λI)r−1vr6= 0, and the remaining vectors v1, . . . , vr−1are then given byvs= (A − λI)r−svr6= 0 (1 ≤ s ≤ r − 1).(b). Show that the vectors {v1, v2, . . . , vr} are linearly independent. (Hint: Supposethat c1v1+c2v2+. . .+crvr= 0. Multiply this equation by (A−λI), (A−λI)2, (A−λI)3,etc... and see what happens.)Suppose that c1v1+ c2v2+ . . . + crvr= 0. Note that(A − λI)kvs= 0 (k ≥ s),and(A − λI)kvs= vs−k(s > k).Therefore if we take the above equation and multiply it by (A−λI)kfor k = 1,