Page 1 of 7Student 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 7Question: 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.(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.Student Net ID: MATH 286 X1 Page 3 of 72. (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.(b) (4 points) Let A be the matrix from part (a). Solve the initial value problemx0(t) = Ax(t); x(0) =123.Student Net ID: MATH 286 X1 Page 4 of 73. (a) (9 points) Find two linearly independent solutions to the systemx0(t) = Ax(t); where A =7 1−4 3(b) (3 points) Write down a fundamental matrix Φ(t) for the system in part (a).(c) (5 points) Compute the matrix exponential etA, where A is the matrix from part(a).Student Net ID: MATH 286 X1 Page 5 of 7(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).Student Net ID: MATH 286 X1 Page 6 of 7(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, 2, . . . , r−1,we get the following system of equations0 = c1v1+ c2v2+ . . . + crvr0 = c2v1+ . . . + crvr−1(mult. by A − λI)0 = c3v1+ . . . + crvr−2(mult. by (A − λI)2). . .0 = cr−1v1+ crv2(mult. by (A − λI)r−2)=⇒ 0 = crv1(mult. by (A − λI)r−1)The last equation implies that cr= 0, the second last then implies that cr−1= 0, andcontinuing up the list of equations, we see that c1= c2= . . . = cr= 0. Therefore thegiven vectors are linearly independent.Student Net ID: MATH 286 X1 Page 7 of 7(Extra work space for your “2