Page 1 of 8Student Name:Student Net ID:MATH 286 SECTION G1 – Introduction to Differential Equations PlusMIDTERM EXAMINATION 3November 15, 2012INSTRUCTOR: M. BRANNANINSTRUCTIONS• This exam 50 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.”Student Net ID: MATH 286 G1 Page 2 of 8Question: 1 2 3 TotalPoints: 24 16 12 52Score:1. Consider the following linear system of first order differential equations:x01= x1+ x2+ x3x02= 2x1+ x2− x3x03= −x2+ x3.(a) (4 points) Write this system in the vector-matrix form x0= Ax.Solution:x0=1 1 12 1 −10 −1 1x, x =x1x2x3(b) (8 points) The eigenvalues of the matrix A in part (a) are λ1= −1 (with multi-plicity 1) and λ2= 2 (with multiplicity 2). Find an eigenvector v1associated toλ1, and determine how many linearly independent eigenvectors v2associated to λ2there are.Solution: Let v1= [a, b, c]Tbe an eigenvector for λ1= −1. Then(A + I)v1= 0 ⇐⇒2 1 12 2 −10 −1 2abc= 0 ⇐⇒2a + b + c = 02a + 2b − c = 0−b + 2c = 0..The third equation says that b = 2c, which implies that the first equation saysthat 2a = −3c. The second equation is simply the first equation minus thethird equation, so it is automatically satisfied once the other two are. Thereforev1= [−3c/2, 2c, c]Tis an eigenvector for all c 6= 0. Taking c = 2, we getv1= [−3, 4, 2]T. Now let v2= [a, b, c]Tbe an eigenvector for λ1= 2. Then(A − 2I)v1= 0 ⇐⇒−1 1 12 −1 −10 −1 −1abc= 0 ⇐⇒−a + b + c = 02a − b − c = 0−b − c = 0..Since the third equation is a linear combination of the first and the second,we really only have the two equations −a + b + c = 0 and b = −c. Thus anysolution to this pair of equations is of the form v2= [0, b, −b] for some b ∈ R.Setting b = 1, we see that λ2= 2 admits (up to scaling) just one eigenvectorv2= [0, 1, −1]T.Student Net ID: MATH 286 G1 Page 3 of 8(c) (2 points) What is the defect of the eigenvalue λ2= 2?Solution: Since λ2= 2 has multiplicity 2, but only admits one linearly inde-pendent eigenvector, the defect is d = 2 − 1 = 1.(d) (10 points) Using the method of (generalized) eigenvectors, write down the generalsolution x(t) for this system.Solution: Based on the above, we know that there are three linearly indepen-dent solutionsx1(t) = e−tv1, x2(t) = e2tv2, x3(t) = e2t(tv2+ w),where {v2, w} is a length 2 chain of generalized eigenvectors based at v2. Tofind w = [a, b, c]T, we must solve(A − 2I)w = v2⇐⇒−1 1 12 −1 −10 −1 −1abc=01−1.Adding the first and last equation, we get a = 1. Then the three equations justsay b + c = 1. Taking b = 1, we get c = 0 and w = [1, 1, 0]T. The generalsolution is then going to be a linear combinationx(t) =3Xi=1cixi(t).Student Net ID: MATH 286 G1 Page 4 of 82. (a) (5 points) Let a and b be real numbers, and consider the matrixB =a b0 a∈ M2(R).Show that etB=eatbteat0 eatfor all t ∈ R.Solution: Write tB = atI + tN, where tN =0 bt0 0. Then (tN)2= 0, and tNand atI commute. ThereforeetB= eatI+tN= eatIetN= eatI(I + tN) = eat1 bt0 1=eatbteat0 eat.(b) (5 points) Compute etA, whereA =1 10 0 0 00 1 0 0 00 0 2 0 00 0 0 4 10 0 0 0 4.(Hint: Use part (a) and the fact that A is block-diagonal.)Solution: Using part (a) and the fact that A is block-diagonal, we haveet10tet0 0 00 et0 0 00 0 e2t0 00 0 0 e4tte4t0 0 0 0 e4tStudent Net ID: MATH 286 G1 Page 5 of 8(c) (6 points) Let A be the matrix from part (b). Solve the initial value problemx0= Ax +et0000; x(0) =00000.Solution: Using the variation of parameters formula,x(t) = etAx(0) + etAZt0e−sAf(s)ds= 0 + etAZt0e−s−10se−s0 0 00 e−s0 0 00 0 e−2s0 00 0 0 e−4sse−4s0 0 0 0 e−4ses0000ds= etAZt010000ds = etAt0000=tet0000.Student Net ID: MATH 286 G1 Page 6 of 83. Let A ∈ Mn(R) be an n × n matrix, and let λ be an eigenvalue of A with associatedeigenvector v ∈ Rn.(a) (5 points) Let ω be a (possibly complex) number such that ω2= λ. Show that thefunction x(t) = eωtv is a solution to the second order systemx00= Ax.Solution:x00(t) =d2dt2(eωtv) = ω2eωtv = eωtλv = eωtAv = Aeωtv = Ax(t).(b) (7 points) Using the method of part (a), find four distinct solutions to the secondorder systemx00=5 33 5x.Solution: Let A be the above matrix. Then the characteristic equation of A isdet(A − λI) = (5 − λ)2− 9 = 0,so A has eigenvalues λ = 5 ± 3 = 8, 2. For λ1= 8, we solve the equation(A − 8I)v1= 0 ⇐⇒−3 33 −3v1= 0, which has a solution v1= [1, 1]T. Forλ2= 2, we solve (A − 2I)v2= 0 ⇐⇒3 33 3v2= 0, which has a solutionv1= [1, −1]T. Therefore the four distinct solutions we get arex1(t) = e√8tv1, x2(t) = e−√8tv1, x3(t) = e√2tv2, x4(t) = e−√2tv2.(c) (Bonus - 3 points) Prove that your four solutions are linearly independent. (Youcan use the fact that distinct eigenvalues yield linearly independent eigenvectors.)Solution: Let c1, c2, c3, c4be such thatP4i=1cixi(t) = 0 for all t. We wantto show that c1= c2= c3= c4= 0. Note that since v1and v2are linearlyindependent, we have0 =4Xi=1cixi(t) = (c1e√8t+ c2e−√8t)v1+ (c3e√2t+ c4e−√2t)v2⇐⇒ 0 = c1e√8t+ c2e−√8t& 0 = c3e√2t+ c4e−√2t(for all t).Student Net ID: MATH 286 G1 Page 7 of 8Now if we note that {e√8t, e−√8t} form a pair of linearly independent functions, itfollows that c1= c2= 0. Similarly, since {e√2t, e−√2t} are linearly independent,we get c3= c4= 0.Student Net ID: MATH 286 G1 Page 8 of 8(Extra work