MATH 286 Sections D1 & X1: Assignment 6Solutions for the graded problems.Additional Problem [4 points] For the eigenvalue λ1= 1 (with multiplicity 2 anddefect 1), there is only one possibility: a chain {v1, v2} of length 2. For the eigenvalueλ2= −1 (with multiplicity 3 and defect 1) the only possibility is two chains, one oflength 1 and one of length 2, say {u1, u2} and {w1}. Finally for λ3= 5 (with multiplicity4 and defect 2), there are two possibilities: In each case we know there will only betwo chains of generalized eigenvectors (since there are only [multiplicity – defect] =4 − 2 = 2 linearly independent eigenvectors). Now, the first possibility is that there is achain of length 3 and a chain of length 1, say { f1, f2, f2} and {g1}. Then the 9 linearlyindependent solutions arex1(t) = etv1, x2(t) = et(tv1+ v2), x3(t) = e−tu1, x4(t) = e−t(tu1+ u2), x5(t) = e−tw1x6(t) = e5tf1, x7(t) = e5t(t f1+ f2), x8(t) = e5t(t22f1+ t f2+ f3), x9(t) = e5tg1.For the second possibility, we could have two chains of length 2, say { f1, f2} and {g1, g2}.Then the 9 solutions would consist of {x1(t), . . . , x5(t)} as above together withx6(t) = e5tf1, x7(t) = e5t(t f1+ f2), x8(t) = e5tg1, x9(t) = e5t(tg1+ g2)Section 5.5, #28 [3 points] WriteA =5 0 010 5 020 30 5= 5I +0 0 010 0 020 30 0| {z }N,and note thatN2=0 0 00 0 0300 0 0and N3= 0.ThusetA= et(5I+N)= e5tIetN(since 5tI and N commute)= e5tI(I + tN +t22N2)= e5t1 0 010t 1 020t + 150t230t 11Now we can solve the IVP x0= Ax, x(0) = [40, 50, 60]T:x(t) = etAx(0) = e5t1 0 010t 1 020t + 150t230t 1405060= e5t4040t + 50600t2+ 230t + 60.Section 5.6, #24 [3 points] We must solve the IVPx0=3 −19 −3x +0t−2; x(1) =37,and we are givenetA=1 + 3t −t9t 1 − 3t.Since Φ( t) = etAis a fundamental matrix for the system, the variation of parametersformula givesx(t) = e(t−1)A| {z }=Φ(t)Φ(1)−1x(1) + etAZt1e−sAf(s)ds=−2 + 3t −t + 19t − 9 4 − 3t37+1 + 3t −t9t 1 − 3tZt11 − 3s s−9s 1 + 3s0s−2ds=2t + 16t + 1+1 + 3t −t9t 1 − 3tZt1s−1s−2+ 3s−1ds=2t + 16t + 1+1 + 3t −t9t 1 − 3tln |t|1 − t−1+ 3 ln |t|=2 + t + ln |t|5 + 3t − t−1+ 3 ln
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