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Unformatted text preview: 286-E1, Spring 2010Practice Exam 2 Solutions11. Solve the initial value problem:x1=x1-x2+x3x2=x2-x3x3=x3with initial conditionsx1(0) = 1,x2(0) = 2,x3(0) = 3.Solution.In matrix-vector notation, the system is:ddt~x=1-111-11~xand we have thatA=I+Bwhere matrixB=A-Iis nilpotent:B=-11-1, B2=0 0 10 0 00 0 0, B3= 0(One way to know in advance thatA-Iis nilpotent is to notice that the characteristic polynomial ofAispA() = (-1)3and every matrix satisfies the equationpA(A) = 0 obtained by setting its characteristicpolynomial to zero.) We now computeetA:etA=et(I+B)=etetB=et(I+tB+t22B2)=et1 0 00 1 00 0 1+-tt-t+0 0t220 00 0=et1-t t+t221-t1The solution with given initial conditions is therefore:~x(t) =etA~x(0) =et1-t t+t221-t1123=et1 +t+3t222-3t3or, given by coordinate equations:x1(t) =et1 +t+3t22x2(t) =et(2-3t)x3(t) = 3et286-E1, Spring 2010Practice Exam 2 Solutions22. Consider the22matrixA=416-1(a) Find eigenvalues and eigenvectors ofA.Solution.The characteristic polynomial is:p() = (4-)(-1-)-6 =2-3-10 = (+ 2)(-5)so the eigenvalues are1=-2 and2= 5. To find corresponding eigenvectors~v1and~v2:(A+ 2I)~v1=~6 16 1~v1=~~v1=1-6(A-5I)~v2=~-116-6~v2=~~v2=11(b) Computeexp(tA).Solution.From part (a) we know thatA=EDE-1whereE=11-6 1, D=-2 05, E-1=171-161etA=EetDE-1=1711-6 1e-2te5t1-161=1711-6 1e-2t-e-2t6e5te5t=17e-2t+ 6e5t-e-2t+e5t-6e-2t+ 6e5t6e-2t+e5t(c) Use your answer in either part (a) or part (b) to find the general solution of the homo-geneous systemddt~x=A~x....
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