Practice 4 Solutions

2 1 1 0 1 1 1 1 1 6 2 3 2 1 1 011 1 1 1 3 find an

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Unformatted text preview: di erential equation y = Ay where 2 −2 3 A = 1 −1 1 , −1 22 1 y(0) = 1 0 λ = 1, 1, 1 −4 8 4 100 1 −2 3 t2 e t −2 4 2 = et 0 1 0 + tet 1 −2 1 + 2 000 001 −1 21 eAt 4 1 −1 2t t + tet −1 + t e 2 1 y=e 2 0 0 1 [5] Solve the di erential equation y = Ay where −2 −1 1 1 −1 , A= 2 2 3 1 0 y ( 0) = 1 2 λ = 0, 0, 0 4 40 100 −2 −1 1 2 t −4 −4 0 1 −1 + = 0 1 0 + t 2 2 4 40 001 2 3 1 eAt 0 1 4 2 t −4 y = 1 + t −1 + 2 2 5 4 [6] Solve the di erential equation y = Ay where 1 3 1 1 1 , A = −1 2 −2 −2 1 y ( 0) = 0 2 , 3 × 3 Exercise Set I (identical roots), November λ = 0, 0, 0 eAt 100 0 4 2 1 3 1 2 t 0 −4 −2 1 1 + = 0 1 0 + t −1 2 001 0 8 4 2 −2 −2 4 1 3 2 t −4 y = 0 + t 1 + 2 8 2 −2 [7] Solve the di erential equation y = Ay where −2 3 2 A = −1 2 −2 , −1 1 3 2 y(0) = 0 1 λ = 1, 1, 1 4 −4 −8 100 −3 3 2 2t te 4 −4 −8 = et 0 1 0 + tet −1 1 −2 + 2 001 −1 1 2 0 0 0 eAt 2 0 −4 t2 e t 0 y = et 0 + tet −4 + 2 1 0 0 [8] Solve the di erential equation y = Ay where 2 1 −1 2 , A = −2 −1 −1 −1 2 1 y ( 0) = 0 1 λ = 1, 1, 1 100 1 1 −1 000 t2 et 0 0 0 2 + = et 0 1 0 + tet −2 −2 2 000 001 −1 −1 1 eAt 1 0 0 2t t + tet 0 + t e 0 0 y=e 2 1 0 0 , 3 × 3 Exercise Set J (symmetric matrices), November , 3 × 3 Exercise Set J (symmetric matrices) Linear Algebra, Dave Bayer, November , [1] Find An where A is the matrix 210 A = 1 1 1 012 1 −2 1 1 0 −1 111 n n 0 2 3 −2 4 −2 + 00 0 + 1 1 1 = 6 2 3 1...
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This document was uploaded on 03/24/2014 for the course MATH V2010 at Barnard College.

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