# Vectory 2 1 6 3 1 3 y 1 y 2 y 3 y 1 y 2 y 3 c 3 1 2 c

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Unformatted text preview: vectory = − 2 1 6 − 3 − 1 − 3 y 1 y 2 y 3 = = ⇒ y 1 y 2 y 3 = c 3 1 2 , c 3 negationslash = 0 Consequently x 1 ( t ) x 2 ( t ) x 3 ( t ) = c 1 1 − 3 e − t + c 2 1 − 1 6 e t + c 3 1 2 e 4 t satsifies the differential equations for all values of the constants c 1 , c 2 and c 3 . To satisfy the initial conditions we also need c 1 1 − 3 + c 2 1 − 1 6 + c 3 1 2 = 1 2 3 or 1 1 1 − 3 − 1 2 6 c 1 c 2 c 3 = 1 2 3 The last equation forces c 2 = 1 2 . Subbing this into the first two equations gives bracketleftbigg 1 1 − 3 2 bracketrightbiggbracketleftbigg c 1 c 3 bracketrightbigg = bracketleftbigg 1 / 2 5 / 2 bracketrightbigg March 31, 2011 Eigenvalues and Eigenvectors 32 Adding three times equation (1) to equation (2) gives 5 c 3 = 4 or c 3 = 4 5 and then equation (1) gives c 1 = − 3 10 . The solution is x 1 ( t ) x 2 ( t ) x 3 ( t ) = − 3 10 1 − 3 e − t + 1 2 1 − 1 6 e t + 4 5 1 2 e 4 t To check, we just sub into the original equations x ′ 1 ( t ) = 3 10 e − t + 1 2 e t + 16 5 e 4 t 2 x 1 ( t ) + x 2 ( t ) = 2 braceleftbig − 3 10 e − t + 1 2 e t + 4 5 e 4 t bracerightbig + braceleftbig 9 10 e − t − 1 2 e t + 8 5 e 4 t bracerightbig = 3 10 e − t + 1 2 e t + 16 5 e 4 t = x ′ 1 ( t ) x 1 (0) = − 3 10 + 1 2 + 4 5 = 1 x ′ 2 ( t ) = − 9 10 e − t − 1 2 e t + 32 5 e 4 t 6 x 1 ( t ) + x 2 ( t ) − x 3 ( t ) = 6 braceleftbig − 3 10 e − t + 1 2 e t + 4 5 e 4 t bracerightbig + braceleftbig 9 10 e − t − 1 2 e t + 8 5 e 4 t bracerightbig − 3 e t = − 9 10 e − t − 1 2 e t + 32 5 e 4 t = x ′ 2 ( t ) x 2 (0) = 9 10 − 1 2 + 8 5 = 2 x ′ 3 ( t ) = 3 e t x 3 ( t ) = 3 e t = x ′ 3 ( t ) x 3 (0) = 3 5) Find the functions x 1 ( t ) , x 2 ( t ) and x 3 ( t ) satisfying x ′ 1 ( t ) = 2 x 1 ( t ) − 6 x 2 ( t ) − 6 x 3 ( t ) x 1 (0) = 0 x ′ 2 ( t ) = − x 1 ( t ) + x 2 ( t ) + 2 x 3 ( t ) x 2 (0) = 1 x ′ 3 ( t ) = 3 x 1 ( t ) − 6 x 2 ( t ) − 7 x 3 ( t ) x 3 (0) = 0 Solution. The system of differential equations is of the form vectorx ′ ( t ) = Avectorx ( t ) with A = 2 − 6 − 6 − 1 1 2 3 − 6 − 7 The eigenvalues of this matrix are the solutions of 0 = det( A − λI ) = det 2 − λ − 6 − 6 − 1 1 − λ 2 3 − 6 − 7 − λ = det − 4 − 3 λ + λ 2 − 2 − 2 λ − 1 1 − λ 2 − 3 − 3 λ − 1 − λ (1) + (2 − λ )(2) (2) (3) + 3(2) = det bracketleftbigg − 4 − 3 λ + λ 2 − 2 − 2 λ − 3 − 3 λ − 1 − λ bracketrightbigg = ( − 4 − 3 λ + λ 2 )( − 1 − λ ) − ( − 2 − 2 λ )( − 3 − 3 λ ) = ( − 4 + λ )(1 + λ )( − 1 − λ ) − 6(1 + λ )(1 + λ ) = (1 + λ ) 2 [4...
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