sys2 - [2-2 i ] b 1 + 8 b 2 = 0 ,-b 1 + [-2-2 i ] b 2 = 0 ,...

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ALGORITHM FOR SOLVING dx/dt = Ax CASE OF COMPLEX CONJUGATE EIGENVALUES 1. Determine the eigenvalues λ and λ * which are roots of the character- istic equation det ( A - λI ) = 0 . 2. Find the corresponding eigenvectors b and b * which are solutions of ( A - λI ) b = 0 , ( A - λ * I ) b * = 0 . 3. Form the REAL vectors k 1 and k 2 according to the definitions: k 1 = 1 2 [ b + b * ] , k 2 = i 2 [ - b + b * ] 4. Two linearly independent REAL solutions are given by x (1) ( t ) = ( k 1 cos βt - k 2 sin βt ) e αt , and x (2) ( t ) = ( k 2 cos βt + k 1 sin βt ) e αt , where λ = α + i β . The general solution is x ( t ) = c 1 x (1) ( t ) + c 2 x (2) ( t ) . 5. If initial data x ( t 0 ) = x 0 are given, find the constants c 1 and c 2 by solving the algebraic system x 0 = c 1 x (1) ( t 0 ) + c 2 x (2) ( t 0 ) . EXAMPLE : Find the solution of the system d x dt = " 2 8 - 1 - 2 # x .
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The eigenvalues are the roots of ± ± ± ± ± 2 - λ 8 - 1 - 2 - λ ± ± ± ± ± = λ 2 + 4 = 0 , so λ = 2 i , λ * = - 2 i . The vector b , found by solving the system of equations
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Unformatted text preview: [2-2 i ] b 1 + 8 b 2 = 0 ,-b 1 + [-2-2 i ] b 2 = 0 , is 2 + 2 i-1 ! = 2-1 ! + i 2 ! . Now, from step (3) above, k 1 = 2-1 ! and k 2 = 2 ! . Since is pure imaginary (and so = 0), the general solution of the system is x ( t ) = c 1 " 2-1 ! cos 2 t- 2 ! sin 2 t # + c 2 " 2 ! cos 2 t + 2-1 ! sin 2 t # or x ( t ) = c 1 2 cos 2 t-2 sin 2 t-cos 2 t ! + c 2 2 cos 2 t + 2 sin 2 t-sin 2 t ! . 1. Find the general solutions of d y dt = " 1 2-1 2 1 # x . 2. Find the solution of the initial value problem d y dt = "-1-1 2-1 # x , x (0) = " 2-2 # ....
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sys2 - [2-2 i ] b 1 + 8 b 2 = 0 ,-b 1 + [-2-2 i ] b 2 = 0 ,...

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