sys1 - x (1) ( t ) = " 1 1 # e-t , x (2) ( t )...

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ALGORITHM FOR SOLVING dx/dt = Ax DISTINCT REAL EIGENVALUE CASE 1. Determine the eigenvalues λ 1 and λ 2 which are roots of the character- istic equation det ( A - λI ) = 0 . 2. Find the corresponding eigenvectors b 1 and b 2 which are solutions of ( A - λ j I ) b = 0 , j = 1 , 2 . 3. Two linearly independent solutions are given by x (1) ( t ) = b 1 e λ 1 t , x (2) ( t ) = b 2 e λ 2 t , and the general solution is x ( t ) = c 1 x (1) ( t ) + c 2 x (2) ( t ) . 4. 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 initial value problem d x dt = " - 5 4 1 - 2 # x , x (0) = " 3 - 2 # . The eigenvalues are the roots of ± ± ± ± ± - 5 - λ 4 1 - 2 - λ ± ± ± ± ± = λ 2 + 7 λ + 6 = 0 , so λ 1 = - 1 , λ 2 = - 6. A vector b 1 , found by solving the system of equations [ - 5 - ( - 1)] b 1 + 4 b 2 = 0 , b 1 + [ - 2 - ( - 1)] b 2 = 0 ,
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is " 1 1 # . Similarly, b 2 = " 4 - 1 # satisfies [ - 5 - ( - 6)] b 1 + 4 b 2 = 0 , b 1 + [ - 2 - ( - 6)] b 2 = 0 . The corresponding solutions are
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Unformatted text preview: x (1) ( t ) = " 1 1 # e-t , x (2) ( t ) = " 4-1 # e-6 t , and the general solution is x ( t ) = c 1 x (1) ( t ) + c 2 x (2) ( t ) = c 1 " 1 1 # e-t + c 2 " 4-1 # e-6 t . To obtain the solution satisfying the initial conditions, c 1 and c 2 must satisfy x (0) = " 3-2 # = c 1 " 1 1 # + c 2 " 4-1 # or 3 = c 1 + 4 c 2 ,-2 = c 1-c 2 . Solving these gives c 1 =-1, c 2 = 1, and the desired solution is x ( t ) = (-1) " 1 1 # e-t + (1) " 4-1 # e-6 t = "-1-1 # e-t + " 4-1 # e-6 t . 1. Find the general solution of d x dt = " 2 3-4 # x . 2. Find the general solution of d x dt = " 2 6-2-5 # x ....
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sys1 - x (1) ( t ) = " 1 1 # e-t , x (2) ( t )...

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