solutions6bd

solutions6bd - values ╬ = 3 ┬▒ 2 i with corresponding...

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Math 33b, Quiz 6bd, March 6, 2008 Name: UCLA ID: 1. Find the solution, with initial condition ~x (0) = ± 2 4 ² , to the system of differential equations d~x dt = ± 2 1 - 5 4 ² ~x. Solution. The characteristic polynomial of the matrix is det ± 2 - λ 1 - 5 4 - λ ² , which is λ 2 - 6 λ + 13. The roots of this polynomial are 3 ± 2 i , so these are the eigenvalues. To find a complex eigenvector associated to λ = (3 + 2 i ), we solve the matrix system ( A - (3 + 2 i ) I ) ~x = ~ 0). Hence we want ± - 1 - 2 i 1 - 5 - 1 - 2 i ²± x 1 x 2 ² = ± 0 0 ² . Writing this out as equations, we want ( - 1 - 2 i ) x 1 + x 2 = 0 - 5 x 1 + (1 - 2 i ) x 2 = 0 One can check that the second equation is simply the first equation multiplied by (1 - 2 i ). Hence we need only solve the first equation ( - 1 - 2 i ) x 1 + x 2 = 0. As usual, there are infinitely many nonzero solutions, and we are free to choose a convenient one. If we choose x 1 = 1, then x 2 = 1 + 2 i . Hence we can take as our eigenvector ± 1 1 + 2 i ² = ± 1 1 ² + i ± 0 2 ² . Since complex eigenvalue-eigenvectors come in conjugate pairs, we have the two eigen-
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Unformatted text preview: values ╬╗ = 3 ┬▒ 2 i with corresponding eigenvectors ┬▒ 1 1 ┬▓ ┬▒ i ┬▒ 2 ┬▓ . We can write down the general solution to the system of di´ČÇerential equations: it is ~x ( t ) = c 1 e 3 t ┬│┬▒ 1 1 ┬▓ cos 2 t-┬▒ 2 ┬▓ sin 2 t ┬┤ + c 2 e 3 t ┬│┬▒ 1 1 ┬▓ sin 2 t + ┬▒ 2 ┬▓ cos 2 t ┬┤ . Using the initial condition ~x (0) = ┬▒ 2 4 ┬▓ , we have that ┬▒ 2 4 ┬▓ = c 1 ┬▒ 1 1 ┬▓ + c 2 ┬▒ 2 ┬▓ . Writing this out as equations, we get c 1 = 2 c 1 + 2 c 2 = 4 1 We thus have c 1 = 2 and c 2 = 1, so our solution is ~x ( t ) = 2 e 3 t ┬▒┬▓ 1 1 ┬│ cos 2 t-┬▓ 2 ┬│ sin 2 t ┬┤ + e 3 t ┬▒┬▓ 1 1 ┬│ sin 2 t + ┬▓ 2 ┬│ cos 2 t ┬┤ = e 3 t ┬▒┬▓ 2 4 ┬│ cos 2 t + ┬▓ 1-3 ┬│ sin 2 t ┬┤ . 2...
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solutions6bd - values ╬ = 3 ┬▒ 2 i with corresponding...

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