solutions6bd

solutions6bd - values λ = 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 λ = 3 ± 2 i with corresponding eigenvectors ± 1 1 ² ± i ± 2 ² . We can write down the general solution to the system of differential equations: it is ~x ( t ) = c 1 e 3 t ³± 1 1 ² cos 2 t-± 2 ² sin 2 t ´ + c 2 e 3 t ³± 1 1 ² sin 2 t + ± 2 ² cos 2 t ´ . Using the initial condition ~x (0) = ± 2 4 ² , we have that ± 2 4 ² = c 1 ± 1 1 ² + c 2 ± 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 ±² 1 1 ³ cos 2 t-² 2 ³ sin 2 t ´ + e 3 t ±² 1 1 ³ sin 2 t + ² 2 ³ cos 2 t ´ = e 3 t ±² 2 4 ³ cos 2 t + ² 1-3 ³ sin 2 t ´ . 2...
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This note was uploaded on 04/02/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.

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solutions6bd - values λ = 3 ± 2 i with corresponding...

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