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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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- Spring '07
- staff
- Equations, initial condition