HW14solutions - Section 9.2 4 We know that e3it = cos(3t i sin(3t Therefore the trajectory is a circle of radius one drawn in the counter-clockwise

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Section 9.2 4. We know that e 3 it = cos(3 t ) + i sin(3 t ). Therefore the trajectory is a circle of radius one, drawn in the counter-clockwise direction and the period is given by 3 T = 2 π , so T = 2 π 3 . 6. The eigenvalues are λ = ± i and the corresponding eigenvectors are ~v = ± 3 ± i 5 . Therefore the general solution is given by: ~x ( t ) = c 1 e it ± 3 + i 5 + c 2 e - it ± 3 - i 5 . If c 1 = c 2 = 1 then, using Euler’s equation we get: ~x ( t ) = (cos( t ) + i sin( t )) ± 3 + i 5 + (cos( t ) - i sin( t )) ± 3 - i 5 = ± 6cos( t ) - 2sin( t ) 10cos( t ) . 16. If A = ± 0 1 a b then the trace is b and the determinant is - a . By Fact 9.2.5, the zero is stable if both a and b are negative. 22. The eigenvalues are λ 1 = 3 and λ 2 = 1 2 . The corresponding eigenvec- tors are v 1 = ± 1 - 1 , v 2 = ± 0 1 and since the system is discrete, it corresponds to phase plane VII. 26. We proceed like above. λ 1 = 1, λ 2 = - 2, v 1 = ± 0 1 , v 2 = ± 1 - 1 1
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and since the system is continuous, it corresponds to phase plane V. 28.
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This homework help was uploaded on 02/23/2008 for the course MATH 2210 taught by Professor Pantano during the Fall '05 term at Cornell University (Engineering School).

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HW14solutions - Section 9.2 4 We know that e3it = cos(3t i sin(3t Therefore the trajectory is a circle of radius one drawn in the counter-clockwise

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