HW13solutions

HW13solutions - Illustrations are on a separate page...

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Illustrations are on a separate page. Section 7.6 8. λ 1 = 0 . 9 , λ 2 = 0 . 8 so -→ 0 is a stable equilibrium. 12. λ = 0 . 6 ± ik, so -→ 0 is a stable equilibrium if | λ | < 1, i.e. 0 . 36 + k 2 < 1 , or | k | < 0 . 8. 22. λ = 4 ± i, r = 17 , θ = arctan(1 / 4), [ -→ w -→ v ] = ! 0 - 5 1 - 3 " , S - 1 -→ x 0 = ! 1 0 " . So -→ x ( t ) = 17 t/ 2 ! - 5 sin( θ t ) cos( θ t ) - 3 sin( θ t ) " . 23. λ = 0 . 4 ± 0 . 3 i, r = 1 / 2 , θ = arctan(0 . 3 / 0 . 4), [ -→ w -→ v ] = ! 0 5 1 3 " , S - 1 -→ x 0 = ! 1 0 " . So -→ x ( t ) = 2 - t ! 5 sin( θ t ) cos( θ t ) + 3 sin( θ t ) " . 24. λ = - 0 . 8 ± 0 . 6 i, r = 1 , θ = π - arctan(0 . 6 / 0 . 8), [ -→ w -→ v ] = ! 0 - 5 1 - 3 " , S - 1 -→ x 0 = ! 1 0 " . So -→ x ( t ) = ! - 5 sin( θ t ) cos( θ t ) - 3 sin( θ t ) " . 26. This is stable because A and A T have the same eigenvalues. 28. Unstable. If A has eigenvalue λ , then ( A - 2 I ) has eigenvalue λ - 2, and | λ - 2 | > 1. Section 9.1 16.–18. See illustrations. 28. λ 1 = 1 , λ 2 = 10 , -→ v 1 = ! - 3 2 " , -→ v 2 = ! 1 2 " , so -→ x ( t ) = - 1 8 e 2 t ! - 3 2 " + 5 8 e 10 t ! 1 2 " .
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HW13solutions - Illustrations are on a separate page...

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