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# problemset9 - 18.03 Problem Set 9 Solutions Part I 34 4 35...

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1 18.03 Problem Set 9: Solutions Part I: 34. 4; 35. 10; 36. 6. (a) [18] A = 0 2 . . 5 25 0 1 . 5 has characteristic polynomial p A ( ) = 2 +2 . 5, and eigenvalues 1 ± 3 i . An eigenvector for 1 = 1+3 i satisfies ( A 1 I ) v 1 = 0 , that is, 3 9 i/ / 4 2 3 i/ 2 v 1 = 0 . 2 2 1 (1+3 i ) t/ 2 1 One choice is v 1 = . The normal mode is then e , which has real and 3 i/ 2 3 i/ 2 t/ 2 cos(3 t/ 2) t/ 2 sin(3 t/ 2) imaginary parts u 1 = e and u 2 = e . (3 / 2) sin(3 t/ 2) (3 / 2) cos(3 t/ 2) 1 The initial condition is u (0) = , which, conveniently, is satisfied by u 1 . Since u ˙ = A u , we 0 1 0 . 5 find u ˙ (0) = A = . So x ˙(0) = 0 . 5, and the weasel population is increasing at 0 2 . 25 t = 0 while the number of voles is decreasing. y ( t ) = 0 occurs next when 3 t/ 2 = , or t = 2 �/ 3. The graphs of x ( t ) = e t/ 2 cos(3 t/ 2) and y ( t ) = (3 / 2) e t/ 2 sin(3 t/ 2) are “anti-damped” sinusoids, with increasing amplitude. The relevant trajec- tory is the one crossing the positive x axis half way out. The values of u ( t ) are u 2 3 = �/ 3 1 �/ 6 0 e , u = e , 3 0 3 / 2 u (0) = 1 , u 3 = e �/ 6 0 , 0 3 / 2 2 �/ 3 1 u = e . 3 0 (b) [8] With A = 1 b , p A ( ) = 2 2 + 1 = ( 1) 2 , so we have a repeated eigenvalue 0 1 0 b 1 = 1. To find an eigenvector form A 1 I = . A nonzero

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problemset9 - 18.03 Problem Set 9 Solutions Part I 34 4 35...

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