problemset9

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
± ± ± ± ± ± ± ± ± ² ³ ± ± ± ± ± ± ± ± 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 satisFes ( 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 satisFed by u 1 . Since u ˙ = A u , we 0 1 0 . 5 Fnd 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 2 3 ³ = ± ± / 3 1 / 6 0 e , u = e , 3 0 3 / 2 u (0) = 1 , u ² 3 ³ = e / 6 0 , 0 3 / 2 ² 2 ³ / 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.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online