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Unformatted text preview: Homework 9  Due on April 12 (Monday)  Solution 1 1. Let Y := u v and A denote the 2 2 matrix for each problem. (a) [10 pt] u v = 3 2 2 2 u v i) General solution: We can easily calculate two eigenvalues r 1 = 2, r 2 = 1 and corresponding eigenvectors 1 = 2 1 , 2 = 1 2 . So the general solution is Y = u v = c 1 2 1 e 2 t + c 2 1 2 e t . ii)iv) 2 1 1 2 v 2 1 1 2 u saddle point unstable v) Initial value problem when Y (0) = 2 . Since c 1 = 4 3 and c 2 = 2 3 , the solution is Y = u v = 4 3 2 1 e 2 t 2 3 1 2 e t . (b) [10 pt] u v = 1 2 3 4 u v i) General solution: Eigenvalues r 1 = 1, r 2 = 2. Eigenvectors 1 = 1 1 , 2 = 2 3 . So the general solution is Y = u v = c 1 1 1 e t + c 2 2 3 e 2 t . ii)iv) 2 1 1 2 v 2 1 1 2 u node asymptotically stable v) Initial value problem when Y (0) = 2 . Since c 1 = 6 and c 2 = 2, the solution is Y = u v = 6 1 1 e t 2 2 3 e 2 t . MATH2400  Introduction to Differential Equations  Spring 2004  JeongRock Yoon Homework 9  Due on April 12 (Monday)  Solution 2 (c) [10 pt] u v = 5 3 3 5 u v i) General solution: Eigenvalues r 1 = 2, r 2 = 8. Eigenvectors 1 = 1 1 , 2 = 1 1 . So the general solution is Y = u v = c 1 1 1 e 2 t + c 2 1 1 e 8 t . ii)iv) 2 1 1 2 v 2 1 1 2 u node unstable v) Initial value problem when Y (0) = 2 . Since c 1 = 1 and c 2 = 1, the solution is Y = u v = 1 1 e 2 t + 1 1 e 8 t . (d) [10 pt] u v = 2 5 1 2 u v i) Eigenvalues r 1 = i , r 2 = i . Eigenvectors 1 = 2 + i 1 , 2 = 2 i 1 . 1 e r 1 t = 2 + i 1 e it = (2 + i )(cos t + i sin t ) cos t + i sin t = (2cos t sin t ) + i (cos t + 2sin t ) cos t + i sin t = 2cos t sin t cos t  {z } Re [ 1 e r 1 t ] + i cos t + 2sin t sin t  {z } Im [ 1 e r 1 t ] So, Y = c 1 Re 1 e r 1 t / + c 2 Im 1 e r 1 t / = c 1 2cos t sin t cos t + c 2 cos t + 2sin t sin t ....
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This homework help was uploaded on 02/24/2008 for the course MATH 2400 taught by Professor Yoon during the Spring '04 term at Rensselaer Polytechnic Institute.
 Spring '04
 Yoon
 Differential Equations, Equations

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