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Unformatted text preview: ACM 95b/100b Solutions for Problem Set 3 Sean Mauch sean@caltech.edu February 3, 2002 1 Problem 1 Problem. Find the general solutions to the following homogeneous linear systems of ODEs. De scribe the behavior of the solutions as t . 1. x = 3 6 1 2 x 2. x = 0 1 1 1 0 1 1 1 0 x Solution. 1. We consider the problem, x = Ax 3 6 1 2 x . The matrix has the distinct eigenvalues 1 = 0, 2 = 1. The corresponding eigenvectors are x 1 = 2 1 , x 2 = 3 1 . The general solution of the system of differential equations is x = c 1 2 1 + c 2 3 1 e t . If c 2 = 0, then the solution is a constant. Otherwise, the solution is dominated by the second term for large time and both coordinates tend to infinity. 2. We consider the problem, x = Ax 0 1 1 1 0 1 1 1 0 x . The matrix has the eigenvalues 1 = 2 = 1, 3 = 2. The corresponding linearly independent eigenvectors are x 1 =  1 1 , x 2 =  1 1 , x 3 = 1 1 1 . 1 The general solution of the system of differential equations is x = c 1  1 1 e t + c 2  1 1 e t + c 3 1 1 1 e 2 t . If c 3 = 0, then the each coordinate of the solution vanishes as t . Otherwise, the solution is dominated by the third term for large time and all the coordinates tend to infinity. x c 3 1 1 1 e 2 t , as t 2 Problem 2 Problem. Find a fundamental matrix for each of the following systems of ODEs using a similarity transformation to decouple the system. 1. x = 5 1 3 1 x 2. x = 1 1 4 3 2 1 2 1 1 x Solution. 1. We consider the problem x = Ax 5 1 3 1 x . The matrix has the distinct eigenvalues 1 = 2, 2 = 4. The corresponding eigenvectors are x 1 = 1 3 , x 2 = 1 1 . The Jordan canonical form of A is J = 2 0 0 4 . A is diagonalized by a similarity transformation. J = S 1 AS , S = 1 1 3 1 The solution of the differential equation is x = e A t c . x = e A t c = S e J t S 1 c = S e J t c = 1 1 3 1 e 2 t e 4 t c x = e 2 t e 4 t 3e 2 t e 4 t c 2 2. We consider the problem x = Ax 1 1 4 3 2 1 2 1 1 x The matrix has the distinct eigenvalues 1 = 3, 2 = ( 1 17) / 2 and 3 = ( 1 + 17) / 2. The corresponding eigenvectors are x 1 = 3 14 5 , x 2 =  1 (5 17) / 2 1 , x 3 =  1 (5 + 17) / 2 1 . The Jordan canonical form of A is J = 3 0 ( 1 17) / 2 ( 1 + 17) / 2 A is diagonalized by a similarity transformation. J = S 1 AS , S = 3 1 1 14 (5 17) / 2 (5 + 17) / 2 5 1 1 The solution of the differential equation is x = e A t c ....
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 Winter '09
 NilesA.Pierce

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