MAP 4305 Exam Practice Answers 1

MAP 4305 Exam Practice Answers 1 - -x 2 ( t ) y 1 ( t )) =...

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Answers practice exam I: MAP 4305 * September 16, 2007. Name : Student ID : This is a closed book exam and the use of calculators is not allowed. 1. A 231 = TD 231 T - 1 , T = ± 1 1 - 1 1 ² , D 231 = ± ( - 1) 231 0 0 3 231 ² , T - 1 = 1 2 ± 1 - 1 1 1 ² System is unstable. 2. P = ± 0 . 5 0 0 . 5 1 ² , Pagerank = ± 0 1 ² 3. Since m ( t ) = x 1 ( t ) y 2 ( t ) - x 2 ( t ) y 1 ( t ), it follows that ˙ m ( t ) = ˙ x 1 ( t ) y 2 ( t ) + x 1 ( t ) ˙ y 2 ( t ) - ˙ x 2 ( t ) y 1 ( t ) - x 2 ( t ) ˙ y 1 ( t ) . Using the fact that ± x 1 ( t ) x 2 ( t ) ² and ± y 1 ( t ) y 2 ( t ) ² are solutions of the system, we can simplify this to: ˙ m ( t ) = ( a + d )( x 1 ( t ) y 2 ( t )
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Unformatted text preview: -x 2 ( t ) y 1 ( t )) = ( a + d ) m ( t ) . Solve this first order differential equation like you were taught in MAP2302, and you get m ( t ) = e ( a + d ) t m (0) . 4. x ( t ) = X ( t ) c, X ( t ) = ± e-t e 2 t 3 e 2 t ² , c = ± e-2 3 e-3 2 3 e-2 ² * Instructor: Patrick De Leenheer....
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This note was uploaded on 04/07/2008 for the course MAP 4305 taught by Professor Deleenheer during the Spring '06 term at University of Florida.

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