s251ex2(su07) - Therefore, it is a proper node (star...

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MATH 251 Exam 2 July 30, 2007 ANSWER KEY 1. (a) k < 12; (b) critically damped; (c) k = 15; (d) T = 2 π ; (e) true. 2, (a) k = 50 (b) 2 u ″ + 50 u = 0, u (0) = 1, u ′(0) = 4 3. D 4. (a) f ( t ) = -1 + 2 e - t - 2 te - t (b) f ( t ) = u 4 ( t )(2 e - t + 4 cos(5 t - 20) - 2 e - t + 4 sin(5 t - 20)) 5. f ( t ) = (1 - u 4 ( t )) e 2 t + u 4 ( t )(8 t - t 2 ) = e 2 t + u 4 ( t )(8 t - t 2 - e 2 t ); - - + - + - = - 2 16 2 ) 2 ( 1 ) ( 8 3 4 s e s s e s s F s 6. - + - = + - - - 18 6 3 3 6 7 1 7 1 ) ( 7 2 7 2 ) ( t t t t e e t u e e t y 7. 4 2 2 4 2 ) 4 2 )( 2 ( 1 ) ( 2 3 2 3 2 3 + + - - + + + - + + + - + = - s s s s s s s e s s s s s Y s π 8. (a) t t e C e C t x 8 2 2 1 1 1 4 6 ) ( + - = - (b) β = -4 9. (a) There are two linearly independent eigenvectors for the eigenvalue 2.
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Unformatted text preview: Therefore, it is a proper node (star point), and it is unstable. (b) -+ + -=--t t t e C t t t e C t x t t 5 cos 2 5 sin 5 cos 5 sin 2 5 sin 5 cos ) ( 4 2 4 1 (c) It is a spiral point ; asymptotically stable. (d) It is a saddle point ; unstable....
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This note was uploaded on 07/23/2008 for the course MATH 251 taught by Professor Chezhongyuan during the Spring '08 term at Pennsylvania State University, University Park.

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