s251ex2(su10)

s251ex2(su10) - 9. (a) t t e t C e C t x + + = 1 1 2 1 2 )...

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MATH 251 Exam 2 July 27, 2010 ANSWER KEY 1. + - + - = = = = t x t x tx x x x x x x x sin 2 3 2 3 4 2 2 1 4 4 3 3 2 2 1 2. g (3) = 10 3. 2 2 ) 10 6 ( 6 2 ) ( + - - = - s s s e s F s 4. f ( t ) = u 3 ( t )( t - 4 + e - t + 3 ) 5. f ( t ) = (1 - u 3 ( t )) (2 t 2 + t ) + u 3 ( t ) e 4 t = 2 t 2 + t + u 3 ( t )( e 4 t - 2 t 2 - t ); - - - - + + = - s s s s e e s s s F s 21 13 4 4 1 4 ) ( 2 3 12 3 2 3 6. (a) L { y ( t )} = 13 4 6 ) 13 4 ( 2 10 2 3 + + - + + + + - - s s e s s s s e s s (b) t e t e t y t t 3 sin 3 3 cos 2 ) ( 2 2 - - - = - - + - + + - + - 13 1 ) 30 3 sin( 39 2 ) 30 3 cos( 13 1 ) ( 20 2 20 2 10 t e t e t u t t 7. (a) t t e e t x 5 2 1 3 4 2 ) ( - - + = (b) α = 0 (c) It is a saddle point , unstable. 8. (a) + = t t t t x 3 sin 2 3 cos 2 3 sin 4 ) (
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(b) It is a center ; (neutrally) stable.
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Unformatted text preview: 9. (a) t t e t C e C t x + + = 1 1 2 1 2 ) ( 2 1 (b) It is an improper node , unstable 10. (a) The other critical points are (0, 0), and (1 / 2, -1 / 2). (b) -= 1 1 1 A (c) It is a saddle point , unstable....
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s251ex2(su10) - 9. (a) t t e t C e C t x + + = 1 1 2 1 2 )...

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