s251ex2(fa02)

s251ex2(fa02) - 4( t-3) ) ; 8. X ( t ) = 3-3 e-3 t + 4 3 e...

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MATH 251 Fall 2002 Exam 2 November 12, 2002 ANSWERS: 1. For y 1 = y and y 2 = y 0 we have: ( x 2 y 0 2 + xy 2 + ( x 2 - λ 2 ) y 1 = 0 y 1 (0) = y 0 , y 2 (0) = y 0 0 y 0 1 = y 2 Simplifying: ( y 0 1 = y 2 y 1 (0) = y 0 , y 2 (0) = y 0 0 y 0 2 = - 1 x 2 ( x 2 - λ 2 ) y 1 - 1 x y 2 2. Y = Ae 2 t + t ( B sin(2 t ) + C cos(2 t )) + D sin t + E cos t + Ft 2 + Gt + H 3.(b) f ( t ) = t 2 + (6 - t - t 2 ) u 2 ( t ) + ( t - 6) u 6 ( t ) 4. It is critically damped, so γ 2 - 4 km = 0 . Therefore, γ = 4 km = 12 . 5. L{ y ( t ) } = s 3 3 ( s - π ) 2 + 9 - 3 s - 6 π 6. y ( t ) = 1 2 e 2 t + 1 2 e - 4 t + ( 1 7 e 3( t - 3) - 1 6 e 2( t - 3) + 1 42 e - 4( t - 3) ) e 9 u 3 ( t ) 7. y ( t ) = e - 4 t + u 3 ( t )( 2 3 e - ( t - 3) - 2 3 e -
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Unformatted text preview: 4( t-3) ) ; 8. X ( t ) = 3-3 e-3 t + 4 3 e 4 t 9.(a) X ( t ) = C 1 1 e t cos(3 t ) + C 2 1 e t sin(3 t ) (b) X ( t ) = C 1 e t 1-1 + C 2 e t t + 1 2-t...
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