answers_1 - Practice Finals Answers December 7, 2010 Fall...

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Practice Finals’ Answers December 7, 2010 Fall 07 1) f ( t ) = 20 + (30 - 2 t ) exp( - t/ 5) 2) λ n = n 2 π 2 4 + 1, y n = C n exp( - 2 t ) sin( n π t ) where n = 1 , 2 , 3 , . . . 3) f ( x ) = 1 + n =1 4 n 2 π 2 (1 - cos n π ) cos 2 x , since f 0 is piecewise continuous we can use term by term differ- entiation, f ( x ) = n =1 2 (1 - cos n π ) sin 2 x 5) u n ( x, t ) = sin nπ x 2 ( A n sinh nπ t 2 + B n cosh nπ t 2 ) 6) y ( x, t ) = n =1 B n sin nπ x L sin nπ at L , B n = 3 ( - 1) n L n 2 ( n +1) Spring 09 2)sine series with coeff. b n = 8 ( - 1) n +1 π (4 n 2 - 1) 6) k = 9 . 8 E 6 N/m, y ( t ) = A cos ω 0 t + B sin ω 0 t + cos ω t +sin ω t k - 2 7) y ( t ) = C 1 exp( - 2 t ) + C 2 exp ( -
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This note was uploaded on 01/21/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell.

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