SampleTest1v1

SampleTest1v1 - MATH 3705* Test 1 - Solutions January 2006...

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MATH 3705* Test 1 - Solutions January 2006 Questions 1-6 are multiple choice. Circle the correct answer. Only the answer will be marked. [12 marks] 1 . L{ e 4 t cos(5 t ) } = ( a ) s +4 ( s +4) 2 +25 ( b ) s s 2 ( c ) s 4 ( s 4) 2 ( d )Noneo ftheabove 2 . L{ t sin(2 t ) } = ( a ) 4 s ( s 2 2 ( b ) s 2 4 ( s 2 2 ( c ) 4 s 2 ( s 2 2 ( d ) None of the above 3 . L F8 t 0 e x sin( t x ) dx k = ( a ) e s s 2 +1 ( b ) 1 ( s 1)( s 2 +1) ( c ) 1 ( s 1) 2 ( d ) None of the above 4 . L 1 F 6 s 2 9 k = ( a ) e 3 t e 3 t ( b ) e 3 t + e 3 t ( c ) sin(3 t )( d ) None of the above 5 . L 1 F 3 s +6 s 2 4 s +13 k = ( a )3 e 2 t cos(3 t )+4 e 2 t sin(3 t b e 2 t cos(3 t )+ 9 4 e 2 t sin(3 t ) ( c e 2 t cos(3 t 9 4 e 2 t sin(3 t d ) None of the above 6 . L 1 F e 2 s s 2 6 s k = ( a ) u ( t 2) e 3( t 2) sin[4( t 2)] ( b ) 1 4 u ( t 2) e 3 t 6 sin(4 t 8) ( c ) 1 4 u ( t 2) e 3 t sin[4( t 2)] ( d ) None of the above Answers: 1.(c), 2.(a), 3.(b), 4.(a), 5.(a), 6.(b),
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2 7. [9 marks] Employ the Laplace transform to solve the initial-value problem y II +2 y I +10 y =0 ,y (0) = 1 I (0) = 2 . Solution: Take the Laplace transform of both parts of the equation: [ s 2 Y ( s ) sy (0) y I (0)] + 2[ sY ( s ) y (0)] + 10 Y ( s )=0 . Substitute y (0) = 1 I (0) = 2 and solve for Y ( s ): ( s 2 s +10) Y ( s )= s +4 , Y ( s s s 2 s = ( s +1)+3 ( s +1) 2 +9 . Using the First Shifting Theorem with a = 1, f nd L 1 { Y ( s ) } = e t cos(3 t )+ e t sin(3 t y ( t ) .
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This note was uploaded on 07/16/2009 for the course MATH 3705 taught by Professor Jaberabdualrahman during the Winter '08 term at Carleton CA.

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SampleTest1v1 - MATH 3705* Test 1 - Solutions January 2006...

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