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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 + 4 s 2 + 25 ( c ) s 4 ( s 4) 2 + 25 ( d ) None of the above 2 . L{ t sin(2 t ) } = ( a ) 4 s ( s 2 + 4) 2 ( b ) s 2 4 ( s 2 + 4) 2 ( c ) 4 s 2 ( s 2 + 4) 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 + 1 ( 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 ) 3 e 2 t cos(3 t ) + 9 4 e 2 t sin(3 t ) ( c ) 3 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 + 25 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 , y 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 , y I (0) = 2 and solve for Y ( s ): ( s 2 + 2 s + 10) Y ( s ) = s + 4 , Y ( s ) = s + 4 s 2 + 2 s + 10 = ( s + 1) + 3 ( s + 1) 2 + 9 . Using the First Shifting Theorem with
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