Test 1 - MATH 3705A Test 1 Solutions January 23, 2009...

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MATH 3705A Test 1 Solutions January 23, 2009 Questions 1-4 are multiple choice. Circle the correct answer. Only the answer will be marked. [Marks] 1. Lf e 2 t cos(3 t ) g = [3] (a) s ( s ¡ 2) 2 +9 (b) s ¡ 2 s 2 (c) s ¡ 2 ( s ¡ 2) 2 (d) s +2 ( s +2) 2 (e) None of these Solution: (c) 2. Lf u ( t ¡ 3) e ¡ 2 t g = [3] (a) e ¡ 3( s +2) s (b) e ¡ 3 s s (c) e 3 s s (d) e ¡ 3( s ¡ 2) s (e) None of these Solution: (a) 3. L ¡ 1 f 6 s ( s 2 +9) 2 g = [3] (a) t sin(3 t )( b ) t cos(3 t c ) ¡ t sin(3 t d ) ¡ t cos(3 t ) (e) None of these Solution: (a) 4. L ¡ 1 f se ¡ s s 2 ¡ 2 s +4 g = [3] (a) e t cos( p 3 t )+ 1 3 e 3 t sin( p 3 t ) (b) u ( t ¡ 1) f e t ¡ 1 cos[ p 3( t ¡ 1)] + e t ¡ 1 sin[ p 3( t ¡ 1)] g (c) u ( t ¡ 1) e t ¡ 1 f cos[ p 3( t ¡ 1)] + 1 p 3 sin[ p 3( t ¡ 1)] g (d) u ( t ¡ 1) f e t ¡ 1 cos[ p 3( t ¡ 1)] + 1 3 e t ¡ 1 sin[ p 3( t ¡ 1)] g (e) None of the above Solution: (c) 5. Solve the initial-value problem y 00 ¡ y 0 +6 y =0 ;y (0) = 0 0 (0) = 5. [9] Solution: [ s 2 Y ( s ) ¡ sy (0) ¡ y 0 (0)] ¡ [ sY ( s ) ¡ y (0)] + 6 Y ( s )=0 ) ( s 2 ¡ s +6) Y ( s ) ¡ 5=0 ) Y ( s )= 5 s 2 ¡ s = 5 ( s ¡ 1 = 2) 2 +23 = 4 ) y ( t 10 p 23 e 1 2 t sin à p 23 2 t ! . For those who guessed correctly that the intended equation was y 00 ¡ y 0 ¡ 6 y ,the solution is below.
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2 [ s 2 Y ( s ) ¡ sy (0) ¡ y 0 (0)] ¡ [ sY ( s ) ¡ y (0)] ¡ 6
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This note was uploaded on 09/21/2009 for the course MATH 3705 taught by Professor Dr.e.devdariani during the Winter '09 term at Carleton.

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Test 1 - MATH 3705A Test 1 Solutions January 23, 2009...

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