2065f09ex2a

2065f09ex2a - Name: Solutions Exam 2 Instructions. Answer...

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Name: Solutions Exam 2 Instructions. Answer each of the questions on your own paper. Put your name on each page of your paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. A table of Laplace transforms and the statement of the main partial fraction decomposition theorem have been appended to the exam. 1. [8 Points] Compute the partial fraction expansion of the following rational function: 7 s + 13 ( s ¡ 1)( s 2 + 9) : I Solution. 7 s + 13 ( s ¡ 1)( s 2 + 9) = A s ¡ 1 + p 1 ( s ) s 2 + 9 ; where A = 7 s + 13 s 2 + 9 s =1 = 20 10 = 2 ; and p 1 ( s ) = 7 s + 13 ¡ 2( s 2 + 9) s ¡ 1 = ¡ 2 s 2 + 7 s ¡ 5 s ¡ 1 = ¡ (2 s ¡ 5)( s ¡ 1) s ¡ 1 = ¡ 2 s + 5 : Hence, 7 s + 13 ( s ¡ 1)( s 2 + 9) = 2 s ¡ 1 + ¡ 2 s + 5 s 2 + 9 : J 2. [20 Points] Compute the inverse Laplace transform of each of the following rational functions. (a) F ( s ) = 5 s ¡ 1 s 2 ¡ 2 s ¡ 15 I Solution. F ( s ) = 5 s ¡ 1 s 2 ¡ 2 s ¡ 15 = 5 s ¡ 1 ( s ¡ 5)( s + 3) = A s ¡ 5 + B s + 3 ; where A = 5 s ¡ 1 s + 3 s =5 = 24 8 = 3 and B = 5 s ¡ 1 s ¡ 5 s = ¡ 3 = ¡ 16 ¡ 8 = 2 : Hence, F ( s ) = 3 s ¡ 5 + 2 s + 3 , and f ( t ) = L ¡ 1 f F ( s ) g = 3 e 5 t + 2 e ¡ 3 t : J Math 2065 Section 1 October 26, 2009 1
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Name: Solutions Exam 2 (b) G ( s ) = s + 4 s 2 ¡ 6 s + 13 I Solution. G ( s ) = s + 4 s 2 ¡ 6 s + 13 = ( s ¡ 3) + 7 ( s ¡ 3) 2 + 4 = ( s ¡ 3) ( s ¡ 3) 2 + 4 + 7 ( s ¡ 3) 2 + 4 : Thus, g ( t ) = L ¡ 1 f G ( s ) g = e 3 t cos2 t + 7 2 e 3 t sin2 t: J 3. [10 Points] Find the Laplace transform Y ( s ) of the solution y ( t ) of the initial value problem 3 y 00 + 2 y = 2 cos 3 t; y (0) = 5 ; y 0 (0) = ¡ 2 : Note that you are only asked to flnd the Laplace transform Y ( s ) of y ( t ), not y ( t ) itself.
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2065f09ex2a - Name: Solutions Exam 2 Instructions. Answer...

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