quiz6-sol - Name: MA 366, Spring 2010, Quiz 6 (1) (5...

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Unformatted text preview: Name: MA 366, Spring 2010, Quiz 6 (1) (5 points) Determine the Laplace transform of f ( t ) = t u ( t ). Solution: To make it match line 13 in the table we write f ( t ) = t u ( t ) = ( t- + ) u ( t ) = ( t- ) u ( t ) + u ( t ) . So we can use lines 13 and 3 (for the Laplace transforms of 1 and t ) to get F ( s ) = e s 1 s 2 + e s 1 s . (5 points) Determine the inverse Laplace transform of F ( s ) = s s 2 + 4 s + 13 . Solution: We cant factor the denominator, so we write F ( s ) = s ( s + 2) 2 + 9 . This looks a bit like lines 9 and 10, but F doesnt have the right numerator for either of them. So we write F ( s ) = s + 2- 2 ( s + 2) 2 + 9 = s + 2 ( s + 2) 2 + 9- 2 ( s + 2) 2 + 9 = s + 2 ( s + 2) 2 + 9- 2 3 3 ( s + 2) 2 + 9 , and now the first term matches line 10 and the second matches line 9, so f ( t ) = e 2 t cos 3 t- 2 3 e 2 t sin 3 t....
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This note was uploaded on 01/18/2012 for the course MATH 366 taught by Professor Edraygoins during the Fall '09 term at Purdue University-West Lafayette.

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quiz6-sol - Name: MA 366, Spring 2010, Quiz 6 (1) (5...

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