s251ex2Laplace(fa03)

# s251ex2Laplace(fa03) - MATH 251 Fall 2003 Exam II Laplace...

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MATH 251 Fall 2003 Exam II Laplace Transforms Make-Up Exam Solutions 6. (14 pts) Write the following function in terms of unit step functions, and ±nd its Laplace trans- form. g ( t ) = t 2 + 1 0 t < 1 e - 3 t + 1 1 t < 2 1 t 2 g ( t ) = (1 - u 1 ( t ))( t 2 + 1) + ( u 1 ( t ) - u 2 ( t ))( e - 3 t + 1) + u 2 ( t ) = t 2 + 1 + u 1 ( t )( e - 3 t + 1 - t 2 - 1) + u 2 ( t )(1 - e - 3 t - 1) = t 2 + 1 + u 1 ( t )( e - 3 t - t 2 ) - u 2 ( t ) e - 3 t G ( s ) = L{ g ( t ) } = 2 s 3 + 1 s + e - s L{ e - 3( t +1) - ( t + 1) 2 } - e - 2 s L{ e - 3( t +2) } = 2 s 3 + 1 s + e - s L{ e - 3 e - 3 t - t 2 - 2 t - 1 } - e - 2 s L{ e - 6 e - 3 t } = 2 s 3 + 1 s + e - s [ e - 3 s + 3 - 2 s 3 - 2 s 2 - 1 s ] - e - 2 s e - 6 s + 3

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8. (12 pts) Find the inverse Laplace transform of: F ( s ) = s 2 - 4 s 3 + 6 s 2 + 9 s First simplify F ( s ) using partial fractions: F ( s ) = s 2 - 4 s ( s 2 + 6 s + 9) = s 2 - 4 s ( s + 3) 2 = a s + b s + 3 + c ( s + 3) 2 = a ( s + 3) 2 + bs ( s + 3) + cs s ( s + 3) 2 . Therefore,
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## This note was uploaded on 07/23/2008 for the course MATH 251 taught by Professor Chezhongyuan during the Spring '08 term at Pennsylvania State University, University Park.

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s251ex2Laplace(fa03) - MATH 251 Fall 2003 Exam II Laplace...

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