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Unit Step Functions - Ordinary and Partial Differential Equations - Solved Exam | DOWNLOAD SOLVED EXAM FOR QUALITY SCORE

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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 find its Laplace transform. g(t) =   � ... � t 2 + 1 0 ≤ t < 1 e −3t + 1 1 ≤ t < 2 1 t ≥ 2 g(t) = (1 − u1(t))(t 2 + 1) + (u1(t) − u2(t))(e −3t + 1) + u2(t) = t 2 + 1 + u1(t)(e −3t + 1 − t 2 − 1) + u2(t)(1 − e −3t − 1) = t 2 + 1 + u1(t)(e −3t − t 2 ) − u2(t)e −3t G(s) = L{g(t)} = 2 s 3 + 1 s + e −sL{e −3(t+1) − (t + 1)2 } − e −2sL{e −3(t+2)} = 2 s 3 + 1 s + e −sL{e −3 e −3t − t 2 − 2t − 1} − e −2sL{e −6 e −3t } = 2 s 3 + 1 s + e −s [ e −3 s + 3 − 2 s 3 − 2 s 2 − 1 s ] − e −2s e −6 8. (12 pts) Find the inverse Laplace transform of: F(s) = s 2 − 4 s 3 + 6s 2 + 9s First simplify F(s) using partial fractions: F(s) = s 2 − 4 s(s 2 + 6s + 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, s 2 − 4 = a(s 2 + 6s + 9) + b(s 2 + 3s) + cs = (a + b)s 2 + (6a + 3b + c)s + 9a. Solving the system 1 = a + b 0 = 6a + 3b + c −4 = 9a a = −4 9 , b = 13 9 , c = −5 3 . Hence, F(s) = − 4 9 1 s + 13 9 1 s + 3 − 5 3 1 (s + 3)2 . Finally, f(t) = L −1 (F(s)) = −4 9 + 13 9 e −3t − 5 3 te−3 [Show More]

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