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Also, setting s −1 B − C + D, or 0 gives 0 B − C. Equating coefﬁcients of s3 and s, respectively, 0 A + C, 0 A − 2B + C. These last two equations imply B 0. Then from the ﬁrst equation, C 1; ﬁnally, the second equation shows A −1. Therefore, 2s2 (s2 + 1)(s − 1)2 L−1 −L−1 + L−1 s s2 + 1 + L−1 1 s−1 1 (s − 1)2 − cos t + et + tet . Simple Poles. Suppose that we have F(t) F(s) P(s) Q (s) L f (t) for P(s) , (s − α1 )(s − α2 ) · · · (s − αn ) αi αj , where P(s) is a polyomial of degree less than n. In the terminology of complex variables (cf.

1) s2 (s Write s+1 − 1) s2 (s A B C + 2+ , s s s−1 or s+1 As(s − 1) + B(s − 1) + Cs2 , which is an identity for all values of s. Setting s 0 gives B −1; setting s 1 gives C 2. Equating the coefﬁcients of s2 gives 0 A + C, and so A −2. Whence L−1 s+1 − 1) s2 (s −2L−1 1 s − L−1 1 s2 + 2L−1 −2 − t + 2et . 41. Find L−1 2s2 (s2 + 1)(s − 1)2 . We have 2s2 (s2 + 1)(s − 1)2 As + B C D + + , 2 s +1 s − 1 (s − 1)2 1 s−1 38 1. Basic Principles or (As + B)(s − 1)2 + C(s2 + 1)(s − 1) + D(s2 + 1). 2s2 1 gives D Setting s 1.

11). n Inﬁnite Series. For an inﬁnite series, ∞ n 0 an t , in general it is not possible to obtain the Laplace transform of the series by taking the transform term by term. 18 1. 17. f (t) e −t ∞ 2 (−1)n t 2n , n! n 0 −∞ < t < ∞. Taking the Laplace transform term by term gives ∞ n 0 (−1)n L(t 2n ) n! ∞ n 0 1 s (−1)n (2n)! n! s2n+1 ∞ n 0 (−1)n (2n) · · · (n + 2)(n + 1) . s2n Applying the ratio test, 2(2n + 1) n→∞ |s|2 un+1 un lim n→∞ ∞, lim and so the series diverges for all values of s. 2 2 However, L(e−t ) does exist since e−t is continuous and bounded on [0, ∞).