18.03SC Practice Problems 28 Inverse Laplace transform Solution suggestions 1. Find the inverse Laplace transform for each of the following. 2s + 1 s2 + 9 s2 + 2 s3 − s , , s2 ( s 2 . − 1) For each of these expressions, we will ﬁrst use the method of partial fractions to break up the expression, and then use formula tables and linearity to read off its inverse transform. The ﬁrst expression breaks up as 2s + 1 s 3 =2 2 + (1/3) 2 , s2 + 9 s +9 s +9 so it has inverse transform � � 1 −1 2s + 1 L = 2 cos(3t) + sin(3t), s2 + 9 3 for t > 0. The denominator in the second expression factors as s3 − s = s(s − 1)(s + 1). So it has a partial fractions decomposition of the form A B C s2 + 2 = + + s(s − 1)(s + 1) s s−1 s+1 . We use the Heaviside cover up method to determine that the coefﬁcients here are A= 02 + 2 = −2, (0 − 1)(0 + 1) B= 12 + 2 3 = , (1)(1 + 1) 2 and C= (−1)2 + 2 3 = , (−1)(−1 − 1) 2 so the inverse transform of the second expression is L −1 � � � � s2 + 2 3/2 3/2 3 3 −1 − 2 = L + + = −2 + e t + e − t , s3 − s s s−1 s+1 2 2 for t > 0. The denominator in the third expression is s2 (s − 1), which has a factor of multi plicity two. So this time the partial fraction decomposition has the slightly more complicated form s2 ( s 2 − 1) = A B C + + . 2 s s s−1 We can uncover the leading coefﬁcients A and C by Heaviside as before, and use them to solve for the remaining coefﬁcient: A= 2 = −2 0−1 and C= 2 = 2, (1)2 so, from the original equation, 2 = −2(s − 1) + Bs(s − 1) + 2s2 , and B = −2. We can now read off that the inverse transform of the third expression is � � � � 2 −2 2 −1 −1 − 2 L =L + 2 + = −2 − 2t + 2et , s2 ( s − 1) s s s−1 for t>0.

18.03SC Practice Problems 28: Inverse Laplace Transform

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