Laplace Inverse by Table Lookup The ﬁrst thing we need to be able to do is to use the Laplace table to ﬁnd the inverse Laplace transform. We will illustrate this entirely by examples. Notation: The inverse Laplace transform will be denoted L−1 . Example 1. Find L−1 (1/(s − 2)). Solution. Use the table entry L(e at ) = 1/(s − a): L−1 (1/(s − 2)) = e2t . Example 2. Find L−1 (1/(s2 + 9)). Solution. Use the table entry L(sin(ωt)) = ω/(s2 + ω 2 ) and linearity: � � � � 1 1 −1 3 1 −1 L = L = sin(3t). 2 2 2 s +9 3 s +3 3 Example 3. Find L−1 (4/s2 ). Solution. Use the table entry L(t) = 1/s2 : L−1 (4/s2 ) = 4t. Example 4. Find L−1 (4/(s − 2)2 ). Solution. Use the s-shift formula L(ezt f (t)) = F (s − 2), where, in this case, F (s) = 4/s2 ⇒ f (t) = 4t by example (3). Therefore, L−1 (4/(s − 2)2 ) = L−1 ( F (s − 2)) = e2t f (t) = e2t 4t. Example 5. Find L −1 � � 1 . s2 + 4s + 13 Solution. We ﬁrst need to complete the square s2 + 4s + 13 = s2 + 4s + 4 + 9 = (s + 2)2 + 9. We have a shifted function F (s + 2), where F (s) = 1/(s2 + 9). Using exam ple (2), we know that f (t) = sin(3t)/3, so using the s-shift rule we get � � 1 sin(3t) −1 L = L−1 ( F (s + 2)) = e−2t . s2 + 4s + 13 3 Laplace Inverse by Table Lookup Example 6. Find L −1 � OCW 18.03SC � s . ( s2 + ω 2 )2 Solution. We haven’t seen this formula yet, but there is a table entry, which t gives: sin(ωt). 2ω Example 7. Find L−1 � 1 ( s2 + ω 2 )2 � . Solution. This is also a table entry, answer: 2 1 (sin(ωt) − ωt cos(ωt)). 2ω 3