Question 5.6.2: Find the inverse Laplace transform of F(s) = 9/(s^2 +1)^2(s...

The partial fraction decomposition of F(s) is

F(s)=− \frac {2/3}{s^{2} +1}+ \frac {1}{(s^{2} +1)^{2}}+ \frac {2/3}{s^{2} +4}+ \frac {1}{(s^{2} +4)^{2}}          (5.6.2)

Two of the terms in (5.6.2) are straightforward to invert, but the two involving squares of irreducible quadratic terms are not among familiar functions from our previous work. In the following subsection, we demonstrate how to use Maple to compute the inverse transform of such functions. These computations reveal that

L^{−1} [\frac {1}{(s^{2} +1)^{2}}]= \frac {1}{2}sin t − \frac {1}{2}t cos t

and

L^{−1} [\frac {1}{(s^{2} +4)^{2}}]= \frac {1}{16}sin2t − \frac {1}{8}t cos2t

From this work and (5.6.2), we find

L^{−1}[F(s)]=−\frac {2}{3}sin t + \frac {1}{2}sin t − \frac {1}{2}t cos t + \frac {1}{3}sin2t + \frac {1}{16}sin2t − \frac {1}{8}t cos2t

=−\frac {1}{6}sin t − \frac {1}{2}t cos t + \frac {19}{48}sin2t − \frac {1}{8}t cos2t