Page 115 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

INTEGRAL TRANSFORMS: THE LAPLACE TRANSFORM 105
Note also that the limit of
s − a
p(s ) (6.54)
q(s )

as s approaches a is simply p(s )/q (s ).

If q(s ) has no repeated roots and is of the form

q(s ) = (s − a 1 )(s − a 2 )(s − a 3 ) ··· (s − a n ) (6.55)

then
n
p(s ) p(a m )
L −1 = e a m t (6.56)
q(s ) q (a m )

m=1
Example 6.6. Find the inverse transform of

4s + 1
F(s ) =
2
2
(s + s )(4s − 1)
First separate out the roots of q(s )
q(s ) = 4s (s + 1)(s + 1/2)(s − 1/2)
4
2
3
q(s ) = 4s + 4s − s − s
2
3

q (s ) = 16s + 12s − 2s − 1
Thus
q (0) =−1 p(0) = 1

q (−1) =−3 p(−1) =−3

q (−1/2) = 1 p(−1/2) =−1

q (1/2) = 3 p(1/2) = 3
f (t) = e −t − e −t/2 + e t/2 − 1

Example 6.7. Solve the differential equation

y − y = 1 − e 3t
subject to initial conditions

y (0) = y(0) = 0