238 Ordinary Differential Equations
Since the linear factor ( s -4) appears to the second power in the denominator
of F(s) and the quadratic factor appears to the first power, the partial fraction
expansion for F(s) has the form
-2s^3 +3s^2 +37s-55 A B C(s-2)+3D
--------= --+ +.(s-4)^2 (s^2 -4s+13) s-4 (s-4)^2 (s-2)^2 +3^2
(13)
Multiplication of (13) by (s - 4)^2 (s^2 - 4s + 13), yields
(14) -2s^3 + 3s^2 + 37s - 55 =
A(s - 4)(s^2 - 4s + 13) + B(s^2 - 4s + 13) + C(s - 2)(s - 4)^2 + 3D(s - 4)^2.Setting s = 4 in (14), we obtain
-2(4)^3 + 3(4)^2 + 37(4) - 55 = B(4^2 - 4(4) + 13) or 13 = 13B.So B = l. Substituting B = 1 into (14), and then subtracting s^2 - 4s + 13
from both sides of the resulting equation, yields
(15) -2s3 +2s^2 +41s-68 = A(s-4)(s^2 -4s+13)+C(s-2)(s-4)^2 +3D(s-4)^2.Lettings= 0 in (15), results in
Lettings= 2 in (15), results in
And letting s = 1 in (15), results in-68 = -52A - 32C + 48D.
6 = -18A + 12D.
-27 = -30A-9C + 27D.Solving the last three equations simultaneously, we find A= -3, C = 1, and
D = -4. Hence,
(1 ) 1 s-2 ( 3 )
F(s)=-^3 s-4 + (s-4)^2 + (x-2)^2 +3^2 -^4 (x-2)^2 +3^2 ·
Since the inverse Laplace transform is a linear operator.c-1[F(s)] = -3.c-1 [-1-] + .c-1 [ 1 ]
s-4 (s-4)^2
-1 [ s - 2 ] -1 [ 3 ]
+ .C (x - 2)^2 + 3^2 -
4
.C (x - 2)^2 + 3^2= -3e^4 x + xe^4 x + e^2 x cos3x - 4e^2 x sin3x.
As we have discovered, manually calculating the Laplace transform of a
function from the definition can involve using various integration techniques
such as integration by parts, substitution, and so forth, or it can involve look-
ing up definite integrals in a table of integrals. If we have a table of Laplace
transforms available, then manually computing the Laplace transform can