Page 36 - 4811

Partial Fractions

fractional expressions:

1 A B
= + ,
(s − 2)(s − 3) s − 2 s − 3

that is,

1 = A(s − 3) + B(s − 2). (2.9)
Since(2.9)equatestwopolynomials[1andA(s−3)+B(s−2)]thatareequalfor

all s in Ω, except possibly for s = 2 and s = 3, the two polynomials are identically
equalforallvaluesofs.Thisfollowsfromthefactthattwopolynomialsofdegreen
that are equal at more than n points are identically equal. Thus, if s = 2, A = −1,
and if s = 3, B = 1, so that

1 −1 1
F(s) = = + .
(s − 2)(s − 3) s − 2 s − 3

Finally,

( ) ( )
1 1
−1 −1 −1 2t 3t
f(t) = L (F(s)) = L − s − 2 + L s − 3 = −e + e .
.

Example 2.27,

1 A B
= +
(s − 2)(s − 3) s − 2 s − 3

or
1 = A(s − 3) + B(s − 2),

as we have already seen. Since this is a polynomial identity
valid for all s, we may equate the coefficients of like powers of
s on each side of the equals sign. Thus, for s, 0 = A + B; and
0
for s , 1 = −3A−2B. Solving these two equations simultaneously,
A = −1, B = 1 as before.
. . . . . .

Example 2.28, Find
( )
s + 1
−1
L .
2
s (s − 1)
. . . . .
.
Solution. Write
s + 1 A B C
= + + ,
2
s (s − 1) s s 2 s − 1
or
2
. . . . s + 1 = As(s − 1) + B(s − 1) + Cs ,

35