Page 140 - Advanced engineering mathematics

120 CHAPTER 3 The Laplace Transform

EXAMPLE 3.24
Decompose
2
x + 2x + 3
2
(x + x + 5)(x − 2) 2
2
into partial fractions. First observe that x + x + 5 has complex roots and so is irreducible. Thus,
use the form
2
x + 2x + 3 A B Cx + D
= + + .
(x + x + 5)(x − 2) 2 x − 2 (x − 2) 2 x + x + 5
2
2
2
If we add the fractions on the right, the numerator must equal x + 2x + 3. Therefore,
2
2
2
2
A(x − 2)(x + x + 5) + B(x + x + 5) + (Cx + D)(x − 2) = x + 2x + 3.
Expand the left side, and collect terms to write this equation as
2
3
(A + C)x + (−A + B − 4C + D)x + (3A + B + 4C − 4D)x − 10A + 5B + 4D
2
= x + 2x + 3.
Equate coefficients of like powers of x to get
A + C = 0,
−A + B − 4C + D = 1,
3A + B + 4C − 4D = 2,
and
−10A + 5B + 4D = 3.
Solve these to obtain A = 1/11, B = 1,C =−1/11, and D =−3/11. The partial fractions
decomposition is
2
x + 2x + 3 1 1 1 1 x + 3
= + − .
2
(x + x + 5)(x − 2) 2 11 x − 2 (x − 2) 2 11 x + x + 5
2

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October 14, 2010 14:14 THM/NEIL Page-120 27410_03_ch03_p77-120