1550078481-Ordinary_Differential_Equations__Roberts_

The Laplace Transform Method 237

EXAMPLE 9 Finding an Inverse Laplace Transform

2

For F(s) = s(s + l), calculate .c-^1 [F(s) ].

SOLUTION

The denominator of F(s), which is Q(s) = s(s + 1), h as two linear factors

of multiplicity 1, so the partial fraction expansion for F(s) has the form

(10)

2 A B

F(s) = s(s + 1) = --; + s + 1

where A and B are constants to be determined. Mult iplying equation (10) by

s(8 + 1), we see A and B must be chosen to satisfy

(11) 2 = A(8 + 1) + B8.

Setting 8 = 0 in equation (11), we find A = 2. And setting 8

equation (11), we see B = -2. Hence,

.C[f(x) ] = 8(8^2 + 1) =--;^2 - 8 +^2 1 =^2 (1) --; -^2 ( 8 + 1 ) 1 ·

Since

£[1] = ~ and .C[e-x] = _2_

1

,

8 8+

(12) .C[f(x) ] = 2£[1] - 2.C[e-x] = £[2 - 2e-x].

-1 in

Taking the inverse Laplace transform of equation (12), we find f (x) = 2-2e-x.

EXAMPLE 10 Finding an Inverse Laplace Transform

-28^3 + 382 + 378 - 55 _ 1

For F(8) = (

8

_

4

) 2 ( 82 _

48

+

13

) , calculate£ [F(8)].

SOLUTION

The quadratic factor 82 - 48 + 13 appearing in the denominator of F(8)

is irreducible, since its discriminant (-4)^2 - 4(1)(13) = 16 - 52 = -36 < 0.

Completing the square, we can rewrite this quadratic factor as

82 - 48 + 13 = (8^2 - 48 + 4) + 9 = (8 - 2)^2 + 32 = (8 - a)^2 + b^2.