Integral Transformations - Heat Conduction: Mathematical Models and Analytical Solutions - page 472

L − 1

by using different methods.

1

( )=

Example 6 .Find f t

s 2 ( 1 + s 2 )

Solution.

1. By the linearity of Laplace transformation,

L − 1 L − 1 1

s 2

L − 1

1 1

f ( t )= = − = t − sin t .

s 2 ( 1 + s 2 ) 1 + s 2

2. By the Heaviside expansion,

e st s = i + e st s = − i +

s 2 e st

1 1 1

f ( t )= lim

s

)] )]

1 +

[ s 2 ( 1 + s 2 [ s 2 ( 1 + s 2

→ 0

s 2 s = 0 =

e i t e − i t

2i

t e st −

= − + t − sin t .

1 +

3. By the convolution theorem,

L − 1 1
s 2 s 2

t
0 (

1

f ( t )= = t ∗ sin t = t − τ ) sin τ d τ

1 +

t
0

t
0

=( − t cos τ ) | − ( − τ cos τ + sin τ ) | = t − sin t .