Matrices and Tensors - Continuum Mechanics of Anisotropic Materials - page 409

Table A.1 A very short table
of Laplace transforms

Transform
function Object function Conditions

1 = s 1 s > 0

e at

1 =ð s a Þ s > 0

s 2 a 2

ð 1 = a Þ sin at a 6¼ 0 ; s > 0

1 =ð þ Þ

s=ðs 2 þ a 2 Þ cos at s > 0

1 =ðsðs þ aÞÞ

ð 1 =aÞð 1 e at Þ s>a

¼ L 1 P

ð s Þ ða 1 Þ

Q 0 ða 1 Þ

P ða n Þ

Q 0 ða n Þ

P

e a 1 t e a n t þ ...þ : (A.216)

Q ð s Þ

The result above was obtained by using the Table A.1 to verify that

a i Þ 1

L 1 e a i t

fð s g¼ :

Example A.16.1

d y
d t

e 3 t

Problem : Solve the differential equation 5 y ¼ þ 4 for y (0) ¼ 0- using

Laplace transforms.

Solution: Noting that the Laplace transform of e at

1

sþa and that the Laplace
transform of the derivative of a function is equal to s times the Laplace transform of
the function minus the value of the function at t

is

¼ 0, the Laplace transform of

d y
d t

e 3 t

1 4

5 y ¼ þ 4is s y ^ ð s Þ 5 y ^ ð s Þ¼

s 3 þ s , thus

5 s 12

y ^ ð s Þ¼

Þ : s ð s 3 Þð s 5

By partial fractions

1 4

5 s

1 13

Þ ¼ Þ þ Þ ;

thus

s ð s 3 Þð s 5 2 ð s 3 10 ð s 5

4
5 s

1 13

y ^ ð s Þ¼

Þ þ Þ :

2 ð s 3 10 ð s 5

L 1 Þ 1 e at , the inverse Laplace transform is Using fð s a g¼

4
5

1
2 e 3 t

13
10 e 5 t

y ð t Þ¼ þ :

The convolution of the functions f ( t ) and g ( t ) is indicated by f ð t Þ g ð t Þ and

defined as the integral