Understanding Engineering Mathematics

method of solution – we do not have to find the general solution first and then apply the
initial conditions afterwards.

Exercises on 17.4

1. Evaluate the Laplace transforms of

(i) 3t^2 +cos 2t (ii) 2t+t^3 + 4 tet

2. Solve the following initial value problem by using the Laplace transform

y′+ 2 y= 3 y( 0 )= 0

Answers

1. (i)

6

s^3

+

s

s^2 + 4

(ii)

2

s^2

+

6

s^4

+

4

(s− 1 )^2

2.^32 ( 1 −e−^2 t)

17.5 The inverse Laplace transform

Iff(s) ̃ is some function of the Laplace transform variables, then that functionf(t)whose
Laplace transform isf(s) ̃ is called theinverse Laplace transformoff(s) ̃ and is denoted:

f(t)=L−^1 [f(s) ̃ ]

Crudely, we may think ofL−^1 as ‘undoing’ the Laplace transform operation.
Table 17.1 gives a number of important inverse transforms, simply by reading it from
right to left.

Problem 17.7

Find the inverse transforms of 3=s^2 ,1=.s− 4 /,4=.s^2 Y 9 /,.s^2 − 9 /=.s^2 Y

9 /^2 ,.s− 1 /=[.s− 1 /^2 Y1], from Table 17.1. Check your results by taking

their transform.

Many more inverse transforms may be obtained by algebraic and mathematical manipula-
tion on the transform to put it into a suitable form for inversion by already known inverses.
Like the Laplace transform itself, the inverse Laplace transform is a linear operator, and
this alone accounts for a large number of inverses. Also completing the square (66
➤
) can
be useful as for example in

L−^1

[

s− 1

s^2 − 2 s+ 2

]

=L−^1

[

s− 1

(s− 1 )^2 + 1

]

=e−tcost

Exercise on 17.5

Find the inverse Laplace transform in each case and check by taking the Laplace transform
of your results.