-
85
Chapter 5: Laplace Transform
Sec 5.1 Introduction
Example: Find Laplace of ctf )( using definition.
Example: Find Laplace of atetf )( .
The Laplace of other function can be evaluated using definition
as in Examples(1) & (2).
Definition:
Let )(tf be a function defined in ),0[ . The Laplace Transform
of )(tf is the function F
defined by the integral
0dttfesF
st)()(
The domain of )(sF is all the values of s for which the integral
in (1) exists. The Laplace
Transform of )(tf is denoted by both F and & f .
-
85
Table of Laplace transforms:
)(xf )(sF & stf )( Domain of )(sF
C
s
C
0s
...3,2,1, ntn 1
!n
s
n
0s
ate
as
1
as
,..3,2,1, nte nat 1
)(
! nas
n
as
btcos
22bs
s
0s
btsin
22bs
b
0s
bteat
cos 22
)( bas
as
as
bteat
sin 22
)( bas
b
as
btcosh
22bs
s
bs
btsinh
22bs
b
bs
-
06
Properties of Laplace:
Proof:
Exercises: 9-12 Sec5.1: Use Theorem 5.1.1 and Table of Laplace
to determine the following laplace
of functions:
# 5.1.10 t
etttf 2sin42)( 3
# 5.1.12 11sin23cosh4)( tttf
Theorem 5.1.1 :Property (1): Linearity
Let 21, ff be two functions for which Laplace exists and let 1c
and 2c be constants then
& 12211 cfcfc & 1f + 1c & 2f
-
06
Sec 5.2 : Continue properties of Laplace
Exercise 5.2.2-5.2.5 use theorem 5.2.1 to evaluate
# 5.2..2 & te t 2sin3
# 5.2.4 & te t sinh2
Theorem 5.2.1 Property 2: Shifting property(Translation
property)
Theorem: If the Laplace Transform & )()( sFsf exist for s
then )()()( asFstfeat for as .
-
06
Proof: using definition
From property 3 ,
& 2)('' ssf & )0(')0( fsff and & 3)(''' ssf &
)0('')0(')0(2 fsffsf and in general,
& nn ssf )()( & )0().........0('')0(')0( )1(321 nnnn
ffsfsfsf
Exercise 5.2.6-5.2.9 use theorem 5.2.2 to evaluate
#5.2.6 & btcos
# 5.2.8 & te4
Theorem 5.2.2: Property 3: Laplace
of the derivative
& ssf )(' & )0(ff
-
06
Derivative of Laplace:
Exercise 5.2.12 – 5.2.15 Use theorem 5.2.4 to evaluate
# 5.2.12 & btt sin
# 5.2.13 & btt cos2
# 5.2.14 tet 23
Theorem 5.2.4:Property 4: Derivative of Laplacen
n
ds
d & nf )1( & )(tft n
which is used mostly to compute & )(tft n n)1(n
n
ds
d & f
-
06
Sec 5.3: Inverse of Laplace:
)(sF is the Laplace of )(tf denoted by & f , )(tf is the
inverse Laplace of )(sF which is denoted by & -1 F . Linearity
of Inverse of Laplace:
& -1 12211 cFcFc & -1
1F + c 2 & -1 2F .
In order to calculate inverse of Laplace . we need to recall the
following Rules of
partial fractions
Now how do we find Partial Fractions of )(
)(
xQ
xP
For example
Next How do we find A,B,C ……
The Rules of Partial fractions are as follows
(1) The numerator )(xP must be of lower degree than that of the
denominator )(xQ . If
it is not, Use long division to ensure that the degree of )(xP
less than the degree of
)(xQ .
(2) Factorize the denominator )(xQ as far as possible. This is
important since the factors
obtained determine the shape of the partial fractions.
(3) This fraction can be broken down into partial fractions,
that is dependent upon the factors of the denominator , there are
four cases:
(4) linear factors dcx
B
bax
A
dcxbax
xP
))((
)(.
(5) Repeated Linear Factors 22 )()(
)(
bax
B
bax
A
bax
xP
.
(6) More Repeated linear Factors 323 )()()(
)(
bax
C
bax
B
bax
A
bax
xP
.
(7) Quadratic factor edx
C
cbxax
BAx
edxcbxax
xP
22 ))((
)(
2223 )22()2)(43(
4
xxxxx
x
-
08
Consider the example above
3132
352
x
B
x
A
xx
x
Multiply both sides by the denominator )3)(1( xx .
)1()3(35 xBxAx
Now we can solve for A and B using one of the following
methods:
Substitution Or Equate coefficients )1()3(35 xBxAx BBxAAxx
335
put 3x )2(33
)1(5
BA
BA
34315 BB solve eq's (1) & (2) 3255
284
BAB
AA
Put 1x
248 AA
So 3
3
1
2
32
352
xxxx
x
Exercise 5.3.1-5.3.4: Find & -1 )(sF for the given
function
# 5.3.1 ss 4
12
-
00
# 5.3.2 )2(
4522
2
sss
ss
# 5.3.3 )4)(16(
12 ss
# 5.3.7 )1)(256(
2 sss
s
-
06
Sec 5.4 Solving Initial value problems using Laplace: Method of
solution will be illustrated using the following two examples
IVP with constant coefficients
(#5.4.6) Solve
2)0(',0)0(06'5'' yyyyy
-
05
(#5.4.6) Solve
2)0(',1)0(4'16'' yyeyy t
Exercise 5.4.10
1)0(''',0)0('')0(')0(11)4( yyyyy
-
05
Example: 4)0(',2)0(05'2'' yyyyy
Answer: teteytt
2sin2cos2
-
66
System of differential equations
# 5.4 12 - # 5.4.13
-
66
-
66
-
66
-
66
Answers to odd-numbered exercises:
