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Introduction:-Laplace Transform Was Named After The Great French

Laplace transforms are used to solve differential equations by converting them into algebraic equations. The Laplace transform of a function F(t) is defined by the integral of e−st F(t) from 0 to infinity. This transforms the differential equation into an algebraic equation that can be more easily solved. Some key properties of Laplace transforms are: 1) Shifting property - The Laplace transform of eatF(t) is equal to the transform of F(t) evaluated at s-a. 2) Scaling property - The transform of F(at) is equal to the transform of F(t) with s/a substituted for s. 3) Multiplication by t property - The transform of tnF(t

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108 views7 pages

Introduction:-Laplace Transform Was Named After The Great French

Laplace transforms are used to solve differential equations by converting them into algebraic equations. The Laplace transform of a function F(t) is defined by the integral of e−st F(t) from 0 to infinity. This transforms the differential equation into an algebraic equation that can be more easily solved. Some key properties of Laplace transforms are: 1) Shifting property - The Laplace transform of eatF(t) is equal to the transform of F(t) evaluated at s-a. 2) Scaling property - The transform of F(at) is equal to the transform of F(t) with s/a substituted for s. 3) Multiplication by t property - The transform of tnF(t

Uploaded by

Manoj Sen
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Mathematics-II (301)

Laplace transforms
Introduction :- Laplace transform was named after the great French
mathematician Pierre De Laplace. This is method has become an
important part for Engineer and Scientist. This method reduce the
problem of solving a differential equation with given boundary condition
of algebraic problem.
Transform Concept:- Let F(t) be a function of t. The integral

K ( s , t ) F (t )dt

Is defined the integral transform of the function F(t), provided the


integral is convergent.
Where

K (s , t )

called the kernel of transformation is a function of two

variables s and t; s is a parameter independent of t.


If we define
K (s, t )

K (s, t )

as :

t< 0
{e0 when
whent 0
st

Laplace Transform:- Let F(t) be a function of t defined for all positive


values of t, then the Laplace transform of F(t) is denoted by

L { F (t) }

f(s) is defined by

f (s)= est F (t) dt


0

(1)

Provided the integral is convergent. The Laplace transform of F(t) is


called to exist if the integral in (1) converges for some values of s.

or

Laplace transform of elementary functions:Now we shall obtain Laplace transform of some elementary functions by
applying definition

1
L {1 } = , s> 0
s

(I)

Proof:- We know that

L { F (t) }= est F (t )dt


0

put

F(t)=1

L {1 } = est 1 dt
0

( II )

1
s

est
)
s 0

, s>0 integral is convergent and s


L {tn }=

n!
s n+1

if n=0,1,2,3, s>0

Proof:-

L { F (t) }= est F (t )dt


0

L { t n } = est t n dt
0

integal is divergent.

est t (n+1)1 dt

( By def. of Gamma function

n!
n+1
s

=
L { eat }=

(III)

or

n+ 1
n +1
s

1
, if s>0
sa

Proof:-

L { F (t) }= est F (t )dt

:-

L { eat }= est e at dt
0

e(sa)t dt
0

=
=

e(sa )t
)
(sa) 0

1
sa

Similarly as proved by definition of Laplace transform


(IV)

L { sinat }=

a
s> 0
2
s +a

(V)

L { cosat }=

s
s >0
2
s +a

(VI)

L { sinhat }=

a
s a2
2

(VII)

L { coshat }=

s
2
s a
2

Properties of Laplace transform:(1) First shifting property:If

L { F (t) }=f ( s )

then

L {e at F (t) }=f ( sa )

Proof:- We have

L { F (t) }= est F (t )dt


0

L {e F (t) }= est {e at F(t ) } dt


at

e(sa)t { F (t) } dt
0

( by def. of Laplace transform

L {e at F (t) }=f ( sa )

(2) Change of scale property:If

L { F (t) }=f ( s )

then

1 s
L { F (at ) }= f ( )
a a

Proof:

L { F (at ) }= est F( at)dt


0

put at=u
( dt=du
/a )

u
s( )
a

F( u)

du
a

( )u
1
e a F(u)du
a0

1 s
L { F (at ) }= f ( )
a a

(3) Multiplication by t:If

L { F (t) }=f ( s )

then

L {t n . F (t ) }=(1)n

dn ( )
f s
ds n

Proof:- We know that

L { F (t) }= est F (t )dt


0

.(1)

Differentiating both sides with respect to s, we get


d
f ( s )=
ds
d
f ( s )=
ds

(t )est F( t)dt
0

est { t . F (t)} dt
0

L { t . F (t) } =(1)

d
f ( s)
ds

This result true for n=1


Again diff. (1) both sides w. r. to s, we get
d
f ( s )=
ds

(t )est { t . F (t)} dt
0

d2
(1) 2 f ( s ) =
ds
2

L {t 2 . F (t) }=(1)2

est {t 2 . F (t)} dt
0

d ( )
f s
2
ds

This proves that the result is true for n=2


Now suppose that the result is true for n=k
(1)k

k
f ( s )=
k
ds

L {t k . F (t ) }=(1)k

est {t k . F (t)} dt
0

d ( )
f s
k
ds

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