Embed Size (px)
Citation preview
UNIT VIII
INVERSE LAPLACE TRANSFORMS
Suppose )()}({ sftfL = then )(tf is called as the inverse Laplace transform of
)(sf and is written as )()}({1 tfsfL =− . Here 1−L denote the inverse Laplace
transform. The inverse Laplace transforms given below follow at once from the
results of Laplace transforms, studied earlier.
(1) 111
=
−
sL (2) at
eas
L =
−
− 11 (3) ate
asL
−−=
+
11
(4) atas
aL sin
22
1=
+
− (5) atas
sL cos
22
1=
+
− (6)
atas
aL sinh
22
1=
−
−
(7) atas
sL cosh
22
1=
−
− (8) .....3,2,1 !1
1==
+
−nt
s
nL
n
n
(9) )(
!1
1 nat
nte
as
nL =
− +
− (10) btebas
bL
atsin
)( 22
1=
+−
−
(11) btebas
asL
atcos
)( 22
1=
+−
−− (12) btebas
bL
atsinh
)( 22
1=
−−
−
(13) btebas
asL
atcosh
)( 22
1=
−−
−−
Partial fractions:
1) bs
B
as
A
bsas ++
+=
++ ))((
1
2) 22 )())((
1
bs
c
bs
B
as
A
bsas ++
++
+=
++
3) cbss
cBs
as
A
cbssas ++
++
+=
+++ 22 ))((
1
Examples: Find the inverse Laplace transform of
1.
+
+−
8
2532
1
s
sL
++
+=
−−
22
1
22
1
)22(
125
)22(3
sL
s
sL
= )22sin(22
1.25)22cos(3 tt +
= )22sin(2
5)22cos(3 tt +
2.
+−
+−
134
22
1
ss
sL
+−
+−=
−
22
1
3)2(
42
s
sL
+−+
+−
−=
−−
22
1
22
1
3)2(
4
3)2(
2
sL
s
sL
)3sin3
43(cos3sin
3
43cos 222 ttetete ttt +=+=
3.
−
−
5
1
)2(
2
sL =
−
+−−
5
1
)2(
22
s
sL
=
−+
−
−
54
1
)2(
2
)2(
1
ssL
=
−+
−
−−
5
1
4
1
)2(
!4
!4
2
)2(
!3
!3
1
sL
sL
= 4232
12
1
6
1tete tt + = )2(
12
1 432 tte t +
4. L-1
− 4)2(
3
s = 3e
2t L
-1
4
1
s
= 3e2t
2!3
233 tett=
5. L-1
++ 204
12 ss
(s2 + 4s + 20= s
2 + 4s + 2
2 + 20 - 2
2
= (s + 2)2 + 16
= L-1
++ 22 4)2(
1
s = (s + 2)
2 + 4
2)
= e-2t
L-1
+ 22 4
1
s
= e-2t
4
4sin t
6.
++
+−
258
22
1
ss
sL
++
−++=
−
22
1
3)4(
442
s
sL
++−
++
+=
−−
22
1
22
1
3)4(
12
3)4(
4
sL
s
sL
+−
+=
−−−−
22
14
22
14
3
12
3 sLe
s
sLe
tt
3
3sin23cos 44 t
ete tt −− −=
−=
−tte
t3sin
3
23cos
4
7. L-1
+−
−
52
322 ss
s
= L-1
+−
−−
4)1(
1222s
s
= L-1
+−
−
4)1(
)1(22s
s - L
-1
+− 4)1(
12s
= et L
-1
+ 22 2
2
s
s - e
t L
-1
+ 22 2
1
s
= et [ 2 cos 2t – ½ sin 2t].
8. L-1
− 3
2
)( as
s = L
-1
−
+−3
2
)(
)(
as
aas
= L-1
−
−++−3
22
)(
)(2)(
as
asaaas
= L-1
−+
−+
− 23
2
)(
2
)(
1
as
a
as
a
as
= eat + e
at a
2
!2
2t + 2ae
at
!1
t
= eat +
2
2a t
2 e
at + 2at e
at
9.
−
+−
2516
1542
1
s
sL
−
+=
−
+
16
2516
154
2516
154 sin
22
s
s
s
sce
−
+
−
=
16
25
1.
16
15
16
25.
4
1
22 ss
s
( ) ( )
−+
−=
−
+∴ −−−
22
1
22
1
2
1
4/5(
1
16
15
4/5(4
1
2516
154
sL
s
sL
s
sL
)4/5sinh(4/5
1
16
15)4/5cosh(
4
1tt +=
)4/5sinh(4
3)4/5cosh(
4
1tt +=
10. Find the inverse Laplace transform of 2)2)(1(
32
+−
+
ss
s
Let 2)2)(1(
32
+−
+
ss
s=
2)2(21 ++
++
− s
c
s
B
s
A
)1()2)(1()2(32 2 −++−++=+∴ sCssBsAs
Putting s=1,-2, and 0 respectively we get 3
1,
9
5,
9
5=
−== CBA
−+
−−
−=∴ −−−
2
111
)2(
1
3
3
2
1
9
5
1
1
9
5f(t)
sL
sL
sL
ttt teee 22
3
1
9
5
9
5 −− +−=
11. Find the inverse Laplace transform of )3)(2)(1(
42 2
−−+
−
sss
s
Let )3)(2)(1(
42 2
−−+
−
sss
s=
)2(21 −+
−+
+ s
c
s
B
s
A
)2)(1()3)(1()3)(2(42 2 −++−++−−=−∴ ssCssBssAs
Putting s=-1, 2, and 3 respectively we get 2
7,
3
4,
6
1=−=−= CBA
−−+
−=∴
−
)3)(2)(1(
42f(t)
21
sss
sL
=
−+
−−
−
−−−
2
111
)2(
1
3
3
2
1
9
5
1
1
9
5
sL
sL
sL
12. Find
−+−
+−−
6116
56223
21
sss
ssL
Let )3(216116
56223
2
−+
−+
−=
−+−
+−
s
c
s
B
s
A
sss
ss
)2)(1()3)(1()3)(2(562 2 −−+−−+−−=−∴ ssCssBssAss
Putting s= 1, 2, and 3 respectively we get 2
5,1,
2
1=−== CBA
−+−
+−=∴
−
6116
562
23
21
sss
ssLf(t)
−+
−−
−=
−−−
)3(
1
2
5
2
1
1
1
2
1 111
sL
sL
sL
ttt eee 32
2
5
2
1+−=
12. Find
+
−
44
1
4as
sL
Let 2222244 4)2(4 saasas −+=+
)22)(22(4 ., 222244 asasasasasie −+++=+
++−
−+=
+ )22(
1
)22(
1
4
1
4 222244 asasasasaas
s
++−
+−=
+
−−−
22
1
22
1
44
1
)(
1
)(
1
4
1
4 aasL
aasL
aas
sL
+−
+=
−−−
22
1
22
1 11
4
1
asLe
asLe
a
atat
−=−
a
ate
a
ate
a
atat sinsin
4
1
2244
1
2
sinhsin
22
sin
4
a
atatee
a
at
as
sLThus
atat
=
−
=
+
−−
Find the Inverse L.T of the following functions:
Evaluation of )}({1 sFeL as−−
By Heaviside shifting theorem we have if )()}({ sFtfL = then
)()}()({ sFeatHatfL as−=−−
)()()}({1 atHatfsFeL as −−=∴ −−
Find the Inverse Laplace Transforms of:
1. Find
−+
−−−
s
e
s
e
sL
ss 2
32
1 323
−
+
=
−−
−−−
s
eL
s
eL
sL
ss 21
3
1
2
132
13 …………………… (1)
We have 11
,2
1 ,
1 12
3
1
2
1 =
=
=
−−−
sL
t
sLt
sL
Equation (1) becomes
)2(1.3)1(2
)1(23
2
−−−−
+= tutut
t
)2(.3)1()1(3323 2
2
32
1−−−−+=
−+∴
−−−
tututts
e
s
e
sL
ss
2. Find
+
+ −−−
22
21
π
π
s
eseL
ss
++
+=
−−−−
22
1
22
21
π
π
π seL
s
seL
ss …………………. (1)
We have ts
sL π
πcos
22
1=
+
− , ts
L ππ
πsin
22
1=
+
−
Hence equation (1) becomes
)1()1(sin)2/1()2/1(cos −−+−−= tuttut ππ
)1(sin)2/1(sin −−−= ttuttu ππ
( ))1()2/1(sin −−−= tututπ
+
+ −−
4
1
)1(
)2(
s
esL
s
4)1(
)2(
+
+=
s
sf(s)Let
We shall first find
)()}({1 tfsfL =∴ −
+
=
+
++=
+
+ −−−−
4
1
4
1
4
1 1
)1(
1)1(
)1(
)2(
s
sLe
s
sL
s
sL
t
+
=
+
+ −−−
43
1
4
1 11
)1(
)2( .,
ssLe
s
sLie
t
get we!
1 sin
1
1
n
t
sLgU
n
n=
+
−
+=
+=
−−
62!3!2
3232 tte
ttef(t)
tt
Next we have [ ] )1( )1()(1 −−=−− tutfsfeL s
)1( 6
)1(
2
)1(
)1(
)2(
32)1(
4
1−
−+
−=
+
+ −−−−tu
tte
s
seLThus
ts
3. L-1
−
−
2
3
)4(s
e s
Here F (s) = 2)4(
1
−s = L
-1 {F(s)} = L
-1
− 2)4(
1
s
= e4t
L-1
2
1
s = - e
4t t = f (t)
L-1
−
−
2
3
)4(s
e s
= L-1
{e-3s
F (s)} = f (t – 3) H (t – 3)
= e4 (t – 3)
(t – 3) H (t – 3).
4. L-1
+
−
12s
e sπ
.
Here F (s) = 1
12 +s
L-1
{F(s)} = L-1
+1
12s
= sin t = f (t)
L-1
+
−
12s
e sπ
= L-1
{eπs
F (s)} = f (t - π) H (t - π)
= sin (t - π) H (t - π)
= - sin (π - t ) H (t - π)
= - sin (π - t) H (t - π)
= - sin t H (t - π)
Evaluation of
−
s
sFL
)(1 &
−
))((1
sFds
dL
n
n
.
� If )()}({ sFtfL =
Then s
sFdttfL
t)(
)(0
=
∫
∫∫−−
==
∴
tt
dtsFLdttfs
sFL
0
1
0
1)}({)(
)(
� If )()}({ sFtfL = then
)}({)1()}({ sFds
dtftL
n
nnn −=
)}({)1()()1()}({11
sFLttftsFds
dL
nnnn
n
n−−
−=−=
∴
Find the inverse Laplace transforms of the following functions:
1. )(
1
ass +
Let F (s) = as +
1
L-1
{F(s)} = e-at
L-1
+ )(
1
ass = ∫
−
t
ate
0
dt
=
tat
a
e
0
−
−
= - a
1 (e
-at – e
-0)
= - a
1 (e
-at – 1)
= a
1 (1 – e
-at).
2.
+
−
)(
122
1
assL
22
1)(
assF
+=
a
attf
sin)( =∴
∫=
−
t
duufs
sFL
0
1)(
)(
+
−
)(
122
1
assL
at
a
at
adt
a
at
00
cos1sin
−== ∫
)1cos(1
2+−= at
a
)cos1(1
2at
a−=
3. Find the inverse Laplace transforms of )1(
123 +ss
1
1)(
2 +=
ssF ttf sin)( =∴
∫=
−
t
duufs
sFL
0
1)(
)(
)cos1(sin)1(
1
0
2
1tdtt
ssL
t
−==
+ ∫−
( )tt
ttdttss
L 0
0
22
1sin)cos1(
)1(
1−=−=
+ ∫−
ttss
L sin)1(
122
1−=
+
−
∫ −=
+
−
t
dtttss
L0
23
1)sin(
)1(
1
t
tt
0
2
cos2
+=
1cos2)1(
1
2
23
1−+=
+
−t
t
ssLThus
4.
−
ssL
1sin
1
1
We know that --------- !5!3
sin53
−+−=xx
xx
−−−−−−−−++−=23 120
1
6
111sin
ssss
−−−+−=
−−
23
11
120
1
6
111sin
sssL
sL
−−−−−+−−=
− t
t
ssL
120
1
!26
11
1sin
1 21
∫
−−−−+−=
−
t
dttt
ssL
0
21
120
1
121
1sin
1
−−−−−−−−+−=
∴ −
24036
1sin
123
1 ttt
ssL
Evaluate:
5. L-1
{log (1+
+
2
11log
s.
Let L-1
+
2
11log
s = f (t)
Then log
+
2
11
s = L {f(t)} = F (s)
Now ds
d { F(s)} =
ds
d [ log (s
2 + 1) – log s
2] =
ss
s 2
1
22
−+
L-1
{ds
d (F(s))} = L
-1
−+ ss
s 2
1
22
- t f(t) = 2 cos t - 2
= - 2 (1 – cos t)
f (t) = t
t)cos1(2 −.
6.
+
+−
bs
asL log
1
We use the formula
−=−
ds
sdFL
ttf
)(1)(
1
+
+−=∴
−
bs
as
ds
dL
ttf log
1)(
1
+−+−=−
))log()(log(1
)(1
bsasds
dL
ttf
+−
+−=
bsast
111
)(1 btat eet
−− −−=
t
eetf
atbt −− −=)(
7.
+
−
+−2
1
1log
1
s
ssL
Sol: )(21
1log
1tf
s
ssLLet =
+
−
+−
21
1log +
−
+==
s
ssf(s))then L{f(t
+
−
+−==∴ 2
1
1log)1(
s
ss
ds
df(s)L{tf(t)
[ ]{ }21log(1log( +−−+−= sssds
d
−
+−
−−
+−=
1
1log
1
1
1
1
s
s
sss
−
+−
−=
1
1log
1
22 s
s
s
s
−
−=−
t
eetL
tt
cosh2
−
−=−
t
eetttf
tt
cosh2)(
−
−=∴−
2
cosh2)(
t
ee
t
ttf
tt
−
−=
+
−
+∴
−−
2
1 cosh22
1
1log
t
ee
t
t
s
ssL
tt
8. Evaluate
+
−
222
1
)( as
sL
We know that )(sin1
)( 22
1tfat
aas
sL ==
+
−
)( 1
ttfds
dLhaheweNow −=
−
ata
t
asds
dL sin
122
1−=
+
−
ata
t
as
sL sin
)(
2222
1−=
+
−⇒ −
ata
t
as
sL sin
2)( 222
1=
+∴
−
9. Prove that ttts
L sinhsin22
tan2
11=
−−
Let )(2
tan2
11tf
sL =
−− then { } )(2
tan)(2
1sF
stfL =
=
−
{ }
=
−
2
1 2tan)(
sds
dsF
ds
d
4
44
21
1432
2
+
−=
−
+
=s
s
s
s
Consider 2224 4)2(4 sss −+=+
)22()22( 22 ssss −+−+=
{ })22)(22(
4)(
22 ++−+−=
ssss
ssF
ds
d
++−
+−=
22
1
22
122 ssss
++−
+−=
1)1(
1
1)1(
122 ss
{ }
++−
+−=
−−
1)1(
1
1)1(
1)(
22
11
ssLsF
ds
dL
++
+−=−
−−−
1
1
1
1)(
2
1
2
1
sLe
sLettf
tt
[ ]tetettf tt sinsin)( −−−=−
−=
−
tt
eetf
tt
sin)(
ttt
tf sinsinh2
)( =
Exercise:
1. ssss
1
15
4
23
1+
++
−
2. 1682
2
++
−
ss
se s
3. 44
32
+s
es s
4.
+
2
2
1logs
a
5.
−
s
a1tan
6.
−
a
s1cot
Convolution Definition: The convolution of two functions )(tf and )(tg usually denoted
)(*)( tgtf is denoted in the form of an integral as follows.
duutguftgtf
t
u
)()()(*)(0
−= ∫=
Property: )(*)()(*)( tftgtgtf =
That is to say that the convolution operation * is commutative.
Convolution theorem
If )()]([ and )()]([ 11 tgsgLtfsfL == −− then
duutgufsgsfL
t
u
)()()]().([0
1−= ∫
=
−
Proof: We shall show that
)().()()(0
sgsfduutgufL
t
u
=
−∫
=
We have L.H.S by the definition
dtduutgufeduutgufL
t
ut
st
t
u
−=
− ∫∫∫
=
∞
=
−
=
)()()()(000
dudtutgufeduutgufL
t
u
st
t
t
u
)()()()(000
−=
− ∫∫∫
=
−
∞
==
----------------(1)
We shall change the order of integration in respect to this double integral.
Existing region:
strip) (Vertical t to0u
strip) l(Horizonta to0
=
∞=t
On changing the order:
strip) l(Horizonta tot
strip) (Vertical to0
∞=
∞=
u
u
On changing the order of integration, (1) becomes
dtduutgufeut
st
u
)()(0
−∫∫∞
=
−
∞
=
----------------- (2)
Now, let us put vut =− where u is fixed dvdt =∴
If ∞=∞=== v, tIf 0; v,ut and hence (2) becomes
dvduvgufev
vus
u
)()(0
)(
0
∫∫∞
=
+−
∞
=
= ∫∫
∞
=
−
∞
=
− dvvgeduufev
sv
u
su )( . )(00
R.H.Sg(s) . )( == sf
Hence we have proved that
)().()()(0
sgsfduutgufL
t
u
=
−∫
=
duutgufsgsfLThus
t
u
)()()]().([ 0
1−= ∫
=
−
This proves the convolution theorem.
Verification of the convolution theorem
Working procedure for problems
We need to verify the theorem in respect of the two given functions )(tf and )(tg
1. We find )]([ g(s) and )]([)( tgLtfLsf ==
2. We evaluate duutguftgtf
t
u
)()()(*)(0
−= ∫=
3. We find )].(*)([ tgtfL
4. If )().()](*)([ sgsftgtfL = then we can conclude that the theorem is verified.
Ex. Verify convolution theorem for the pair of the function
-teg(t) and sin)( == ttf
Sol:
1
1)L(eg(s) ,
1
1)(sin)( t-
2 +==
+==
sstLsf
duuee
t
u
ut
∫=
−=
0
sin
t
u
ut uu
ee
0
)cos(sin11
=
−
−
+=
22
cossin1)cos(sin
2)(*)(
ttt etttte
etgtf
−−
+−
=
+−=
duuduutguftgtf
t
u
t
u
∫∫==
=−=0
u)-(t-
0
e sin)()()(*)(
++
+−
+=∴
1
1
11
1
2
1)](*)([
22 ss
s
stgtfL
)1)(1(
12 ++
=ss
Also )1)(1(
1)().(
2 ++=
sssgsf
)().()](*)([ sgsftgtfLThus = . The theorem is verified.
Ex. Verify convolution theorem for the pair of the function
btg(t)attf cos and cos)( ==
Sol. 222
cos ,
)(cos)(bs
sbt)L(g(s)
as
satLsf
+==
+==
duauduutguftgtf
t
u
t
u
∫∫==
=−=00
bu)-cos(bt cos)()()(*)(
dubu)(au-bt bt-bu)(au tgtf
t
u
]cos[cos2
1)(*)(
0
∫=
+++=
t
uba
btbubuau
ba
bubtautgtf
0
)sin()sin(
2
1)(*)(
=
+
−+
−
−+=
{ } { }
+
++−
−= btat
babtat
basinsin
1sinsin
1
2
1
−−
++
++
−=
bababt
babaat
11sin
11sin
2
1
−
−+
−=
2222
2sin
2sin
2
1)(*)(
ba
bbt
ba
aattgtf
ba ),sinsin(1
22≠−
−= btbata
ba
[ ]
+−
+−=
222222..
1)(*)(
bs
bb
as
aa
batgtfL
[ ]))(())((
)(.
1)(*)( .,
2222
2
2222
222
22 bsas
s
bsas
bas
batgtfLie
++=
++
−
−=
))(()().(
2222
2
bsas
ssgsfAlso
++=
)().()](*)([ sgsftgtfLThus = . The theorem is verified.
Type-2 Computation of the inverse transform by using convolution theorem
Working procedure for problems 1. The given function is expressed as the products of two functions say f(s) and g(s)
2. We find )()]([ and )()]([ 11 tgsgLtfsfL == −−
3. We apply the convolution theorem in one of the form:
duutgufsgsfL
t
u
)()()]().([ 0
1−= ∫
=
−
4. We evaluate the convolution integral to obtain the required inverse.
Ex. Using convolution theorem obtain the inverse Laplace transform of the
function
)(
122 ass +
Sol: Let 22s
1g(s) ;
1)(
assf
+==
Taking inverse,
a
at
aL
sLtf
sin
s
1g(t) ; 1
1)(
22
11=
+==
=
−−
We have convolution theorem,
duutgufsgsfL
t
u
)()()]().([ 0
1−= ∫
=
−
dua
auat
aL
t
u
∫=
− −=
+∴
0
22
1 )sin(.1
)s(s
1
)cos1(1)cos(
2
0
2at
aa
auatt
−=
−=
)cos1(1
)s(s
1
222
1at
aaLThus −=
+
−
Ex. Using convolution theorem obtain the inverse Laplace transform of the
function
22 )54(
2
++
+
ss
s
Sol: Let 54
1g(s) ;
54
2)(
22 ++=
++
+=
ssss
ssf
++=
++
+=
−−
1)2(s
1g(t) ;
1)2(
2)(
2
1
2
1L
s
sLtf
+=
+=
−−−−
1s
1g(t) ;
1)(
2
12
2
12Le
s
sLetf
tt
teetf tt sing(t) ;cost )( 22 −− ==∴
Now by applying convolution theorem we have,
∫=
−−−−−=
++
+t
u
utuduutuee
ss
sL
0
)(22
22
1)sin(cos
)54(
2
∫=
−−−=
++
+t
u
tduuute
ss
sL
0
2
22
1 cos )sin(
)54(
2
∫=
−
−−++−=
t
u
t
duuutuute
0
2
)]sin()[sin(2
∫=
−
−+=
t
u
t
duutte
0
2
)]2sin([sin2
−+=
++
+ −−
t
tt ut
ute
ss
sL
0
0
2
22
1
2
)2cos(][sin
2)54(
2
−+=−
)cos(cos2
1sin
2
2
tttte t
2
sin
)54(
2
2
22
1 tte
ss
sLThus
t−−
=
++
+
Ex. Using convolution theorem obtain the inverse Laplace transform of the
function
222 )( as
s
+
Sol:
++=
+
−−
)(
1.
)(
1
)( 2222
1
222
1
asasL
as
sL
2222
s ;
1)(
asg(s)
assf
+=
+=
atas
Lg(t)as
Ltf coss
;sinat a
1
1)(
22
1
22
1=
+==
+=
−−
Now by applying convolution theorem we have,
∫=
−−=
+
t
u
duauatauaas
sL
0
222
1)cos(sin
1
)(
∫=
+−+−+=
t
u
duauatauauataua
0
)sin()sin(2
1
∫=
−+=
t
u
duatauata
0
)2sin(sin2
1
t
a
atauatu
aas
sL
0
222
1
2
)2cos(sin
2
1
)(
−−=
+
−
+−=
a
at
a
atatt
a 2
)cos(
2
)cos(sin
2
1
attaas
sL sin
2
1
)( 222
1=
+∴
−
Type-3 Laplace transform of the convolution integral and solution of integral
equations
Working procedure for problems
1. Laplace transform of the convolution integral by using the result
−==
− ∫∫
==
duugutfLsgsfduutgufL
t
u
t
u
)()()().()()( 00
in the appropriate form.
2. Given an equation for f (t) involving the convolution integral we first take Laplace
transform on both sides
3. We evaluate the convolution integral and simplify to obtain
)()]([1 tfsfL =− as a function of s
4. Taking inverse we obtain f (t)
Ex: Find )(tf from the equation )(21)(2
∫−
−+=
t
o
udueutftf
Sol: Taking Laplace transform on both sides we have,
[ ] [ ] )(21)( 2
−+= ∫
−
t
o
u dueutfLLtfL
)(21
)( 2
−+= ∫
−
t
o
u dueutfLs
sf
-(1)----------- )().(21
)( sgsfs
sf +=
[ ]2
1)()()( 2
+=== −
seLugLsg u
Hence equation (1) becomes
2
1).(2
1)(
++=
ssf
ssf
1
2)(or
1
2
21)(
ss
ssf
sssf =
+=
+−
s
2
12)(
22+=
+=∴
ss
ssf
By taking inverse we have,
[ ]
+=
−−
s
2
1)(
2
11
sLsfL
2t1 )( +=tfThus
Ex: Solve the integral equation )sin()(1)( ∫ −+=
t
o
duutuftf
Sol: Taking Laplace transform on both sides we have,
[ ] [ ] )sin()(1)(
−+= ∫
t
o
duutufLLtfL
-(1)----------- )().(1
)( sgsfs
sf +=
1s
1g(s)sint g(t)or )sin()(
2 +=∴=−=− ututgwhere
Hence equation (1) becomes
sssf
ssf
ssf
1
1
11)(or
2
1).(
1)(
22=
+−
++=
33
2
2
2 111or
1
1)( .,
sss
sf(s)
ss
ssfie +=
+==
+
[ ]
+=
−−
s
1
1)(
3
11
sLsfLNow
21 )(
2ttfThus +=
APPLICATION OF LAPLACE TRANSFORMS:
SOLUTION OF DIFFERENTIAL EQUATIONS USING LAPLACE
TRANSFORMS:
The Laplace transform method of solving differential equations given particular
solution without the necessity of first finding the general solution and then evaluating
the arbitrary constants. This method is especially useful for solving liner differential
equations with constants coefficients.
Working Procedure
Working Procedure to solve liner differential equations with constants coefficients by
transform method.
Laplace transform has to be taken on both sides of the differential equation, using the
formula of derivative and the given initial conditions.
All the terms with negative sign are transposed to right.
Divide by the coefficient of [ ])(tyL , getting [ ])(tyL as known function of s.
Resolve this function of s
Resole this function of s into partial fractions
Take the inverse Laplace transforms on both sides.
Then we get y as a function of t which is the required solution satisfying the given
condition.
Note: We use the following the formula of derivative in every example.
[ ] )0(y- -------- )0()0()()( 1-n21 −′−−= −− ysyssystyL nnnn
)0(y)()}({ )1( −=′ ssytyL
)0()0(y)()}({ )2( 2 yssystyL ′−−=′′
)0()0()0(y)()}({ )3( 23 yysssystyL ′′−′−−=′′′
Ex: Solve by the method of Laplace transforms the equation
-1.(0)y 2,y(0)given 22
2
=′==+− teydt
dy
dt
yd
Sol: The given differential equation is
tetytyty =+′−′′ )()(2)(
Taking Laplace transforms on both sides
}{)}({)}({2)}({ teLtyLtyLtyL =+′−′′
1
1)}({)]0(y)([2)0()0(y)(2
−=+−−′−−∴
styLssyyssys
521
1)()12( 2 −+
−=+−∴ s
ssyss
)12(
52
)12)(1(
1)(
22 +−
−+
+−−=∴
ss
s
ssssy
23 )1(
52
)1(
1)(
−
−+
−=∴
s
s
ssy
223 )1(
3
)1(
)1(2
)1(
1
−−
−
−+
−=
ss
s
s
23 )1(
3
)1(
2
)1(
1
−−
−+
−=
sss
Taking Inverse Laplace transforms on both sides
−−
−+
−=
−−−−
2
11
3
11
)1(
13
)1(
12
)1(
1)}({
sL
sL
sLsyL
teete ttt 322
1 2 −+=
+−=∴ 23
2)(
2
tt
etyt
Ex: Solve by the method of Laplace transforms the equation
02
y ,0y(0) given that 1892
2
=
==+
πty
dt
yd
Sol: Given differential equation is
18)(9)( ttyty =+′′
Taking Laplace transforms on both sides
}{18)}({9)}({ tLtyLtyL =+′′
2
2 18)(9)0()0(y)(
ssyyssys =+′−−∴
0y(0) and (0)y Take ==′ k
ks
sys +=+∴2
2 18)()9(
)9()9(
18)(
222 ++
+=∴
s
k
sssy
Taking Inverse Laplace transforms on both sides
++
+=
−−−
)9(
1
)9(
118)}({
2
1
22
11
skL
ssLsyL
++
+−
=∴
−−−
)9(
1
9
1
9
1.18
1
9
1.18)(
2
1
2
1
2
1
skL
sL
sLty
tk
ttty 3sin3
3sin3
22)( +−=
02
y 2
t =
=
ππwhen
23sin
323sin
3
2
220
πππ k+−=∴
23k +=∴ π
tttty 3sin3
233sin
3
22)(
++−=∴
π
ttty 3sin2)( π+= is the required solution
Ex: 0at x 2 2, , 1ygiven 02y- 2 2
2
2
2
3
3
=====−+dx
yd
dx
dy
dx
dy
dx
yd
dx
ydSolve
by Laplace transforms method.
Sol: Given differential equation is
0)(2)()(2)( =−′−′′+′′′ xyxyxyxy
Taking Laplace transforms on both sides
0)}({2)}({)}({2)}({ =−′−′′+′′′ xyLxyLxyLxyL
0)}({2)]0(y)([
)]0()0(y)([2
)0()0()0(y)(
2
23
=−−−
′−−+
′′−′−−∴
tyLssy
yssys
yysssys
54
14222)()22(
2
223
++=
−++++=−−+∴
ss
ssssysss
)22(
54)(
23
2
−−+
++=∴
sss
sssy
)1)(1)(2(
542
−++
++=
sss
ss
Taking Inverse Laplace transforms on both sides
−++
++=∴
−−
)1)(1)(2(
54)}({
211
sss
ssLsyL
)1()1()2()1)(1)(2(
54
2
−+
++
+=
−++
++
s
C
s
B
s
A
sss
ssLet
)()1)(2()1)(2()1)(1(542 issCssBssAss −−−−+++−++−+=++∴
Putting S = -2, -1, 1 in (i) we obtain
A=1/3 , B=-1 and C= 5/3
−+
+−
+=
−−−−
)1(
1
3
5
)1(
1
)2(
1
3
1)}({
1111
sL
sL
sLsyL
xxx eeexy3
5
3
1)( 2 +−=∴ −−
SIMULTANEOUS DIFFRENIAL EQUATIONS
Engineering experiments involving two or more dependent variables and only one
independent variable results in simultaneous differential equations which can be
solved by applying Laplace transform.
Ex: Solve the simultaneous equations.
txdt
dyty
dt
dxcos ; sin =+=+
0.y(0) 2, x(0) ==that given
Sol: The given equations are
cos)()(y ; sin)()( ttxtttytx =+′=+′
Taking the Laplace transform on both sides
1)()0()(
1
1)()0()(
2
2
+=+−
+=+−
s
ssxyssy
ssyxssx
Using the given initial conditions we obtain,
1)(0)(
1
1)(2)(
2
2
+=+−
+=+−
s
ssxssy
ssyssx
(2) --------- 1
)()(
(1) 21
1)()(
2
2
+=+
−−−−−−++
=+
s
ssxssy
sssxsy
Multiplying (1) by s and then subtract from (2)
ssxs 2)()1( 2 =−
)1(
2)(
2 −=∴
s
ssx
Taking the inverse Laplace transform on both sides
−=
−−
)1(
2)}({
2
11
s
sLsxL
ttx cosh2)( =∴
Then substitute )1(
2)(
2 −=
s
ssx in equation (2)
)1(
2
)1(
1)(
22 −−
+=∴
sssy
Taking the inverse Laplace transform on both sides
−−
+= −−−
)1(
2
)1(
1)}({
2
1
2
11
sL
sLsyL
ttty sinh2sin)( −=∴
tttyttx sinh2sin)( , cosh2)( −==∴
Ex: Solve the equations.
0222 ; =+++−=++ − yxdt
dy
dt
dxex
dt
dy
dt
dx t
1y(0) -1, x(0) ==ditions to the conSubject
Sol: Given equations are
0)(2)(2)(2)(
)()()(
=++′+′
−=+′+′ −
tytxtytx
etxtytx t
Taking the Laplace transform on both sides
1
1)()0()()0()(
+−=+−+−
ssxyssyxssx
0)(2)(2)]0()([2)0()( =++−+− sysxyssyxssx
Using the given initial conditions we obtain,
1
1)()(1)(
+−=+++
ssxssyssx
0)(2)(2)(21)( =++++ sysxssyssx
(1) ------ 1
2)()()1(
+
+−=++
s
sssysxs
-(2)----- 1)()1(2)()2( −=+++ syssxs
Multiplying (1) by 2(s+1) and (2) by s
(3) ------ )2(2)()1(2)()1(2 2 +−=+++ ssysssxs
-(4)----- )()1(2)()2( ssysssxss −=+++
Subtracting,
4)()]2()1(2[ 2 −−=+−+ ssxsss
)4()()22( 2 +−=++∴ ssxss
)22(
)4()(
2 ++
+−=∴
ss
ssx
Taking the inverse Laplace transform on both sides
++
+−=
−−
)22(
)4()}({
2
11
ss
sLsxL
+++
++
+−=∴
−
2222
1
1)1(
3
1)1(
1)(
ss
sLtx
)sin3(cos)( ttetx t +−=∴ −
)sin1()( tety t += −
SOME APPLICATION TO ENGINEERING PROBLEMS
In this section, we illustrate the use of Laplace transform by considering some
examples connected with the vibrations of string, deflection of beams and LRC
circuits.
Vibration of string:
Let an elastic string be attached to a support and hanging downward. Let a
body of mass m be attached to the lower end of an elastic string. When the spring and
the body are rest in the equilibrium position E, suppose the weight m is pulled
downwards and released, then the string vibrates in the vertical direction freely and
there is a resistance due to the medium opposing the movement resulting in damped
vibrations.
If y denotes the downward displacement of the body of m from the
equilibrium position at time t, the differential equation for this model is
-(1)------- )(2
2
tfkydx
dyc
dt
ymd=++
where dt
dyis the velocity,
2
2
dt
yd is the acceleration, c is the damping coefficient, k
is the spring modulus and f(t) is the driving force.
If 0y be the distance by which the body m is pulled down then the equilibrium
position before releasing it, and 0v be the velocity with which the body m is released.
Then (2)------- 0at t dt
dy and 00 === vyy
These are called as initial conditions for vibrations. The equation (1) and (2)
together form an initial value problem. The displacement y can be obtained by solving
(1) under the conditions (2).
If the medium does not cause any resistance to the vibrations, then we have
“undamped vibrations” and in such a case, c=0
Therefore (1) becomes
-(3)------- )(2
2
tfkydt
ymd=++
If c>0 the we have “resonance damped vibrations”
The driving force (extra force) 0)( ≠tf then the vibrations are called “forced
vibrations” otherwise 0)( =tf , they are called simple harmonic vibrations (motions).
Deflection of beams: Consider a horizontal uniform beam supported at both the ends, which is
slightly bent (deflect) by a downward vertical load w from a horizontal position. Let x
denote the horizontal distance of a point P of the beam from the one end O of the
beam. Then the deflection ‘y’ at the point P at a distance x from one end of the beam
experienced due to bending satisfies the differential equation
wdx
ydEI =
4
4
Where E is the modules of elasticity and I is a moment of inertia. E and I are
supposed to be constants. The constant EI is called as “flexural rigidity of the beam”.
LRC Circuits: A simple electric consisting of an inductance ‘L’(henry) resistance ‘R’(Ohms)
and capacitance ‘C’ (farads) connected in a series is called an LRC-circuit.
If a constant electromotive force (emf) E (volts) is applied to an LRC- circuit
then the current I (amperes) in the circuit at time t is given by the differential
equation, which depends upon the Kirchoff’s law i.e, the algebraic sum of the voltage
dropsaround any closed circuits is equal to the resultant electcteromotive force in the
circuit using the following relations.
∫== idtqordt
dqi )1(
RiR resistance across drop Voltage )2( =
dt
diL=L inductance across drop Voltage )3(
C
q=C ecapacitanc across drop Voltage )4(
)1()( −−−−−−−−=++ tEC
qRi
dt
diL
Therefore (1) also can be written as
)2()(2
2
−−−−−−−−=++ tEC
q
dt
dqR
dt
qdL
From (2) we can find q
Differentiate (1) w.r.t ‘t’ we get
)(1
2
2
tEdt
dq
Cdt
diR
dt
idL ′=++
)3()(1
2
2
−−−−−−−−=++ tEiCdt
diR
dt
idLor
From (3) we can find i.
The differential corresponding to LR – circuit is given by
)4()( −−−−−−−−=+ tERidt
diL
Ex: Spring can extended 20 cm when 0.5 kg mass is attached to it. It is suspended
vertically from a support and set into vibration by putting it down 10 cms and
imposing a velocity 5 cm/sec vertically upwards. Find the displacement from its
equilibrium position at time t.
Sol: When m = 0.5 kg = 500 grams. B= 20cm, taking g=980cm/sec2 we have
dynes/cm. 500,2420
980500===
X
b
mgk
Equation of motion is
02
2
=+ xm
k
dt
xd
0500
500,24 ,.
2
2
=+ xdt
xdei
049 2
2
=+∴ xdt
xd
Taking Laplace transform on both sides.
0)(49)0()0()(2 =+′−− sxxsxsxs
Initially, the mass at a depth of 10 cm below the equilibrium position
.10)0( =∴ x
Also the initial velocity was 5 cm/sec. vertically upwards
5)0( −=′∴ x
510)()49( 2 −=+∴ ssxs
)49(
510)(
2 +
−=∴
s
ssx
+−
+=∴
−−
)49(
15
)49(10)(
2
1
2
1
sL
s
sLtx
tttx 7sin7
57cos10)( −=∴
Ex: A beam is simply supported at its end x=0 and is clamped at the other end x=l. It
carries a load at x=l/4. Find the resulting deflection at any point.
Sol: The differential equation of deflection is
)4/(4
4
lxEI
w
dt
xd−= δ
Taking the Laplace transform, we get
)}4/({)}({ lxLEI
wtyL −=′′′′ δ
)1()0()0()0()0()( 4/234 −−−−−=′′′−′′−′−− −sleEI
wyysysyssys
Using the given condition
,)1(,)0(,)0( assuming and )0(0)0( 21 becomescycyyy =′′′=′′′==
4
4/
4
2
2
1)(s
e
EI
w
s
c
s
csy
sl−
++=
Taking inverse transform we get
4/0,6
1 3
21 lxxcxcy <<+=
)2(4/,462
13
3
21 −−−−<<
−++= lxl
lx
EL
wxcxcyand
Using the conditions,
,)2(,,0)( ,0)( becomeslyly =′=
EL
wllclc
128
9
6
10
33
21 ++=
EL
wllccand
32
9
6
10
22
21 ++=
EL
wc
EL
wlc
128
81,
256
92
2
1 ==⇒
Substituting these values of C1 and C2 in (2) we get the deflection at any point.
Ex: A volt atEe
− is applied at t=0 to a circuit of Inductance L and Resistance R, show
by Laplace transform method the current at time t is
[ ]Ltratee
aLR
E /−− −−
Sol: Required differential equation is
)1(−−−−−−−−=+ ERidt
diL
Since it is given that atEeE
−= , equation (1) becomes
)2(−−−−−−−−=+ −ateL
E
L
Ri
dt
di
Since the voltage is applied for t >0, we suppose that i=0 at t=0
i.e., i(0)=0
Taking Laplace transform on both the sides at (2), we have
asL
Esi
L
Rissi
+=+−
1.)()0()(
Putting i(0)=0, we get
asL
E
L
Rssi
+=
+
1.)(
)(
1.)(
asL
Rs
L
Esi
+
+
=⇒
+−
+−=
asRLsLRaL
Esi
11
)/(
1.)(
+−
+−=
LRsasaLR
Esi
/
11.)(
Taking inverse transform we get
[ ] 0.)( / >−−
= −−twhereee
aLR
Eti
LtRat
Ex: A particle is moving with damped motion according to the law
.02562
2
=++ xdt
dx
dt
xd If the initial position of the particle is at x=20 and the initial
speed is 10, find the displacement of the particle at any time t using Laplace
transforms.
Sol: The given equation is 0)(25)(6)( =+′+′′ txtxtx
Initial conditions are 10)0(,20)0( =′= xx
Thus )sin.4/74cos2(10)( 3ttetx
t += −
Ex: The differential equation of a uniformly loaded beam with simple support is
)(22
2
xlwx
dx
ydEI −−=
Where EI and w are constants. Solve by Laplace transform method given that
.0)0()0( =′= yy
Ex: An LR circuit carries an emf of voltage wtEE sin0= where 0E and 0w are
constants. Find the current I in the circuit, if initially there is no current in the circuit.
