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Transcript of Chapter 3 math 3
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8/10/2019 Chapter 3 math 3
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1
Arab Academy for Science, Technology and Maritime Transport
College of Engineering and Technology
Department of Basic and Applied Science
Mathematics (3)
BA223Chapter (3)Laplace Transforms
Prepared by:
Hossam Shawky
Associate Prof. of Engineering Mathematics
Fall 2014-2015
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Chapter (3)
Laplace Transforms
Laplace: Newton of France [1749-1827]
Definition:
Suppose that )t(f is a piecewise continuous function, defined for 0t . TheLaplace transform of )t(f is denoted )t(fL and defined as
0s),s(Fdt)t(fe)t(fL
0
st
,
provided that 0)s(FLims
.
Time Domain L.T Frequency Domain
)t(f )s(F I.L.T
Sufficient conditions for existence of Laplace transform:
1) A function )t(f is a piecewise continuous function on ),0[ .2) A function )t(f is of exponential order for Tt ,
i.e 0T,0c,0M,Me
)t(fct
tLim
.
So, neither
t1L nor 2teL exists.
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A function is called piecewise continuous on an interval if the interval can be brokeninto a finite number of subintervals on which the function is continuous on each open
subinterval and has a finite limit at the endpoints of each subinterval. Below is a
sketch of a piecewise continuous function.
Recall
(1) 1e0 (2) e (3) 0e (4) cke
dtekt
kt
Laplace transforms for some basic functions:
s
K
s
10K
s
eKdteKKL]1[
0st
st
0
as,
as
1
)as(
10
)as(
e
dtedteeeL]2[0
t)as(
0
t)as(
0
atstat
Result:
as
1
eL at
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Examples:
2s
1eL t2
,
2s
1eL t2
dtettL]3[ st0
nn
dtsdxst xLet
0x0 tat , x tas
1n
x
0
n
1n
x
0
nn
s
)1n(dxex
s
1dxs
1e
s
xtL
where is the Gamma function.
!n)1n( (n is a positive integer number)
1n
n
s
!ntL
(n is a positive integer number)
Example
6
5
s
!5tL
22
atat
as
s
)as)(as(
asas
2
1
as
1
as
1
2
1eeL
2
1atcoshL]4[
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22
atat
as
a
)as)(as(
asas
2
1
as
1
as
1
2
1eeL
2
1atsinhL]5[
22
iatiat
as
s
)ias)(ias(
iasias
2
1
ias
1
ias
1
2
1eeL
2
1atcosL]6[
22
iatiat
as
a
)ias)(ias(
iasias
i2
1
ias
1
ias
1
i2
1eeL
i2
1atsinL]7[
Properties of Laplace Transforms
Property No.1: Linearity of the transformation
If )s(F)t(fL 11 and )s(F)t(fL 22 , then
)s(FC)s(FC)t(fLC)t(fLC)t(fC)t(fCL 221122112211
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Sheet (15)
F ind the Laplace transform of each of the following functions
No.1 t3sinh4t3cos5t3e23)t(f 4t2
Solution:
3s
34
9s
s5
s
72
2s
2
s
3
3s
34
9s
s5
s
!43
2s
12
s
3)s(F
225
225
No.22t2 )3e5()t(f
Solution:
s
9
2s
30
4s
25)s(F
9e30e25)t(f t2t4
No.32)t2cost2(sin)t(f
Solution:
16s
4
s
1)s(F
t4sin1t2cost2cost2sin2t2sin)t(f
2
22
No.4 t3cos)t(f 2
Solution:
36s
s
s
1
2
1)s(F
)t6cos1(2
1)t(f
2
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Recall:
(1) BsinAsinBcosAcos)BAcos(
(2)
BsinAsinBcosAcos)BAcos(
(3) BsinAcosBcosAsin)BAsin(
(4) BsinAcosBcosAsin)BAsin(
From the above four equations, we get
(5) )BAcos()BAcos(2
1BcosAcos by adding (1) +(2)
(6) )BAcos()BAcos(2
1BsinAsin by subtracting (2) -(1)
(7) )BAsin()BAsin(2
1BcosAsin by adding (3) +(4)
(8) )xcos()xcos( , since )xcos( is an even function
(9) )xsin()xsin( , since )xsin( is an odd function
(10) )xcosh()xcosh( , since )xcosh( is an even function
(11) )xsinh()xsinh( , since )xsinh( is an odd function
(12) 1at2cosh2
1atcosh2
(13) 1at2cosh2
1atsinh 2
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Sheet (15)
F ind the Laplace transform of each of the following functions
No.5 )3t3cos()t(f
Solution:
9s
3
2
3
9s
s
2
1)s(F
t3sin
2
3t3cos
2
1
3sint3sin
3cost3cos)t(f
22
No.6 t2sinh)t(f 2
Solution:
1t4cosh2
1)t(f
s
1
16s
s
2
1)s(F
2
No.7 t6cost3sin)t(f
Solution:
]t3sint9[sin2
1)t6t3sin()t6t3sin(
2
1)t(f
9s
3
81s
9
2
1)s(F
22
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No.8 t5sint3sin)t(f
Solution:
]t8cost2[cos2
1
)]t5t3cos()t5t3[cos(21)t(f
64s
s
4s
s
2
1)s(F
22
No.9 t6cost3cos)t(f
Solution:
81s
s
9s
s
2
1)s(F
]t9cost3[cos2
1
)]t6t3cos()t6t3[cos(2
1)t(f
22
Property No.2: F irst Shi f ting Theorem [s-shi f ting]
If )s(F)t(fL , then )as(F)}t(fe{L at
Proof:
)s(Fdt)t(fe)t(fL0
st
)as(Fdt)t(fedt)t(fee)t(feL0
t)as(
0
atstat
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No.10 t4sine)t(f t2
Solution:
16)2s(
4}t4sine{L
16s
4}t4{sinL
2
t2
2
No.11 t4sine)t(f 2t
Solution:
64)1s(
1s
1s
1
2
1}t4sine{L
64s
s
s
1
2
1}t4{sinL
]t8cos1[2
1t4sin
2
2t
2
2
t2
No.12 )t6cos5t3sin4(e)t(f t3
Solution:
36)3s()3s(5
9)3s(12)}t(f{L
36s
s5
9s
34}t6cos5t3sin4{L
22
22
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No.132t )t2sinhe()t(f
Solution:
s
1
16s
s
2
1
4)1s(
4
2s
1)}t(f{L
s
1
16s
s
2
1}t2{sinhL
]1t4[cosh2
1t2sinh
4)1s(
2}t2sinhe{L
4s
2}t2{sinhL
2s
1}e{L
t2sinht2sinhe2e)t(f
22
2
2
2
2
t
2
t2
2tt2
No.14 t4sint2sine)t(f t2
Solution:
36)2s(
2s
4)2s(
2s
2
1)}t(f{L
36s
s
4s
s
2
1}t4sint2{sinL
]t6cost2[cos2
1
)]t4t2cos()t4t2[cos(2
1t4sint2sin
22
22
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No.15 t6cost3sine)t(f t3
Solution:
9)3s(
3
81)3s(
9
2
1
)}t(f{L
9s
3
81s
9
2
1}t6cost3{sinL
]t3sint9[sin2
1
)]t6t3sin()t6t3[sin(2
1
t6cost3sin
22
22
No.16t5 3.t)t(f
Solution:
6
t5
65
5t)3(ln
)3lns(
!5}3t{L
s!5}t{L
tet3lne5t)t(f
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Property No.3: Derivatives of Transforms
If )s(F)t(fL , then )s(Fsd
d)t(ftL
Proof:
)s(Fdt)t(fe)t(fL0
st
)t(ftLdt)t(ftedt)t(fet
dt)t(fes
dt)t(fesdd)s(F
sdd
0
st
0
st
0
st
0
st
)s(Fsdd
)t(ftL
Result:
)s(Fsd
d)t(ftL
2
22
In general:
)s(Fsd
d)1()t(ftLn
nnn
Recall:
v
uy 2v
vuuvy
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No.17 t2sint)t(f
Solution:
22222
2
)4s(
s4
)4s(
s220
4s
2
ds
d}t2sint{L
4s2}t2{sinL
No.18 t3cost)t(f 2
Solution:
42
22222
22
2
22
22
22
22
2
)9s(
s2)9s(2)s9(]s2[)9s(}t3cost{L
)9s(s9
dsd
)9s()s2s(1)9s(
dsd}t3cost{L
9s
s
ds
d}t3cost{L
9s
s}t3{cosL
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No.19 t3coste)t(f t2
Solution:
222t2
22
2
22
2
22
2
2
2
9)2s(
9)2s(}t3coste{L
)9s(
9s
)9s(
s9
)9s(
]s2.s[1).9s(
9s
s
ds
d}t3cost{L
9s
s
}t3{cosL
No.20 t2sinte)t(f 2t3
Solution:
22
2
2
2t3
222
2
22
2
22
2
2
2
2
]16)3s[(
)3s(16
)3s(
1
2
1}t2sinte{L
)16s(
)s16(
s
121
)16s(
]s2.s[1).16s(
s
1
2
1
16s
s
s
1
ds
d
2
1}t2sint{L
16s
s
s
1
2
1}t2{sinL
]t4cos1[2
1
t2sin
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No.21t37et)t(f
Solution:
8
t37
8
7
)3s(
!7
}e.t{Ls
!7
}t{L
No.22t42
e)1t()t(f
Solution:
)4s(
1
)4s(
2
)4s(
!2)}t(f{L
s
1
s
2
s
!2}1t2t{L
e]1t2t[)t(f
23
23
2
t42
No.23 )3
t3cos(t)t(f
Solution:
2222
2
2222
2
22
22
)9s(
s.33
)9s(
s9
2
1
)9s(
]s2.3[0.
2
3
)9s(
]s2.s[1).9s(
2
1
9s
3
2
3
9s
s
2
1
ds
d)}3t3cos(.t{L
9s
3
2
3
9s
s
2
1)}3
t3{cos(L
t3sin2
3t3cos
2
13
sint3sin3
cost3cos)3
t3cos(
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No.24 t3cosht2cost)t(f
Solution:
22
2
22
2
22
2
2
t3t3
]4)3s[(
4)3s(
]4)3s[(
4)3s(
2
1)}t(f{L
)4s(
4s}t2cost{L
4s
s}t2{cosL
)]ee)(t2cos.t(
2
1)t(f
Property No.4: Transform I ntegration
If )s(F)t(fL , thens
)s(Fdx)x(fL
t
0
No.25 }dx)x3sinx({L
t
0
Solution:
9s
3
s
1
s
1}dx)x3sinx({L
9s
3
s
1)s(F
t3sint)t(f
22
t
0
22
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No.26 }dx)x2sinhe({L
t
0
2x
Solution:
t2sinht2sinhe2e)t2sinhe()t(f 2tt22t
From No.13
s
1
16s
s
2
1
4)1s(
4
2s
1)s(F
22
s
1
16s
s
2
1
4)1s(
4
2s
1
s
1}dx)x2sinhe({L
22
t
0
2x
No.27 }dxx3cosxe{L
t
0
x2
Solution:
t3coset)t(f t2
From No.19
22
2
]9)2s[(9)2s()s(F
22
2t
0
x2
]9)2s[(
9)2s(
s
1}dx)x3cos.e.x({L
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No.28
t
0
x2t2dx)x3cose(e)t(f
Solution:
dx)x3cos.e({L)}t(f{L
t
0
x2
9s
s}t3{cosL
2
9)2s(
2s
}t3cos.e{L 2
t2
9)2s(
2s
s
1dx)x3cos.e({L
2
t
0
x2
9s
s
)2s(
1)}t(f{L
2
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Property No.5:
If )s(F)t(fL , then sd)s(F
t
)t(fL
s
,
Provided thatt
)t(fLim
0t exists
Recall:
(1)
C
a
stansd
as
a 122
(2)
2
1tan
(3)
11
2cottan
No.1t
t4sin)t(f
Solution:
4
scot
4
stan
24
stantan
4
stands
16s
4}
t
t4sin{L
16s
4}t4{sinL
1111
s
1
s
2
2
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No.2t
tsine)t(f
t
Solution:
)1s(cot)1s(tan2
)1s(tands
1)1s(
1}
t
tsin.e{L
1)1s(
1}tsin.e{L
1s
1}t{sinL
11
s1
s
2
t
2
t
2
No.3t
ee)t(f
t3t2
Solution:
2s
3sln
3s
2sln1ln
3s
2sln
3s
2slnLim
3s
2sln]3sln2s[ln
ds3s
1
2s
1}
t
ee{L
3s
1
2s
1}ee{L
s
ss
s
t3t2
t3t2
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No.4t
t4cost6cos)t(f
Solution:
36s
16sln
16s
36sln1ln
16s
36sln
16s
36slnLim
16s
36sln
16sln
2
136sln
2
1
ds16s
s
36s
s}
t
t4cost6cos{L
16ss
36ss}t4cost6{cosL
2
2
2
2
2
2
2
2
ss
2
2
s
22
s
22
s2
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No.5t
t4cose)t(f
t3
Solution:
3s
16sln
16s
3sln1ln
16s
3s
ln16s
3s
lnLim16s
3s
ln
16sln2
13slnds
16s
s
3s
1)}t(f{L
16s
s3s
1}t4cose{L
2
2
22ss
2
s
2
s
2
2
t3
Unit Step Function:
ot0
ot1)t(u
s
1)}t(u{L
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24
at0
at1)at(u
ases
1)}at(u{L
Property No.6: Second Shi f ting Theorem (t-shi f ting)
If )s(F)t(fL , then )s(Fe)at(u)at(fL as ,
No.6 s34
3 es!3)}3t(u)3t{(L
No.7s
2 e4s
2)}t(u)t(2{sinL
No.8s3)3t(4 e
4s
1)}3t(ue{L
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Expressing the function )t(f in terms of a Unit step function:
Let,
ctb)t(f
bta)t(fat0)t(f
)t(f3
2
1
, then
......)bt(U)t(f)t(f)at(U)t(f)t(f)t(U)t(f)t(f 23121
No.9
5t1
5t20
2t01
)t(f
Solution:
s5s2
s5s2sst2
0st
2
0
5
2 5
t5st
0
st
ee1
s
1
e0s
11e
s
1
s
]e[
s
]e[
dte0dtedt)t(fe)}t(f{L
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Another solution
s5s2s5s2 ee1s
1es
1es
1
s
1)s(F
)5t(u)2t(u)t(u
)5t(u]01[)2t(u]10[)t(u1)t(f
No.10 )}2t(u)3t2{(L
Solution:
7)2t(23]2)2t[(23t2 s2
2e.
s
7
s
1.2)}2t(u]7)2t(2{[L
No.11 )}3t(u)1t3{(L 2
Solution:
8)3t(31]3)3t[(3)1t3(
64)3t(48)3t(9)1t3( 22
s3
23
2 e.s
64
s
1.48
s
!2.9)}3t(u.]64)3t(48)3t(9{[L
No.12)}2t(u)t2{(L
2
Solution:
2]2t[)t(
4)2t(4)2t(t 22
)2t(4)2t(2]4)2t(4)2t[(2t2 222
s2
23
2 e.s
1.4
s
!2
s
2)}2t(u)]2t(4)2t(2{[L
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No.13 )}3t(u)4t4t{(L 2
Solution:
)}3t(u)2t{(L)}3t(u)4t4t{(L 22 5)3t(2]3)3t[(2t
25)3t.(10)3t()2t( 22
s3
23
2 e.s
25
s
1.10
s
!2)}3t(u]25)3t.(10)3t{[(L
No.14 )}2t(u]t5e2{[L
2t3
Solution:
2)2t(t
6)2t(3]2)2t[(3t3
4)2t(4)2t(t 22
s2
23
6
2)2t(36
es
20
s
1.20
s
!2.5
3s
1.e2
)}2t(u]20)2t(20)2t(5e.e2{[L
No.15 )}t(ut2{sinL
Solution:
2)t(2])t[(2t2
s
2e.
4s
2)}t(u)t(2{sinL)}t(u]2)t(2{sin[L
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No.16
)2
t(u)et(sinL t
Solution:
)2
t(ue2
)2
t(sin)2
t(u).et(sin 2)2
t(t
)2
t(ue.e2
sin).2
tcos(2
cos).2
tsin( 2t
2
2tue.e
2tcos)t(f 2
t2
s22
2e
1s
1.e
1s
s)s(F
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29
No.17
1t4
1t02)t(g
Solution:
)1t(u2)t(u2
)1t(u].24[)t(u.2)t(g
(i) ss e1s
2e.
s
2
s
2)}t(g{L
(ii) )3s(t3 e1)3s(
2)}t(g.e{L
(iii)
)e1(s2
s1}dx)x(g{L s
t
0
(iv)
s
22
s
2
ss
s
e.s
2
s
2e.
s
2
s
2
).e1()e(s
2
)e1(s
2
ds
d)}t(g.t{L
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Inverse Laplace Transforms
)s(F)t(fL )t(f)s(FL 1
Sheet (17)
F ind the I nverse Laplace transform of each of the following
functions
No.1s
3)s(F
Solution: 3s
3
L1
No.22s
1)s(F
Solution:t21 e
2s
1L
No.3 2s1)s(F
Solution: ts
1L
2
1
No.44s
s)s(F
2
Solution: t2cos4s
sL
2
1
No.59s
3)s(F
2
Solution:
t3sin9s
3L 21
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31
No.64s
s)s(F
2
Solution:
t2cosh4s
sL
2
1
No.79s
3)s(F
2
Solution:
t3sinh9s
3L
2
1
No.84s
1)s(F
Solution:
!3
t
s
1L
3
4
1
No.99s
1)s(F2
Solution:
t3sin3
1
9s
1L
2
1
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32
No.107s
1)s(F
2
Solution:
t7sinh7
1
7s
1L
2
1
Recall: )as(F)}t(fe{L at
No.119)2s(
)2s()s(F2
Solution:
t3cos.e9)2s(
2sL t2
2
1
No.129)4s(
1)s(F
2
Solution: t3sin.e.3
1
9)4s(
1L t4
2
1
Completion Square
9)2s(13s4s 22
16)3s(25s6s 22
7)2s(3s4s 22
4/1)2/1s(ss 22
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No.1325s8s
)2s()s(F
2
Solution:
]t3sin2t3[cose
t3sine31.6t3cos.e
9)4s(
1L6}
9)4s(
4s{L
9)4s(
6)4s(L
25s8s
2sL
t4
t4t4
2
1
2
1
2
1
2
1
No.145s2s
)3s2()s(F
2
Solution:
]t2sin21t2cos2[et2sine
21t2co.e2
4)1s(
1L
4)1s(
)1s(L2
4)1s(
1)1s(2L
4)1s(
3]1)1s[(2L
5s2s
3s2L
ttt
2
1
2
1
2
1
2
1
2
1
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Recall: )s(Fe)at(u)at(fL as
No.15
s
2 e13s4s
)3s2(
)s(F
Solution:
)t(u)t(3sin3
1)t(3cos2e
e.13s4s
)3s2(L
]t3sin3
1t3cos2[et3sine
3
1t3cose2
9)2s(
1
L9)2s(
)2s(
L2
9)2s(
1)2s(2L
9)2s(
3]2)2s[(2L
13s4s
3s2L
)t(2
s
2
1
t2t2t2
2
1
2
1
2
1
2
1
2
1
No.16s3
4e
)5s(
s)s(F
Solution:
t6
5
2
1et
!3
t.e.5
!2
t.e
)5s(
1L5
)5s(
1L
)5s(
5)5s(L
)5s(
sL
t523
t52
t5
4
1
3
1
4
1
4
1
)3t(u)3t(6
5
2
1e)3t(e.
)5s(
sL )3t(52s3
4
1
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35
I nverse Laplace Transform using Partial fractions:
)s(QofdegreesPofdegree
sQ
sP
Purpose:
To transform
basA
bassQ
basB
bas
A
bassQ 22
cbsas
BAs)cbsas(sQ
2
2
cbsas
DCs
cbsas
BAscbsassQ
222
22
Find the unknowns A,B,C,. Using the Heaviside Method:
We can get all the unknowns if 1bassQ
No.17)2s)(1s(
s)s(F
Solution:
)2s(
B
)1s(
A
)2s)(1s(
s
3
2
2s)1s(
sB,
3
1
1s)2s(
sA
2s1s
3s2
2s
1
1s
1
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36
t2t111 e3
2e
3
1
)2s(
3
2
L)1s(
3
1
L)2s)(1s(
sL
No.18)3s)(2s)(1s(
s)s(F
2
Solution:
3s
C
2s
B
1s
A
)3s)(2s)(1s(
s2
,2
1
1s)3s)(2s(
sA
2
41
4
2s)3s)(1s(
sB
2
2
9
)1)(2(
9
3s)2s)(1s(
sC
2
t3t2t111
21
e2
9e4e
2
1
3s2
9
L2s
4L
1s2
1
L
)3s)(2s)(1s(
sL
Find the unknowns A,B,C,. using the general Method:
Steps:
(1) * Q(s)
(2) Sub. By zeros of Q(s)
(3) Compare Coefficients
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37
Sheet (17)
F ind the I nverse Laplace transform of each of the following
functions
No.192)4s)(3s(
s)s(F
Solution:
122 )4s(
C
)4s(
B
3s
A
)4s)(3s(
s
)4s)(3s(C)3s(B)4s(As
)4s)(3s(*
2
2
A33s 4s 4BB4
Coeff2s 3CCA0
t4t4t3
1
2
11
2
1
e3te4e3
)4s(
3L
)4s(
4L
3s
3L
)4s)(3s(
sL
No.20)4s)(2s(
s)s(F
2
Solution:
4s
CBs
)2s(
A
)4s)(2s(
s22
)4s)(2s(* 2
)2s)(CBs()4s(As 2
2s 4
1AA82
Coeff2
s 4
1
BBA0
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38
Coeff0s
2
1CC210C2A40
t2sin2
1.
2
1t2cos
4
1e
4
1
4s
1L
2
1
4s
sL
4
1e
4
1
4s
2
1
s4
1
L2s4
1
L)4s)(2s(
sL
t2
2
1
2
1t2
211
21
No.21)3s(s
1)s(F
Solution:
3s
B
s
A
)3s(s
1
3
1A ,
3
1B
t3t3111 e13
1e
3
1
3
1
3s
3
1
Ls
3
1
L)3s(s
1L
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39
No.22)9s(s
1)s(F
2
Solution:
9s
CBs
s
A
)9s(s
122
s)cBs()9s(A1
)9s(s*
2
2
0s 9
1AA91
Coeff :s2 9
1BBA0
Coeff :s1 C0
t3cos19
1t3cos
9
1
9
1
9s
s9
1
Ls
9
1
L)9s(s
1L
2
11
2
1
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40
By Inspection Method:
(1)
bs
1
as
1
)ab(
K
)bs)(as(
K2222
(2)
bs
BAs
as
BAs
)ab(
1
)bs)(as(
BAs2222
(3)
bs
s
as
s
)ab(
K
)bs)(as(
Ks2
2
2
2
22
2
as
a
bs
b
)ab(
K
bs
b1
as
a1
)ab(
K
bsb)bs(
asa)as(
)ab(K
22
22
2
2
2
2
(4)
bs
1
as
1
)ab(
K
)bs)(as(
K
(5)as
a
bs
b
)ab(
K
)bs)(as(
Ks
No.23)4s)(1s(
1)s(F 22
Solution:
4s
1
1s
1
3
1
)4s)(1s(
12222
t2sin2
1
tsin3
1
)4s)(1s(
1
L 221
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41
No.24)4s)(2s(
s)s(F
22
Solution:
4s
s
2s
s
2
1
)4s)(2s(
s2222
t2cost2cos2
1
)4s)(2s(
sL
221
No.25)4s)(2s(
4s3)s(F22
Solution:
t2sin2
1.4t2cos3
t2sin2
1.4t2cos3
2
1
)4s)(2s(
4s3L
4s
4
4s
s3
2s
4
2s
s3
2
1
4s
4s3
2s
4s3
2
1
)4s)(2s(
4s3
22
1
2222
2222
No.26 )4s)(2s(
s
)s(F 22
2
Solution:
2s
2
4s
4
2
1
)4s)(2s(
s2222
2
t2sin2
1.2t2sin
2
1.4
2
1
)4s)(2s(
sL
22
21
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No.2716s
s)s(F
4
Solution:
4s
s
4s
s81
)4s)(4s(
s
16s
s22224
t2cost2cosh8
1
)4s)(4s(
sL
22
1
No.28 )4s2s)(1s(
7s7s3
)s(F 2
2
Solution:
4s2s
CBs
1s
A
)4s2s)(1s(
7s7s322
2
)4s2s)(1s(* 2
)1s)(CBs()4s2s(A7s7s3
22 1s 1AA3773
Coeff2s : 2BBA3
Coeff0s : 3CCA47
4s2s
3s2L
1s
1L
)4s2s)(1s(
7s7s3L
2
11
2
21
t3sine3
1t3cose2e
3)1s(1L
3)1s()1s(2Le
3)1s(1)1s(2Le
ttt
21
21t
21t
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43
No.29)ss2s(
5s2s)s(F
23
2
Solution:
1s
C
)1s(
B
s
A
)1s(s
5s2s
)1s2s(s
5s2s22
2
2
2
2)1s(s*
)1s(CsBs)1s(A5s2s 22 0s A5 1s 4BB4
Coeff2s 4cCA1
tt
1
2
11
3
21
e4te45
1s
4L
)1s(
4L
s
5L
ss2s
5s2sL
No.30 2
3
)1s(
2s
)s(F
Solution:
)s(Flims
I.L.T doesnt exist
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44
Applications
Solution of Ordinary Differential Equations by means
of Laplace Transform
Property No.7: Transform Differentiation
If )s(F)t(fL , then
)0(f)s(Fstd
)t(fdL
,
)0(f)0(fs)s(Fstd
)t(fdL 2
2
2
Sheet (18)
Using Laplace transform of differentiation, find the transformation of thefunctions
No.1 atcos
Solution:
22 as
a}at{sinL
0)0(f
0as
a.s)}at(sindtd{L
22
22 as
as}atcosa{L
22 as
s}at{cosL
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No.2 atsinh Solution:
22
as
s}at{coshL
1)0(f
1as
s.s)}at(cosh
dt
d{L
22
22
222
as
ass}atsinha{L
22
as
a}at{sinhL
Sheet (18)
Solve the following D.E. using Laplace transforms
No.3t4
e)t(y2td
yd , 0)0(y
Solution:
Take L.T for both sides
4s
1)s(Y2)]o(y)s(sY[
4s
1)s(Y2]o)s(sY[
4s
1]2s)[s(Y
)2s)(4s(1)s(Y
By Inspection
2s
1
4s
1
6
1
t2t4 ee6
1)t(y
Check 0]11[6
1)0(y
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46
No.4t2
2
2
e3)t(y3td
yd4
td
yd 0)0(y)0(y
Solution:
Take L.T for both sides
2s
1.3)s(Y3)]0(y)s(sY[4)]0('y)0(sy)s(Ys[ 2
2s
3]3s4s)[s(Y 2
2s3)]3s)(1s)[(s(Y
2s
C
3s
B
1s
A
)2s)(3s)(1s(
3)s(Y
Using Heaviside method
2
1
)3)(2(
3A
10
3
)5)(2(
3B
5
1
)5)(3(
3C
2s
51
3s
103
1s
21
)s(Y
t2t3t e
5
1e
10
3e
2
1)t(y
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47
No.5t32
2
2
et)t(y9dt
dy6
dt
yd
2)0(y
6)0('y
Solution:
Take L.T for both sides
3
2
)3s(
!2)s(Y9)]0(y)s(sY[6)]0('y)0(sy)s(Ys[
3
2
)3s(
!2)s(Y9]2)s(sY[6]6s2)s(Ys[
32
)3s(2126s2]9s6s)[s(Y
3
2
)3s(
2126s2])3s)[(s(Y
25
25
25
)3s(
12
)3s(
2
)3s(
2
)3s(
18]3)3s[(2
)3s(
2
)3s(
18s2
)3s(
2)s(Y
t3t34
t3 te12e2!4
t.e2)t(y
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No.6 )2t(u)t(y6dt
dy5
dt
yd2
2
0)0(y)0(y
Solution:
Take L.T for both sides
s22 es
1)s(Y6)]0(y)s(sY[5)]0('y)0(sy)s(Ys[
s22 es
1)s(Y6]0)s(sY[5]00)s(Ys[
s22 e
s
1]6s5s)[s(Y
s2es
1)]3s)(2s)[(s(Y
s2e)3s)(2s(s
1)s(Y
3s
C
2s
B
s
A
)3s)(2s(s
1
6
1A
2
1B
3
1C
s2e3s
3
1
2s2
1
s
6
1
)s(Y
)2t(ue31e
21
61)t(y )2t(3)2t(2
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No.7 )2t(u)2t()t(ytd
yd2
2
0)0(y)0(y
Solution:
s2
2
2 es
1)s(Y)]0('y)0(sy)s(Ys[
s2
2
2 es
1)s(Y]00)s(Ys[
s2
2
2 es
1]1s)[s(Y
s222
e)1s(s
1)s(Y
1s
1
s
1
1
1
)1s(s
12222
s222
e1s
1
s
1)s(Y
)2t(u)]2tsin()2t[()t(y
0)5707.1(y)2(y
]2)22
sin[(]2)22
{[()22(y
12
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No.8 The differential equation of the R-L circuit connected in series is given by
t2cos)t(Ritd
idL
Determine the electric current i(t) where L=1 Henry, R=2 ohms and i(0)=0.
Solution:
t2cos)t(i2dt
di
Take L.T for both sides
4ss)s(I2)]0(i)s(sI[
2
4s
s)s(I2]0)s(sI[
2
4s
s]2s)[s(I
2
)4s)(2s(
s)s(I
2 See sheet 17 Problem No. 20 Lec.12
t2sin2
1.2
1t2cos
4
1e
4
1)t(i t2