chapter 3 math 3

8/10/2019 Chapter 3 math 3

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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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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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28

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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33

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


34/50

34

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


35/50

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


36/50

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


37/50

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


38/50

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


39/50

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


40/50

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


41/50

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


42/50

42

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


43/50

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


44/50

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


45/50

45

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


46/50

46

No.4t2

2

2

e3)t(y3td

yd4

td

yd 0)0(y)0(y

Solution:


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


47/50

47

No.5t32

2

2

et)t(y9dt

dy6

dt

yd

2)0(y

6)0('y

Solution:


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


48/50

48

No.6 )2t(u)t(y6dt

dy5

dt

yd2

2

0)0(y)0(y

Solution:


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


49/50

49

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


50/50

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


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

chapter 3 math 3

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