unit 8 geotechnical engineering

8/12/2019 UNIT 8 geotechnical engineering

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UNIT 8 SECOND-ORDER LINEAR ORDINARY

DIFFERENTIAL EQUATIONS

Unit Structure

8.0 Overview

8.1 Learning Objectives

8.2 Introduction

8.3 Homogeneous Equations with Constant Coefficients

8.4 Differential Operators

8.5 Solving the Inhomogeneous Equation

8.6 Summary

8.7 Answers to Activities

8.0 OVERVIEW

In this Unit, we shall study second-order ordinary differential equations. First we

consider the solution of homogeneous equations, and then show how to solve

inhomogeneous equations.

8.1 LEARNING OBJECTIVES

By the end of this unit, you should be able to do the following:

1.

Find the general solution of homogeneous equations.2. Use D-operators.3. Obtain the complementary function and particular integral of inhomogeneous

second-order linear differential equations.


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8.2 INTRODUCTION

We shall study 2nd

-order constant-coefficientordinary differential equations of the form

)(2

2

xfycdx

dyb

dx

yda =++ , (1)

where the coefficients a, b, care real constants and 0a . is called the free term

or the forcing function.

)(xf

If , then (1)is called a Homogeneousdifferential equation. If , then

(1)is called an Inhomogeneous differential equation.

0)( xf 0)( xf

8.3 HOMOGENEOUS EQUATIONS WITH CONSTANT

COEFFICIENTS

The simplest 2nd

-order differential equation is 02

2

=dx

yd. To find its general solution we

simply integrate twice w.r.t.x. Now,

002

2

=

=dx

dy

dx

d

dx

yd.

Integrating w.r.t.x, we have Adxdx

dy == 0 ;

Integrating again w.r.t. x gives +== BAxdxAy .

Hence the general solution of 02

2

=dx

yd is BAxy += , where A and B are arbitrary

constants, which can be fixed if we are given two conditions onxandy.


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Example 1

Solve 02

2

=dx

yd, given that 3)1(',2)1( == yy .

The general solution is BAxy += . The condition 2)1( =y (i.e., when2=y 1=x )

yields

BA += 12

The second condition (i.e.,3)1(' =y 3'=y when 1=x ) gives [since ]Axy =)('

A=3 .

Hence we have ; therefore theparticular solutionis1,3 == BA

13 = xy .

In general, to solve the equation )(xfdx

ydn

n

= , we just integrate ntimes w.r.t.x.


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Example 2

Solve xexxdx

yd 322

2

sin ++= .

Integrating once, we have

.3

cos3

)sin(

33

32

Ae

xx

dxexxdx

dy

x

x

++=

++=

Integrating once more now yields the general solution

BAxexxy x +++= 3

914

121 sin .

[N.B.Since we are not given any conditions onxandy, we cant determineAandB.]

Activity 1

Solve the following differential equations:

(i) 1)2(',3)2(,02

2

=== yydx

yd;

(ii) 2)0(',1)0(,02

2

=== yydx

yd;

(iii) 3)2(,2)1(,02

2

=== yydx

yd;

(iv) 2)1(,3)1(,02

2

=== yydx

yd;

(v) 252

2

= xdx

yd;

(vi) 4cosh3sinh22

2

++= xxdxyd

;

(vii) 3)0(',2)0(,1322

2

==+= yyxedx

yd x ;

(viii) 5)2(,3)1(,762

2

=== yyxdx

yd;


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We shall now learn how to solve the homogeneous equation

02

2

=++ ycdx

dyb

dx

yda . (2)

Let us assume that is a solution of (2). ThenmxAey=

mxmAe

dx

dy= ,

mxAemdx

yd 22

2

= .

On substituting into (2), we obtain

02 =++ mxmxmx cAebmAeAeam .

Since , this simplifies to0mxAe

02 =++ cbmam .

This quadratic equation is called the Auxiliary Equation (A.E.) of Eqn (2).

The auxiliary equation can easily be written down. We simply replaceyby 1,dx

dyby m,

and2

2

dx

ydby .2m

Now, the quadratic equation can have either 2 distinct roots, or 2 equal roots, or 2

complex roots. Each case gives a different type of solution to Eqn (2), which is given in

the table below.

General Solution of Eqn (2)

Distinct Roots21

, mm xmxm eBeAy 21 +=

Equal Roots00 ,mm

xmeBAxy 0)( +=

Complex Roots iqp )sincos( qxBqxAey px +=

Note that the arbitrary constants have been calledAandB here; other symbols could be

used.

We shall now consider a few examples to illustrate.

Example 3


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Consider the differential equation

d y

dxy

2

2 = , (3)

The A.E. is , which has distinct roots012 =m 1,1=m . Hence, from the table,

is the m

xample 4

xx BeAey += ost general solution of (3).

E

olve the differential equationS

0652

2

=+ ydxdy

dx

yd

.

Now the A.E. is , whose roots are0652 =+ mm 3,2=m . Therefore, the general

.

Example 5

solution is

xx BeAey32 +=

differential equationConsider the

0962

2

=+ ydxdy

dx

yd

.

The A.E. gives0962 =+ mm 3,3=m , two repeated roots.

general solution

.

Example 6

Hence the is

xeBxAy 3)( +=

uationSolve the eq

01342

2

=+ ydx

dy

dx

yd

.

The A.E. is given by , whose roots are01342 =+ mm im 32 = . The general solution

.

Obtaining Particular Solutions.

is

)3sin3cos(2 xBxAey x +=


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First we solve the A.E. and obtain the general solution. Using the 2 conditions given, we

then fix the 2 arbitrary constants.

Example 7

Solve the following differential equation

14'19''3 5)0(',1)0(,0 == yyy =yy .

The A.E. is , whose roots are014193 2 = mm 7,32=m . Hence the general solution

is xx

Aey 32

=

No

Be7+ .

w yields1)0( =y 1=+BA .

xxBexy 7)(' 3

2

+

at gives 5732 =+ BA .Ae 7

32= , so th 5)0(' =y

Solving simultaneously fo , we seerA andB that2317

236 , == B .A

Hence our particular solution is

xxeey

7

2317

236 3

2

+=

.

Example 8

ferential equationSolve the dif

6'' 5)0(',2)0(,025'+ yy = = =yyy .

The A.E. is , whose roots are02562 =+ mm im 43 = . The general solution is

by

.

We now determineAandB.

in == A .

, Bexy x ++= on using product rule.

0 ++=

therefore given

)4sin4cos(3

xBxAey x +=

s0cos(2)0(0 += BAey 22)0

So, )4sin4cos2(3 xBxey x +=

Now x ]4sin)83(4cos)64[()('3

xB

5' =]0sin)83(0cos)64[(5)0( BBe

i.e.

y

564 =+B

B41=

Hence, the particular solution is

)4sin4cos2(413 xxey= x .

Activity 2


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i)

(i) 02'3'' =+ yyy ;

(i 1)0(',0)0(,04'4'' ===+ yyyyy ;

(,0'5''

(iii) 02'2'' =+ yyy ;

(iv) 06''' =+ yyy ;

(v) )0 1)0(',0 === yyy ;y

(vi) 0 ;'3''2 =+ yyy

(vii) 0'2''4 = yyy ;

(viii) 0'6''5 =+ yy ;

(ix) 1)0(',0)0(,05'4'' ===++ yyyyy ;

2++ kyk .(x) ' =yy 0'2'


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8.4 DIFFERENTIAL OPERATORS

The D-operator

LetDdenote the differentiation operatord

dx . Then

dy

dxDy= , ( ) yD

dx

ydyDDyD

dx

dy

dx

d

dx

yd nn

n

===

= ,,22

2

L .

We can now express our differential equations in terms of theDoperator. Thus Eqn (1)

can be written as

)()( 2 xfycbDaD =++ .

Similarly, the differential equation

xyyy 3sin2'3''5 =+

can be written as

. (*)xyDD 3sin)235(2 =+

Inverse of theD operator

SinceDdenotes differentiation w.r.t.x, its inverse will represent integration w.r.t.x.

In general will mean integrate ntimes w.r.t.x. The symbol may also be written

as

1D

nD 1D

1

D. In general, it is customary to write asmD

1

Dmwhen mis a positive integer.

Thus,

DD D

=

=2

21 1

D

1, i.e., integrate twice.

So,

( )

.246

)(

;62

43213

32

2

===

===

xdx

xxDDxD

xdxxdxxxD

While evaluating , the arbitrary constant of integration may be omitted.)(1 xfD


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Some Properties of theDoperator

We shall denote by a function of theDoperator, e.g., in the above Eqn (*),)(DL

235)( 2 += DDDL .

Symbolically we can write a differential equation in the form)()( xfyDL = ,

or,

).()(

1xf

DLy=

Our task is to find ways of evaluating the RHS for different functions ).(xf

We first consider some properties of the D operator. You can easily verify all these

properties.

(i) ;)()( aLeeDL axax =

(ii) ;)()()]([)( xfaDLexfeDL axax +=

(iii) ;axaLaxDL cos)(cos)(22 =

(iv) ;axaLaxDL sin)(sin)(22 =

(v) ;axaLaxDL cosh)(cosh)(22 =

(vi) .axaLaxDL sinh)(sinh)(22 =

We note that there are no simple formulae for , and the

hyperbolics.

axDLaxDL cos)(,sin)(

Lets see how to use the above properties.

Example 9

(a) Suppose we wish to evaluate . Of course we could do it from

first principles; however, using Property (i), we find that here

xeDD 32 )725( +

725)( 2 += DDDL

and .3=a

447)3(2)3(5)3()( 2 =+== LaL .

Hence, .xx eeDD 332 44)725( =+


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(b) Prove that

( )( ) ( )1432122 233223 ++=+ xxeexDDD xx .

Here,

DDDDL += 23 2)( , . Then, Property (ii)gives2)(,3 xxfa ==

( )( ) ( ) ( ) ( ) 22333223 33232 xDDDeexDDD xx ++++=+

( )( ) ( )( )

( ).14321212216270

12167

23

23

2233

++=

+++=

+++=

xxe

xxe

xDDDe

x

x

x

(c) Evaluate .xDD 7cos)253(2 +

We use Property (iii). We have , and253)( 2 += DDDL 7=a .

We therefore replace by . [ Note: ]2D 27 4972 =

xx

xdx

dx

xDx

xD

xDxDD

7sin357cos145

7cos57cos145

)7(cos57cos145

7cos)5145(

7cos]25)49(3[7cos)253(2

+=

=

=

=

+=+


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8.5 SOLVING THE INHOMOGENEOUS EQUATION

The method of solution of the inhomogeneous equation

)(2

2

xfycdx

dyb

dx

yda =++

consists of 3 parts:

1.Find the Complementary Function Cy

2.Find the Particular Integral Py

3. The general solution is then given by

PC yyy += .

The complementary function (C.F.) is obtained by solving the corresponding

homogeneous differential equation, i.e.,

02

2

=++ ycdx

dyb

dx

yda , or 0)( 2 =++ ycbDaD

as we have done above.

The second step, finding the particular integral (P.I), is the hardest part. The P.I. is

obtained by the following result

)()(

1xf

DLyP = .

The following table summarizes the main results for different forcing functions .)(xf


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Table of Inverse Operator Techniques

1.ax

keDL )(

1, k :constant 0)(

)(aL

aL

ek ax

2. (i) )cos()(

12

baxkDL

+

(ii) )sin()(

12

baxkDL

+

)cos()( 2

baxaL

k+

0)(2 aL

)sin()( 2

baxaL

k+

3. )()(

1xP

DL, where is a)(xP

polynomial of degree m

Expand)(

1

DLin ascending powers ofDby

the Binomial Theorem as far as the term in

. Then operate on .m

D )(xP

4. )()(

1xe

DL

ax )()(

1x

aDLeax

+ [Shift Theorem]

5. )(1

xfmD

dxxfee mxmx )(

6. axre

aD )(

1

K,2,1,

!

=re

r

x axr

7. )cos(1

22 baxk

aD+

+

)sin(1

22 baxk

aD+

+

)sin(2

baxa

kx+

)cos(2

baxa

kx+

We shall now consider a few examples to illustrate.

Example 10

Solve the differential equation

912'5''2 = yyy .In terms of theDoperator, our equation is

9)1252(2 = yDD .

If we compare with the entries in the table, we find that here we need to use the first entry

where


14/29

1252)( 2 = DDDL

9,0 == ka We now find the solution to our equation.

STEP 1. Obtain the Complementary Function .

.

Cy

Now, the A.E. is

01252 2 = mm

4,23= m .

Hence

STEP 2. Find the Particular Integral

rom the table,

xx

C eBeAy42/3 += .

Py

F

4

3

91200

1

91252

1=y2

=

=

DDP

[Putting 0==aD ]

TEP 3. Write down the General SolutionS PC yyy +=

4342/3 += xx eBeAy .

Example 11


.x

eyyy 3410'3'' =+

Now we have

103)( 2 = DDDL

3,4 == ak

5,201032 == mmm A.E.

C.F. is

xxC eBeAy52 += .

P.I. is


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x

x

x

P

e

e

eDD

y

3

2

1

2

3

3

2

10)3(3)3(

4

4103

1

=

=

=

[Putting 3==aD ]

General solution is

xxxeeBeAy 3

2152 ++= .

Example 12


xx eeydx

dy

dx

yd 22

2

5296 =++ .

7)0(',0)0( == yy .Here we note that the forcing function consists of 2 terms, and . To find the

P.I., we use entry 1 of the table twice, and add the results, i.e.,

xe xe 25 2

)5()(

12

)(

1 2xxP e

DLe

DLy += ,

where .

0962

==++ mmm

.F.

22 )3(96)( +=++= DDDDL

A.E. 3 3,

xC eBAxy3)(

+= C

P.I.

xx

xx

xx

P

ee

ee

eD

eD

y

2

81

2

2

2

2

22

5

)32(

5

)31(

2

)5()3(

12

)3(

1

=

+

+

+=

+

++

=

General solution is

xxxeeeBAxy

2

813 5)( ++= .

es ofAandB.We now fix the valu

839

81 050)0( ==+= BBy


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Also, we have on differentiating and simplifying

)]248(8024[)(' 381 eBexy

xx ++= 4 xAe x +

0]818024[0)0('81 =+++= ABy

29= A after putting

Hence, th lutionis

8/39=B .

e Particular So

xxxeeexy

2

813

839

29 5)( ++= .

Example 13


.xyDD 2cos5)1( 2 =++

mm2 01 =++ imA.E.

2

3

21 =

+

=

xBxAey

x

C 2

3

sin2

3

cos

2/

C.F.

We now use entry 2(i)in the table with 0,2,5 === bak . Note we replace only

y . TheDstays as it is.

2D

2ab

xD

x

D

xDD

P

2cos53

1

2cos5

12

1

2cos51

1

2

2y

=

++

=

++

Since the result works only for , the trick is to obtain in the denominator. This is

achieved by multiplying top and bottom by

=

2D 2D 3+D . This then gives


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)2cos32sin2(13

5

2cos)3(13

5

2cos592

3

2cos59

3

2cos53

1

D

xD

yP

+

2

2

xx

xD

xD

xD

+=

+=

+

=

=

=

Hence the General solution is

.13

2cos152sin10

2

3

sin2

3

cos2

1x

xxxBxAey

+

+

=

Example 14


xyDD 3sin4)65( 2 =++ ,

0)0(',0)0( == yy .

A.E.

C.F.

the Particular Integral. We use entry 2(ii) in the table with

.

3,20652 ==++ mmm

xx

C eBAey32 +=

Next, we find

k 0,3,4 === ba

P.I.


18/29

)3sin3cos5(39

2

3sin)35(117

2

3sin49)3(25

35

3sin4925

35

3sin435

1

3sin4653

1

3sin465

1

2

2

2

2

xx

xD

xD

xD

D

xD

xD

xDD

yP

+=

+=

+

=

+

=

=

++=

++=

Hence, the General Solution is

)3sin3cos5(39232 xxeBAey

xx ++= .

We now determine the arbitrary constants.

00)0(3910 =+= BAy ----- (i)

Also,

)3cos3sin5()32()('13

232 xxeBeAxy xx ++=

0320)0('132 =++= BAy -----(ii)

Solving (i)and (ii)simultaneously, we obtain32

1312 , == BA .

Hence, the Particular Solution is given by

)3sin3cos5(3923

3

22

13

12

xxeey xx

+=

.


19/29

Evaluating )()(

1xP

DLwhen is a polynomial of degreem.)(xP

In this case we expand the operator)(

1

DLby the binomial theorem in ascending powers

ofD as far as the term in .mD

If is factorisable, use partial fractions and then expand.)(DL

For this purpose, the following binomial expansions are useful:

( )

( ) L

L

+++++=

++=+

4321

4321

11

11

DDDDD

DDDDD

( )( ) L

L

+++++=++=+

4322

4322

543211

543211

DDDDD

DDDDD

Note that we need to have our expression in the right form before carrying out the

expansion. Thus to expand we must first express it as1)53( +D 1531 )1(5 + D .

Likewise, must first be written as1)72( D 1721 )1(7 D . Also, we must express

as2

)45(

D2

542

)1(5

D and then expand.

Example 15

Evaluate 4

1

1x

D.

Here the degree of the polynomial is 4; we therefore expand1

1

Dup to .4D

).2424124(

)1(

)1(

)1(

1

1

1

234

4432

41

44

++++=

+++++=

=

=

xxxx

xDDDD

xD

xD

xD

L

Note:We just differentiate the previous term as we go along.


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Example 16


323672'3'' xxxyyy +=+ .

A.E. 2,10232 ==+ mmm

C.F. xxC eBeAy2+=

P.I.

]367[23

1 322

xxxDD

yP ++

=

Here the degree of the polynomial is 3, therefore expand up to . Also, the operator is

factorisable and so we split it into partial fractions before expanding. Thus

3D

1

1

2

1

)2)(1(

1

23

12

=

=+ DDDDDD

.

On using the binomial theorem we obtain

)1()( 3216842

132

LL DDDDDD

which simplifies to

)1514128( 32161 DDD +++

Hence, is given byPy

]51864[

)]6(15)66(14

)366(12)367(8[

]367)[1514128(]367[23

1

23

81

232

161

3232

16132

2

+=

+++

+++=

++++=++

xxx

x

xxxxx

xxxDDDxxxDD

Note again, we just differentiate previous brackets, while paying attention to the coefficients of the in

the first brackets.

iD

General Solution: )51864( 23812 +++= xxxeBeAy xx .


21/29

Example 17


9725'3'' 3 +=+ xxyyy .

A.E. immm2

11

232 053 ==+

C.F. ]sincos[2

11

2

112/3 xBxAey x

C +=

P.I.

)972(53

1 32

++

= xxDD

yP .

In this case, the degree of the polynomial is 3, so that we need to expand up to . But

now the operator does not factorize, so that we cant use partial fractions. We therefore

proceed as follows:

3D

L

L

+++=

+

+

=

+=+=

+

32

32

222

12

12

2

625

3

125

4

25

3

5

1

5

3

5

3

5

31

5

1

5

31

5

1)53(

53

1

DDD

DDDDDD

DDDD

DD

Hence, the P.I. is

].6361115450250[

)12()12()76()972(

)972(625

3

125

4

25

3

5

1)972(

53

1

23

625

1

6253

12542

2533

51

3323

2

++=

++++=

+

++=++

xxx

xxxx

xxDDDxxDD

Then, the general solution is given by

]6361115450250[]sincos[ 236251

2

11

2

112/3 ++++= xxxxBxAey x .


22/29

The Shift Theorem

Evaluating )()(

1xe

DL

ax . [Here, ])()( xexf ax=

We use the formula

)()(

1)(

)(

1x

aDLexe

DL

axax +

=

TheDis shiftedto .aD +

Example 18

Evaluate ).(

12

1 32

xe

DD

x

+

Here, ( ) .)(,1,12 32 xxaDDDL ==+=

1

2 1

1

123

2

3

D De x

De x

x x

+ =

( ) 3

2]1)1[(

1x

Dex

+=

3

2

1x

Dex= .

20

5x

ex= (Integrating twice).

Example 19

Solve the differential equation .xexxyy 32 )976(4'5'' +=+

A.E. .0452 =+ mm 4,1= m

C.F. .xxC BeAey4+=

We now need to find the particular integral.

P.I. xP exxDD

y 322

)976(45

1+

+=

Here .976)(,3,45)( 22 +==+= xxxaDDDL

The Shift Theoremgives


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)976(12

)976()1)(2(

1

)976(

2

1

)976(4)3(5)3(

1)976(

45

1

231

31

3

23

2

2

3

2

2

332

2

+

++

=

++

=

+

+

=

++++

=++

xxDD

e

xxDD

e

xx

DD

e

xxDD

eexxDD

x

x

x

xx

We now proceed as for Case 3; we expand the operators by the binomial theorem up to

the term in since we have a 2nd

degree polynomial. Thus2D

[ ]

.8

3

42

1

)1()2/1(3

1

122

11

213

131

+++=

++=

++

L

DD

DDDD

Now,429

2122

2

3)976(8

3

42

1+=+

++ xxxx

DD.

)3(429

2123 += xxey xP .

The general solution is hence given by

)3(4

2921234 +++= xxeBeAey xxx .

Example 20

Solve the equation .0)0(',1)0(,sin2'3'' 2 ===++ yyxeyyy x

A.E. , so that0232 =++ mm 2,1 =m .

xx

C BeAey2 += .

Next we find theP.I. We have here

Then, by the Shift Theorem( ) .sin)(,2,232 xxaDDDL ==++=


24/29

xDD

exeDD

y xx

P sin2)2(3)2(

1sin

23

12

22

2 ++++=

++=

=+ +

eD D

xx22

1

7 12sin

= + +

eD

xx2 1

1 7 12sin (replacing by2D 21 )

( )

=+

=

eD

x

e DD

x

x

x

2

2

2

1

11 7

11 71

121 49

sin

sin

( )= +

e Dx2 11 71

121 49sinx (replacing by )2D 21

( )

( ).cos7sin11170

1

sin711170

1

2

2

xxe

xDe

x

x

=

=

Therefore, the general solution is

170/)cos7sin11(22 xxeBeAey xxx ++= .

We now fix the arbitrary constantsAandBby using the given data .0)0(',1)0( == yy

1170/71)0( =+= BAy . ----(i)

Now, , so that170/)cos3sin29(2)(' 22 xxeeBAexy xxx +=

0170/320)0(' == BAy . -----(ii)

Solving (i) and (ii) simultaneously forAandB, we obtain 17/18,170/357 == BA .

Hence, the particular solution is

( )}cos7sin11{180357170

1 22 xxeeey xxx += .


25/29

Forcing Function Consisting of Several Functions

When the forcing function consists of several terms, the P.I. is given by the sum of

the P.I.s of each term. Thus,

)(xf

)]([)(

1)]([

)(

1)]([

)(

1)]()()([

)(

1xr

DLxq

DLxp

DLxrxqxp

DL++=++ .

Example 21

Evaluate ]2sin[23

1 3322

xexeDD

xx +++

.

Letting ]2sin[23

1 3322

xexeDD

y xxP +++= , we have 321 yyyyP ++= , where

)cos7sin11(170

1]sin[

23

1 2221

xxexeDD

y xx =++

= ,

xxee

DDy 33

22 2

1][

23

1 =++

= ,

445

2212

2933

23]2[

231 +=

++= xxxx

DDy .

Hence,

4

45

2

21

2

9

2

1)cos7sin11(

170

1 2332 ++= xxxexxey xxP .

SPECIAL CASES


26/29

ofax

keDL )(

11. Evaluation when 0)( =aL

In this case we can proceed in two ways. One is to use the Shift theorem, the other is to

use Case 6 in the table.

Example 22

Lets find the P.I. for the equationx

eyyy252'3'' =++ .

x

x

P

eDD

eDD

y

2

2

2

5)1)(2(

1

523

1

++=

++=

Clearly, the denominator is zero when we replace D by 2 . We therefore proceed as

by in the factor

follows.

ReplaceD only. We then have2 )1( +D

x

P eD

y25

)1)(2(

1=

+ ------ (#)

Now we use the Shift Theorem with . So theDis shifted to . Thus)2( D5)( =x

.5

)5(

)5(1

5

)1](2)2[(

1

2

2

2

2x

x

x

x

P

xe

xe

De

D

=

=

=+

[Recall means Integrate]

Hence the P.I. is .

lternatively, we can use Result 6 in the table to Eqn. (#) above, where

ey =

D/1

xxe 25

A 1=r . This

yields againx

P xey25 = .


27/29

2. Evaluation of )cos()(

12

baxkDL

+ and )sin()(

12

baxkDL

+ when 0)( 2 =aL

We use Result 7 in the table.

Example 23

To find the P.I of the equation xyy 2cos34'' =+ , we proceed as follows:

xx

xx

xD

yP

2sin

2sin)2(2

3

2cos34

1

4

3

2

=

=

+=

[Using Result 7 with 0,2,3 === bak ]

Activity 3


(i) ;725'6'' 3 =++ xeyyy

(ii) ;18)0(',1)0(,549'' 3 === yyeyy x

(iii) ;xyy cosh2''' =

(iv) ;xeyy 28'4'' =

(v) ;xyy 2sin'' =+

(vi) ;xyyy 4sin1006'5'' =+

(vii) xxyyy sin16cos4825'8'' =++ ;

(viii) ;xxyyy 20cos4020sin401'2'' +=++

(ix) ;0)0(',0)0(,''

3

===+ yyxyy(x) ;xeyy x 2sin'2'' =+

(xi) ;24876442''' xxyyy =

(xii) ;18549'6'' +=+ xyyy

(xiii) 1)0(',2)0(,cos4'' ===+ yyxyy .


28/29

8.6 SUMMARY

In this unit, you have studied how to solve homogeneous linear second-order ordinary

differential equations and inhomogeneous equations by finding the complementary

functions and particular integrals usingD-operators.

8.7 ANSWERS TO ACTIVITIES

Activity 1

(i) ;xy = 5

(ii) ;12 += xy

(iii) ;1+=xy

(iv) ;2/)5( xy =

(v) BAxxxy ++= 2365 ;

(vi) BAxxxxy ++++= 22sinh2cosh3 ;

(vii) 3212

212 xxxey x += ;

(viii) ;2/)711(3 xxxy +=

Activity 2

(i) ;xx

BeAey2+=

(ii) ;xxey 2=

(iii) ;)sincos( xBxAey x +=

(iv) ;xx BeAey 32 +=

(v) )4( 551 xey += ; (vi) ;2/xx BeAey +=

(vii) )4/5sin4/5cos(4/ xBxAey x += ;

(viii) ;5/6x

BeAy +=

(ix) ;xey x sin2=

(x)kx

eBxAy += )( .


29/29

Activity 3

(i)573

1615 ++= xxx eBeAey ;

(ii) ;

xx

eexy

33

2)91(

+=(iii) xx exCexBAy ++++= )()(

21

21 ;

(iv) xx eBAey 2324 ++= ;

(v) xxBxAy 2sinsincos31+= ;

(vi) ;xxBeAey xx 4sin24cos432 ++=

(vii) ;xxBxAey x cos2)3sin3cos(4 ++=

(viii) ;xxBxAey x 20sin)20sin20cos( ++=

(ix) ;xxxy sin663 +=

(x) xeBeAy xx 2sin512 += ;

(xi) ;51424 22 +++= xxBeAey xx

(xii) ;66)( 3 +++= xeBxAy x

(xiii) .xxxy sin)12(cos2 +=

unit 8 geotechnical engineering

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