MATHEMATICS- II S.NO CONTENTS PAGE NO UNIT-1 VECTOR ...€¦ · integrals of transforms -...

MA6251 MATHEMATICS-II

S.NO CONTENTS PAGE NO

UNIT-1 VECTOR CALCULUS

1.1 Gradient-Directional Derivative 1

Problems

1.2. Divergence And Curl –Irrotational And Solenoidal Vector 4

Fields Divergence

Problems

1.3 Vector Integration 7

Problems

1.4 Green’s Theorem In A Plane;(Excluding proof) 9

Problems

1.5 Gauss Divergence Theorem:(Excluding proof) 10

Problems

1.6 Stoke’s Theorem(Excluding proof) 12

Problems

UNIT –II ORDINARY DIFFERENTIAL EQUATIONS

2.1 Higher Order Linear Differential Equations With 14

Constant Coefficients

Problems

2.2 Method of Variation of Parameters 22

Problems

2.3 Differential Equations for the Variable Coefficients 24

Problems

2.4 Simultaneous First Order Linear Equations with Constant Coefficients26

Problems

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UNIT –III LAPLACE TRANSFORMATION

3.1 Laplace transformation-Conditions and existence 28

Problems

3.2 Transforms of Elementary functions-Basic Properties 29

Problems

3.3 (a) Transforms of derivatives 31

(b)Derivatives and integrals of Transforms

(c)Integrals of Transforms

Problems

3.4 Transforms of the unit step functions and impulse function 34

Problems

3.5 Transforms of periodic functions 35

Problems

3.6 Inverse Laplace Transform 36

Problems

3.7 Convolution theorem 42

Problems

3.8 Initial and final value theorems 45

3.9 Solution of linear ODE of Second Order with constant coefficients 47

Problems

UNIT-IV ANALYTIC FUNCTIONS

4.1 Introduction –Function of A Complex Variable 49

4.2 Analytic Functions(C-R Equations) 49

Problems

4.3 Harmonic and Orthogonal Properties Of Analytic Functions 51

Problems

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4.4 Construction of Analytic Functions 56

Problems

4.5 Conformal Mapping 58

Problems

4.6 Bilinear Transformation 61

Problems

UNIT V- COMPLEX INTEGRATION

5.1 Prerequisite 62

5.2 Introduction 62

5.3 Cauchy’s Theorem 62

Problems

5.4 Taylor’s and Laurent’s Series Expansion. 64

Problems

5.5 Singularities 67

Problems

5.6 Residues 69

Problems

5.7 Evaluation of real definite Integrals as contour integrals 72

Problems

5.8 Applications 79

APPENDICES

A Question Bank

B University Questions

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MA6251 MATHEMATICS – II REGULATION 2013 SYLLABUS

MA6251 MATHEMATICS – II L T P C

3 1 0 4

OBJECTIVES: • To make the student acquire sound knowledge of techniques in solving ordinary differential equations that model engineering problems. • To acquaint the student with the concepts of vector calculus, needed for problems in all engineering disciplines. • To develop an understanding of the standard techniques of complex variable theory so as to enable the student to apply them with confidence, in application areas such as heat conduction, elasticity, fluid dynamics and flow the of electric current. • To make the student appreciate the purpose of using transforms to create a new domain in which it is easier to handle the problem that is being investigated.

UNIT I VECTOR CALCULUS 9+3 Gradient, divergence and curl – Directional derivative – Irrotational and solenoidal vector fields

– Vector integration – Green’s theorem in a plane, Gauss divergence theorem and Stokes’ theorem (excluding proofs) – Simple applications involving cubes and rectangular parallelopipeds.

UNIT II ORDINARY DIFFERENTIAL EQUATIONS 9+3 Higher order linear differential equations with constant coefficients – Method of variation of parameters – Cauchy’s and Legendre’s linear equations – Simultaneous first order linear equations with constant coefficients.

UNIT III LAPLACE TRANSFORM 9+3

Laplace transform – Sufficient condition for existence – Transform of elementary functions – Basic properties – Transforms of derivatives and integrals of functions - Derivatives and integrals of transforms - Transforms of unit step function and impulse functions – Transform of periodic functions. Inverse Laplace transform -Statement of Convolution theorem – Initial and final value theorems – Solution of linear ODE of second order with constant coefficients using Laplace transformation techniques.

UNIT IV ANALYTIC FUNCTIONS 9+3

Functions of a complex variable – Analytic functions: Necessary conditions – Cauchy-Riemann equations and sufficient conditions (excluding proofs) – Harmonic and orthogonal properties of analytic function – Harmonic conjugate – Construction of analytic functions – Conformal mapping: w = z+k, kz, 1/z, z2, ez and bilinear transformation.

UNIT V COMPLEX INTEGRATION 9+3 Complex integration – Statement and applications of Cauchy’s integral theorem and Cauchy’s integral formula – Taylor’s and Laurent’s series expansions – Singular points – Residues – Cauchy’s residue theorem – Evaluation of real definite integrals as contour integrals around unit circle and semi-circle (excluding poles on the real axis).

TOTAL: 60 PERIODS

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TEXT BOOKS:

1. Bali N. P and Manish Goyal, “A Text book of Engineering Mathematics”, Eighth Edition, Laxmi Publications Pvt Ltd.,(2011). 2. Grewal. B.S, “Higher Engineering Mathematics”, 41 (2011). Edition, Khanna Publications,

Delhi,

REFERENCES: 1. Dass, H.K., and Er. Rajnish Verma,” Higher Engineering Mathematics”, S. Chand Private Ltd., (2011) 2. Glyn James, “Advanced Modern Engineering Mathematics”, 3rd Edition, Pearson Education, (2012). 3. Peter V. O’Neil,” Advanced Engineering Mathematics”, 7th Edition, Cengage learning, (2012). 4. Ramana B.V, “Higher Engineering Mathematics”, Tata McGraw Hill Publishing Company, New Delhi, (2008).

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UNIT-1

VECTOR CALCULUS

1.1Gradient-Directional Derivative

1.1.1. Gradient

1.1(a) The Vector Differential Operator The differential operator (read as del) is defined as ≡ + + where

, , are unit vectors along the three rectangular axes OX, OY, OZ.

1.1(b) The Gradient (Or Slope Of A Scalar Point Function)

continuously differentiable then the vector Let ( , , ) be a scalar point function and is

is called the gradient of the scalar function

= +

+

=

+

+

and is written as = .

Note: 1.1.1 is a vector differential operator and also it is a vector. Note: 1.1.2 ≡ + + Note: 1.1.3 If is a constant ,then = 0. Note: 1.1.4 If is a vector whose three components are , , Note: 1.1.5 ) + φ .

1. Find r ,

Solution:

We know that ,→ = + + ,

r = → = + + , = + +

= ; = ; =

(i) r = + + = + +

= =

(ii) = ∑ ı =∑ ı =∑ x ı → = .

2. Prove that )= Solution: )=∑ ( )=∑

=∑ = + +

= SCE 1 Dept of S&H

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1.1.2. Directional Derivative

Directional derivative= . | | 1. Find the directional derivative of = + 4 + (1,2,3) in the direction of

2 + − Solution:

Given: = + 4 +

=

+

+

=(2 + 4 + ) + ( + ) + ( + 8 + ) =54 + 6 + 28 Given: = + − | | = √

D.D= . | | = (54 + 6 + 28 ). √ =√ (86)

1.1.3. Unit Tangent Vector

Unit Tangent vector=

1.Find a unit tangent vector to the following surfaces at the specified

points= + , = − , = − = 2.

Solution:

= + +

=( + 1) + (4 − + (2 − ) = 4 + 4 + 2

= 6

Unit tangent vector= =

=

1.1.4 Normal Derivative

Normal derivative =| |

1. What is the greatest rate of increase of = (1,0,3)

Solution: Given: =

=

= + + SCE 2 Dept of S&H

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= ( ) + ( ) + (2 ) ==9 ∴ Greatest rate of increase = | | = 9

1.1.5. Unit Normal Vector: Unit normal vector = | |

1. Find a unit vector normal to the surface + − = 10. (1,1,1)

Solution:

Given: = + − − = + +

=2 + 2 − ( , , ) = + −

| |=3

Unit normal vector = | | =

1.1.6Angle Between The Surfaces . = | | | |

1. Find a and b such that the surfaces − = ( + 2) and 4 + = 4 cut orthogonally at (1,-1,2)

Solution:

Let =− − ( + 2

) …………….. (1)

=

+

+

= [2 − + ] + − + − ) , − , = − − + = 4 + = 4

= + +

=8 + 4 + 3 , − , = −8 + + Given: . = 0.

( − 2) − 2 + . −8 + + = . −8 − − 8 + 12 = 0 − − = … … … … … . Since the points (1,-1, 2) lies on the surface ( , , ) = 0.

− − = + 2)(1) = 1 ∴ … . > − − =

=

SCE 3 Dept of S&H

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1.1.7 .Scalar Potential : ,then find the value of .

1. If = + +

Solution: Given: =

+ +

+ += 2 ++

Equating the coefficients of , , ,we get

= … . . , = … . .

= … . .

Integrating (1) p.w.r.to ‘x’ we get

= + ( , … … … … . . Integrating (2) p.w.r.to ‘y’ we get

= + ( , … … … … . . Integrating (3) p.w.r.to ‘z’ we get

= + ( , … … … … . . Combining (4),(5),(6) we get

= + , where c is the arbitrary constant.

1.1. 8.The Vector Equation Of The Tangent Plane And Normal Line To The Surface:

(i) Equation of the tangent plane is ( − . = 0.

(ii) Equation of the normal line is ( − × = 0.

Tutorial Problems:

1. Find the values of a and b so that the surfaces

− = ( + 3 ) 4 − = 1 may cut orthogonally at (2,-1,-3)

2. If = ( + + ) find

3. In what direction from the point (2,1,-1) is the DD of = a maximum? What is the magnitude of this maximum?

4. Find the angle between the surfaces = − = − (1,1,1)

1.2 Divergence And Curl –Irotational And Solenoidal Vector Fields:

1.2.1 Divergence and curl

= . =

+

+

Note:

= × = ×

+ ×

+ ×

,then

If = + +

× =

1. If = + + then find = . and ×

Solution:

Given: = + +

= . =

+

+

=

SCE 4 Dept of S&H

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=2

=

( ) +

( ) +

( )

+ 2 + 2

= × = =

=0

2. Find .

Solution:

We know that, = + +

=

+

+

. (

. =

+

+

+

+

)

=

+

+

( )

( ) ( )

=

+

+

=

3. Prove that ( ) = 0

Proof: = =

+

+

( = ×

=

=∑ − = 0

3. Find where = ( + + − ) Solution:

Given: = + + −

= − + − + −

. = − + − + −

= 6 + 6 + 6

× = (3 − ) (3 − ) (3

− ) = − + 3 − − + 3 + − + 3 )= 0

SCE 5 Dept of S&H

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1.2.2 SOLENOIDAL VECTOR,IRROTATIONAL VECTOR: Solenoidal vector formula: . = 0

Irrotational vector formula: × = 0 1. Show that = + +

Solution: = + + To prove: × = 0 × =

=0

2. Prove = ( + ) +(2ysinx-4) + 3 is irrotational and find its scalar potential Solution:

= ( + ) +(2ysinx-4) + 3 × =

(( + ) ysinx − (3 ) × = 0.

Hence, is irrotational =

( + ) + (2ysinx-4) + 3 = + + Equating the coefficients of , , we get,

= + … … = 2 … …

= … …

Integrating (1) p.w.r.to ‘ We get = + , … …

Integrating (2) p.w.r.to ‘ We get = − 4 + ( , … …

Integrating (3) p.w.r.to ‘ We get = + ( , … …

Combining (4),(5),(6) we get, Where c is a constant

= + − +

3. Show that is an irrotational Vector for any value of but is solenoidal only if = −

Solution:

Let = = ( + + )

= ++

SCE 6 Dept of S&H

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MA6251 MATHEMATICS-II × = =∑ − =0.

For all values of n is irrotational. . = ( ) = + =3 + ( + + ) =(3 + )

When n= -3 we get . = 0.

Laplace Operator: = + +

= 0 is called the laplace equation.

Tutorial Problems:

1. Prove ( ) = + ( ). ( ) 2. Prove that ( ) = 0.

3. Determine ( ) so that the vector ( ) is solenoidal.

4. Show that = (6 + + − ) + (3 − )

5. Prove that . × )= . × − . × 1.3 Vector Integration:

Conservative Vector Field: The line integral . depends not only on the path C but also the terminal

points A and B. If the integral depends only on the end points but not on the path C, then

is called the conservative vector field.

1.3.1. Line Integral:

1. If = ( 3 + 6 ) − 14 + 20 , evaluate . from (0, 0,0) to (1,1,1) along the curve

= , = , =

Solution: The end points are (0,0,0) and (1,1,1)

= 1,

N The points corresponds to = 0

∴ = , = 2 = 3

. = + − + 20

= + − + )

= 5.

2.If = − ,evaluate the line integral . from (0,0) to (1,1) along the path = .

Solution: = ………….(1)

Given:

SCE 7 Dept of S&H

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⇒ = … … … … . .

Given: = − We know that = +dy +

. = +

( by (1)&(2) )

= + ) =

. = +

1.3.2. Surface Integrals:

Definition: Consider a surface S .Let n denote the unit outward normal to the surface S. Let R be

the projection of the surface x on xy plane. Let be a vector function defined in some region

containing the surface S, then the surface integral of is defined to be

. = .

.

.

Note: We can define surface integral by considering the projection of the surface on the yz plane

or zx plane and we get

. =

. .

| . |

. =

. .

| . |

1. Evaluate . where = +x - and S is the surface of the cylinder + = 1

included in the first octant between the planes = 0 = 2.

Solution:

+x -

Given: =

= + − 1

= 2 + 2

| | = 4 + 4 =2.

Unit normal vector = | | = +

. = ( +x - ) . ( + ) = + .

Now, .

. = . | . |

= + ) =3.

1.3.3. Volume Integral: The volume integral of ( , , ) over a region enclosing a volume V is given

by ( , , ) ( , , )

SCE 8 Dept of S&H

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1.3.4 Tutorial Problems:

1. If = + , evaluate . along the curve C in the XY plane =from the point

(0,0) to (1,1)

2. Evaluate . where = ( + − + 2 and S is the surface of the plane

2 + + 2 = 6 in the first octant.

( − 3 ) − 2 − 4 , evaluate

( , ,

) where V is the region bounded by

3. If = 2

= 0, = 0 2 + 2 + = 4.

1.4 .Green’s Theorem In A Plane:

Statement: If , ,

,

are continuous and one-valued functions in the region R enclosed by the

curve C, then

+ ) =

−

.

1. Verify Green’s theorem in the xy plane for + ) + where C is the closed curve of

the region bounded by = = .

Solution:

Green’s theorem in the xy plane is

( + ) = − .

Here = + =

= + 2

= 2

Evaluation of + ) To evaluate + ),we shall take C in two different paths viz..,

(i) Along ( = ), ( ) ( = ) + = +

(i) Along ( = , = 2 )

= [{ ( ) + ( ) } + . 2 ]

= [ + ) + 2 ]

(ii)

=

( = , = )

= [( + ) + ]

=

= -1.

Hence, + ) +

= + =

− = −

……………..(1)

Evaluation of −

−= [ − + 2 )] SCE 9 Dept of S&H

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=√ ( − ) √

= − 2 = ………………..(2)

From (1) & (2) we get,

+ = − . Hence, Green’s theorem is verified.


1. Verify Green’s theorem in the xy plane for { − 8 ) + (4 − ) } where C is the closed curve of the region bounded by = 0, = 0, + = 1..

2. Prove that the area bounded by a simple closed curve C is given by − ).

Hence find the area of the ellipse = , = .

1.5 Gauss Divergence Theorem:

Statement: The surface integral of the normal component of a vector function F over a closed surface

S enclosing volume V is equal to the volume integral of the divergence of F taken throughout the

volume V . = .

1. If =+ + ,a,b,c are constants, Show that . = ( + + )

Proof:

We know that,

The Gauss-divergence theorem is,

. = .

=

( ) +

( ) +

( )

=( + + )

(Note: V is the volume of the unit sphere; sphere volume =

)

(

+ + )

= ( +

( )

= + ) 1

. =

( +

+ )

2 .Verify G.D.T for = ( − ) + ( − ) + ( − ) taken over the rectangular parallelepiped ≤ ≤ , ≤ ≤ , ≤ ≤ .

Solution: The Gauss-divergence theorem is,

. = .

Given: = ( − ) + ( − ) + ( − )

SCE 10 Dept of S&H

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. = 2 + 2 + 2 = 2( + + ) R.H.S = .

= + + ) = + + ) = + + = + +

= abc (a+b+c).

L.H..S: . = + + + + +

= − + − ) + ( − ) ,=unit outward normal vector

Face . Equation . on S .

( − ) = − ( − )

(

− )

= 0

− -

( − ) = − ( − )

− (

−

)

= 0

-

( − ) = − ( − )

− (

−

)

= 0

-

(i) . + . = ( − ) +

= )

=

(ii) . + . = − ) +

= ( )

(iii) . =

+ . = − ) +

= )

=

SCE 11 Dept of S&H

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∴ . = + + + + +

= ( + + ) ( ), ( ), ( )

L.H.S=R.H.S

. = .

Hence, Gauss-divergence theorem is verified.

Tutorial Problems:

over the cube bounded by =

1. Verify the GDT for = 4 − +

0, = 1, = 0, = 1, = 0, = 1.

over the cube bounded by = ±1, =

2.Verify the GDT for = + +

±1, = ±1.

1.6. Stoke’s Theorem

Statement:

The surface integral of the normal component of the curl of a vector function F over an open surface S

is equal to the line integral of the tangential component of F around the closed curve C bounding S.

. = × .

1. Verify Stokes theorem for = ( + ) − 2 taken around the rectangle bounded by the lines = ± , = 0, = . Solution:

Stoke’s theorem is

. = × .

Given: = ( + ) − 2 × =

( + − 0 =−4

R.H.S = × . = −4 . =−

=−4 Given: = ( + ) − 2

. = + +

. = + ) − L.H.S: = + + + (i) = + ) −

= 0, = 0.

SCE 12 Dept of S&H

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= =

(ii) = + ) −

= −2 = − . (iii) = + ) − 2

= , = 0.

= , = 0. = ( + ) = − − 2 ..

(iv) = + ) −

= 2 = − ..

= − , = 0.

∴ = + + + =−4

L.H.S=R.H.S

( ),( ),( )

. = × .

Hence, Stoke’s theorem is verified.


1. Evaluate by Stokes’s theorem + 2 − ) where C is the curve + = 4, = 2

2. Using Stoke’s theorem, prove that = 0 and ( ) = 0.

SCE 13 Dept of S&H

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UNIT-II ORDINARY DIFFERENTIAL EQUATIONS

2.1 Introduction:

The study of a differential equation in applied mathematics consists of three phases. (i) Formation of differential equation from the given physical

situation, called modeling. (ii) Solutions of this differential equation, evaluating the

arbitrary constants from the given conditions, and (iii)Physical interpretation of the solution.

2.1.1. Higher Order Linear Differential Equations with Constant Coefficients

General form of a linear differential equation of the nth order with constant

coefficients is d y d y

+ K

+ + K y = X

dx dx

Where , K , K … K are constants.

D is the operator of differential. (i. e) Dy =

dy

,

dx

=

,etc

Generally =

.

2.1.2 Note:

1. X =∫ X dx.

2. X = ∫ X dx

X = ∫ X dx 2.1.3.Solution of the linear differential equation

Working rule: i. The general form of the linear differential equation of second order is

+P +Qy = R. where P and Q are constants and R is a function of x or constant.

ii. Differential operators

The symbol D stands for the operation of differential ( . ) = ,

=

stands for operation of integration. stands for the operation of integration twice.

+P +Qy = R. It can be written in the operator form

SCE 14 Dept of S&H

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+ + = (or) ( + + ) = iii.

iii. Complete solution is Y= complementary function + particular integral

To ind the complementary functions

Roots of A.E C.F Roots are real and different

1. A +B , ( ≠ )

Root are real and equal

2. (Ax+B) (or) = = (A+Bx)

3. Roots are imaginary α±iß (Acos ßx + B sin ßx)

iv. To find particular integral: P.I =

X

( )

X P.I

P.I =

(

)

=

,f(a)≠0

( )

1.

=x

, f(a)=0, ≠

( )

0

=

, f(a)=0,

( )

( ) = 0, ≠

2.

P.I =

= [f(D)]

( )

3.

sin ax (or) cos ax P.I =

[sin ax (or) cos ax]

( )

4.

P.I=

( )

( )

=

( )

2.1.4. Problems Based On R.H.S Of The Given Differential Equation Is Zero.

1. Solve ( − D + D − y = 0 Solution: :

Given: ( − D + D − y = 0

The auxiliary equation is − + 3 − = . ( − 1) = 0.

= 1,1,1.

SCE 15 Dept of S&H

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MA6251 MATHEMATICS-II The general solution is given by y = C.F

y = [ + Bx + Cx2 ]

2. Solve − + 13y = 0. Solution:

Given: − 6 + 13y = 0. (i.e) (D + 6D + 13)y = 0.

The auxiliary equation is − + 13 = 0 m = ±√

= ±√ m = ±

= 3 ± 2i

The roots are imaginary and occur in conjugate pairs. Hence, the solution is y = (A cos 2 + B sin 2 ).

2.1.5 Problems Based On P.I. = ( ) => Replace D By a

1. Solve (D 2

4D 13)y

Solution:

Given: (D 2

4D 13)y

The auxiliary equation is − + 13 = 0 m = ±√ = ±√

= 2 ±3i

C.F (A cos 3 + B sin 3 )

= =

y = C.F + P.I.

y = (A cos 3 + B sin 3 ) + 2. Solve + 4 + 4y =

Solution: Given: (D + 4D + 4) y =

The auxiliary equation is + 4 + 4 = 0

(m + 2)2 = 0

m = -2,-2.

C.F = (Ax + B)

SCE 16 Dept of S&H

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P.I. =

=

=

= x

=

= x2 (Ax + B)

y = C.F + P.I.

= (Ax + B) +

2.1.6 Problems Based On P.I. = ( ) ( ) ( ) => Replace D2 By –a

2

1. Find the P.I. of (D2 +1)y = sin x

Solution: Given: (D

2 +1)y = sin x

P.I. = sin x

= sin x = x sin x

= sin dx

= (− cos )

P.I. = 2.Find the P.I. of + 4 = sin 2x

Solution: Given: (D

3+4D) y = sin 2x

P.I. = sin 2x = ( ) sin 2x

= ( ) sin 2x

= x

sin 2x

= x

sin 2x

= − sin 2x

2.1.7 Problems Based On R.H.S = + Sin Ax (Or) + Cos Ax 1.Solve (D − D + y) = + cos 2x

Solution: Given: (D − D + )y = + cos 2x

The auxiliary equation is m2- 4m + 4 = 0

(m - 2)2 = 0

m = 2,2

C.F. = ( + ) SCE 17 Dept of S&H

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MA6251 MATHEMATICS-II P.I1 =

=

=

= x

= x

= x

=

P.I2 = cos 2x = cos 2x

= cos 2x

= [ cos 2x]

= [] =

y = C.F. P.I1+ P.I2

= ( + ) +

Problems Based On R.H.S =

Solution: Given:

The auxiliary equation is m2- 1= 0

m = ± 1

C.F Aex Be

-x

P.I. = x

= -[1- D2] x

= -x

y = C.F + P.I.

y = Aex Be

-x – x

2. Solve (D2 – D) y = x

Solution: Given:

The auxiliary equation is m2- m= 0

m(m-1) = 0 m =0,1

C.F Ae0x

Bex

C.F = A + Bex

P.I. =

= x

= [1 + D + D + D + ]x = [x + 1]

= - + ) SCE 18 Dept of S&H

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= - −

y = A + Bex - −

2.1.9 Problems Based On R.H.S =

P.I. = ( ) = ( ) 1. Obtain the P.I. of ( -2D +1) y = (3 − 2)

Solution:

Given: ( -2D +1) y = (3 −

P.I. =

(3 − 2)

( )

=

(3 − 2)

( )

=

(3 − 2)

=

[ −

]

P.I. = [ − ]

2. Solve (D+2)2 y = sin

Solution: Given: (D+2)

2 y = sin

The auxiliary equation is (m + 2)2 = 0

m = -2,-2 C.F. = (

+ )

P.I. =

sin

( )

=

sin

( )

=

sin

( )

= sin

=− sin x

sin x

y = (+ ) −

2.1.10 Problems Based On f(x) = in ax (Or) os ax

1.Solve (D2 +4) y = x sin

Solution: Given: (D

2 +4) y = x sin

The auxiliary equation is + 4 = 0 m = ± 2i

C.F. = (A cos 2 + B sin 2 ) P.I. = ( ) x sin x

= x () sin x - ( ) sin x

= sin x - cos y = C.F + P.I.

y = (A cos 2 + B sin 2 ) + sin x - cos . SCE 19 Dept of S&H

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2. Solve (D2 -2D +1)y = x sin x

Solution:

Given: (D2 -2D +1)y = x sin x

The auxiliary equation is − 2 + 1 = 0

m = 1,1

x

C.F. = (Ax + B)

P.I. =

sin x

= x sin x

) ( )

(

= x sin x

= [ sin x ] - ( ) sin x = - x sin x -2 cos = - (x sin x + 2 cos )

y = C.F + P.I.

y = (Ax + B) - (x sin x + 2 cos )

2.1.11 Problems Based On

1. Solve (D2 + a

2 ) y = sec

Solution:

Given: (D2 + a

2 ) y = sec The auxiliary equation is (m

2 +a

2) =

0

m =± ∴ C. F. = A cos ax + B sin ax

P.I. = sec

=

sec

( )( )

= (

-

) sec

=

sec dx -

sec dx

= − itan ) -

+ tan ) dx

=

(x -

log sec ) -

(x +

log sec )

=

(

) -

log sec (

)

= sin ax − cos ax log sec

y = C.F + P.I.

y = A cos ax + B sin ax + sin ax − cos ax log sec

2.1.12 General ODE Problems

1.Solve (D2 -6D +13)y = 2

x

Solution: Given: (D

2 -6D +13)y = 2

x

(D2 -6D +13)y =

= = ( )

The auxillary equation is − + 13 = 0

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MA6251 MATHEMATICS-II m = ±√

= ±√ m = ±

C.F (A cos 3 + B sin 3 ) P.I. = ( )

= ( ) =

y = C.F + P.I.

y = (A cos 3 + B sin 3 ) +

2.1.13 Problems Based On R.H.S = + + cos ax

1.Solve (D2 +1)y = + sin hx

Solution: Given: (D

2 +1)y = +

The auxillary equation is (m2 +1)

2 = 0

m = ±1 C .F. = (A+Bx) cos +(C+Dx)sin

P.I1 = ( ) = [1+ D ] -2 x

= [ 1- 2D + 3D ] x = − 24 + 72

P.I2 = ( ) =

=

P.I3 = ( ) − ) = − = −

y = C.F + P.I1 + P.I2 + P.I3

y = (A+Bx) cos +(C+Dx)sin + − + 72 + − 2.1.14 Tutorial Problems

1. Solve (D2 + 1) y = 0 given y(0) = 1, y( 2) = 0

2. Solve (D3 –D

2 – D – 1)y =o

3. Solve (D2 –4D + 4)y = 2

x

4. Solve (4D2 –4D + 1)y = 4

5. Solve (D2 – 4) y =e + e

6. Solve + 3

+ 2y = sin 3x

7. Find the particular integral of (D2 + 1) y = sin x sin 2x

8. Solve (D − 4D + 4 y) = + cos2 x

9. Solve

+ 2

+

= + sin 2x

10. Solve

− 5

+ 6y = + 3

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11. Solve D2 (D

2 + 4) y = 96 x

2

12. Obtain the P.I. of (D -2D +5) y = cos 2x

13. Solve (D + 5D + 4) y = sin 2

14. Solve (D2 +1) y = x sin

15. Solve − 6 + 9y = 6 + 7e - log 2

16. Solve (D2 –4D + 4)y = + cos 4x +

2.2 Method Of Variation Of Parameters

This method is very useful in finding the general solution of the second order equation.

+ a

+ y = X ….(1)

The complementary function of (1)

C.F = c1f1 + c2f2 where c1, c2 are constants and f , f are functions of x.

then P.I = P f +Q f

P = -

dx

Q =

dx

∴ = c1f1 + c2f2 + P.I

2.2.1 Problems Based On Method Of Variation Of Parameters

1. Solve +y = cosec x by using method of variation of parameters. Solution:

Given: (D2 +1) y = cosec x

The auxiliary equation is (m2 +1) = 0

m = ±i ∴ C. F. = c cos x + c sin x

= c1f1 + c2f2 Here, f1 = cos

= − sin f2 = sin

= cos x X = cosec x

f f − f f = cos x + sin x = 1 P = - dx

= -

= -x Q = dx

=

= log (sin x)

P.I. = Pf1 + Qf2

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MA6251 MATHEMATICS-II = -x cos x + log (sin x)

y = c cos x + c sin x − x cos x + log (sin x)

2. Solve (D2 +1) y = cosec x cot x by the method of variation of parameters.

Solution: Given: (D

2 +1) y = cosec x

The auxiliary equation is (m2 +1) = 0

m = ±i ∴ C. F. = C cos x + C sin x

= c1f1 + c2f2 Here, f1 = cos

= − sin f2 = sin

= cos x X = cosec x cot x

f f − f f = cos x + sin x = 1 P = - dx

= -

= - cot

= - log (sin x) Q =

=

= = (

dx

dx dx

) dx = cosec x − dx

= - cot x –x P.I. = Pf1 + Qf2

= - log (sin x) cos x + [-cot x – x] sin x = - cos x log (sin x) –[cot x + x] sin

x y = C.F. + P.I y = C cos x + C sin x - cos x log (sin x) –[cot x + x] sin x

2.2.2. Tutorial Problems

1. Solve + 4y = 4 tan 2x by using method of variation of parameters

2. Solve (D2 + 4) y = sec 2x by the method of variation of parameters

3. Solve (D2 –4D + 4)y = by the method of variation of parameters

2.3 Differential Equations For The Variable Coefficients

(Cauchy’s Homogeneous Linear Equation) An equation of the form

x + a x + a x + … + a y = f(x) where a , a , … a are constants and f(x) is a function of x.

It can be reduced to linear differential equation with constant coefficients by putting the substitution

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x = (or) z = log x

now,

=

x

=

=

x

= D’y where D’ =

2.3.1 Problems Based On Cauchy’s Type

.1. Solve − x

+ y = 0

Solution: Given: [x

2 D

2 –xD +1] y =0

Put x= ez

log x = z

xD = D’

x2D

2 = D

’ (D

’ – 1)

[D’ (D

’ – 1) - D

’ + 1] y =0

[(D’)2 -2D

’ +1] y =0

auxiliary equation is − + 1 = 0 (m-1)

2 =0

m = 1,1

y= (Az +B) ez

= x(A logx +B) 2. Solve x + = 0. Solution: Given:

xD2 y + Dy = 0

[ xD2 + D] y = 0

Put x= ez

log x = z

xD = D’

x2D

2 = D

’ (D

’ – 1)

[D’ (D

’ – 1) - D

’ ] y =0

[(D’)2 -D

’ +D’] y =0 ( ,)

2 y = 0 The auxiliary equation is =0

m = 0,0.

y = (Az +B) e0z

y = (Az +B) y = (A logx +B)

2.3.2 Problems Based On Legendre’s Linear Differential Equation

(Equation Reducible To Linear Form )

An equation of the form

(ax +b)n + k1 (ax +b)

n-1 + … kn y =0

Where k’s are constants and Q is a function of x is called Legendre’s linear differential equations. Such equations can be reduced to linear equations with constant coefficients by

putting ax + b = ez

z = log(ax +b)

If D’ = , Then (ax+b) D = aD’

( ax+b)2 D

2 = a

2 D’ (D’ -1) and so on.

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1.Transform the equation

(2x +3)2

- 2(2x+3)

-12y =6x

into a differential equation with constant coefficients. Solution :

Given: (2x +3)2

- 2(2x+3)

-12y =6x

(2x +3)2 D

2 - 2(2x+3) D -12y =6x

Let 2x +3 = log(2x +3) = z

2x = ez – 3

x = − let (2x+3) D =2D’

(2x+3)2 D

2 = 2

2 D’ (D’ -1)

= 4 D’ (D’ -1)

[4 D’ (D’ -1) – 4D’ -12] y = 6[ − ]

[(D ) − − ] =

[3 − 9]

2. Solve (1+x)2 + (1+x)

+y = 2 sin [log (1+x)]

Solution : D + (1+x) D + y] = 2 sin [log (1+x)]

Given: [(1+x)2

Put 1+x = ez

log(1+ x) = z

(1+x) D = D’

(1+x) D = D’(D’ -1)

[D’ (D’ -1) + D’ +1] y = 2sin z

[ D − + + 1] = 2sin z

((D’)2 +1)y = 2sin z

The auxiliary equation is + 1 = 0

C.F = A cos z + B sin z m = ± i

= A cos (log(1+ x)) + B sin (log(1+ x) )

P.I. = 2 sin z

( )

= 2 sin z ( )

= 2

sin z

( )

= 2z

sin z

= z sin

= -z cos z = - (log(1+ x)) cos (log(1+ x))

y = C.F + P.I.

y = A cos z + B sin z - (log(1+ x)) cos (log(1+ x))


1. Solve − x + y = log x

2. Solve [x2 D

2 – 2xD - 4] y = x

2 + log x

3. (x + 2)2

- (x + 2) +y = 3x +4

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2.4 Simultaneous First Order Linear Equations With Constant Coefficients

2.4.1 Simultaneous linear equations

Linear differential equations in which there are two (or) more dependent variables and a single independen Variable. such equations are known as Simultaneous linear equations. Consider the Simultaneous Equation in two dependent variables x and y and one independent variable.

f1(D) x + g1(D) y = h1(t) … (1) f2(D) x + g2(D) y = h2(t) … (2)

where f1 ,f2,g1,g2 are polynomials in the operator D.

The number of independent arbitrary constants appearing in the general solution of the system of Differential equation (1) & (2) is equal to the degree of D in the coefficient determinant

∆ = f1(D) g1(D) provied ∆ ≠ f2(D) g2(D)

2.4.2 Problems Based Simultaneous First Order Linear Equations With Constant Coefficients 1. If D = , how many arbitrary constants are involved in the solution of the Simultaneous

Equations + 2x + 3y = 2e dy dt + 3x + 2y = 0

Solution : Given: Dx+2x + 3y = 2e

(D+2)x +3y = 2e …(1) Dy + 3x + 2y = 0

3x + (D+2)y =0 …(2) The coefficient determinant of degree (1) & (2) is

∆ = D + 2 3

3 D + 2

= (D+2)2 – 9

It is an expression in D of degree 2, ∴ The number of arbitrary constants in the solution is 2.

2. Eliminate y from the system + y = − sin t , − 2 = cos .

Solution : Given: + y = − sin t

Dx +2y = - sin t … 2) × D => D2 x + 2Dy = -D(sin t)

…(3)

(1) D

2 x + 2Dy = - cos t

× 2 => 2Dy – 4x = 2 cos t

…(4)

-4x +2Dy =2cos t

(2) – (4) => (D2 +4) x = -3 cos t

If we eliminate y we get (D2 +4) x = -3 cos t


1. Solve Dx + y = sin 2t ; -x +Dy = cos 2t 2. Solve the simultaneous equations

+ 2x + 3y = 2e ,

+ 3x + 2y = 0 given that x(0) =0, and y(0) = -1

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UNIT III LAPLACE TRANSFORM

Introduction

Laplace Transformation named after a Great French mathematician PIERRE SIMON DE LAPLACE

(1749-1827) who used such transformations in his researches related to “Theory of Probability”.

The powerful practical Laplace transformation techniques were developed over a century later by the

English electrical Engineer OLIVER HEAVISIDE (1850-1925) and were often called “Heaviside -Calculus”.

3.1 Laplace Transform

Definitions

1. Transformation

A “Transformation” is an operation which converts a mathematical expression to a different but

equivalent form

2. Laplace Transformation

Let a function f(t) be continuous and defined for positive values of ‘t’. The Laplace transformation of

f(t) associates a function s defined by the equation

L[ f ( t )] ( s ) F ( s )

e st

f ( t ) dt

0

Here, F(s) is said to be the Laplace transform of f(t) and it is written as L[f(t)] or L[f]. Thus F(s) = L(f(t))

L[ f ( t )] st

f (t ) dt , t 0

e

0

3. Exponential Order

A function f(t) is said to be of exponential order if

Lt e st

f (t) 0 t

3.1.1 Laplace Transform – Sufficient Conditions For Existence

i) f(t) should be continuous or piecewise continuous in the given closed interval [a, b] where a > 0.

ii) f(t) should be of exponential order.

Example:

1. L[tan t] does not exist since tan t is not piecewise continuous. i.e., tan t has infinite number of

infinite discontinuities at , 3

, 5

,... 2 2 2

3.1.2 Problems Based On Laplace Transform – Sufficient Conditions For Existence

1. Show that t2 is of exponential order.

Solution:

Given f(t) = t2

By the definition of exponential order,

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Lt e st

f (t ) Lt e st

t2

t t t2

" Apply L'Hospital rule

Lt i.e.,Indeterminant form#

t est

$

2t " Apply L'Hospital rule

Lt i.e.,Indeterminant form#

t sest

$

Lt 2

t s 2

est

2

t

Hence t2 is of exponential order.

2. Show that the function the following function is not of exponential order f (t ) et2

Solution:

Given f(t) = et2

By the definition of exponential order,

Lt e st et 2 Lt e st t2 n n

e

Lt e st et2 n

3.1.3 Define function of class A

A function which is sectionally continuous over any finite interval and is of exponential order is

known as a function of class A

3.2 Transforms Of Elementary Functions- Basic Properties

Important Results

1. L[1] 1

where s 0

S

2. L[t n ]

n!

where n 0,1,2,...

sn1

3. L[tn ]

n 1

where n is not a integer

sn1

4. L[ eat

] 1 where s > a or s - a > 0

s a

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MA6251 1

MATHEMATICS-II

5. L[ eat

]

where s + a > 0

s a

6. L[sin at] a

where s > 0

s 2 a

2

7. L[cos at]

s

where s > 0

s 2 a

2

8. L[sinh at]

a

wheres >

a

or s2 > a

2

s 2 a

2

9. L[cosh at] s where s2 > a

2

a2

s 2

10. Linearity property L[ af (t) bg(t)] a L[f(t)] bL[g(t)]

3.2.1 Problems Based On Transforms Of Elementary Functions- Basic Properties

1. Find L[t5 + e

3t + 5e

-2t]

Solution:

L[t 5 e

3 t e

2 t ] L[t

5 ] L[ e

3 t ] L[5e

2t ]

5! 1

s 5 1 s 3

L[t 5 e

3 t e

2t ] 5! 1

s 3

s 6

5 s

2

5 s 2

2. Find L[sin 2t+ cos πt – 8cosh 7t +sinh bt]

Solution:

L[sin 2t cos t 8cosh 7 t sinh bt]

2

s

8s

b

s 2 2

2 s

2

2 s

2 7

2 s

2 b

2

2

s

8s

b

s 2 4 s

2

2 s

2 49 s

2 b

2

1 "

3. Find L #

t $

Solution: 1 "

L t 1 "

Given L # 2

t $ #

$

1 2

1

s1

21

1

2

s 1

2

s

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MA6251 1 "

MATHEMATICS-II

Hence L #

s

t $

3.2.2 First Shifting Theorem

If L f t " F s , then L[ eat

f ( t )] F ( s a)

$

If L f t " F s , then L[ e at

f ( t )] F ( s a)

$

3.2.3 Second Shifting Theorem

If L f t " F s and G(t) f (t a), t a

$ 0, t a

then L[G(t)] e as

F ( s)

3.2.4 Problems Based On First And Second Shifting Theorem

1. Find L[tn e

-at]

Solution:

L[t n

e

at n "

] L[t ]$s (s a)

n! "

#

n1

s $s (s a)

L[t n eat ] n! "

n1 # ( s a) $

2. Find L[eat

sinh bt]

Solution

L[ e at

sinh bt]

L[ eat

sinh bt]

L[sinh bt] s (s a) b "

#

2

b

2

s $

s(s a)

b

(s a)

2 b

2

3.2.5 Tutorial Problems: 1. Find L[cos 4t sin 2t]

2. Find L[sinh22t]

3. Find L[cos(3t-4)

4. Find L[e-t

t9]

3.3 Transforms Of Derivatives And Integrals Of Functions

Properties:

L[ f '( t )] s L[ f ( t )] f (0)

L[ f ''(t )] s2 L[ f (t )] s f (0) f '(0)

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3.3.1 Transform of integrals

1

t "

If L[ f (t )] F (s), then L f (u) du # L[ f (t)]

0 $ s

3.3.2 Derivatives of transform

If L[ f (t )] F (s) then L[ t f ( t )]

d F ( s ) F '(s)

ds

L[ f (t )] F (s) then L[t n

If f (t )] (1)n F

( n)

(s)

3.3.3 Problems Based On Derivatives Of Transform

1. Find L[t sin at]

Solution:

We know that d

If L[ f ( t )] F (s) then L[ t f ( t )] F ( s ) F '(s)

d

ds

L[t sin at ] L[sin at]

ds

d a "

#

ds 2

a 2

s $

(s

2 a

2 )(0) a(2s)

s

2

a

2

2

!

a(2 s)

s

2 a

2

2

!

L[t sin at]

2 a s

s 2 a2 2

2. Show that e t t cos t dt 0

0

Solution:

Given e t t cos t dt [ L[t cos t]]s 1

0

d

L(cos t)

"

#

ds

$s1

d s "

#

2

ds s 1

!$s1

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( s

2 1)(1) s(2s)

" "

# #

s2 12 # #

$ $s1

" " 1 s

2

" "

s2 1 2s

2 # # # # [ 0]

s2 12

s 2 12 # # # #

$ $ s 1 $ $s1

e t t cos t dt 0

0


1. Find L[t sin2t]

2. Find L[t cos at] 3. Find L[t sin3t cos2t]

3.3.5 Problems Based On Integrals Of Transform

sin 3t cos t "

1. Find L #

t

$

Solution:

sin 4t sin 2t "

sin 3t cos t "

Given L # L #

t

2t

$ $

1 sin 4t sin 2t "

L

#

2

t

$

1 sin 4t " sin 2t " "

L

# L

# #

2

t

t

$ $ $

1 1 s 1 s sin at " 1 s

cot

cot

L

# cot

2

!

! !

4 2 t $ a ! !

sin 3t cos t " 1 1 s 1 s

L

#

cot

cot

t 2

!

! !

$ 4 2

cos at cos bt " 2. Find L #

t $ Solution:

cos at cos bt Given L

t

"

# L[cos at

$ s

s

2 a

2

s s

1

log s 2

2

cos bt ] ds

s "

# ds

s

2 b

2

$

a 2 log s 2 b2 " $s

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MA6251

1 2 a 2 "

log s

#

2 b

2

2 s $s

1 2 a 2 "

0 log s

#

2 b

2

2 s $

cos at cos bt " 1 s 2 a

2 "

L

#

log

#

t

s

2

b

2

$ 2 $

3.3.6 Tutorial Problems: t 1 "

1. Find L e (cosh 2t

sinh 2t #

2

$

1 cos 2t

2. Using Laplace transform prove that

dt

t 2

0

3.4 Transforms Of Unit Step Function And Impulse Function

MATHEMATICS-II

log 1 0

3.4.1 Problems Based On Unit Step Function (Or) Heaviside’s Unit Step Function

1. Define the unit step function.

Solution:

The unit step function, also called Heaviside’s unit function is defined as 0 for t a

U (t a)

1 for t a

This is the unit step functions at t = a. It can be also denoted by H(t-a).

2. Give the L.T of the unit step function.

Solution:

The L.T. of the unit step function is given by

L[U ( t a )] e st

U (t a) dt 0 a

e st

(0)dt e st

(1)dt 0 a

e st

dt

a

e

st "

#

s

$s sa "

e

0

#

s $

e

as "

L[U (t a)] #

s

$

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1. Give the L.T. of the Dirac Delta function

2. Find the L.T. of t U(t-9).

3. Find L-1

[1]

3.5 Transform Of Periodic Functions

Definition: (Periodic)

A function f(x) is said to be “periodic” if and only if f(x+p) = f(x) is true for some value of p and

every value of x. The smallest positive value of p for which this equation is true for every value of x will

be called the period of the function. The Laplace Transformation of a periodic function f(t) with period p given by

p

11e ps e

st f (t) dt

0

3.5.1 Problems Based On Transform Of Periodic Functions

1. Find the Laplace Transform of the Half-sine wave rectifier function

sin t , 0 t

f (t)

2

0, t

Solution:

We know that

1 2

L[ f (t)] e

st f (t) dt

1 e

2 s

0

1

st

L[sin t ]

e

sin t dt 0

1

2 s

e

0 !

1

e

st [ s sin t

2 s

1 e s

2 2

1 e s

1 e 2 s

s2

2 !

1 e

s

1 e 1 e (s2 2 )

s s

1 e

s

(s2 2 )

L[sin t]

( s 2

2 ) 1 e

s

cos t]

!0

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1. Find the Laplace Transform of triangular wave function

t , 0 t a

f (t)

f (t 2a) f (t)

2a t , a t 2a with

2. Find the Laplace transform of the square wave function (Meoander function) of period ‘a’ defined as

a

1 , 0 t

2

f (t)

a

1,

2

t a

3.6 Inverse Laplace Transform

a. If L[f(t)] = F(s), then L–1

[F(s)] = f(t) where L–1

is called the inverse Laplace transform operator.

b. If F1(s) and F2(s) are L.T. of f(t) and g(t) respectively then

L 1 [C1 F1 (s) C 2 F2 (s)] C1 L

1 [ F1 (s)] C 2 L

1[ F2 (s)]

Important Formulas

1. L1 1"

# 1

s $

2. L1 1

"

tn1

# s n $ n 1

3. 1 1 " at

L

# e

s a $

4. 1 s " cosh at

L

#

2 a

2

s $

5. 1 1 " 1 sinh at

L

#

2 a

2

a

s $

6. 1 1 " 1

sin at

L

#

2

a

2

a

s $

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MA6251

s

MATHEMATICS-II

7. 1 "

cos at

L

#

2 a

2

s $

8. L1[F ( s a )] e

at f (t)

9. 1 1 " 1 at

sin bt

L

#

e

2 b

2

b

( s a ) $

10. 1 s a " at

cos bt

L

# e

( s a ) 2 b

2

$

11. 1 1 " 1 at sinh bt

L

# e

( s a ) 2

b

2

b

$

12. 1 s a " at cosh bt

L

# e

( s a )

2

b

2

$

13.

s "

1

t sin at

L1 #

s 2 a2 2

2a

$#

14.

s2 "

1

[sin at at cos at]

L1 #

s 2 a2 2

$# 2a

15.

1 "

1

(sin at at cos at)

L1 #

s 2 a2 2

2a3

$#

16.

s 2 a

2

"

t cos at

L1 #

s 2 a2 2

$#

17. L1[1] (t)

3.6.1 Problems based on Inverse Laplace Transform

1 2s "

1. Fi nd L

#

2

16

s $

Solution:

Given 1

2s "

s "

1

L

#

2L

#

2 16

2 16

s $ s $

2cosh 4t

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2. Find

s 3 "

1

L

#

2

4 s 13

Solution: s $

s 3 "

s 3

"

1 1

L

# L

#

2 4s 13

(s 2) 2

13 4

s $ $

1 s 3 "

L

#

(s 2) 2 9

$

1 s 2 5 "

L

#

(s 2) 2 9

$

1 s 2 " 1 1 "

L

# 5L

#

(s 2) 2 3

2 (s 2)

2 2

$ 3 $

2 t 1 s

e

L

2

2

s 3

" 5 1 3 "

#

L

#

3 (s 2) 2

3 2

$

$

2 t cos3t

5 2 t 1 3 "

e

e

L

#

3

2

2

s 3 $

1 s 3 " 2 t cos 3t

5 2t sin 3t

L

# e

e

2

4 s 13

3

s $

3.6.2 Inverse Laplace Transforms of derivatives of F(s)

If L1 [ F ( s )] f (t ), then L

1 [ F '( s )] t f (t )

t L1 [ F ( s)]

3.6.3 Problems based on Inverse Laplace Transforms of derivatives of F(s)

1 s "

1. Find L

#

(s

2

a

2

) 2

$

Solution:

s

"

Let F '(s)

#

2 a

2 )

2

(s $

F '(s)ds

s "

# ds

(s 2

a 2 )

2

$

F (s)

s "

# ds

(s 2

a 2 )

2

$

Put s2 a

2 t

2s ds dt

sds dt

2

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1 dt

1

t 2 2 2

1

2t 1

F (s)

2(s 2 a2 )

MATHEMATICS-II

1" t $#

3.6.4 Inverse Laplace Transform of Integrals

1 " 1 1 1

L F (s)ds#

f (t)

L [F (s)]

t

t

s $

(or)

L1 [F (s)] t L

1 "

F (s)ds#

s $

3.6.5 Problems based on Inverse Laplace Transform of Integrals

1. Find 1 2s "

L

#

(s

2

1)

2

$

Solution:

We know that

"

L

1 [ F (s)] t L

1

F (s) ds#

s $

1 2s " 1 2s "

L

# t L

ds#

2

1)

2

(s

2

1)

2

(s $ s $

1 1 "

t L

#

(s2 1) !

s

#

$

1 1 "

t L 0

#

s 2

1

1$

1 "

t L

#

2

s 1$

1 2s " t sinh t

L

#

2

1)

2

(s $

3.6.6 Problems based on Partial fractions method

1. Find L1

5

s 2

15

s

11"

3 # (s 1)(s 2) $

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Solution:

Consider

5s 2 15s 11

A

B

C

D

(s 1)(s 2)3

s 1 S 2 (S 2)

2 (S 2)

3

5s 2 15s 11 A(S 2)

3 B (S 1)(S 2)

2 C(S 1)(S 2) D(S1)

Put s = -1, we get

5 15 11 A( 1 2)3

9 27 A

A 1

3

Equating the coefficients of s3 on both sides, we get

0 A B B

A

B 1

3 Put s = 2, we get

21 D D 7

Put s = 0, we get

11 8A 4B 2C D 1

1

8 4 2C 7

4 8 4 2C

3 3

8 2C

C 4

5 s

2 15 s 11

1 1

4

7

3 3

(s 1)(s 2) 3 s 1 S 2 (S 2)

2 (S 2)

3

1 5s 2 15s 11" 1 1 1 " 1 1 1 " 1 1 " 1 1 "

L

#

L

#

L

# 4L

# 7L

#

(s1)(s 2) 3

3

3

2 2

(s 2) 3

$ s 1$ s $ (s 2) $ $

1 t 1 2 t 4e

2 t 1 1 " 7e

2 t 1 1 "

e

e

L

#

L

#

3

3 2 3

s $ s $

1 t 1 2 t 4e

2 t 7 2 t 1 2 "

e

e

t

e

L

#

3

3

2

3

5 s

2 15 s 11 " 1

1

7 s $

1 t 2 t 4e

2 t 2 t 2

L

#

e

e

t

e

t

3

3

3

2

(s 1)(s 2) $

3.6.7 Second Shifting property

L1[e

as F (s)] f (t a ) U (t a)

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3.6.8 Problems based on second shifting property

1. Find L1

e s

"

s 3#$

Solution:

Consider

1 1 " 3t

L

# e

s 3 $

1 e s " 3( t )

L

# e U (t )

s 3 $


a. Find the inverse L.T of Derivatives.

1. log 1 1

s2 !

2. tan1 (s 1)

b. Partial Fraction Method

s 1.

s 4 4a

4

3s 1 2.

(s 1)(s2 1)

3.6.10 Change of scale property

If L[ f (t)] F(s), then L[ f ( at ) 1 s "

F #

a

a $

1 1 1 t "

If f (t) L [F (s)], then L [F ( cs )] f

#

c

c $

3.6.11 Problems based on Change of scale property

1. If L[ f (t)] F(s) find L t "

f

#

Solution: a !$

We know that

L[ f (t)] e st

f (t) dt

0

t

t "

st

dt

L f

#

e

f

a !$ 0 a !

Put u =

t as t 0 u 0

a

du dt t u

a

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L f t a

"$ e

s(au) f (u)a du

0

a e s au

f (u) du 0

a e s at f (t) dt

0 a F [ as]


1. If L[ f (t)] F(s) then L[F(t 2) 2 F(2s)

2. 1 s "

Find L

#

2

a

2

b

2

s $

3.7 Convolution Theorem

If f (t) and g(t) are functions defined for t 0,

then L[ f ( t ) g ( t )] L[ f ( t )] L[ g ( t)]

3.7.1 Problems on Convolution Theorem

1.Define convolution

The convolution of two functions f(t) and g(t) is defined as

f (t) g(t) t f (u) g(t u) du

0

Note: Convolution Integral or Falting integral

1 1 "

2. Using convolution theorem find L

#

(s a)(s b) $

Solution:

We know that L1 [F (s) G(s)] L

1 [F (s)] L

1[G(s)]

1 1 1 " 1 1 " 1 1 "

L

# L

# L

#

s a s b $ s a $ s b $

e at ebt

Here ( ) =

( ) =

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f (t) g(t) t f (u)g(t u)du

0

t e au e b(t u) du

0

t e au e bt ebu du

0

e bt t e ( a b ) u du

0

e

bt e ( a b ) u "t

#

(a b)

$0

1

e

bt e ( a b) t

"

#

(a b) (a b) $

ebt e

at

e

bt "

a b 1 $

1 1 1 " 1

bt at "

L

#

e

e

$

a b

∴ s a s b $

3. Using convolution theorem find

1 1 "

L

#

2

1)

s(s $

Solution:

We know that L1 [F (s) G(s)] L

1 [F (s)] L

1[G(s)]

1 1 " 1 1 " 1 1 "

L

# L

# L

#

2 2

s (s 1) $ s $ s 1$

1sin t sin t 1 ( f (t) g(t) g(t) f(t))

t sin u du

0

cos uu t

u0

1

( cos t) ( 1)

1 " 1 cos t

L

#

2 1)

s(s $

4. Using convolution theorem find

1 s "

L

#

(s

2

a

2

) 2

$

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Solution:

1 s " 1 s " 1 1 "

L

# L

# L

#

(s 2

a 2 )

2 2 a

2 2 a

2 )

$ (s ) $ (s $

1 s " 1 1 a "

L

#

L

#

2

a

2

a (s

2

a

2

(s ) $

) $

cos at 1

sin

at a

1a cos at sin at

1

t cos au sin a (t u)du

a 0

1 t

cos au sin (a t au)du

a

0

1 t sin (a t au au ) sin (a t au au) "

# du

a 2

0 $

1

t sin (a t) sin a ( t 2u)du

2a 0

1

sin (a t) sin a ( t 2u)du 2a

0 1 (sin at) u cos a (t 2u) "

t

#

2a 2a!$0t

1 cos a (t 2u) "t

(sin at) u #

2a

2a

$0

1 cos at cos at "

t sin at

0 #

2a

2a !

2a ! $

1 s " 1 t sin at

L

#

(s

2

a

2

) 2

2a

$


Find the inverse Laplace Transform using convolution theorem.

4 1.

(s2 2s 5)

2

1 2.

(s2 4)(s

2 4)

s 3.

(s2 a

2 )

3

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3.8 Initial and final value theorems

3.8.1 Initial value theorem

If L[ f (t)] F(s), then Lt f (t) Lt sF (s) t 0 s

3.8.2 Final value theorem

If L[ f (t)] F(s), then Lt f (t) Lt sF (s) t s0

3.8.3 Problems based on initial value and final value theorems

1. If L[ f (t)] 1 , find Lt f (t) and Lt f (t)

s(s a)

t t0

Solution:

We know that Lt f (t) Lt sF (s)

t 0 s

Lt s 1

s(s a)

s

Lt 1

s (s a)

1

Lt f (t) 0 t0

We know that Lt f (t) Lt sF (s)

t s0

Lt s 1

s(s a)

s0

Lt 1

(s a)

s0

Lt f (t) 1

a

t

2. Verify the initial and final value theorem for the

function f (t) 1 et (sin t cos t)

Solution:

Initial value theorem states that

Lt f (t) Lt sF (s) t 0 s

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MA6251 1

MATHEMATICS-II

L[ f (t)] F(s) L[sin t cos t]s s 1

s

1 1 s 1

1)2 1

s (s 1)2 1 (s

1 s 2

(s 1)2 1

s

L.H.S Lt f (t) 1 1 2

t0

1

s 2

R .H.S Lt s

"

#

2

s s (s 1) 1$

Lt 1 s ( s 2) "

#

(s 1) 2

s 1$

s

2 (1 2 )

"

#

Lt s #

1

2

2 #

s 2

s

1

#

s

s

2 !

$

(1 2 )

"

#

Lt s #

1

2

2

#

s

1

#

s

s

2

! $

11

R.H.S 2 L.H.S

= R.H.S

Initial value theorem verified.

Final value theorem states that Lt f (t) Lt sF (s)

t s0

t "

L. H .S tLt 1 e (sin t cos t)$

1 0 1

R. H .S Lt 1 s ( s 2) "

#

1) 2

1

s0 (s $

1 0 1

L.H.S = R.H.S

Final value theorem verified.


Verify the initial and final value theorems for the functions

1. f (t) t2 e

3t

2. If F (s) s

2 5s 2

find f (0) and f ( )

s 3 4s

2 2s

3.

f (t) aebt

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3.9. Problems based on solution of linear ODE of second order with constant coefficients

1. Using L.T solve y '' 3y ' 2 y et given y(0) =1,

y '(0)

0

Solution:

y '' 3 y' 2 y e t and y(0) 1, y' 0 0

Taking L.T on bothsides, L y ''(t) 3L y'(t) 2L y(t) L e

t "

$

s 2 L[ y (t)] sy(0) y '(0) 3[sL[ y(t)] y(0)] 2L[y(t)]

1

s 1

s

2 L[ y (t)] s 0 3sL[ y(t)] 3 2L[y(t)]

1

s 1

(s

2 3s 2) L[y(t)]

1 s 3

s 1

(s 1)(s 2) L[y(t)]

s 2 2s 2

s 1

L[ y(t)] s 2 2s 2 A B C

(s 1)(s 1)(s 2) s

s 2

1 s 1

s 2 2s 2 A(s 1)(s 2) B(s 1)(s 2) C(s 1)(s1) Put s 1, we get

1 2 2 2B

3 2B

B 3

2 Put s 2, we get

4 4 2 3C

C 2

3 Put s 1, we get

1 2 2 6A

A 1

6

L[ y(t)] 1 6 3 2 2 3

s 1

s 2

s 1

1 1 3 1 2 1

6 s 1

2 s 1 3 s 2

y (t) 1 1 1 " 3

1 1 " 2 1 1 "

L # L

# L

#

6

2

3

2

s 1$ s 1$ s $

1 e t

3 e t

2 e2t

623

SCE 46 Dept of S&H

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1. Solve : y t ydt t

2 2t

0

2. Solve (D2 5D 6) y 2, given y(0) 0, y'(0) 0

3. Solve d 2 y 2 dy 2 y 0 given that y dy 1 at x 0 using L.T . method

dx 2 dx dx

SCE 47 Dept of S&H

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MA6251

UNIT-IV 4.1 Introduction: Analytic Functions

MATHEMATICS-II

ANALYTIC FUNCTIONS

4.1.1 Function of Complex Variable Many complicated integrals of real functions are solved with the help of complex

variable. They are very useful in solving large number of engineering and science problems

4.1.2 Complex Variable:

z=x +iy is a complex variable where i=√−

4.1.3 Function of Complex Variable: z=x+ i y and w=u+ iv are two complex variable. If for each value of z in a given region R of

the complex plane there corresponds one or more values of w, then w is called a function of z and it is denoted by w=f(z)=u(x, y)+iv(x, y)where u(x, y) ,v(x ,y) are real functions of the real variable x and y.

4.1.4 Single Valued Function If for each value of z in R, there is correspondingly only one value of w, the w is called

a single valued function of z.

Example:

= , =

4.1.5 Multiple Valued Function

If for each value of z in R, there is correspondingly more than one value of w, the w is called

a multiple valued function of z.

Example:

=

Neighbourhood of a Point :

4.1.6 Neighbourhood of a point is a small circular region excluding he points on the boundary

with centre at .

i.e.,| − | < ,Here is a small positive number.

4.1.7 Note:

i. The distance between two points z and is is

ii. The circle C of radius with centre at the

point

iii. | − |.

represents the interior of the circle

excluding is circumference

| − | <

| − | =

iv. represents the interior of the circle including is circumference

v. | − | >

of the circle excluding is circumference

| − | represents the exterior

vi. A circle of radius 1 with centre at the origin can be represented by| | =

4.2 Analytic function

4.2.1 Limit of The Function: Let f(z) be a single valued function defined at all points in some neighbourhood of the point Then the limit of f(z) as z approaches is .

lim = →

4.2.2 Continuity:

then lim → =

If f(z) is said to be continuous a z=

.

4.2.3 Note:

i. If two functions are continuous at the point , then their sum, product are also continuous at the point and their quotient is also continuous provided that

≠

at that point.

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ii. In real variable implies that x approaches along with x-axis (or) a line parallel to

the X-axis but in complex variable implies that z approaches along any path

→ the

joining the points z and that lie in →

z-plane .

4.2.4 Differentiability at the Point

A function is said to be differentiable at a point z = if the limit

= lim∆ → ∆

′ +∆ − exists. This limit is called derivative of f(z) at z=

4.2.5 Note:

If f(z) is differentiable at ,then f(z) is continuous at that point but converse need not true.

4.2.6 Analytic (Or) Holomorphic (Or) Regular Function A function is said to be analytic at a point if its derivative exists not only at the point but

also in some neighbourhood of that point.

4.2.7 Entire Function: A function which is analytic everywhere in the finite plane is called an entire function.

Example: , , , ℎ , ℎ .

4.2.8 The Necessary Condition For f(z) To Be Analytic:(Cauchy-Riemann Equations) i. Cartesian form: The necessary condition for a complex function

f(z)=u(x,y)+iv(x,y) to be analytic in region R are = and

. i.e. =

and = −

ii. Polar f orm

: If

is differentiable at

= −

= −

= = = − =

,then

=

and , +

,

. i.e.

and

4.2.9 Problems Based on Analytic Function-Necessary Conditions (C-R Equations)

1. Show that the function f(z)=xy+iy is continuous everywhere but not differentiable anywhere.

Solution: = +

Given

x = , =

= , and y are continuous functions ,therefore u and v are also continuous.

But, =

= , =

= , =

≠ −

≠ and

C-R equations are not satisfied. Hence f(z) is not differentiable anywhere though it is continuous everywhere

2. Show that the function f(z)= is differentiable everywhere in the complex plane.

Solution: , = +

=

Given, f(z)= = + = =

= , =

= − , =

= and = −

SCE 49 Dept of S&H

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MA6251

C -R equations are satisfied.

MATHEMATICS-II

= + = +

= +

= = +

= 3. Test the analyticity of the function f(z)=

Solution:

Let =

= u= = +

=

, =

− = −

. =

− and ,

=

4. = = −

2.10 Tutorial problems

1. Check whether is analytic everywhere.

2. of the function .

Test the analyticity=

analytic function and find its derivative

3. Prove that is an =

4.3 Harmonic and Orthogonal Propert ies Of A nal ytic Func ti ons

= ℎ

4.3.1 Laplace Equation: + = is known as Laplace equation 4.3.2 Properties Of Analytic Functions And Harmonic Conjugate

1. The real and imaginary part of an analytic function w=u+iv satisfy the Laplace Equation in two dimensions i.e. ∇ = , ∇ =

2. The real and imaginary part of an analytic function w=u(r, )+iv(r, ) satisfy the Laplace equation in polar coordinates

3. If w=u(x,y)+iv(x,y) is analytic function the curves of the family u(x,y)=a and the curves of

the family v(x,y)=b cut orthogonally where a and b are varying constants. 4. If w=u(r, )+iv(r, ) is analytic function the curves of the family u(r, )=a and the curves of the

family v(r, ) =b cut orthogonally where a and b are varying constants.

5. An analytic function with constant modulus is constant. 6. An analytic function whose real part is constant must itself be a constant

7. If f(z) and are analytic in a region D, then f(z) is constant in that region D

4.3.3 Problems Based On Properties

1. If f(z)= u+iv is a regular function of z in a domain D the following relations hold in D

i.

∇ |

|

=4

|

′

ii. =0 ,if| f(z)f’(z)

≠

in D

iii. ∇ | =p(p

|-1)

− | −

′

′|

iv. ∇ |

= | | |

v. ∇ | |

∇ |

| =0 |

′

|

= |

|

vii. + ′

vi.

|

=

| ′

|

viii. ∇ |

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MA6251

Solution:

i. Let f(z)=u+iv Here f(z) is an analytic function = and | | = | − |

=√ + | | = +

+

| |

=

+

+

=

+ +

MATHEMATICS-II

= −

+

=

+

+

+

=

=

(

) =

+ (

)

= ( ) = + ( )

+

+

= (

+

) (

+ )

=2

+

+

=2 ′ +

| ∇

=

+ =2| |

| ii. | + | =0 ,if f(z)f’(z)≠ in D

|

| =

+

+

|

=

+

) =

| (

+

+

+

+

+

|

|

=( + ) + −

+

|

| =( + )( + + + )−( + )( + )

( + )(

+

+

+ + + +

+)

+

|

|

+

= − ( + ) − +

+

= ( + ) + + + − (− + ) − ( + )

+

∇

= ( + ) + + + − (− + ) − ( + ) =0

Let

=p(p-1)

−

|

′

|

+

iii.

f(z)=u+iv is an analytic function

SCE 51 Dept of S&H

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= and

= −

+

=

+ = + =

f ’ +

| ’(z)=

=

+

|

= −

= − + −

−

−

= − + −

= ( − + − − )

+ = − ( + ) + −

− ( + )

= − + − − + −

= − − | ’ |

∇ | |

|

| − | ′

|

iv. =

Let f(z)=u+iv is an analytic function

| | = +

|

| =

+

|

| =

+

+

=

+

−

++

= + −

−

+

=+

+

+

+

+

−

+

=

++

−

− +

+ + | ′ |

+

+ − + + ′ + −

+ = + − + + | |

+ − + −

+ + − ′

+

+

=

+

+

+ + + |

|

−

+ + + +

+

−

= 2 ′ +

+

−

|

|

+

−

+

−

+ ( + )

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= +

−

2

|

′

| + − +

−

+

2 ′ ′

= + − | ′ | + − + − | |

= + − | ′ | + −

= + − − | ′ |

= | | | |

v. ∇ =0

= − = −

−

=

+

−

= +

= +

−

−

=

(

+

)

+

+

− − +

− − +

=

= + + −

− + − − +

= + − − + − + −

−

(

) =

+ ( − ) − ( − ) +

+

+ ( + )

( + ) − ( + ) − − (+ )

+

= − ++ − =

vi. |

| +

|

| = | ′ |

Let

| f(z)=u+iv is an analytic function

| = +

|

| =

+

|

| =

+

√ +

= √ + +

| | = +

√ +

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| | = +

√ +

| | +

| |

= + + +

= √ + + + √ + + + +

= + | | = +

Let ∇ | | = | ′ | + | |

vii.

f(z)=u+iv is an analytic function =

| | =

=

=

+

=

= ( + )

=

+

| ′ + + + =2

∇ |

|

|

viii. | = | ′

=

| | =

=

=

+

=

= = ( + )

+

+ +

=2

| ′

+

|

4.3.3 Problems Based On Harmonic Conjugate

1. If f(z)= then show that u and v are harmonic function.

Solution: +

=

u =

=

Given f(z)=

=

= =

=-

=

= =

= −

−

+

= =0

Type equation here. − 54 Dept of S&H

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functions. =0

They are harmonic

+ =−

4.4 Constructions of Analytic Functions (Milne- Thomson Method) i. To find f(z) when u given,

f’(z )= + = −

, =

, , =

, f’(z)=, − ,

∫ f’ z dz = ∫, − ∫,

+ c

= ∫, − ∫,

ii. To find f(z) when v given,

f’(z)= + , =

, , =

, f’(z)=, + ,

∫ f’ z dz = ∫, + ∫,

+ c

= ∫, + ∫,

4.4.1 Problems Based on Constructions Of Analytic Functions (Milne- Thomson Method)

1. Find an analytic function whose real part is

−

Solution: −

Given, =

= + −

, =

, = +

, = = − − +

, =

− ∫, + c

= ∫,

=∫ + − ∫ + c

=

+

∫

+−+

2. Find an analytic function whose real part is = −

Solution: =

−

+ − +

Given

, = = − +

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= + 55 Dept of S&H

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MA6251

MATHEMATICS-II

, =

= − −

, =

− ∫, + c

= ∫,

= ∫ + − ∫ + c

= + +

Find an analytic function f(z)=u+iv whose + =

3. +

Solution: = +

Given

−

= −

+ + = + F(z)= +

= − = −

Given = +

, = ,

=

= ( + ) −

= −

+ +

=0

,

, = =( + ) −

= −

= =

− + +

,

, ,

+ ,

F’(z)=

∫ F’ z dz = ∫, + ∫,

=0+i

∫ − dz

= +c

(1+i)f(z)= +c

= +

+

4.4.5 Tutorial problems

=

+

1.

Show that the function is harmonic and determine its conjugate. Also

find f(z)

−

2. Find the analytic function given

function whose real part is

3. Determine the analytic = + = −

ℎ −

4.5 Conformal Mapping

4.5.1 Definition: The transformation w=f(z) is called as conformal mapping if

it preserves angle between every pair of curves through a point, both in magnitude and sense

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The transformation w=f(z) is called as Isogonal mapping if it preserves angle between every pair of curves through a point in magnitude but altered in sense

4.5.2 Standard Transformations

1. Translation:

= + , where C is a complex constant ,represents

The transformation

a translation 2. Magnification:

The transformation

=

, where C is a real constant ,represents

magnification

3. Magnification And Rotation:

The transformation

=

, where C is a complex constant ,represents

magnification and Rotation

4. Magnification , Rotation And Translation:

The transformation , where C,D are complex constant ,represents

Magnification, Rotation and Translation

= +

=

represents inversion w.r.to the unit circle | | = ,

5. Inversion And Reflection:

The transformation

followed by reflection in the real axis

4.5.3 Problems Based on Transformation

1. Find the image of the circle | | = by the transformation = + +

Solution:

+

= + +

Given

= + + +

= + =

= + + +

+

= − = −

Given, | | =

+ = − + − =

Hence, the circle + = is mapped into

in W-plane which is also a circle with centre (2,4)

and − + − =

radius 1

2. Determine the region D of the w-plane which the triangular region D enclosed by the lines x=0,y=0,x+y=1 transformed under the transformation w=2z

= +

Given , =

+

+ =

+ = +

= =

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= =

In the z plane, the line x=0 is transferred into u=0 in the w-plane

In the z plane, the line y=0 is transferred into v=0 in the w-plane

In the z plane, the line x+y=1 is transferred into u+v=2 in the w-plane

3. Find the image of | − | = under the transformation = .

Solution:

Given =

z= Here, z=x+iy and w=u+iv

z = = −i

=

+i

+

x+iy =

−i

x=

+

y= −

+ | −

|

=2

=2 +

given:

|

+

|

−

+ − = + − + =

+ − =

u

+

−v

=

−v

++ = −+ +

+ +

= −

= − which is a straight line

4. Discuss the conformal transformation w=sinz

Solution: Given, w=sinz

+ = +

+

=

=

ℎ ,

= ℎ

Elimination=

of y usingℎ+the formula,ℎ

ℎ − ℎ =

=

If

Case I: −

x=c

−

= which is a hyperbola with foci at (1,0) and (-1,0)

∴ The lines parallel to y axis in the z-plane are transformed into confocal hyperbolas. Type equation here.

Elimination of x using the formula, + =

( ℎ ) + ( ℎ ) =

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Case II: If y=c

ℎ + ℎ = which is a hyperbola ellipse with foci at (1,0) and (-1,0)

∴ The lines parallel to x axis in the z-plane are transformed into confocal ellipses. Case III: When x=0 u=0,v=sinhy If y is –ve, v is –ve

y is +ve, v is +ve If y varies from then v is varies from

Hence the image of the line x=0 in the z-plane is the complete imaginary axis in the plane.

−∞ ∞ −∞ ∞

Case IV:

When y=0 u is u=sinx ,v=0

– ve

If x is –ve,

x is +ve, u is +ve

x=0, u=0 −∞ ∞ −∞ ∞

w-plane.

then u varies from

If x varies from

Hence the image of the line y=0 in the z-plane is the complete real axis in the

Case V:

When x= , which is given by u=cushy and v=0 ∴u 1 and v=0.

Similarly th map of line x=

− is given by

= − ℎ

and

=

u and v=0

Thus the part of the u axis for which u and th part of the u axis for which u

Are the images of the lines x= and x=− respectively

−

4.5.4 Tutorial problems

1. Find the critical points of the transformation

2. Find the image of the region y the transformation

= −

=

3. Discuss the transformation >

ℎ

−

4.6 Bilinear Transformation =

4.6.1 Definition:

+

− ≠

The transformation , where a,b,c,d are complex numbers

is called a bilinear

transformation

+

=

This transformation was first introduced by A.F.Mobius. So it is also called as

Mobius transformation

4.6.2 Note:

i. Under bilinear transformation, no two points in z-plane go to the same point in w-plane ii. Bilinear transformation has atmost two fixed point or invariant point

iii. The bilinear transformation which transforms , , into , , is

− − − −

− − −

−

−

ratio −

iv. Given four points

,

in this order ,the =

− −

is called the cross

ratio of the points.,,

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v. = ++

can be expressed as cwz+dw-(az+b)=0 .It is linear in both w and z

∴ It is called as bilinear transformation. vi. Bilinear transformation is conformal only when ≠

−

≠ +

− ≠ − = very point in the z-plane is a critical point

vii. The inverse of the bilinear transformation

=

+

=

− + which is also a bilinear

transformation except

+

−

except z= corresponds to a unique point i n the w-plane

viii. Each point in the plane = −

The point z= − corresponds to a infinity in the w-plane

ix. The cross ratio of four points − −

=

− − is invariant under bilinear

transformation − − − −

x. If one of the point is the point at infinity the quotient of those difference which involve this point is replaced by 1

4.6.3 Problems based on Bilinear Transformation = −

1. Find the fixed point of +

Solution:

= +

− −

= +

− − =

± √ − +

=

= , Given: ∞, ,

onto , , ∞

respectively.

2. Find the bilinear transformation that maps the points

Solution:

=

= ,

= , = ∞

Let= ∞, = , &

−

−

the transformation be,

− − =

− − − −

−

−

=

− −

− − − −

−

−

=

− −

− −

−

−

−

=

−− −

− − − −

=

=

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= −

Solution: , , − onto

, , − respectively.

3. Find the bilinear transformation that maps the points

Given:

= ,

= −

& = , = , = −

Let= ,

1

−

the transformation be,

− − =

−

−

−

− −

+

−

+ =

−

+ − + −

− =

− +

+ − + −

+

− − =

− +

+ + − +

− − =

−

+ + −

− − =

− −

+ +

−

=

−

+ +

−

=

−

+ +

+ + − =+ − −

+ + − =− − +

= − + +

4.6.4 Tutorial problems−

+ +

+ , − , −

1.

Find the bilinear transformation that maps the points of the Z-plane into the

points of the W-plane. as the

2. bilinear transfo rmation wh ich maps onto and has

Find the, ,

invariant points. Also show that under this transformation the upper half of the Z-plane maps

onto the interior of the unit circle in the w-plane=

= − −

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CHAPTER 5 COMPLEX INTEGRATION

5.1 Prerequisites Before starting this topic students should be able to carry out

integration of simple real-valued functions and be familiar with the basic ideas of functions of a complex variable. The students should also familiar with line integrals.

5.2 Introduction Complex integration is an intuitive extension of real integration. Since

a complex number represents a point on a plane while a real number is a

number on the real line, the analog of a single real integral in the complex

domain is always a path integral. For some special functions and domains,

the integration is path independent, but this should not be taken to be the

case in general. Given the sensitivity of the path taken for a given integral

and its result, parametrization is often the most convenient way to evaluate

such integrals.Complex variable techniques have been used in a wide

variety of areas of engineering. This has been particularly true in areas such

as electromagnetic field theory, fluid dynamics, aerodynamics and elasticity.

5.3 Cauchy’s Theorem

5.3.1 Definitions

5.3.1.1Connected Region A connected region is one which any two points in it can be

connected by a curve which lies entirely with in the region.

5.3.1.2 Simply connected region A curve which does not cross itself is called a simple closed curve.

A region in which every closed curve in it encloses points of the region only is called a simply connected region.

5.3.1.3 Contour integral An integral along a simple closed curve is called a contour integral.

5.3.1.4 Cauchy’s Integral Theorem

If a function f(z) is analytic and its derivative Rf0(z) is continuous at

all points inside and on a simple closed curve c, then c f(z)dz = 0.

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5.3.1.5 Cauchy’s Integral formula

If f(z) is analytic inside and on a closed curve c of a simply connected region R and if a is any point with in c, then

f(a) = 1 R

f(z) dz

2 i c z−a

the integration around c being taken in the positive direction.

5.3.1.6 Cauchy’s integral formula for derivative

If a function f(z) is analytic within and on a simple closed curve c and a is any point lying in it, then

f (a) = 1

R

f(z)

5.3.2 Worked out examples0

2 i c (z−a)2

dz

1. Evaluate

R z

dz where c is a circle

|

z

|

= 1.

C z−2

Solution.

Let f(z) =

z = 2 lies outside c.

) f(z) is analytic inside and on c .

f0(z) is continuous inside c

R

Hence by Cauchy’s theorem

c

f(z)dz = 0

1

2. Evaluate


z

= 1.

Solution. 1

RC 2z−31 1 3 | |

Given c dz = dz z =

lies outside c.

2z−3 2 z−23 2

) f(z)

Ris analytic inside and on c .

f0(z) is continuous inside c

R

Hence by Cauchy’s theorem

c

f(z)dz = 0

1

3. Evaluate


z

= 2.

RC 2z+3 | |

Solution.

dz = Z

c

1 1

2z + 3 2(z +

3 )dz

2

1 Z 1

= 2 c z +

32 dz

= 1

22 if (−

23

)

1

= i2z + 3dz

= i.

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z

z−2

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4. Evaluate

R

sin 3z dz where c is a circle

|

z

|

= 5.

C z+

Solution. 2

R

W.K.T Cauchy’s Integral formula is f(z) dz = 2 if (a)

Given sin 3z

dz c z−a

C z+

2

−

|

|

Here f(z) = sin 3z, a = lies inside z = 5.

R 2

f(a) = sin( −23 ) = 1

) By Cauchy’s Integral formula, we get

ZC

sin 3z

dz = 2 i

z +

2

5.3.3 Tutorial Problems

R

3z2+7z+1 1

1. Evaluate

dz where c is a circle |z| =

.

C 1 z+1 2

2. Evaluate C e

dz where c is a circle | z |

= 1.

2

R 1

3. Evaluate C e


|

z

|

= 3.

z2+4

R

5.4 Taylor’s and Laurent’s Series Expansion.

5.4.1 Taylor’s Series. A function f(z), analytic inside a circle C with center at a, can be

expanded in the series

f(z) = f(a)+f0(a)(z−a)+

f00

(a)

(z−a)2+

f000

(a)

(z−a)3+· · ·+

f(n)(a)

(z−a)n+. . .

2! 3! n!

5.4.2 Laurent’s Series.

Let C1, C2 be two concentric circles |z − a| = R1 and |z − a| = R2

where R2 < R1. Let f(z) be analytic on C1andC2 and in the annular region R between them. Then, for any point z in R,

1 1 bn

f(z) = X

X

(z −

a)n

n=0 an(z − a)n

+

n=0

where

Z

1 f(z)

an =

c1

dz

2 i (z − a)n+1

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1

Z

f(z)

bn =

c2

dz

2 i (z − a)1−n

where the integrals being taken anticlockwise.

5.4.3 Worked out examples

1. Expand ez in a Taylor’s series about z =

0 Solution.

Function Value at z = 0

f(z) = ez f(0) = 1

f0(z) = e

z f’(0) = 1

f00

(z) = ez f”(0) = 1

f000

(z) = ez f”’(0) = 1

. . . . . .

) Taylor’s series about z = 0 is

f(z) = f(0) + f0(0)(z) +

f00(0)(z)2 +

f000(0)(z)3 + . . .

2! 3!

= 1 + 1(z) + 2!1

(z)2 + 3!

1(z)

3 + . . .

2. Expand z−12 at z = 1 in a Taylor’s

series. Solution.

Function Value at z = 1

f(z) = 1 f(1) = -1

z−2

f0(z) = −1 2 f’(1) = -1

(z−2)

f00

(z) = 2

f”(1) = -2

(z−2)3

f000

(z) = −6 4 f”’(1) = -6

. (z−2) .

. .

. .

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) Taylor’s series about z = 1 is

f(z) = f(1) + f0(1)(z) +

f00(1)(z)2 +

f000(1)(z)3 + . . .

2! 3!

= −1 + (−1)(z − 1) + −

2!2

(z − 1)2 + −

3!6

(z − 1)3 + . . .

= −1 − (z − 1) − (z − 1)2 − (z − 1)3 − . . .

3. Expand f(z) = 1

as a Laurent’s series in powers of z and state the

z(z−1)

respective region of validity.

Solution. Given f(z) =

1 .

z(z−1) f(z) is not analytic at z = 0 and z = 1. But it is analytic in the region

1)0 < |z| < 1(Deleted disc) 2) |z| > 1

Case i) For all z in 0 < |z| < 1

1 = −1

z(z − 1)

z(1 − z)

= −z1 [1 + z + z2 + . . . ]

1

+ z + z2 + . . . ]

= −[

z

Case ii)

For all z in |z| > 1 we have | 1

| < 1.

z

1

=

1 1

z(z − 1) z z − 1

1 1

]−1

=

[1 −

z2

z

1 1 1

=

[1 +

+

+ . . . ]

z2 z z2


1. Expand f(z) = cos z about z = in Taylor’s series.

3

2. Expand f(z) = sin z about z = in Taylor’s series.

4 1

in the region 1 < |z| < 2

3. Find Laurent’s series expansion of

z2+3z+2

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5.5 Singularities

5.5.1 Definitions

5.5.1.2 Zeros of an analytic function:

If a function f(z) analytic in a region R is zero at a point z = z0 in R

then z0 is called a zero of f(z).

5.5.1.3 Simple Zero:

If f(z0) = 0 and f0(z0) 6= 0 then z = z0 is called a simple zero of f(z)

or a zero of the first order.

5.5.1.4 Zero of order n: An analytic function f(z) is said to have a zero of order n if f(z) can be

expressed as f(z) = (z − z0)mφ(z) where φ(z) is analytic and φ(z0) 6= 0

5.5.1.5 Singular Points:

A point z = z0 at which a function f(z) fails to be analytic is called a singular point. 5.5.1.6 Entire function

A function f(z) which is analytic everywhere in the finite plane is called an entire funcction. 5.5.1.7 Meromorphic function

A function f(z) which is analytic everywhere in the finite plane except at finite number of poles is called a meromorphic function.

5.5.2 Types of Singularities

5.5.2.1. Isolated Singularity

A point z = z0 is said to be isolated singularity of f(z) if 1. f(z) is not analytic at z = z0

2. There exist a neighbourhood of z = z0 containing no other singularity. 5. 5.2.2. Removable singularity:

If the principal part of f(z) in Laurent series expansion of f(z) about

the point z0 is zero then the point z = z0 is called removable singularity. 5. 5.2.3. Pole:

If we can find a positive integer n such that limz!a(z − a)nf(z) 6= 0 then z = a is called a pole of order n for f(z). 5. 5.2.4. Essential singularity:

If the principal part of f(z) in Laurent series expansion of f(z) about

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the point z0 contains infinite number of non zero terms then the point z =

z0 is called essential singularity.

5.5.3 Worked out Examples

1. Find the zeros of zz

33−

+11 . Solution.

The zeros of f(z) are given by f(z) = 0

That is z3 − 1 = 0.

) z = 1, ! , !2

2. Find the zeros of sin z3−z .

Solution. z

The zeros of f(z) are given by f(z) = 0 That is

f(z) = sin z − z

z3

−z

3 + z

5 − . . .

= 3! 5! z

3

= −1

+ z2

− . . . 3! 5!

Now,limz!0 sin

zz

3−z

6= 0

3. What is the nature of the singularity z = 0 of the function Solution.

sin z−z

z3

Given f(z) = sin z3−z z

the function f(z) is not defined at z = 0.

By, L’Hospital rule

lim sin z − z = lim cos z − 1 z!0 z

3 z!0 3z

2

= lim −sin z z!0 6z

= −1

6

Since the limit exist and is finite, the singularity at z = 0 is a removable singularity.

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4. Find the nature of the singularity at z = 0 of f(z) = sin z

z

Solution.

Given f(z) = sin z

z

the function f(z) is not defined at z = 0.

By, L’Hospital rule

lim sin z = lim cos z

z 1

z!0 z!0

= 1

Since the limit exist and is finite, the singularity at z = 0 is a removable singularity.

5. Classify the singularity of the function f(z) = zz−

22

sin(z−11 ) Solution.

Poles of f(z) are z = 0, 0. That is z = 0 is a pole of order 2.

Zeros of f(z) are obtained by, (z − 2) sin(z−11 = 0

=) z − 2 = 0 and sin(z−11 ) = 0

=) z = 2andz = n1

+ 1, n = 0, 1, 2, . . . The limits of zeros is 1. Therefore z = 1 is isolated and essential singularity.


1. Find the singular point of f(z) = sin(z−1

a )the nature of the singularity.

2. Find the singular points of f(z) = 1

z(ez−1)

e z

3. Classify the nature of singularities of the functions

2

z +4

5.6 Residues

5.6.1 Definitions: If z = z0 is an isolated singular point of f(z), we can find the

Laurent’s series of f(z)aboutz = z0 1 n 1 bn

f(z) = n=0 an(z − z0) + n=1 (z−z0)n

fficient of 1 in the above expansion is called the residue of f(z) at

The coeP z−z0 P

z = z0

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5.6.1.2 Worked out Examples

e2z

1. Calculate the residue of f(z) = 1−

z3

Solution.

e2z

Given f(z) = 1−

z3

Here z = 0 is a pole of order 3.

1 lim d2 z3 1 e2z

Res at(z = 0) =

dz2

−z3 &

2! z!0

1 ! d

−2z

= 1 lim 2e2z

dz

2! z 0

=

lim

− 4e

2! z!0

= −2

2. Find the residue of f(z) = Solution

Given f(z) = z 2

(z−1) Here z = 1 is a pole of order 2.

Res at(z = 1) = 1 lim d (z

−

1)2 z

&

dz (z − 1)2

1! z!1

= lim d

z z!1 dz

= 1

5.6.1.3 Tutorial Problems

1. Calculate the residue of f(z) = e2z

at its pole.

(z+1) 2

2. Find the residue of

z3

(z−1)4(z−2)(z−3)

5.6.2 Cauchy Residue Theorem

If f(z) be analytic at all points inside and on a simple closed curve c, except for a finite number of isolated singularities z , z , z , . . . then

R 1 2 3

c f(z)dz = 2 i [sum of residues].

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z

2 at its pole. (z−1)


5.6.2.1Worked out Examples

R

1. Evaluate c z2ez

1 where c is |z| = 1. Solution.

Let f(z) = z2ez

1

Here z = 0 is the only singular point which lies inside c.

1 1 ( 1 )

2

2 2

z

z

f(z) = z

ez = z 1 +

+

+ . . . &

1! 2!

= z

2 + z +

1 +

1 1 + . . .

2 6 z

The residue at z = 0 is 1

6

c f(z)dz = 2 i(sum of residues)

By Cauchy residue theorem,

R z2

R

c f(z)dz = i

3

2. Evaluate R

e dz where c is |z| = 1.

c cos z

Solution R ez

2

Let f(z) = dz

c cos z

The singular points are obtained by

cos z = 0

z = (2n + 1)

, n = ±1, ±2, . . .

2

1 3 1

= ±

, ±

, ±

, . . .

2 2 2

1

1

1

φ( 1 )

2

Here z = 2 and z = −2 lies inside c. R1 f(z), 2 = 1

f(z) = ez

2

= φ(z)

0 ( 2 )

cos z (z)

R1 f(z),

= 1

1

e

4

R2 f(z), − 1 =−e 4

2

2

1

By Cauchy residue theorem, c f(z)dz = 2 i(sum of residues)

R

residues) = 0.

c f(z)dz = 2 i(sum of

R


1. Evaluate

c dz

where c is |z| = 4.

sin z

2. If c is a

circle | z | = 3, evaluate R tan zdz.

c

R

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5.7 Evaluation of real definite Integrals as contour integrals.

5.7.1 Contour Integration:

The complex integration along the scro curve used in evaluating the definite integral is called contour integration. Here we are going to see under three types. They are

1. Type I - Integrals of the form

R

2

0

function in cos ✓ and sin ✓.

2. Type II - Integrals of the form R 1

3. Type III - Integrals of the form

−1

R1−1

5.7.2 Type I

5.7.2.1 Worked out examples

f(cos ✓, sin ✓)d✓ where f is rational

P (x) dx.

R−

11 f(x) sin(nx)dx.

Q(x)

f(x) cos(nx)dx or

1. Evaluate R

2 d✓

.

0 2+cos ✓

Solution.

Let z = ei✓

dz = izd✓

cos ✓ = 1 z + 1 and sin ✓ = 1 z 1

z

− z

Now, 2 2i

Z

2 d✓

Zc

1

dz

=

where c is z

= 1. 1 z

2+1

| |

0

2 + cos ✓

2 +

iz

2 z

2

Zc

dz

=

i z2 + 4z + 1

p z

2 + 4z + 1 = 0 =) z = −2 ± 3

p ↵ = −2 + 3 is simple pole and lies inside c

and β = −2 p

− 3 is simple pole and lies outside c.

Z

2 ✓ 2

Zc

dz

0

d

=

2 + cos ✓ i (z − ↵)(z − β)

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Res[f(z), ↵] =

lim(z −

↵) 1

(z −

↵)(z −

β)

z ! ↵

1

=

↵ − β

1

= 2 p

3

Hence by Cauchy Residue theorem,

c f(z)dz = 2 i (sum of residues) = i

p

3

) R

2 d✓

=

2

R

p

0 2+cos ✓

3

2 d✓

2. Using contour integration evaluate

Rc

13+5 sin ✓.

Solution.

Let z = ei✓

dz = izd✓ cos ✓ =

12 z

+ Now,

Z 2

0

1

and sin ✓ =

1

z −

1

z 2i z

d✓ = 1 dz where c is z = 1. 2

2

| |

13 + 5 sin ✓ Zc 13 + 521i z z−1 iz

=

Z

dz

5 c z2 +

26

iz − 1

5

z2 +

26 iz − 1 = 0 =) z = −5i, −5i

5

↵ = −5i is simple pole and lies inside c and β = −5i is simple pole and lies

outside c.

2 d✓ 2 dz

R0

=

Rc

13+5 sin ✓ 5 (z−↵)(z−β)

Res[f(z), β] = lim(z

− β)

1

(z −

↵)(z −

β)

z ! β

1

=

β − ↵

=

5

24i

Hence by Cauchy Residue theorem, f(z)dz = 2 i (sum of residues) = 5

2 d✓ 2 5 R

c 12

) R

0

=

(

) =

13+5 sin ✓ 5 12 6

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1. Evaluate the integral 2

sin2 ✓ d✓.

R 0 5+4 cos ✓

d✓

2. Evaluate

, aR

> 0.

0 a2+sin

2 ✓

5.7.3 Type II


1. Evaluate R

−1 1

x2dx

using contour integration.

(x2+1)(x

2+4)

Solution.

Let us consider R

c f(z)dz = R

c

z2dz

(z2+1)(z

2+4)

where c consist of the semi circle": |z| = R and the bounding di-

ameter [−R, R].

Now, c f(z)dz = −R

R f(x)dx + ! f(z)dz + 4) = 0

The poles of f (z) are obt ained by (z2 + 1 )(z 2

R R R

i.e., z = i, −i, 2i, −2i. where z = i, 2i are simple poles lie inside"and z = −I, −2i are simple

poles lie outside"

R1[f(z), i] = lim(z − i)f(z) z!i

z2

= lim(z − i)

z!i(z + i)(z − i)(z2 + 4)

z2

= lim

z!i (z + i)(z2

+ 4) i2

= (i + i)(i

2 + 4)

= −1 6i

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R2[f(z), 2i] =

lim (z

−

2i)f(z)

z!2i z2

= lim(z −

2i)

(z + 2i)(z −

2i)(z2 + 1)

z !

i

z

2

= lim

(z + 2i)(z2 + 1)

z!i

(2i)2

= (2i + 2i)((2i)

2 + 1)

= 31

i

Hence by Cauchy’s Residue theorem, Z

f(z)dz = 2 i[R1 + R2] c 6i 3i

&

= 2 i −1 +

1

= 2 ✓ 6 ◆

1

=

3

i.e., R

−R

R f(x)dx + R

! f(z)dz = 3 .

when R ! 1, the semi-circle becomes very large and the real and imagi-nary parts of any point lying on the semi-circle becomes very large so that |z| ! 1

) R ! 1 then ! f(z)dz ! 0.

1 = .

3

i.e., −1 f(x)dx R2

Hence

R

1 x dx =

+4) 3

R −1 (x2+1)(x

2

1 dx 3

2. Show that R

−1

=

.

(x2+1)

3 8

Solution.

R

R dz

Consider c f(z)dz = c

(z2+1)

3

where c is the upper half of the semi circle"with the bounding diam-

eter [−R, R].

R−

By Cauchys Residue Theorem we have, c f(z)dz = R Rf(x)dx + ! f(z)dz

The poles of f(z) are obtained by (z2 +

R1)

3 = 0 R

z = −i

i.e., z = i, −i,. where z = i is a pole of order 3 which lies inside"and

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is a pole of order 3 which lies outside".

Res [f(z), i] = lim 1 d

2 [(z

− i)

3f(z)]

2! dz2

z!i

&

=

lim 1 d2 (z

−

i)3

1

(z + i)3(z − i)3

z!i 2! dz2

lim 1 d2

1

= 2 dz2 (z + i)3

&

z!i

= lim 1 12

2 (z + i)5

z!i

= 3

16i

Hence by Cauchy’s Residue theorem, Z

f(z)dz = 2 i [sum of residues]

Zc

f(z)dz = 2 i[R1 + R2] c 3 &

= 2 i 16i

= 3

8

i.e., R

−R

R f(x)dx + R

! f(z)dz = 38 .

when R ! 1, the semi-circle becomes very large and the real and

imaginary parts of any point lying on the semi-circle becomes very large so that |z| ! 1

) R ! 1 then ! f(z)dz ! 0.

1 = 3 .

8

i.e., −1 f(x)dx R

Hence

R

1 dx dx = 3

R −1 (x2+1)

3 8


1. Evaluate 1 dx

, a > 0, b > 0.

R 0 (x2+a

2)(x

2+b

2)

dx

2. Evaluate R

01

.

x4+10x

2+9

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5.7.4 Type III


1. Evaluate R

1 cos(ax)

dx, a > 0.

0 x2+1

Solution.

W.K.T 1 cos(ax)

dx = 1 1 cos(ax)

dx

0 x2+1 2 −1 x

2+1

To find R

1 cos(ax)

dx

R

x2+1

considerR

−1

Z

Z

1 cos(az)

c f(z)dz =

1

−1

dz

2 z2 + 1

= R.P Z

eiaz

c

dz

z2 + 1

Where c is the upper half of the semi-circle"with the bounding

diam-eter [−R, R] By Cauchy’s residue theorem,

R R R R

c f(z)dz = −R f(x)dx + ! f(z)dz The

poles of f(z) are obtained by (z2 + 1) = 0

i.e., z = i, −i,. where z = i is a simple pole which lies inside"and z = −i is a simple

pole which lies outside"

Res [f(z), i] =

lim(z

−

i)f(z)

z!i eiaz

= lim(z

−

i)

z!i z2 + 1

= lim (z − i)(ia)eiaz + e

iaz

z!i 2z

= e−a

2i

Hence by Cauchy’s Residue theorem,

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Z

f(z)dz = R.P 2 i [sum of residues] c

= R.P 2 ie−a

2i

= R.P e−a

R

) i.e., Z

−R f(x)dx + Z

! f(z)dz = e−a

If R ! 1, then

Z! f(z)dz ! 0

1

Z

−1 f(x)dx = e−a

1 cos(ax)

Z0

dx =

e−a.

x2 + 1 2

2. Show that R

01

sin x

dx =

x 2

Solution.

W.K.T 1 sin(x) dx = 1 1 sin(x) dx

0 x 2 −1 x

To find R

1 sin(x)

R

x

considerR

−1

sin(z)

Zc f(z)dz =

Zc

dz

z

I.P Z

c

iz

=

e

dz

z

Where c is the upper half of the semi circle"with the bounding diameter R R R R

[−R, R]. i.e., c f(z)dz = I.P −R f(x)dx + ! f(z)dz The poles of f(z) are obtained by z = 0

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z = 0 is a simple pole lies on the real axis inside " .

Res [f(z), 0] = lim(z)f(z) z!0

= lim(z)eiz

z!0 z

= lim eiz

z!0

= 1

Hence by Residue theorem

Z

f(z)dz = 2 i(0) + i(1) c

1

I.P Z

ix

e

dx = I.P i

−1 x

1 sin x

Z−1

dx =

x

i.e., Z

1 sin x

dx =

0 x 2


1 cos(ax) [e−

ab

1. Show that

dx =

(1 + ab)]

0 (x2 +b

2)2 4b

3

R (1+x2)

Prove that R

01

2.

log

dx = log2

1+x2

5.8 Applications:

Blasius Theorem. The following figure shows a cross-section of a cylinder (not necessarily cir-

cular), whose boundary is C,placed in a steady non-viscous flow of an ideal

fluid; the flow takes place in planes parallel to the xy plane. The cylinder is

out of the plane of the paper. The flow of the fluid exerts forces and turning

moments upon the cylinder. Let X, Y be the components, in the x and y

directions respectively, of the force on the cylinder and let M be the

anticlockwise moment (on the cylinder) about the origin.

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R R

Blasius theorem states that X−iY = 12 i c(dw

dz ) 2dz and M =

−2

1 c z(

dwdz )

2dz.

where Re denotes the real part, is the (constant) density of the fluid and w = u + iv is the complex potential for the flow both of which are presumed known. We shall find X, Y and M if the cylinder has a circular cross-section and the boundary is specified by |z| = a. Let the flow be a uniform stream with speed U.

Now, using a standard result, the complex potential describing this situation is:

a2 dw a2

w = U ✓

z +

◆ so that

= U ✓ 1 −

◆

z dz z2

dw 2 2a2 a4

✓

◆ = U2 ✓

1 −

+

◆

dz z2

z4

Zc ✓

◆ dz = 0.

1 dw 1 2a2 a

4

Now X − iY =

i

Zc(

)2dz =

i U 2 1 −

+

2 dz 2 z2 z4

dw 2 2a2 a

4

Hence X = Y = 0. Also z ✓

◆

= U2

✓ z −

+

◆

dz z z3

The only term to contribute to M is −2a2U2

z

Again using the Key Point above this leads to 4 a2U

2i and this has zero

real part. Hence M = 0, also. The implication is that no net force or

moment acts on the cylinder. This is not so in practice. The discrepancy

arises from neglecting the viscosity of the fluid.

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MATHEMATICS-II

UNIT-I VECTOR CALCULUS

Part-A

1. If = + .Evaluate . d along the curve ‘C’ in the XY plane y= from the point (0,0) and (1,1). 2. Prove that = + +

3. Is the position vector =x + +z is irrotational?Justify.

4. Find the constants a,b,c so that =(x+2y+az) +(bx-3y-z) +(4x+cy+2z) may be irrotational

5. If are irrotational, prove that × is solenoidal.

6. Find ‘a’ such that (3x-2y+z) +(4x+ay-z) +(x-y+2z) is solenoidal. 7. Prove that Gradient of a constant is zero. 8. Find grad(r”) where =x + +z and r=| |

9. Find a unit normal to the surface x2y+2xz

2=8 at the point (1,0,2).

10. Prove that div = 3 =0

11. Find the gradient of Ø where Ø=xy+yz+zx at (1,1,1).

12. If =x2+y

2+z

2,find div and curl

13. Prove that Curl (Curl )=grad(div )-

14. If is a constant vector prove that (i) .( × )= 0 (ii) ×( × )= 2 . 15. State the physical interpretation of the line integral

16. Statement Green’s, Stoke’s and Gauss divergence theorem.

17. Prove by Green’s theorem that the area bounded by a simple closed C curve is

− ) PART-B

1.Show that = ( − + 3 − ) + (3 + 2 ) + xy − xz + z is conservative vector field. Find (i) the scalar potential (ii) the work down by in moving a particle from (1,0,1) to (2,1,3).

2.Find the Directional Derivative 0f Ø=x2yz+4xz

2 at (1,-2,-1) in the direction of

2 − −2

3. Find the angle between the normal’s to the surface xy = z2 at the points (1,4,2) and

(-3,-3,3)

4. Find the angle between the normal’s to the surface xy3 z

2= 4 at the points (-1,-1,2) and

(4,1,-1)

5. Prove that =(2x+yz) +(4y+zx) - (6z-xy) is solenoidal as well as irrotational.

Also find the scalar potential of .

6. Prove that is irrotational. Also find the scalar potential.

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7.Prove that curl( × = . − ( . + −

8. Prove that ) = ( + 1)

. If r = | | and find

.

9. Verify the Green’s theorem in the plane for − 8 ) + (4 − 6 ) where

C is the boundary of the region given by x=0,y=0,x+y=1.

10. Verify Green’s theorem for = ( + − taken around the rectangle

bounded by the lines x=± , = 0 =

11. Verify the Green’s theorem in the XY plane for + ) + where C is the

closed curve of the region bounded by y=x and y= .

12. Verify the Green’s theorem in the plane for − 8 ) + (4 − 6 ) where

C is the boundary of the region defined by y= and x= .

13. Verify the stoke’s theorem for a vector field defined by = ( − ) + 2 in the rectangular region in the XOY plane bounded by the line x=0,x=a,y=0 and y=b.

14. Verify the stoke’s theorem for a vector field defined by = xy − 2 − zx in the rectangular parallelepiped formed by the planes the line x=0,x=1,y=0,y=2 and z=3 above the XY line.

15. Verify the stoke’s theorem for = ( − ) + − where S is the surface bounded by the plane in the rectangular region in the XOY plane bounded by the line x=0,x=a,y=0 and y=b.

16. Evaluate − + )by using Stoke’s theorem,where C is the boundary of the rectangle defined by 0≤ ≤ , ≤ ≤ , = 3

17. Verify Stoke’s theorem when = (2 − ) − ( − ) nd C is the boundary of the region enclosed by the parabolas y2=x and x

2=y

18. Evaluate + ) + ( + ) where C is the square bounded by the lines x=0,x=1 and y=0,y=1.

19. Verify the G.D.T for =4xz − +yz over the cube bounded by x=0,x=1,y=0,y=1,z=0 and z=1

20. Verify the G.D.T for =(x3-yz) -2x

2 y + 2k over the cube bounded by x=0,y=0,z=0 and x=a, y=a, z=a

21. Verify the G.D.T for =x2

+ +z where S is the surface of the cuboid formed by the planes x=0,x=a,y=0,y=b,z=0 and z=c

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UNIT-II ORDINARY DIFFERENTIAL EQUATIONS

Part-A

1. Find the particular integral of (D2- 4D+4)y = 2

x

2. Find the particular integral of (D-1)2y = e

x sinx

3. Find the P.I of (D2- 4)y = cosh2x

4. Find the P.I of (D2+1)y = sinx

5. Find the P.I of (D2+2D+1)y = e

-x cosx

6. Find the P.I of (D2+4)y = sin2x

7. Find the P.I of (D+1)2y = e

-x cosx

8. Find the P.I of (D2-2D+2)y = e

x cosx

9. Solve: (D2-6D+13)y = 0

10. Transform the equation (2x+3)2

− + 3) − = 6

coefficients

11. Transform the equation x2y”+xy ‘= x

coefficients

12. Reduce the equation (x2D

2 +xD +1)y = logx into an ordinary differential equation with

constant coefficients

Part-B

1. Solve the equation (D2+5D+4)y = e

-x sin2x

2. Solve (D2-4D+3)y = e

x cos2x

3. Solve: (D2-3D+2)y = 2cos(2x+3)+ 2e

x

4. Solve (D2+5D+4)y = e

–x sin2x

5. Solve (D2+4)y = x

2 cos2x

6. Solve (D2+16)y = cos

3x

7. Solve: (D2+4D+3)y = e

-x sinx

8. Solve (D2+a

2)y = Secax by the method of variation of parameters

9. Solve + = by the method of variation of parameters

10. Apply the method of variation of parameters to solve (D2+4)y = cot2x

11. Solve by the method of variation of parameters to solve (D2+a

2)y = tanax

12. Solve + 4 = 2 by the method of variation of parameters

13. Solve: (x2D

2 +3xD +5)y = x cos(logx)

14. Solve: (x2D

2 - xD +4)y = x

2 sin(logx)

15. Solve (x2D

2-3xD+4)y = x

2 cos(logx)

16. Solve: (x2D

2 - 2xD -4)y = x

2 + 2logx

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17. Solve (1+x)2

+ (1 + ) + = 2sin[log(1 + )] 18. Solve (1+x)

2 + (1 + ) + = 4cos [log(1 + )]

19. Solve + = , − =

20. Solve the equation + =

21. Solve the simultaneous differential equations + 2 = 2 , − = 2

22. Solve: − = , + = given x(0)=y(0)=2

23. Solve: + = , + = given x=2 and y=0 at t=0

24. Solve +2 = − , − = given x=1 and y=0 at t=0

25. Solve + 2 + 3 = 2 , + 3 + 2 = 0.

UNIT – III LAPLACE TRANSFORM

PART-A

1. Find L

2. Is the linearity property applicable to L ? Reason out.

3. Find the Laplace transform of 4. Find the Laplace transform of 2

5. State the conditions under which Laplace transform of ( ) exists

6. Find L[ ] 7. State the first shifting theorem on Laplace transforms.

0, < 8. Find the Laplace transform of ( ) =

cos − , > 9. Find the Laplace transform of unit step function.

10. Find the inverse Laplace transform of 11. Find [ ]

12. Find the inverse Laplace transform of


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MA6251 MATHEMATICS-II 14. Find the inverse Laplace transform of ( )( )

15. Evaluate ∞

3 .

16. What in the Laplace Transform of Periodic function f(s) of period T? 17. If L= them find

∞ 18. State the convolution theorem

19. State and prove initial value theorem. 20. State and prove final value theorem 21. Verify the initial and final value theorems for f(t) = 3

22. Verify initial value theorem for 1 + ( + ).

PART-B

1. Find the Laplace transform of 3

2. Find L

3. Find

( )( )

4. Find

5. Find the Laplace transform of .

6. Solve by using Laplace transform + 9 =

2 given that (0) = 1, = − .

7. Solve the differential equation

+ = 2 with y (0) = 0 and (0) = 0 by using

8.

Laplace transforms method.

differential equation

Using Laplace Transform solve the − 3 − 4 = 2

9.

with (0) = 1 = (0)

differential equation

Using Laplace Transform solve the − 3 + 2 = 4

with = − , (0) = 5

10. Solve − 3

+ 2 = 2 given x= 0 and

= 5 at t = 0 using Laplace transform.

11. Solve

+ 4

+ 4 = if

= 0

= 2

= 0 using Laplace transforms.

12. Solve the differential equation − 3

+ 2 =

with (0) = 1 = (0) using

Laplace transforms.


using convolution theorem.

( )( )

14. Using convolution theorem, find

( )(

)

15. Find

using convolution theorem.

( )

16. Apply convolution theorem to evaluate

( )

17. Using convolution theorem

( )( )

18. Using convolution theorem find the inverse Laplace transform of

( )( )

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19. Find the Laplace transform of square wave function defined by

( ) =

1 0 < < with period 2

20.

−1 < < 2

Find the Laplace transform of the following triangular wave function given by

, ≤ ≤

( + 2 ) = ( ).

( ) = − , ≤ ≤

21. Find the Laplace transform of ( )

= , 0 < < 1

( + 2) = ( )

> 0.

0 ,1 < < 2

22. Find the Laplace transform of

23. Find the Laplace transform of the periodic function

( ) =

, ≤ ≤ ( + 2 ) = ( )

24.

− , < ≤

).

Verify initial and final value theorem for the function ( ) = 1 + ( +

25. Find the Laplace transform of ( ) = , ≤ ≤

( + 2 ) = ( ) for all

− ,0 < ≤

26. Find the Laplace transform of square wave function given by

≤ ≤

( ) = 2

( + ) = ( )

−

≤ ≤

27.

2

Find the Laplace transform of the Half wave

rectifier ( ) = , 0 < <

+

= ( )

0 ,

< <

28. Evaluate ∞ using Laplace transform

UNIT-IV ANALYTIC FUNCTIONS

Part-A

1. Show that the function f(z)=

2. Find the map of the circle | | = 3 under the transformation w=2z

3. Prove that a bilinear transformation has atmost two fixed points

4. State the basic difference between the limit of a function of a real variable and that of

a complex variable

5. Show that an analytic function with constant imaginary part is constant 6. Find the invariant points of the transformation w=

7. Show that u=2x-x3+3xy

2 is harmonic

8. Verify whether the function u=x3-3xy

2+3x

2-3y

2+1 is harmonic

9. Find the constants a,b,c if f(z)=x + ay + i(bx+cy) is analytic

10. Find the invariant points of the transformation w =

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11. State the Cauchy-Riemann equation in polar coordinates satisfied by an

analytic function

12. Verify whether f(z)= is analytic function or not

PART-B

1. Determine the analytic function whose real part is

2. Find the bilinear transformation that maps the points z= ∞, , 0 = 0, , ∞

3. Find the image of the hyperbola - = 1 =

4. Prove that the transformation w = the upper half of z-plane onto the upper half of w-plane. what is the image of | | = 1

5. Prove that every analytic function w = u+iv can be expressed as a function z alone, not as a function of

6. Find the bilinear transformation that maps the points z = , , ∞ = , , −

7. Show that the image of the hyperbola x2-y

2=1 under the transformation = is the lemniscates r

2=cos2

8. If f(z) is an analytic function of z, prove that

+

| ( )| = 0

9. Find the analytic function w=u+iv when = ( 2 + 2 ) and find u

10. Show that the map = maps the totality of circles and straight lines as circles or straight lines

11. Prove that the transformation = maps the family of circles and straight lines into the family of circles or straight lines

12. If u(x,y) and v(x,y) are harmonic functions in a region R, Prove that the function − + + is an analytic function of z = x+iy

13. If w=f(z) is analytic, prove that = = −

14. Show that = log( + ) is harmonic. Determine its analytic function. Find also its conjugate.

15. If f(z) is an analytic function prove that + | ( )|²= 4| ( )|²

16. Find the image of = 2 under the mapping (i) w = z+3+2i, (ii) w = 3z

17. Find the bilinear transformation that maps the points z = , − , − = , 1,0

18. If f(z) is a regular function of z, prove that + | ( )|²= 4| ( )|²

SCE 1 Dept of S&H

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19. Find the image of the half plane x > c, when c > 0 under the transformation = Show the regions graphically. Also find the fixed point of w.

20. Find the analytic function f(z) = P+iQ, if P-Q =

21. When the function f(z) = u+iv is analytic, prove that the curves u=constant and

v=constant are orthogonal

22. Find the image of the circle | − | in the complex plane under the mapping =

23. Verify that the families of curves u = c1 and v = c2 cut orthogonally, when u+iv=z3

24. Find the analytic function u+iv, if u = (x-y)(x2+4xy=y

2). Also find the

conjugate harmonic function v.

25. Prove that u = x2-y

2 and = are harmonic but u+iv is not regular.

26. Find the bilinear transformation that transforms 1,i,-1 of the z-plane onto 0,1, ∞ of the w-plane. Also show that the transformation maps interior of the unit circle of the z-plane onto upper half of the w-plane

27. Prove that = ( − ) is harmonic and hence find the analytic function f(z)=u+iv

UNIT-V COMPLEX INTEGRATION

PART-A

1. Evaluate

,

C is the circle | | =

( )( )

2. If f(z) = − [ + − + − 1) + ],

( ) = 1

3. Define Singular point

4. Find the residue of ( ) =

at a simple pole

( )

5. Find the residue of at Z=0

6. Find the residue of ( ) =

at its pole

( )

7. Calculate the residue of f(z) =

at its pole

( )

8. Find the singular point of

and hence find its residue.

( )

9. Identify the type of singularities of the following function f(z) =

10. Determine the residue at the simple pole of

( ) ( )

11. Expand f(z)=sinz in a Taylor series about origin

12. Expand f(z)=sinz in a Taylor series about Z=

13. Evaluate | | = 2

14. Evaluate

, where C is | | =

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15. Evaluate ∮ where C is the circle | | = 2 in the Z-plane.

16. Using Cauchy’s integral formula, evaluate

, where C is the circle | |

− 2 =

17. Evaluate

if C is | | = 2

18. Evaluate

where C is | | = ⁄ .

( )( )

19. State Cauchy’s integral theorem

20. Use Cauchy’s integral formula to evaluate

, where C is the circle with centre 1

and radius=1. [AU 2000]

21. Obtain the expansion of log(1 + ) when | | < 1

PART-B

1. Evaluate ∮

where C is the circle | | = 3.5

2. Evaluate ∮ ( ) where C is the circle

| + 1 + | = 2,

3. Find the residues of f(x) =

at its isolated singularities using Laurent’s

( ) ( )

series expansions. Also state the valid region

4. Evaluate

.

5. Evaluate the integral

6. Evaluate the integral

, a>b>0

| |

7. Evaluate ( )( ) where C is

by using Cauchy’s integral formula.

− 2 =

8. Using Cauchy’s Integral formula, evaluate ∮ ( )

, where C is the circle

( )( )

| | =

9. Evaluate ∞

, > 0

10. Evaluate ∞

,

11. Evaluate ∮ ( )

where C is the circle

( ) ( )

| − | = 2,

12. Evaluate f(z) =

Laurent’s series valid for the region | | > 3 and

( )( )

1< | |

< 3

13. Evaluate ∮ where C is the circle

( )( )

| − 2| =

,

14. Obtain Taylor’s series to represent the function

in the region | | < 2.

( )( )

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15. Obtain Laurent’s series to represent the function f(z) =

in the region

( )( )

2 < | | < 3.

16. Evaluate ∮ ( ) where C is

( ) ( )

| | = 3, 17. Using contour integration, evaluate

∞

∞ ( )()

18. Using contour integration, evaluate ∞

∞ ()

19. Evaluate

by contour integration

20. Evaluate ∞

( ) by contour integration

21. Obtain Laurent’s series to represent the function f(z) = ( )( ) in the region 1 < | + 1| < 3.

22. Evaluate

, (0 < < 1)

23. Prove that ∞

= , > > 0

∞ ( )()

24. Obtain Laurent’s series to represent the function f(z) = ( )( ) in the region | | > 3 1 < | | < 3.

25. Show that ∞

=

[ (1 + )]

( )

26. ∞ , > 0

)

(

27. Evaluate ∞

,

∞ ()()

28. Obtain Laurent’s expansion for ( ) = ( )( ) Valid in the regions.

(i) | − | < (ii)1 < | | < 2

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B.E./B.Tech. DEGREE EXAMINATION, MAY/JUNE 2009

Second Semester

Civil Engineering

MA2161 – MATHEMATICS – II

(Common to all branches of B.E./B.Tech. )

(Regulation 2008)

Time : Three hours Maximum : 100 marks

Answer ALL questions

PART A – (10 x 2 = 20 marks)

1. Find the particular integral of D2

2D 1 y e x cos x .

2. Solve the equation x 2y cc xyc y 0 .

3. Find the values of a ,b c, so that the vector F x y az i bx 2 y z j

x cy 2z k may be irrotational.

4. State Green s theorem in a plane.

5. State the Cauchy-Riemann equation in polar coordinates satisfied by an analytic function.

6. Find the invariant points of the transformation w

2 z 6

.

z 7

7. Evaluate ³tan where C is

z

2 .

c

8. Find the Taylor series for f (z ) sin z about z

S .

4

1 cos t

9. Find the Laplace transform of .

t

1

k

s

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PART B – (5 x 16 = 80 marks)

11. (a) (i) Solve the equation D2

4 y x2 cos 2x .

(ii) Solve the equation D2

a2

y tan ax by the method of variation of

parameters.

Or

(b) (i) Solve the equation x 2D

2 3xD 5 y x cos log x.

(ii) Solve dx y sin t , x dy cos t given that x 2 and y 0 at t 0 .

dt

dt

12. (a) (i) Find the angle between the normals to the surface xy 3z

2 4 at the points

1, 1, 2and 4,1, 1 .

(ii) Verity Stoke s theorem for F xyi 2 yzj zxk where S is the open surface of the

rectangular parallelepiped formed by the planes x 0, x 1, y 0, y 2 and z 3 above the

XY plane.

Or

(b) (i) Find the directional derivative of I 2 xy z2 at the point 1, 1, 3 in the direction

of i 2 j 2k .

(ii) Verify Gauss divergence theorem for F x 2i y

2j z

2k where S is the surface of

the cuboid formed by the planes x 0, x a ,y 0, y b,z 0 and z c .

f (z ) P iQ , if P Q

sin 2 x

13. (a) (i) Find the analytic function .

cosh 2 y cos 2 x

(ii) Find the bilinear transformation which maps the points z 0, i,

w i,1,0 respectively.

Or

w 2 w 2

f (z ) 2 (b) (i) If f (z) is a regular function of z , prove that

wx2

wy2

1into

4 f c( z) 2 .

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(ii) Find the image of the half plane x ! c , when c ! 0 under the transformation w 1 .

z

Show the regions graphically.

Evaluate ³c

zdz

c is the circle

z 2

1

14. (a) (i)

where

using Cauchy s integral

z 1 z 22

2

formula.

2S dT

(ii) Evaluate ³0

, 0 x 1 , using contour integration.

1 2 x sinT x2

Or

(b) (i)

Find the Laurent s series of f ( z)

z

2 1

valid in the region 2

z

3 .

z 2 5 z 6

f x

2dx

(ii) Evaluate

³ x2 a2 x2 b2 , using contour integration, where

a ! b ! 0 .f

15. (a) (i) Find the Laplace transform of te2t

cos 3t .

1

(ii) Find the inverse Laplace transform of s 1s2 4 .

(iii) Solve the equation ycc 9 y cos 2t, y(0) 1 and S

1using Laplace

y

2

transform. Or

(b) (i) Find the Laplace transform of f (t) t, in 0 d t d a

f (t 2a) f (t ) .

and

2a t, in a d t d 2a

(ii) Find the Laplace transform of e4t

³t t sin 3 t dt .

0

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B.E./B.Tech. DEGREE EXAMINATIONS, MAY/JUNE 2010

Regulations 2008

Time: Three Hours

Second Semester

Common to all branches

MA2161 – Mathematics II

Answer ALL Questions

PART A – (10 x 2 = 20 Marks)

Maximum: 100 Marks

1. Transform the equation x 2y cc xyc x into a linear differential equation with constant

coefficients.

2. Find the particular integral of D2

1 y sin x .

3. Is the position vector r xi yj zk irrotational? Justify.

4. State Gauss divergence theorem.

5. Verify whether the function u x3

3 xy2

3 x2 3 y

2 1 is harmonic.

6. Find the constants a ,b c, if f (z ) x ay i (bx cy) is analytic.

3z

2 7z 1

1

7. What is the value of the integral ³ dz where C is z ?

1

c

z 1

2

2

8. If f (z )

2

(z 1) (z 1) ...

f (z) at z 1 .

z 1

1 ¼ , find the residue of

9. Find the Laplace transform of unit step function.

10. Find L1

^cot 1

( s)`.

PART B – (5 x 16 = 80 marks)

11. (a) (i) Solve the equation D2

4D 3 y e x sin x .

(ii) Solve the equation D2

1 y x sin x by the method of variation of

parameters.

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OR

(b) (i) Solve x 2D

2 2xD 4 y x

2 2log x .

(ii) Solve

dx 2 x 3 y 2e2t

,

dt

dy 3 x 2 y 0.

dt

12. (a) (i) Prove that F 6xy z3 i 3x2 z j 3xz

2 y k is irrotational vector and

find the scalar potential such that F �M .

(ii) Verity Green s theorem for ³3x2 8 y

2 dx 4 y 6xy dy where C is the

C

boundary of the region defined by x y2 , y x

2 .

OR

(b) Verify Gauss – divergence theorem for the vector function

f x3 yz i 2 x 2

yj 2k over the cube bounded by x 0, y 0, z 0 and

x a,y a,z a .

13. (a) (i) Prove that every analytic function w u iv can be expressed as a function z alone,

not as a function of z .

0,1, f into

(ii) Find the bilinear transformation which maps the points z

w i,1, i respectively.

OR

(b) (i) find the image of the hyperbola x

2 y

2 1 under the transformation w

1 .

z

(ii) Prove that the transformation w

z

maps the upper half of z - plane on to the

1 z

upper half of w - plane. What is the image of z 1 under this transformation?

f (z)

7z 2 1

z 1

3 .

14. (a) (i) Find the Laurent s series of in

z (z 1)(z 2)

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(ii) Using Cauchy s integral formula, evaluate ³

4 3z dz , Where C is the circle

z (z 1)(z 2)

3

C

z

.

2

OR

f x

2 x 2

(b) (i) Evaluate ³ x 4 10 x2 9 dx using contour integration. f

2S dT

Evaluate ³0

(ii)

using contour integration.

2 cosT

ª s

º

15. (a) (i) Apply convolution theorem to evaluate L1 « »

a2 2

« s2 »

¬ ¼

(ii) Find the Laplace transform of the following triangular wave function given by

f (t) -t , 0 d t d S

and f (t 2S ) f (t ) .

®

S d t d 2S

¯2S t ,

OR

(b) (i) Verify initial and final value theorems for the function f (t ) 1 et (sin t cos t ) .

(ii) Using Laplace transform solve the differential equation ycc 3 yc 4 y 2et with

y(0) 1 yc(0) .

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MATHEMATICS- II S.NO CONTENTS PAGE NO UNIT-1 VECTOR ...€¦ · integrals of transforms -...

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