S.NO CONTENTS PAGE NO UNIT-1 VECTOR …...MA6251 MATHEMATICS-II S.NO CONTENTS PAGE NO UNIT-1 VECTOR...

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 Coefficients 26

Problems

www.studentsfocus.com


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



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


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


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).



SCE 1 Dept of S&H

UNIT-1VECTOR CALCULUS

1.1Gradient-Directional Derivative

1.1.1. Gradient

1.1(a) The Vector Differential OperatorThe 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)Let ( , , ) be a scalar point function and is

continuously differentiable then the vector∇ = + + = + + is called the gradient of the scalar functionand 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 2 Dept of S&H

1.1.2. Directional DerivativeDirectional derivative=∇ . | |

1. Find the directional derivative of = + 4 + (1,2,3) in the direction of2 + −

Solution:Given: = + 4 +

∇ = + +

=(2 + 4 + ) + ( + ) + ( + 8 + )=54 + 6 + 28

Given: = 2 + −| | = √6

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 = + 1 , = 4 − 3, = 2 − 6 = 2.

Solution:= + +

=( + 1) + (4 − 3) + (2 − 6 )= 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 3 Dept of S&H

= ( ) + ( ) + (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: = + − − 10∇ = + +

=2 + 2 −∇ ( , , ) = 2 + 2 −

|∇ |=3Unit 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 − ( + 2)] + (− ) + (− )∇ (1,−1,2) = ( − 2) − 2 +

= 4 + = 4

∇ = + +

=8 + 4 + 3∇ (1,−1,2) = −8 + 4 + 12

Given: ∇ . ∇ = 0.( − 2) − 2 + . −8 + 4 + 12 = 0.

−8( − 2) − 8 + 12 = 02 − − 4 = 0…………… . (2)

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

= 1∴ (2)… .> 2 − (1) − 4 = 0

=



SCE 4 Dept of S&H

1.1.7 .Scalar Potential :1. If ∇ = 2 + + ,then find the value of .

Solution:Given: ∇ = 2 + +

+ + = 2 + +

Equating the coefficients of , , ,we get= 2 … . . (1) , = … . . (2) = … . . (3)

Integrating (1) p.w.r.to ‘x’ we get= + ( , )………… . . (4)

Integrating (2) p.w.r.to ‘y’ we get= + ( , )………… . . (5)

Integrating (3) p.w.r.to ‘z’ we get= + ( , )………… . . (6)

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 themagnitude of this maximum?4. Find the angle between the surfaces = − 1 = 2 − (1,1,1)

1.2 Divergence And Curl –Irotational And Solenoidal Vector Fields:

1.2.1 Divergence and curl= ∇. = + +

= ∇ × = × + × + ×Note:If = + + ,then

∇ × =

1. If = + + then find = ∇. and ∇ ×Solution:

Given: = + +

= ∇. = + + =



SCE 5 Dept of S&H

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

= ∇ × = = =0

2. Find ∇.

Solution:We know that, = + +

= + +

∇. = + + . ( + + )

= + +

=( )

+( )

+( )

=

3. Prove that ( ) = 0

Proof: = ∇ = + +( ) = ∇ × (∇ )

=

=∑ −

= 0

3. Find where = ( + + − 3 )

Solution:Given: = ∇ ( + + − 3 )

= (3 − 3 ) + (3 − 3 ) + (3 − 3 )∇. = (3 − 3 ) + (3 − 3 ) + (3 − 3 )

= 6 + 6 + 6

∇ × =

(3 − 3 ) (3 − 3 ) (3 − 3 )= (−3 + 3 ) − (−3 + 3 ) + (−3 + 3 )= 0



SCE 6 Dept of S&H

1.2.2 SOLENOIDAL VECTOR,IRROTATIONAL VECTOR:

Solenoidal vector formula: ∇. = 0Irrotational 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

∇ × =

(( + ) (2ysinx − 4) (3 )∇ × = 0.

Hence, is irrotational

= ∇( + ) + (2ysinx-4) + 3 = + +

Equating the coefficients of , , we get,

= + ……(1) = 2 …… (2)

= 3 ……(3)Integrating (1) p.w.r.to ‘ We get = + ( , )…… (4)Integrating (2) p.w.r.to ‘ We get = − 4 + ( , )…… (5)Integrating (3) p.w.r.to ‘ We get = + ( , )…… (6)

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

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

Solution:Let =

= ( + + )= + +



SCE 7 Dept of S&H

∇ × =

=∑ −=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 − ) + (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)N The points corresponds to = 0 = 1,∴ = , = 2 = 3

. = (3 + 6 ) − 14 + 20

=∫ (3 + 6 ) − 14 (2 ) + 20 (3 )= 5.

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

Given: = ………….(1)



SCE 8 Dept of S&H

⇒ = ………… . . (2)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 theprojection of the surface x on xy plane. Let be a vector function defined in some regioncontaining 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 orzx plane and we get

. =.

| . |.

. =.

| . |.

1. Evaluate ∬ . where = +x - and S is the surface of the cylinder + = 1included in the first octant between the planes = 0 = 2.

Solution:Given: = +x -

= + − 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 givenby ( , , ) ( , , )



SCE 9 Dept of S&H

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 + 2 and S is the surface of the plane2 + + 2 = 6 in the first octant.3. If = (2 − 3 ) − 2 − 4 , evaluate ( , , ) where V is the region bounded by= 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 ofthe 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) ( = , = )

= [( + ) + ]

=∫ 3= -1.

Hence, ∫ ( + ) + =∫ +∫ = − 1 = − ……………..(1)

Evaluation of ∬ −

∬ − =∬ [2 − ( + 2 )]



SCE 10 Dept of S&H

=∫ ∫ ( − 2 )√

=∫ − 2√

= ………………..(2)From (1) & (2) we get,

∫ ( + ) = ∬ − .

Hence, Green’s theorem is verified.

1.4.1 Tutorial Problems:

1. Verify Green’s theorem in the xy plane for ∫ {(3 − 8 ) + (4 − 6 ) } where C isthe 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 Senclosing volume V is equal to the volume integral of the divergence of F taken throughout thevolume 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 rectangularparallelepiped 0 ≤ ≤ ,0 ≤ ≤ ,0 ≤ ≤ .Solution:

The Gauss-divergence theorem is,

. = ∇.

Given: = ( − ) + ( − ) + ( − )



SCE 11 Dept of S&H

∇. = 2 + 2 + 2 = 2( + + )R.H.S = ∇.

=∫ ∫ ∫ 2( + + )

=∫ ∫ 2( + + )

=∫ 2 + += + += abc (a+b+c).

L.H..S:∬ . = ∬ +∬ +∬ +∬ +∬ +∬= ( − ) + ( − ) + ( − ) , =unit outward normal vector

Face . Equation . on S.

( − ) = −( − )

−-( − ) = 0

( − ) = −( − )

− -( − ) = 0

( − ) = −( − )

− -( − ) = 0

(i) ∬ . +∬ . =∫ ∫ ( − ) +∫ ∫

=∫ ∫ ( )=

(ii) ∬ . +∬ . =∫ ∫ ( − ) +∫ ∫

= ( )

=(iii)∬ . +∬ . = ∫ ∫ ( − ) + ∫ ∫

=∫ ∫ ( )=



SCE 12 Dept of S&H

∴ . = + + + + +

= ( + + ) ( ), ( ), ( )L.H.S=R.H.S

. = ∇.

Hence, Gauss-divergence theorem is verified.Tutorial Problems:

1. Verify the GDT for = 4 − + over the cube bounded by =0, = 1, = 0, = 1, = 0, = 1.

2.Verify the GDT for = + + over the cube bounded by = ±1, =±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 surfaceS 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 bythe lines = ± , = 0, = .

Solution:

Stoke’s theorem is

. = ∇ × .

Given: = ( + ) − 2

∇ × =

( + ) −2 0=−4

R.H.S =∬ ∇ × . =∬ −4 .=−4∫ ∫

=−4Given: = ( + ) − 2

. = + +∫ . = ∫ ( + ) − 2

L.H.S:∫ = ∫ +∫ +∫ +∫

(i) ∫ = ∫ ( + ) − 2= 0, = 0.



SCE 13 Dept of S&H

=∫ =

(ii) ∫ = ∫ ( + ) − 2= , = 0.

=∫ −2 = − .

(iii) ∫ = ∫ ( + ) − 2= , = 0.

=∫ ( + ) = − − 2 ..

(iv) ∫ = ∫ ( + ) − 2= − , = 0.

=∫ 2 = − ..

∴ = + + +

=−4 ( ),( ),( )

L.H.S=R.H.S

. = ∇ × .

Hence, Stoke’s theorem is verified.

1.6.1 Tutorial Problems:1. Evaluate by Stokes’s theorem ∫ ( + 2 − ) where C is the curve +

= 4, = 22. Using Stoke’s theorem, prove that = 0 and ( ) = 0.



SCE 14 Dept of S&H

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 physicalsituation, called modeling.

(ii) Solutions of this differential equation, evaluating thearbitrary constants from the given conditions, and

(iii)Physical interpretation of the solution.

2.1.1. Higher Order Linear Differential Equations with Constant CoefficientsGeneral form of a linear differential equation of the nth order with constant

coefficients isd ydx

+ Kd ydx

+ + K y = X

Where , K , K … K are constants.

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

,

= ,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 equationWorking 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 operatorsThe 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 15 Dept of S&H

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

iii. Complete solution isY= complementary function + particular integral

To ind the complementary functions

Roots of A.E C.F

1.

2.

3.

Roots are real and different, ( ≠ )

Root are real and equal= =

Roots are imaginary α±iß

A +B

(Ax+B) (or)(A+Bx)

(Acos ßx + B sin ßx)

iv. To find particular integral: P.I =( )

X

X P.I

1.

P.I =( )

=( )

,f(a)≠0

=x ( ), f(a)=0, ( ) ≠0

= ( ), f(a)=0,( ) = 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 ( − 3D + 3D − 1)y = 0Solution::

Given: ( − 3D + 3D − 1)y = 0

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

( − 1) = 0.= 1,1,1.



SCE 16 Dept of S&H

The general solution is given by y = C.Fy = [ + Bx + Cx2 ]

2. Solve − 6 + 13y = 0.Solution:

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

The auxiliary equation is − 6 + 13 = 0m = ±√

= ±√

m = ±

= 3 ± 2iThe 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 − 4 + 13 = 0m = ±√

= ±√

= 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 = 0m = -2,-2.

C.F = (Ax + B)



SCE 17 Dept of S&H

P.I. =

=

=

= x== x2 (Ax + B)

��y = C.F + P.I.= (Ax + B) +

2.1.6 Problems Based On P.I. =( )

( )( )

=> Replace D2 By –a2

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

Given: (D2 +1)y = sin xP.I. = sin x

= sin x

= x sin x= ∫ sin dx= (− cos )

P.I. =

2.Find the P.I. of + 4 = sin 2xSolution:

Given: (D3+4D) y = sin 2xP.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 Ax1.Solve (D − 4D + 4 y) = + cos 2x

Solution:Given: (D − 4D + 4 )y = + cos 2x

The auxiliary equation is m2- 4m + 4 = 0(m - 2)2 = 0

m = 2,2C.F. = ( + )



SCE 18 Dept of S&H

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= 0m = ± 1

��C.F � Aex� Be-x

P.I. = x= -[1- D2] x= -x

��y = C.F + P.I.y = Aex

� Be-x – x2. Solve (D2 – D) y = x

Solution:Given: ��

��

The auxiliary equation is m2- m= 0m(m-1) = 0

m =0,1C.F � Ae0x

� Bex

C.F = A + Bex

P.I. == x= [1 + D + D + D + ]x

= [x + 1]= - ∫(1 + )



SCE 19 Dept of S&H

= - −��

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 − 2)P.I. =

( )(3 − 2)

=( )

(3 − 2)

= (3 − 2)

= [ − ]

P.I. = [ − ]2. Solve (D+2)2 y = sinSolution:

Given: (D+2)2 y = sinThe auxiliary equation is (m + 2)2 = 0

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

P.I. =( )

sin

=( )

sin

=( )

sin

= sin=− sin x

y = ( + ) − sin x

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

1.Solve (D2 +4) y = x sinSolution:

Given: (D2 +4) y = x sinThe auxiliary equation is + 4 = 0

m = ± 2iC.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 20 Dept of S&H

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

Given: (D2 -2D +1)y = x sin xThe auxiliary equation is − 2 + 1 = 0

m = 1,1C.F. = (Ax + B)

P.I. = x 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 + a2 ) y = secSolution:

Given: (D2 + a2 ) y = secThe auxiliary equation is (m2 +a2) = 0

m =±∴ C. F. = A cos ax + B sin axP.I. = sec

= ( )( )sec

= ( - ) sec

= ∫ sec dx - ∫ sec dx= ∫(1 − itan ) - ∫(1 + 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 Problems1.Solve (D2 -6D +13)y = 2x

Solution:Given: (D2 -6D +13)y = 2x

(D2 -6D +13)y === ( )

The auxillary equation is − 6 + 13 = 0



SCE 21 Dept of S&H

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 ax1.Solve (D2 +1)y = + sin hx

Solution:Given: (D2 +1)y = +The auxillary equation is (m2 +1)2 = 0

m = ±1C .F. = (A+Bx) cos +(C+Dx)sin

P.I1 =( )

= [1+ D ] -2 x= [ 1- 2D + 3D ] x

= − 24 + 72P.I2 =

( )

=

=

P.I3 =( )

(− )

= −

= −��y = C.F + P.I1 + P.I2 + P.I3

y = (A+Bx) cos +(C+Dx)sin + − 24 + 72 + −2.1.14 Tutorial Problems

1. Solve (D2 + 1) y = 0 given y(0) = 1, y( 2) = 02. Solve (D3 –D2 – D – 1)y =o3. Solve (D2 –4D + 4)y = 2x

4. Solve (4D2 –4D + 1)y = 45. Solve (D2 – 4) y =e + e6. Solve + 3 + 2y = sin 3x7. Find the particular integral of (D2 + 1) y = sin x sin 2x8. Solve (D − 4D + 4 y) = + cos2 x9. Solve + 2 + = + sin 2x

10. Solve − 5 + 6y = + 3



SCE 22 Dept of S&H

11. Solve D2 (D2 + 4) y = 96 x2

12. Obtain the P.I. of (D -2D +5) y = cos 2x13. Solve (D + 5D + 4) y = sin 214. Solve (D2 +1) y = x sin15. Solve − 6 + 9y = 6 + 7e - log 216. Solve (D2 –4D + 4)y = + cos 4x +

2.2 Method Of Variation Of ParametersThis 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) = 0m = ±i

∴ C. F. = c cos x + c sin x= c1f1 + c2f2

Here, f1 = cos = − sin

f2 = sin = cos x

X = cosec xf f − f f = cos x + sin x = 1

P = -∫ dx

= - ∫= -x

Q = ∫ dx

= ∫= log (sin x)

P.I. = Pf1 + Qf2



SCE 23 Dept of S&H

= -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: (D2 +1) y = cosec x

The auxiliary equation is (m2 +1) = 0m = ±i

∴ C. F. = C cos x + C sin x= c1f1 + c2f2

Here, f1 = cos= − sin

f2 = sin= cos x

X = cosec x cot xf f − f f = cos x + sin x = 1

P = -∫ dx

= - ∫= -∫ cot= - log (sin x)

Q = ∫ dx

= ∫ dx

= ∫ dx

= ∫ ( ) dx= ∫(cosec x − 1)dx

= - cot x –xP.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.Iy = C cos x + C sin x - cos x log (sin x) –[cot x + x] sin x

2.2.2. Tutorial Problems1. Solve + 4y = 4 tan 2x by using method of variation of parameters2. Solve (D2 + 4) y = sec 2x by the method of variation of parameters3. 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



SCE 24 Dept of S&H

x = (or) z = log xnow, =

=

x =

x = D’y where D’ =2.3.1 Problems Based On Cauchy’s Type

.1. Solve − x + y = 0Solution:

Given: [x2 D2 –xD +1] y =0Put x= ez

log x = zxD = D’

x2D2 = D’ (D’ – 1)[D’ (D’ – 1) - D’ + 1] y =0

[(D’)2 -2D’ +1] y =0auxiliary equation is − 2 + 1 = 0

(m-1)2 =0m = 1,1

y= (Az +B) ez

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

Solution: Given: xD2 y + Dy = 0

[ xD2 + D] y = 0Put x= ez

log x = zxD = D’

x2D2 = D’ (D’ – 1)[D’ (D’ – 1) - D’ ] y =0

[(D’)2 -D’ +D’] y =0( ,)2 y = 0

The auxiliary equation is =0m = 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 D2 = a2 D’ (D’ -1) and so on.



SCE 25 Dept of S&H

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 D2 - 2(2x+3) D -12y =6x

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

2x = ez – 3x = −

let (2x+3) D =2D’(2x+3)2 D2 = 22 D’ (D’ -1)

= 4 D’ (D’ -1)[4 D’ (D’ -1) – 4D’ -12] y = 6[ − ]

[(D ) − 2 − 3] = [3 − 9]

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

Given: [(1+x)2 D + (1+x) D + y] = 2 sin [log (1+x)]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 = 0m = ± i

C.F = A cos z + B sin z= 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))

2.3.3. Tutorial Problems1. Solve − x + y = log x2. Solve [x2 D2 – 2xD - 4] y = x2 + log x

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



SCE 26 Dept of S&H

2.4 Simultaneous First Order Linear Equations With Constant Coefficients2.4.1 Simultaneous linear equations

Linear differential equations in which there are two (or) more dependent variables and a single independentVariable. such equations are known as Simultaneous linear equations. Consider the SimultaneousEquation 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)f2(D) g2(D) provied ∆ ≠ 0

2.4.2 Problems Based Simultaneous First Order Linear Equations With Constant Coefficients1. If D = , how many arbitrary constants are involved in the solution of the Simultaneous

Equations + 2x + 3y = 2edydt

+ 3x + 2y = 0Solution :

Given: Dx+2x + 3y = 2e(D+2)x +3y = 2e …(1)

Dy + 3x + 2y = 03x + (D+2)y =0 …(2)

The coefficient determinant of degree (1) & (2) is

∆ = D + 2 33 D + 2

= (D+2)2 – 9It is an expression in D of degree 2,

∴ The number of arbitrary constants in the solution is 2.

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

Given: + 2y = − sin tDx +2y = - sin t … (2)

× D => D2 x + 2Dy = -D(sin t)D2 x + 2Dy = - cos t …(3)

(1) × 2 => 2Dy – 4x = 2 cos t-4x +2Dy =2cos t …(4)

(2) – (4) => (D2 +4) x = -3 cos tIf we eliminate y we get (D2 +4) x = -3 cos t

2.4.3. Tutorial Problems1. Solve Dx + y = sin 2t ; -x +Dy = cos 2t2. Solve the simultaneous equations

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



SCE 27 Dept of S&H

UNIT IIILAPLACE 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 theEnglish 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 equivalentform

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

0[ ( )] ( ) ( ) ( )stL f t s F s e f t dt�

� ��

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))

0[ ( )] ( ) , 0stL f t e f t dt t

� ��

3. Exponential Order

A function f(t) is said to be of exponential order if( ) 0

t

stLt e f 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,



SCE 28 Dept of S&H

�

�

2

2

2

2

( )

. ., Indeterminant form Apply L'Hospital rule

2 i.e., Indeterminant form Apply L'Hospital rule

2

2

0

t t

t

t

t

t

st stLt e f t Lt e t

tLt i estetLt stse

Lt sts e

stLt e t

��

��

��

��

��

� ��

�� "� � #�� $�� "� � #�� $

�

��

� �

Hence t2 is of exponential order.

2. Show that the function the following function is not of exponential order ( ) tf t e�2

Solution:

Given f(t) =te

2

By the definition of exponential order,

2

n n

n

st t st tLt e e Lt e

est tLt e e

��

�

��

� � ��

��

2 2

3.1.3 Define function of class A

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

3.2 Transforms Of Elementary Functions- Basic Properties

Important Results

1

1

11. [1] where 0

n!2. [ ] wher 0,1,2,...

13. [ ] where n is not a integer

14. [ ] where s > a or s - a > 0

nn

nn

at

L sS

L t e nsnL ts

L es a

�

�

� �

� �

��

��



SCE 29 Dept of S&H

2 2

2 2

2 22 2

15. [ ] where s + a > 0

6. [sin ] where s > 0

7. [cos ] where s > 0

8. [sinh ] wheres > a or s > a

atL es a

aL ats asL at

s aaL at

s a

� ��

��

��

��

2 22 29. [cosh ] where s > a

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

sL ats a

Linearity propertyL af t

��

� � �3.2.1 Problems Based On Transforms Of Elementary Functions- Basic Properties

1. Find L[t5 + e3t + 5e-2t]

Solution:

5 3 2 5 3 2

5 1

[ ] [ ] [ ] [5 ]5! 1 5

3 2

t t t tL t e e L t L e L e

s s s

� �

�

� � � � �

� � ��

5 3 26

5! 1 5[ ]3 2

t tL t e es s s

��

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

Solution:

2 2 2 2 2 2 2 2

2 2 2 2 2 2

[sin 2 cos 8cosh 7 sinh ]2 8

2 72 8

4 49

L t t t bts s b

s s s s bs s b

s s s s b

�

�

�

� � �

� � � ��

� � � ��

13. Find Lt

� "� #� $

Solution:1

21L L tGt

iven �� " � "�� # � #� $� $

� �1 12

1 12s

� �

� ��

12

12

s

s�

�

�



SCE 30 Dept of S&H

1Hence Lst�� "

�� #� $

3.2.2 First Shifting Theorem

� � � � , the [ ( )] ( )n atIf L f t F s L e f t F s a� � �� "� $

� � � � , then [ ( )] ( )atIf L f t F s L e f t F s a� �� "$

3.2.3 Second Shifting Theorem

� � � �(t a), t a

and G(t,0

)If L f tt a

F sf � ��

�� " �� $ ��

�[G(t)t (hen ] )ase F sL ��

3.2.4 Problems Based On First And Second Shifting Theorem

1. Find L[tn e-at]

Solution:

(s a)

1(s a)

[ ] [ ]

!

n at ns

ns

L t e L t

ns

�

� �

��

� "� � $

� "� � #� $

1![ ]

( )n at

nnL t e

s a�

�

� "� � � #�� $

2. Find L[eat sinh bt]

Solution � (s a)[ sinh bt] [sinh bt]at

sL e L

� ��

2 2

(s a)s

bs b

� �

� "� � #�� $

2 2[ sinh bt](s a)

at bL eb

� ��


1. Find L[cos 4t sin 2t]2. Find L[sinh22t]3. Find L[cos(3t-4)4. Find L[e-tt9]

3.3 Transforms Of Derivatives And Integrals Of Functions

Properties:

[ '( )] s L[ ( )] (0)L f t f t f� �2[ ''( )] s L[ ( )] (0) '(0)L f t f t s f f� � �



SCE 31 Dept of S&H

3.3.1 Transform of integrals

0

1If [ ( )] (s), then L (u)du [ (t)]t

L f t F f L fs

� "� �� #

� $�

3.3.2 Derivatives of transform

[ ( )] (s) then [ ( )] ( ) '(s)dIf L f t F L t f t F s Fds

� � � � �

( )[ ( )] (s) then [ ( )] ( 1) (s)n n nIf L f t F L t f t F� � �

3.3.3 Problems Based On Derivatives Of Transform

1. Find L[t sin at]

Solution:We know that

[ ( )] (s) then [ ( )] ( ) '(s)dIf L f t F L t f t F s Fds

� � � � �

� �

2 2

2 2

22 2

[ sin ] [sin ]

( )(0) (2s)

dL t at L atdsd ads s a

s a a

s a

� �

� "�� #�� $

� �� !

� �22 2

(2s)a

s a

� �� !

� �22 2

2 s[ sin ] aL t ats a

� ��

0

2. cos 0tShow that e t t dt�

� ��Solution:

10

cos [ [ cos ]]tsGiven e t t dt L t t

��

��

1

21

(cos t)

1

s

s

d Ldsd sds s

�

�

� "� �� #� $

� "� �� #�� !� $



SCE 32 Dept of S&H

� �

� � � �

2

22

1

2 2 2

2 22 2

1 1

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

1 2s 1 s [ 0]1 1

s

s s

s

s

s

s s

�

� �

� "� "� �� #� #� �� #� #�� $� $

� " � "� " � "� � �� # � #� # � #� � � � � �� # � #� # � #� �� $ � $� $ � $

0

cos 0te t t dt�

��


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 Transformsin 3 cos1. t tFind L

t� "� #� $

Solution:

1 1 1

sin 3 cos sin 4 sin 2Given2

1 sin 4 sin 221 sin 4 sin 22

1 sincot cot cot2 4 2

t t t tL Lt t

t tLtt tL L

t t

s s at sLt a

� � �

�� " � "�� # � #� $ � $�� "� � #� $

� "� " � "� �� #� # � #� $ � $� $

� � � �� " � �� #� ! � ! � $ � !� ! � !�

1 1sin 3 cos 1 cot cot2 4 2

t t s sLt

� �� " � � � �� #� $ � ! � !� !

cos cos2. at btFind Lt�� "

� #� $Solution:

� � � �

2 2 2 2

2 2 2 2

cos cosGiven [cos cos ]

1 log log2

s

s

s

at btL L at bt dst

s s dss a s b

s a s b

�

�

�

�� " � �� #� $

� "� �� #� �� $

� "� � � �� $

�

�



SCE 33 Dept of S&H

� �

2 2

2 2

2 2

2 2

1 log2

1 0 log log 1 02

s

s as b

s as b

�� "�

� � #�� $

� "�� #�� $

�

2 2

2 2cos cos 1 log

2at bt s aLt s b

� "� �� "� � � #� # �� $ � $

3.3.6 Tutorial Problems:11. (cosh 2 sinh 22

tFind L e t t� "�� #� $

20

1 cos2 2.Using Laplace transform p t dtt

rove that ��

��

3.4 Transforms Of Unit Step Function And Impulse Function

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 as0

(t )1for t a

U afor 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

0

[ ( )] (t a) dtstL U t a e U�

��

0

(0)dt (1)dt

dt

0

ast st

a

st

a

st

s

sa

e e

e

es

es

��

��

��

�

� �

�

� "� � #�� $

� "� � � #�� $

� �

�

[ ( )]aseL U t as

�� "� � � � #

� $



SCE 34 Dept of S&H


1. Give the L.T. of the Dirac Delta function2. 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 everyvalue of x. The smallest positive value of p for which this equation is true for every value of x will be calledthe period of the function.

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

0

1 (t)dt1

pst

ps e fe

��

3.5.1 Problems Based On Transform Of Periodic Functions

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

sin , 0(t)

20,

t tf

t

��

�

� �

� �

� � ��

� � ��

Solution:

We know that

� ��

� �

2

2

2

2

2

0

0

2 20

2 2

2 2

2 2

1[ (t)] (t) dt1

1[sin ] sin dt 01

1 [ s sin cos ]1

11

1

1 1 ( )

1 ( )

s

s

s

s

s

s

s s

s

st

st

st

L f e fe

L t e te

e t tse

ese

e

e e s

e s

��

��

��

��

��

��

��

��

��

� ��

��

� �

� � ��

� �

�

�

�

�

�

�

�

�

�

�

�

� �

�

�

�

�

��

� �� !

� �� !

� ��

�� !

��

� � �

��

�

�

� �2 2[sin ]

( ) 1sL t

s e��

��

��

� ��



SCE 35 Dept of S&H


1. Find the Laplace Transform of triangular wave function, 0 t a

(t)2 , 2 (t 2a) (t)t

fa t a t a with f f

� ��

� � � � ��

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

1 , 02(t)

1,2

atf

a 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 then1 1 1

1 1 2 2 1 1 2 2[C (s) C (s)] C [ (s)] C [ (s)]L F F L F L F� � ��

Important Formulas

1

11

1

12 2

12 2

12 2

11. 1

12.1

13.

4. cosh

1 15. sinh

1 16. sin

n

n

at

Ls

tLs n

L es asL at

s a

L ats a a

L ats a a

�

��

�

�

�

�

� " �� #� $

� " �� # �� $

� " �� #�� $

� " �� #�� $

� " �� #�� $

� " �� #�� $



SCE 36 Dept of S&H

� �

12 2

1

12 2

12 2

12 2

12 2

122 2

7. cos

8. [ ( )] (t)1 19. sin

( )

10. cos( )

1 111. sinh( )

12. cosh( )

113.

at

at

at

at

at

sL ats a

L F s a e f

L e bts a b b

s aL e bts a b

L e bts a b b

s aL e bts a b

sLs a

�

�

�

�

�

�

�

� " �� #�� $

� �

� "�� #� �� $

� "�� #� �� $

� "�� #� �� $

� "�� #� �� $

� "� # �� #�� $

� �

� �

� �

21

22 2

12 32 2

2 21

22 2

1

t sin2

114. [sin cos ]2

1 115. (sin cos )2

16. t cos

17. [1] (t)

ata

sL at at atas a

L at at atas a

s aL ats a

L �

�

�

�

�

� "� # � �� #�� $

� "� # � �� #�� $

� "�� # �� #�� $

�

3.6.1 Problems based on Inverse Laplace Transform

12.

161 2sFi d

sn L� � "

� #�� $

Solution:

Given1 1

2 22 2

16 162cosh 4

s sL Ls s

t

� �� " � "�� # � #� �� $ � $�



SCE 37 Dept of S&H

2. Find1

23

4 13sL

s s� �� "

� #� �� $Solution:

1 12 2

12

12

1 12 2 2 2

2 1 12 2 2 2

2 2 12 2

3 34 13 (s 2) 13 4

3(s 2) 9

2 5(s 2) 9

2 15(s 2) 3 (s 2) 3

5 3s 3 3 (s 2) 3

5 3cos33 s 3

t

t t

s sL Ls s

sL

sL

sL L

se L L

e t e L

� �

�

�

� �

� � �

� � �

� "� �� " � � #� #� � � � �� $ � $

� "�� #� �� $

� "� �� #� �� $

� " � "�� # � #� � � �� $ � $

� "� "� � � #� #� � �� $ � $

��

"� #� $

1 2 22

3 5cos3 sin 34 13 3

t tsL e t e ts s

� � �� "� � �� #� �� $

3.6.2 Inverse Laplace Transforms of derivatives of F(s)

1 1

1

[ ( )] ( ), then [ '( )] ( )[ ( )]

L F s f tIf L F s t f tt L F s

� �

�

� � �

� �

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

12 2 2(s a

1.)

sFind L� � "� #�� $

Solution:

2 2 2'(s)(s a )

sLet F � "� � #�� $

2 2 2'(s)ds(s a )

sF ds� "� � #�� $

� �

2 2 2(s)(s a )

sF ds� "� � #�� $�

2 2s a2

2

Put ts ds dt

dtsds

� �

�

�



SCE 38 Dept of S&H

2

2 2

1 1 12 2

12

1(s)2(s a )

dtt t

t

F

�� "� � � #� $

� �

� � ��

�

3.6.4 Inverse Laplace Transform of Integrals

1 1

1 1

1 1(s)ds (t) [ (s)]

(or)

[ (s)] (s)ds

s

s

L F f L Ft t

L F t L F

��

��

� "� �� #

� $

� "� � #

� $

�

�

3.6.5 Problems based on Inverse Laplace Transform of Integrals

12 221.

(s 1)sFind L� � "

� #�� $

Solution:

We know that

1 1[ (s)] (s)dss

L F t L F�

� � � "� � #

� $�

1 12 2 2 2

12

2 2(s 1) (s 1)

1(s 1)

s

s

s sL t L ds

t L

��

�

�

� "� "� � #� #� �� $ � $

� "� �� #� �� #� !� $

�

12

12

101

11

t Ls

t Ls

�

�

� "� �� #�� $

� "� � #�� $

3.6.6 Problems based on Partial fractions method

21

35 15 111.(s 1)(s 2)s sFind L� � "� �

� #� �� $

12 22 sinh

(s 1)sL t t� � "

� �� #�� $



SCE 39 Dept of S&H

Solution:

Consider2

3 2 3

2 3 2

5 15 11(s 1)(s 2) 1 2 (S 2) (S 2)5 15 11 (S 2) (S 1)(S 2) (S 1)(S 2) D(S 1)

s s A B C Ds S

s s A B C

� ��

� � � � � �

� � � � � � � � � � � �

3

Put s = -1, we get5 15 11 ( 1 2)

9 271

3

AA

A

� � � � � �

��

��

Equating the coefficients of s3 on both sides, we get0

13

A BB A

B

� �

� �

�

Put s = 2, we get21

7D

D� �

� �Put s = 0, we get

11 8 4 21 18 4 2 7

3 38 44 23 3

8 24

A B C D

C

C

CC

� ��

�� ! � !

� � � �

� � �

�

2

3 2 3

1 15 15 11 4 73 3(s 1)(s 2) 1 2 (S 2) (S 2)s s

s S

��

� � � � ��

21 1 1 1 1

3 2 3

2 2 1 2 12 3

2 2 2 13

5 15 11 1 1 1 1 1 14 7(s 1)(s 2) 3 1 3 2 (s 2) (s 2)

1 1 1 14 73 31 1 7 24

3 3 2

t t t t

t t t t

s sL L L L Ls s

e e e L e Ls s

e e e t e Ls

� � � � �

� � �

� �

� " � " � "� � � � " � "� � � �� # � # � #� # � #� � � � � �� $ � $ � $ � $� $� � " � "� � � �� # � #� $ � $� � "� � � � � #� $

21 2 2 2 2

35 15 11 1 1 74(s 1)(s 2) 3 3 2

t t t ts sL e e e t e t� �� "� � �� #� �� $

3.6.7 Second Shifting property

1[e (s)] ( ) ( )asL F f t a U t a� � � � �



SCE 40 Dept of S&H

3.6.8 Problems based on second shifting property

11.3

seFind Ls

�� "

� #�� $

Solution:

Consider

1 3

1 3( )

13

(t )3

t

st

L eseL e Us

��

� �

��

� " �� #�� $

� "� �� #�� $


a. Find the inverse L.T of Derivatives.

211. log 1s

� �� !

12. tan ( 1)s� �b. Partial Fraction Method

4 4

2

1.4

3 12.(s 1)(s 1)

ss a

s�

��

3.6.10 Change of scale property

1 1

1[ (t)] F(s), then L[ ( )

1(t) L [ (s)], then L [ ( )]

sIf L f f at Fa a

tIf f F F cs fc c

� �

� "� � � #� $

� "� � � #� $

3.6.11 Problems based on Change of scale property

1. [ (t)] F(s) find L tIf L f fa

� "� �� #� !� $Solution:

We know that 0

[ (t)] (t) dtstL f e f�

��

0

t dt

Put u = 0 0

sttL f e fa a

t as t uadtdu t ua

�� "� � � �� #� ! � !� $

� � �

� ��

�



SCE 41 Dept of S&H

� � (au)

0

au

0

at

0

(u) a du

(u) du

(t) dt

[ ]

s

s

s

tL f e fa

a e f

a e f

a F as

��

��

��

� " �� $

�

�

�

�

�

�


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

12 2 22. sFind Ls a b

� � "� #�� $

3.7 Convolution Theorem

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

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

0

(t) g(t) (u) g(t u) dut

f f� � ��

Note: Convolution Integral or Falting integral

1 1(s a)(s b

2.)

Using convolution theorem fi Lnd � � "� #� �� $

Solution:

1 1 1[ (s) G(s)] [ (s)] [G(s)]L F L FWe know tha Lt � � ��

1 1 11 1 1 1

at bt

L L Ls a s b s a s b

e e

� � �

� �

� " � " � "� � �� # � # � #� � � �� $ � $ � $

� �

Here ( ) =

( ) =



SCE 42 Dept of S&H

∴

0

(t u)

0

0

( )

0

( )

0

( ) t

1

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

du

du

du

(a b)

1(a b) (a b)

1

1 1 1

t

tau b

tau bt bu

tbt a b u

ta b ubt

a bbt

btat bt

bt at

f f

e e

e e e

e e

ee

ee

e e ea b

L e es a s b a b

� � �

� �

� � �

� ��

� ��

��

� � �

� � �

�

�

�

� "� � #� �� $

� "� �� #� � � �� $

� "� �� $�

� " � "� � �� $� #� � �� $

�

�

�

�

3. Using convolution theorem find

121

(s 1)L

s� � "

� #�� $

Solution:

We know that 1 1 1[ (s) G(s)] [ (s)] [G(s)]L F L F L� � ��

�

1 1 12 2

0

0

1 1 1(s 1) 1

1 sinsin 1 ( (t) g(t) g(t) f(t))

sin

cos( cos t) ( 1)

t

u t

u

L L Ls s s

tt f

u du

u

� � �

�

�

� " � " � "� �� # � # � #� �� $ � $� $� �

� � � � �

�

� �

� � � �

�

�

121 1 cos t

(s 1)L

s� � "

� � �� #�� $4. Using convolution theorem find

12 2 2(s a )sL� � "

� #�� $



SCE 43 Dept of S&H

Solution:

� �

�

1 1 12 2 2 2 2 2 2

1 12 2 2 2

0

0

1(s a ) (s a ) (s a )

1(s a ) (s a )

1cos sin

1 cos sin

1 cos sin (t u)du

1 cos sin ( t ) du

1 sin ( t ) sin ( t )2

t

t

s sL L L

s aL La

at ata

at ata

au aa

au a aua

a au au a au aua

� � �

� �

� " � " � "� �� # � # � #� � �� $ � $ � $

� " � "� �� # � #� �� $ � $

� �

� �

� �

� �

� � � � �� "� ��

�

�

0

dut

#$�

�

�

0

0

0

0

1 sin ( t) sin ( t 2 ) du2

1 sin ( t) sin ( t 2 ) du2

1 cos (t 2u)(sin at) u2 2

1 cos (t 2u)(sin at) u2 21 cos cossin at 0

2 2 2

t

t

t

t

a a ua

a a ua

aa a

aa a

at atta a a

� � �

� � �

� � � "� �� #�� !� $

�� "� �� #� $

� "� � � �� #� ! � !� $

�

�

12 2 2

1 sin at(s a ) 2

sL ta

� � "� �� #�� $


Find the inverse Laplace Transform using convolution theorem.

2 2

2 2

2 2 3

41.(s 2s 5)

12.(s 4)(s 4)

3.(s a )

s

� �

� �

�



SCE 44 Dept of S&H

3.8 Initial and final value theorems3.8.1 Initial value theorem

0[ (t)] F(s), then (t) (s)

t sIf L f Lt f Lt sF

� ��

3.8.2 Final value theorem

0[ (t)] F(s), then (t) (s)

t sIf L f Lt f Lt sF

��

3.8.3 Problems based on initial value and final value theorems

0

11. If [ (t)] , find (t) and (t)(s a) t t

L f Lt f Lt fs ��

��

Solution:

We know that

0(t) (s)

t sLt f Lt sF� ��

�

1s(s a)1

(s a)1

s

s

Lt s

Lt

��

��

��

��

��

0(t) 0

tLt f�

� �

We know that

0(t) (s)

t sLt f Lt sF��

�

0

0

1(s a)1

(s a)

s

s

Lt ss

Lt

�

�

��

��

1(t)tLt f

a��

2. Verify the initial and final value theorem for the function(t) 1 e (sin t cos t)tf ��

Solution:

Initial value theorem states that

0(t) (s)

t sLt f Lt sF� ��

�



SCE 45 Dept of S&H

1

2 2

2

0

2

2

2

22

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

1 1 1( 1) 1 ( 1) 1

1 2( 1) 1

L.H.S (t) 1 1 2

1 2R .H.S( 1) 1

( 2)1( 1) 1

2(1 )1

2 2s 1

s s

t

s

s

s

L f L t ts

ss s s

ss sLt f

sLt ss s

s sLts

ssLt

s s

� �

�

��

��

��

� � � �

��

� � � �

��

� �

� � � �

� "�� #� �� $

� "�� #� �� $

� "�� #

� #� �� #� �� #� !� $

2

2(1 )1

2 21

1 1.H.S 2

L.H.S = R.H.S

s

sLt

s s

R

��

� "�� #

� #� �� #� �� #� !� $

� �

�

Initial value theorem verified.

Final value theorem states that

0(t) (s)

t sLt f Lt sF��

�

20

. .S 1 (sin t cos t)

1 0 1( 2). . 1

( 1) 11 0 1

L.H.S = R.H.S

t

t

s

L H Lt e

s sR H S Lts

�

��

�

� "� � �� $

� � �

� "�� #� �� $� � �

Final value theorem verified.


Verify the initial and final value theorems for the functions2 31. (t) t tf e��

2

3 25 22. (s)

4 2s sIf Fs s s

� ��

� �(0) and ( )find f f �

3. (t) btf ae��



SCE 46 Dept of S&H

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

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

Solution:� �'' 3y' 2 y e (0) 1, y' 0 0

Taking L.T on bothsides,

ty and y��

� � �2

2

2

2

''(t) 3 y'(t) 2 y(t) e

1[ (t)] sy(0) '(0) 3[ [ (t)] y(0)] 2 L[y(t)]1

1[ (t)] s 0 3 [ (t)] 3 2 L[y(t)]1

1(s 3s 2) L[y(t)] 31

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

tL y L L L

s L y y sL ys

s L y sL ys

sss ss

�� "� � � � $

� � � � � ��

� � � � � ��

� � � � ��

� � ��

2

2

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

2 2 (s 1)(s 2) B(s 1)(s 2) C(s 1)(s 1)1,

1 2 2 23 2

322,

4 4 2 32

31,

1 2 2 6161 6 3 2 2 3[ (t)]

1 1 21 16

s s A B CL ys s s

s s APut s we get

BB

B

Put s we getC

C

Put s we getA

A

L ys s s

s

� ��

� � � � � �

� � � � � � � � � � �

�

� � � �

� � �

�

�

� � �

��

� �

� � �

�

� � ��

��

1 1 1

2

3 1 2 11 2 1 3 2

1 1 3 1 2 1(t)6 1 2 1 3 21 3 26 2 3

t t t

s s

y L L Ls s s

e e e

� � �

�

� ��

� " � " � "� � �� # � # � #� � �� $ � $ � $

� � �



SCE 47 Dept of S&H

3.9.2 Tutorial Problems:2

0

1. Solve : 2t

y ydt t t� � ��22. Solve (D 5D 6) y 2, given y(0) 0, y'(0) 0� � � � �2

23. 2 2 0d y dySolve ydx d

given hatx

t� � � 0 . .1 using LTdyy at xdx

method� � �



SCE 48 Dept of S&H

UNIT-IV ANALYTIC FUNCTIONS 4.1 Introduction: 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: = 4.1.6 Neighbourhood of a Point ��: 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 | − |. ii. The circle C of radius � with centre at the point is | − | = �

iii. | − | < � represents the interior of the circle excluding is circumference iv. | − | � represents the interior of the circle including is circumference v. | − | > � represents the exterior of the circle excluding is circumference

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: If f(z) is said to be continuous a z= then lim → = .

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.



SCE 49 Dept of S&H

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 joining the points z and that lie in the 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 form: 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 50 Dept of S&H

C-R equations are satisfied. = + �

= + � � = + � �

= � = +� = 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. Test the analyticity of the function = � . 3. Prove that = ℎ is an analytic function and find its derivative

4.3 Harmonic and Orthogonal Properties Of Analytic Functions 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

vi. �� | | + �

� | | = | ′ |

vii. ∇ |� | = | ′ | viii. ∇ |� | = | ′ |



SCE 51 Dept of S&H

Solution: i. Let f(z)=u+iv

Here f(z) is an analytic function = and = −

| | = | − � | =√ + | | = +

�� + �

� | | =�

� + �� +

= �

� + + �� +

=

�� + �

� + �� + �

�

� � = �

�

�� = �

� ( � � ) = �

� + (� � )

�� = �

� ( � � ) = �

� + (� � )

�

� + �� = ( �

� + � � )+ ( �

� + � � )

=2� � + �

� + �� + �

�

=2 + =2| ′ |

ii. ∇ | | =0 ,if f(z)f’(z)≠ in D

| | = +

| | = +

� � | | = �

� ( + )=++

�

� | | =( + ) + + + − + +

+

�

� | | =( + )( + + + )−( + )( + )

+

�� + �

� | | =

( + )( + + + + + + + )

− ( + ) − ++

=( + ) + + + − (− + ) − ( + )

+

=( + ) + + + − (− + ) − ( + )

+ =0

iii. ∇ =p(p-1) − | ′ |

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



SCE 52 Dept of S&H

= and = −

+ = + = + =

f ’(z)= + � | ’ | = + �

� = ��

−

�� = − + − −

= − + − − �

� = ( − + − − )

�� + �

� = − ( + ) + − − ( + )

= − + − − + − = − − | ’ |

iv. ∇ | | = | | − | ′ |

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

| | = +

| | = + �

� + = + − +

= +�− +

�� + = + − + + +

+ + − +�− +

�� + = + − + + | ′ |

+ − + + +�−

�� + = + − + + | ′ |

+ − + + +�−

�� + �

� + = + − + + + + | ′ |

− + + + ++ + −

= +�− 2| ′ | + − +

�−

+ ( + )



SCE 53 Dept of S&H

= +�− 2| ′ | + − +

�− +

= +�− 2| ′ | + − +

�− | ′ |

= + − | ′ | + −

= + − | ′ |

= | | − | ′ |

v. ∇ =0

= −

�� =

��

−

=+

−

= +−

= + −

�

� ( )=�

�−+

= + + − − − − ++

= + + − − − − ++

= + − − − + −+

�� ( ) = + ( − ) − ( − ) + −

+

�� + �

� ( )

=

+ ( + ) − ( + ) − − ( + )+

− + −+ =

vi. �� | | + �

� | | = | ′ |

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

| | = + �

� | | =√ +

+

=√ +

+

�� | | =

+√ +



SCE 54 Dept of S&H

�� | | =

+√ +

�� | | + �

� | | = +√ + +

+√ +

=+ + + + +

+

=+ |� |

+ =| |

vii. ∇ |� | = | ′ | Let f(z)=u+iv is an analytic function � = |� | = �

� =

�� = �

�

= + �

� = ( + )

�� + �

� = + + +

=2| ′ | viii. ∇ |� | = | ′ | Let f(z)=u+iv is an analytic function � = |� | = �

� =

�� = �

�

= + �

� = ( + )

�� + �

� = + + +

=2| ′ | 4.3.3 Problems Based On Harmonic Conjugate

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

Solution: Given f(z)= = + � � u = = �

= = � = = �

=- � =

=− = − �

Type equation here. + = − =0



SCE 55 Dept of S&H

+ = � − � =0

They are harmonic functions. 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 = − + − +

� , = �� = − +

� , = +



SCE 56 Dept of S&H

� , = �� = − −

� , =

= ∫ � , − � ∫ � , + c

= ∫ + − � ∫ + c

= + +

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

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 = − � −

3. Determine the analytic function whose real part is �

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



SCE 57 Dept of S&H

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: The transformation = + , where C is a complex constant ,represents

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

5. Inversion And Reflection:

The transformation = represents inversion w.r.to the unit circle | | = , 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 Solution: Let, = + � = + � Given , =

+ � = + � + � = + �

= ⟹ =



SCE 58 Dept of S&H

= ⟹ =

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=−+

given: | − �|=2 | + � − �|=2 + − =

+ − + = + − =

u+ + −v

+ = −v+

++ = −

+

= − = − which is a straight line

4. Discuss the conformal transformation w=sinz

Solution: Given, w=sinz

+ � = � + � = � � + � � = � ℎ + � � ℎ

= � ℎ , = � � ℎ Elimination of y using the formula,

ℎ − � ℎ =

� − =

Case I: If 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 ) =



SCE 59 Dept of S&H

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=sinx ,v=0 If x is –ve, u is –ve x is +ve, u is +ve x=0, u=0 If x varies from −∞ ∞ then u varies from −∞ ∞ Hence the image of the line y=0 in the z-plane is the complete real axis in the w-plane. 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


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 − −− − = − −

− −

iv. Given four points , , , in this order ,the ratio − −− − is called the cross

ratio of the points.



SCE 60 Dept of S&H

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 =

viii. Each point in the plane except z= − corresponds to a unique point in the w-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:

= �+− �

− � = � + − � − =

= � ± √− +

= �, �

2. Find the bilinear transformation that maps the points ∞, �, onto , �, ∞ respectively. Solution: Given:

= ∞, = �, = & = , = �, = ∞ Let the transformation be, − −

− − = − −− −

− −

− −=

− −

− −

− −

− −=

− −

− −

− −

− �− = − �−− −

� = �

= �



SCE 61 Dept of S&H

= −

3. Find the bilinear transformation that maps the points , �, − onto �, , −� respectively.

Solution: Given:

= , = �, = −1 & = �, = , = −� Let the transformation be,

− −− − = − −

− −

− � + �+ � − � = − � +

+ � −

− � � + � −� = − � +

+ � −

− − �+ � = − � + � +

+ � − � +

− − �+ � = − �

+ −

− − �+ � = − −�

+

− �+ � = − �

+

− �+ � = � − �

+

+ + � − � = � + � − � − � + + � − � = � − − � +

= − + � +− � + + �


1. Find the bilinear transformation that maps the points + �, −�, − � of the Z-plane into the points , , � of the W-plane.

2. Find the bilinear transformation which maps = onto = − � and has − as 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



CHAPTER 5COMPLEX INTEGRATION

5.1 PrerequisitesBefore starting this topic students should be able to carry out

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

5.2 IntroductionComplex integration is an intuitive extension of real integration. Since

a complex number represents a point on a plane while a real number is anumber on the real line, the analog of a single real integral in the complexdomain is always a path integral. For some special functions and domains,the integration is path independent, but this should not be taken to be thecase in general. Given the sensitivity of the path taken for a given integraland its result, parametrization is often the most convenient way to evaluatesuch integrals.Complex variable techniques have been used in a wide varietyof areas of engineering. This has been particularly true in areas such aselectromagnetic field theory, fluid dynamics, aerodynamics and elasticity.

5.3 Cauchy’s Theorem

5.3.1 Definitions

5.3.1.1Connected RegionA 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 regionA 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 iscalled a simply connected region.

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

5.3.1.4 Cauchy’s Integral TheoremIf a function f(z) is analytic and its derivative f

0(z) is continuous atall points inside and on a simple closed curve c, then

Rc

f(z)dz = 0.

SCE 62 Dept of S&H



5.3.1.5 Cauchy’s Integral formulaIf 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

2⇡i

Rc

f(z)

z�a

dz

the integration around c being taken in the positive direction.

5.3.1.6 Cauchy’s integral formula for derivativeIf a function f(z) is analytic within and on a simple closed curve c and

a is any point lying in it, then

f

0(a) = 1

2⇡i

Rc

f(z)

(z�a)

2dz

5.3.2 Worked out examples

1. EvaluateRC

z

z�2

dz where c is a circle |z| = 1.Solution.Let f(z) = z

z�2

z = 2 lies outside c.) f(z) is analytic inside and on c .f

0(z) is continuous inside c

Hence by Cauchy’s theoremRc

f(z)dz = 0

2. EvaluateRC

1

2z�3

dz where c is a circle |z| = 1.Solution.Given

Rc

1

2z�3

dz = 1

2

1

z� 32

dz z = 3

2

lies outside c.

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

0(z) is continuous inside c

Hence by Cauchy’s theoremRc

f(z)dz = 0

3. EvaluateRC

1

2z+3

dz where c is a circle |z| = 2.Solution.

1

2z + 3dz =

Z

c

1

2(z + 3

2

)dz

=1

2

Z

c

1

z + 3

2

dz

=1

22⇡if(

�3

2)

= ⇡i

1

2z + 3dz

= ⇡i.

SCE 63 Dept of S&H



4. EvaluateRC

sin 3z

z+

⇡2dz where c is a circle |z| = 5.

Solution.W.K.T Cauchy’s Integral formula is

Rc

f(z)

z�a

dz = 2⇡if(a)

GivenRC

sin 3z

z+

⇡2dz

Here f(z) = sin 3z, a = �⇡2

lies inside |z| = 5.

f(a) = sin(�3⇡

2) = 1

) By Cauchy’s Integral formula, we getZ

C

sin 3z

z + ⇡

2

dz = 2⇡i

5.3.3 Tutorial Problems

1. EvaluateRC

3z

2+7z+1

z+1

dz where c is a circle |z| = 1

2

.

2. EvaluateRC

e

12dz where c is a circle |z| = 1.

3. EvaluateRC

e

1z2+4

dz where c is a circle |z| = 3.

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)+f

0(a)(z�a)+ f

00(a)

2!

(z�a)2+ f

000(a)

3!

(z�a)3+· · ·+ f

(n)(a)

n!

(z�a)n+. . .

5.4.2 Laurent’s Series.Let C

1

, C

2

be two concentric circles |z � a| = R

1

and |z � a| = R

2

where R

2

< R

1

. Let f(z) be analytic on C

1

andC2

and in the annular regionR between them. Then, for any point z in R,

f(z) =1X

n=0

a

n

(z � a)n +1X

n=0

b

n

(z � a)n

where

a

n

=1

2⇡i

Z

c1

f(z)

(z � a)n+1

dz

SCE 64 Dept of S&H



b

n

=1

2⇡i

Z

c2

f(z)

(z � a)1�n

dz

where the integrals being taken anticlockwise.

5.4.3 Worked out examples

1. Expand e

z in a Taylor’s series about z = 0Solution.

Function Value at z = 0f(z) = e

z f(0) = 1f

0(z) = e

z f’(0) = 1f

00(z) = e

z f”(0) = 1f

000(z) = e

z f”’(0) = 1...

...

) Taylor’s series about z = 0 is

f(z) = f(0) + f

0(0)(z) +f

00(0)

2!(z)2 +

f

000(0)

3!(z)3 + . . .

= 1 + 1(z) +1

2!(z)2 +

1

3!(z)3 + . . .

2. Expand 1

z�2

at z = 1 in a Taylor’s series.Solution.

Function Value at z = 1f(z) = 1

z�2

f(1) = -1

f

0(z) = �1

(z�2)

2 f’(1) = -1

f

00(z) = 2

(z�2)

3 f”(1) = -2

f

000(z) = �6

(z�2)

4 f”’(1) = -6...

...

SCE 65 Dept of S&H



) Taylor’s series about z = 1 is

f(z) = f(1) + f

0(1)(z) +f

00(1)

2!(z)2 +

f

000(1)

3!(z)3 + . . .

= �1 + (�1)(z � 1) +�2

2!(z � 1)2 +

�6

3!(z � 1)3 + . . .

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

3. Expand f(z) = 1

z(z�1)

as a Laurent’s series in powers of z and state therespective 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 region1)0 < |z| < 1(Deleted disc)2) |z| > 1Case i)For all z in 0 < |z| < 1

1

z(z � 1)=

�1

z(1� z)

=�1

z

[1 + z + z

2 + . . . ]

= �[1

z

+ z + z

2 + . . . ]

Case ii)For all z in |z| > 1 we have |1

z

| < 1.

1

z(z � 1)=

1

z

1

z � 1

=1

z

2

[1� 1

z

]�1

=1

z

2

[1 +1

z

+1

z

2

+ . . . ]


1. Expand f(z) = cos z about z = ⇡

3

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

4

in Taylor’s series.3. Find Laurent’s series expansion of 1

z

2+3z+2

in the region 1 < |z| < 2

SCE 66 Dept of S&H



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 = z

0

in R

then z

0

is called a zero of f(z).

5.5.1.3 Simple Zero:If f(z

0

) = 0 and f

0(z0

) 6= 0 then z = z

0

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 � z

0

)m�(z) where �(z) is analytic and �(z0

) 6= 0

5.5.1.5 Singular Points:A point z = z

0

at which a function f(z) fails to be analytic is called asingular point.5.5.1.6 Entire function

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

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

5.5.2 Types of Singularities

5.5.2.1. Isolated SingularityA point z = z

0

is said to be isolated singularity of f(z) if1. f(z) is not analytic at z = z

0

2. There exist a neighbourhood of z = z

0

containing no other singularity.5.5.2.2. Removable singularity:

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

0

is zero then the point z = z

0

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= 0then 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

SCE 67 Dept of S&H



the point z0

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

0

is called essential singularity.

5.5.3 Worked out Examples

1. Find the zeros of z

3�1

z

3+1

.Solution.The zeros of f(z) are given by f(z) = 0That is z3 � 1 = 0.) z = 1,!,!2

2. Find the zeros of sin z�z

z

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

f(z) =sin z � z

z

3

=�z

3

3!

+ z

5

5!

� . . .

z

3

=�1

3!+

z

2

5!� . . .

Now,limz!0

sin z�z

z

3 6= 0

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

z

3

Solution.Given f(z) = sin z�z

z

3

the function f(z) is not defined at z = 0.By, L’Hospital rule

limz!0

sin z � z

z

3

= limz!0

cos z � 1

3z2

= limz!0

� sin z

6z

=�1

6

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

SCE 68 Dept of S&H



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

limz!0

sin z

z

= limz!0

cos z

1= 1

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

5. Classify the singularity of the function f(z) = z�2

z

2 sin( 1

z�1

)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( 1

z�1

= 0

=) z � 2 = 0 and sin( 1

z�1

) = 0

=) z = 2andz = 1

n⇡

+ 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( 1

z�a

)the nature of thesingularity.

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

z(e

z�1)

3. Classify the nature of singularities of the functions e

z

z

2+4

5.6 Residues

5.6.1 Definitions:If z = z

0

is an isolated singular point of f(z), we can find the Laurent’sseries of f(z)aboutz = z

0

f(z) =P1

n=0

a

n

(z � z

0

)n +P1

n=1

bn(z�z0)

n

The coe�cient of 1

z�z0in the above expansion is called the residue of f(z) at

z = z

0

SCE 69 Dept of S&H



5.6.1.2 Worked out Examples

1. Calculate the residue of f(z) = 1�e

2z

z

3

Solution.Given f(z) = 1�e

2z

z

3

Here z = 0 is a pole of order 3.

Res at(z = 0) =1

2!limz!0

d

2

dz

2

z

3

1� e

2z

z

3

�

=1

2!limz!0

d

dz

⇥�2e2z

⇤

=1

2!limz!0

⇥�4e2z

⇤

= �2

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

(z�1)

2 at its pole.SolutionGiven f(z) = z

(z�1)

2

Here z = 1 is a pole of order 2.

Res at(z = 1) =1

1!limz!1

d

dz

(z � 1)2

z

(z � 1)2

�

= limz!1

d

dz

z

= 1

5.6.1.3 Tutorial Problems

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

2z

(z+1)

2 at its pole.

2. Find the residue of z

3

(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 curvec, except for a finite number of isolated singularities z

1

, z

2

, z

3

, . . . thenRc

f(z)dz = 2⇡i [sum of residues].

SCE 70 Dept of S&H



5.6.2.1Worked out Examples

1. EvaluateRc

z

2

e

1z where c is |z| = 1.

Solution.

Let f(z) = z

2

e

1z

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

f(z) = z

2

e

1z = z

2

1 +

1

z

1!+

(1z

)2

2!+ . . .

�

= z

2 + z +1

2+

1

6

1

z

+ . . .

The residue at z = 0 is 1

6

By Cauchy residue theorem,Rc

f(z)dz = 2⇡i(sum of residues)Rc

f(z)dz = ⇡i

3

2. EvaluateRc

e

z2

cos⇡z

dz where c is |z| = 1.SolutionLet f(z) =

Rc

e

z2

cos⇡z

dz

The singular points are obtained by

cos ⇡z = 0

⇡z = (2n+ 1)⇡

2, n = ±1,±2, . . .

= ±1

2,±3

2,±1

2, . . .

Here z = 1

2

and z = �1

2

lies inside c. R1

⇥f(z), 1

2

⇤=

�(

12 )

0(

12 )

f(z) = e

z2

cos⇡z

= �(z)

(z)

R

1

⇥f(z), 1

2

⇤= e

14

�⇡

R

2

⇥f(z), �1

2

⇤= e

14

⇡

By Cauchy residue theorem,Rc

f(z)dz = 2⇡i(sum of residues)Rc

f(z)dz = 2⇡i(sum of residues) = 0.


1. EvaluateRc

dz

sin z

where c is |z| = 4.2. If c is a circle |z| = 3, evaluate

Rc

tan zdz.

SCE 71 Dept of S&H



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 thedefinite integral is called contour integration. Here we are going to see underthree types. They are

1. Type I - Integrals of the formR

2

0

⇡f(cos ✓, sin ✓)d✓ where f is rationalfunction in cos ✓ and sin ✓.2. Type II - Integrals of the form

R1�1

P (x)

Q(x)

dx.

3. Type III - Integrals of the formR1�1 f(x) cos(nx)dx or

R1�1 f(x) sin(nx)dx.

5.7.2 Type I

5.7.2.1 Worked out examples

1. EvaluateR

2⇡

0

d✓

2+cos ✓

.

Solution.Let z = e

i✓

dz = izd✓

cos ✓ = 1

2

⇥z + 1

z

⇤and sin ✓ = 1

2i

⇥z � 1

z

⇤

Now,

Z2⇡

0

d✓

2 + cos ✓=

Z

c

1

2 + 1

2

⇥z

2+1

z

⇤ dziz

where c is |z| = 1.

=2

i

Z

c

dz

z

2 + 4z + 1

z

2 + 4z + 1 = 0 =) z = �2±p3

↵ = �2 +p3 is simple pole and lies inside c

and � = �2�p3 is simple pole and lies outside c.

Z2⇡

0

d✓

2 + cos ✓=

2

i

Z

c

dz

(z � ↵)(z � �)

SCE 72 Dept of S&H



Res[f(z),↵] = limz!↵

(z � ↵)1

(z � ↵)(z � �)

=1

↵� �

=1

2p3

Hence by Cauchy Residue theorem,Rc

f(z)dz = 2⇡i (sum of residues) = ⇡ip3

)R2⇡

0

d✓

2+cos ✓

= 2⇡p3

2. Using contour integration evaluateR

2⇡

c

d✓

13+5 sin ✓.

Solution.

Let z = e

i✓

dz = izd✓

cos ✓ = 1

2

⇥z + 1

z

⇤and sin ✓ = 1

2i

⇥z � 1

z

⇤

Now,

Z2⇡

0

d✓

13 + 5 sin ✓=

Z

c

1

13 + 5 1

2i

⇥z

2�1

z

⇤ dziz

where c is |z| = 1.

=2

5

Z

c

dz

z

2 + 26

5

iz � 1

z

2 + 26

5

iz � 1 = 0 =) z = �5i, �i

5

↵ = �i

5

is simple pole and lies inside c and � = �5i is simple pole and liesoutside c.R2⇡

0

d✓

13+5 sin ✓

= 2

5

Rc

dz

(z�↵)(z��)

Res[f(z), �] = limz!�

(z � �)1

(z � ↵)(z � �)

=1

� � ↵

=5

24i

Hence by Cauchy Residue theorem,Rc

f(z)dz = 2⇡i (sum of residues) = 5

12

⇡

)R2⇡

0

d✓

13+5 sin ✓

= 2

5

( 5

12

⇡) = ⇡

6

SCE 73 Dept of S&H




1. Evaluate the integralR

2⇡

0

sin

2✓

5+4 cos ✓

d✓.

2. EvaluateR⇡

0

d✓

a

2+sin

2✓

, a > 0.

5.7.3 Type II


1. EvaluateR�1 1 x

2dx

(x

2+1)(x

2+4)

using contour integration.

Solution.

Let us considerRc

f(z)dz =Rc

z

2dz

(z

2+1)(z

2+4)

where c consist of the semi circle � : |z| = R and the bounding di-ameter [�R,R].

Now,Rc

f(z)dz =R

R

�R

f(x)dx+R�

f(z)dz

The poles of f(z) are obtained by (z2 + 1)(z2 + 4) = 0i.e., z = i,�i, 2i,�2i.

where z = i, 2i are simple poles lie inside � and z = �I,�2i are simplepoles lie outside �

R

1

[f(z), i] = limz!i

(z � i)f(z)

= limz!i

(z � i)z

2

(z + i)(z � i)(z2 + 4)

= limz!i

z

2

(z + i)(z2 + 4)

=i

2

(i+ i)(i2 + 4)

=�1

6i

SCE 74 Dept of S&H



R

2

[f(z), 2i] = limz!2i

(z � 2i)f(z)

= limz!i

(z � 2i)z

2

(z + 2i)(z � 2i)(z2 + 1)

= limz!i

z

2

(z + 2i)(z2 + 1)

=(2i)2

(2i+ 2i)((2i)2 + 1)

=1

3i

Hence by Cauchy’s Residue theorem,

Z

c

f(z)dz = 2⇡i[R1

+R

2

]

= 2⇡i

�1

6i+

1

3i

�

= 2⇡

✓1

6

◆

=⇡

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

R�

f(z)dz ! 0.i.e.,

R1�1 f(x)dx = ⇡

3

.

HenceR1�1

x

2

(x

2+1)(x

2+4)

dx = ⇡

3

2. Show thatR1�1

dx

(x

2+1)

3 = 3⇡

8

.

Solution.

ConsiderRc

f(z)dz =Rc

dz

(z

2+1)

3

where c is the upper half of the semi circle � with the bounding diam-eter [�R,R].By Cauchys Residue Theorem we have,

Rc

f(z)dz =R�R

Rf(x)dx+R�

f(z)dz

The poles of f(z) are obtained by (z2 + 1)3 = 0i.e., z = i,�i,. where z = i is a pole of order 3 which lies inside � and z = �i

SCE 75 Dept of S&H



is a pole of order 3 which lies outside � .

Res [f(z), i] = limz!i

1

2!

d

2

dz

2

[(z � i)3f(z)]

= limz!i

1

2!

d

2

dz

2

(z � i)3

1

(z + i)3(z � i)3

�

= limz!i

1

2

d

2

dz

2

1

(z + i)3

�

= limz!i

1

2

12

(z + i)5

=3

16i


Z

c

f(z)dz = 2⇡i [sum of residues]Z

c

f(z)dz = 2⇡i[R1

+R

2

]

= 2⇡i

3

16i

�

=3⇡

8

i.e.,R

R

�R

f(x)dx+R�

f(z)dz = 3⇡

8

.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 sothat |z| ! 1) R ! 1 then

R�

f(z)dz ! 0.i.e.,

R1�1 f(x)dx = 3⇡

8

.

HenceR1�1

dx

(x

2+1)

3dx = 3⇡

8


1. EvaluateR10

dx

(x

2+a

2)(x

2+b

2)

, a > 0, b > 0.

2. EvaluateR10

dx

x

4+10x

2+9

.

SCE 76 Dept of S&H



5.7.4 Type III


1. EvaluateR10

cos(ax)

x

2+1

dx, a > 0.

Solution.

W.K.TR10

cos(ax)

x

2+1

dx = 1

2

R1�1

cos(ax)

x

2+1

dx

To findR1�1

cos(ax)

x

2+1

dx

considerZ

c

f(z)dz =1

2

Z 1

�1

cos(az)

z

2 + 1dz

= R.P

Z

c

e

iaz

z

2 + 1dz

Where c is the upper half of the semi-circle � with the bounding diam-eter [�R,R]By Cauchy’s residue theorem,R

c

f(z)dz =R

R

�R

f(x)dx+R�

f(z)dz

The poles of f(z) are obtained by (z2 + 1) = 0i.e., z = i,�i,.

where z = i is a simple pole which lies inside � and z = �i is a simplepole which lies outside �

Res [f(z), i] = limz!i

(z � i)f(z)

= limz!i

(z � i)e

iaz

z

2 + 1

= limz!i

(z � i)(ia)eiaz + e

iaz

2z

=e

�a

2i


SCE 77 Dept of S&H



Z

c

f(z)dz = R.P 2⇡i [sum of residues]

= R.P 2⇡ie

�a

2i= R.P ⇡e

�a

) i.e.,

ZR

�R

f(x)dx+

Z

�

f(z)dz = ⇡e

�a

If R ! 1, then

Z

�

f(z)dz ! 0

Z 1

�1f(x)dx = ⇡e

�a

Z 1

0

cos(ax)

x

2 + 1dx =

⇡

2e

�a

.

2. Show thatR10

sinx

x

dx = ⇡

2

Solution.

W.K.TR10

sin(x)

x

dx = 1

2

R1�1

sin(x)

x

dx

To findR1�1

sin(x)

x

considerZ

c

f(z)dz =

Z

c

sin(z)

z

dz

= I.P

Z

c

e

iz

z

dz

Where c is the upper half of the semi circle � with the bounding diameter[�R,R]. i.e.,

Rc

f(z)dz = I.PR

R

�R

f(x)dx+R�

f(z)dzThe poles of f(z) are obtained by z = 0

SCE 78 Dept of S&H



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

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

(z)f(z)

= limz!0

(z)e

iz

z

= limz!0

e

iz

= 1

Hence by Residue theorem

Z

c

f(z)dz = 2⇡i(0) + ⇡i(1)

I.P

Z 1

�1

e

ix

x

dx = I.P ⇡i

Z 1

�1

sin x

x

dx = ⇡

i.e.,

Z 1

0

sin x

x

dx =⇡

2


1. Show thatR10

cos(ax)

(x

2+b

2)

2dx = ⇡

4b

3 [e�ab(1 + ab)]

2. Prove thatR10

log(1+x

2)

1+x

2 dx = ⇡ log2

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 idealfluid; the flow takes place in planes parallel to the xy plane. The cylinderis out of the plane of the paper. The flow of the fluid exerts forces andturning moments upon the cylinder. Let X, Y be the components, in the xand y directions respectively, of the force on the cylinder and let M be theanticlockwise moment (on the cylinder) about the origin.

SCE 79 Dept of S&H



Blasius theorem states thatX�iY = 1

2

i⇢

Rc

(dwdz

)2dz andM = �1

2

⇢

Rc

z(dwdz

)2dz.where Re denotes the real part, ⇢ is the (constant) density of the fluid andw = u + iv is the complex potential for the flow both of which are presumedknown. We shall find X, Y and M if the cylinder has a circular cross-sectionand the boundary is specified by |z| = a. Let the flow be a uniform streamwith speed U.

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

w = U

✓z +

a

2

z

◆so that

dw

dz

= U

✓1� a

2

z

2

◆

✓dw

dz

◆2

= U

2

✓1� 2a2

z

2

+a

4

z

4

◆

Now X � iY =1

2i⇢

Z

c

(dw

dz

)2dz =1

2i⇢U

2

Z

c

✓1� 2a2

z

2

+a

4

z

4

◆dz = 0.

Hence X = Y = 0. Also z

✓dw

dz

◆2

= U

2

✓z � 2a2

z

+a

4

z

3

◆

The only term to contribute to M is �2a

2U

2

z

Again using the Key Point above this leads to 4⇡a2U2

i and this has zero realpart. Hence M = 0, also. The implication is that no net force or momentacts on the cylinder. This is not so in practice. The discrepancy arises fromneglecting the viscosity of the fluid.

SCE 80 Dept of S&H



SCE 1 Dept of S&H

MATHEMATICS-II

UNIT-I VECTOR CALCULUS

Part-A

1. If = + .Evaluate ∫ . d along the curve ‘C’ in the XY plane y= from thepoint (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

irrotational5. 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+2xz2=8 at the point (1,0,2).10. Prove that div = 3 =011. Find the gradient of Ø where Ø=xy+yz+zx at (1,1,1).12. If =x2+y2+z2,find div and curl13. 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 − 2 ) + (3 + 2 ) +(3xy − 2xz + 2z) isconservative vector field. Find (i) the scalar potential (ii) the work down by inmoving a particle from (1,0,1) to (2,1,3).

2.Find the Directional Derivative 0f Ø=x2yz+4xz2 at (1,-2,-1) in the direction of2 − −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 z2= 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. Alsofind the scalar potential of .

6. Prove that � �� is irrotational. Also find the scalarpotential.



SCE 1 Dept of S&H

7.Prove that curl( × ) = ( . ∇) − ( . ∇) + −8. Prove that ∇ ( ) = ( + 1) . If r = | | and find ∇ .

9. Verify the Green’s theorem in the plane for ∫(3 − 8 ) + (4 − 6 ) whereC is the boundary of the region given by x=0,y=0,x+y=1.

10. Verify Green’s theorem for = ( + ) − 2 taken around the rectanglebounded by the lines x=± , = 0 =

11. Verify the Green’s theorem in the XY plane for ∫( + ) + where C is theclosed curve of the region bounded by y=x and y= .

12. Verify the Green’s theorem in the plane for ∫(3 − 8 ) + (4 − 6 ) whereC 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 therectangular 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 therectangular parallelepiped formed by the planes the line x=0,x=1,y=0,y=2 and z=3above the XY line.

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

16. Evaluate ∫ − + )by using Stoke’s theorem,where C is theboundary of the rectangle defined by 0≤ ≤ , 0 ≤ ≤ 1, = 3

17. Verify Stoke’s theorem when = (2 − ) − ( − ) nd C is the boundary ofthe region enclosed by the parabolas y2=x and x2=y

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

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

20. Verify the G.D.T for =(x3-yz) -2x2 y + 2k over the cube bounded by x=0,y=0,z=0and x=a, y=a, z=a

21. Verify the G.D.T for =x2 + +z where S is the surface of the cuboid formedby the planes x=0,x=a,y=0,y=b,z=0 and z=c



SCE 1 Dept of S&H

UNIT-II ORDINARY DIFFERENTIAL EQUATIONS

Part-A

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

2. Find the particular integral of (D-1)2y = ex sinx3. Find the P.I of (D2- 4)y = cosh2x4. Find the P.I of (D2+1)y = sinx5. Find the P.I of (D2+2D+1)y = e-x cosx6. Find the P.I of (D2+4)y = sin2x7. Find the P.I of (D+1)2y = e-x cosx8. Find the P.I of (D2-2D+2)y = ex cosx9. Solve: (D2-6D+13)y = 010. Transform the equation (2x+3)2 − 2(2 + 3) − 12 = 6

coefficients

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

12. Reduce the equation (x2D2 +xD +1)y = logx into an ordinary differential equation withconstant coefficients

Part-B1. Solve the equation (D2+5D+4)y = e-x sin2x2. Solve (D2-4D+3)y = ex cos2x3. Solve: (D2-3D+2)y = 2cos(2x+3)+ 2ex

4. Solve (D2+5D+4)y = e –x sin2x5. Solve (D2+4)y = x2 cos2x6. Solve (D2+16)y = cos3x7. Solve: (D2+4D+3)y = e-x sinx8. Solve (D2+a2)y = Secax by the method of variation of parameters9. Solve + = by the method of variation of parameters

10. Apply the method of variation of parameters to solve (D2+4)y = cot2x11. Solve by the method of variation of parameters to solve (D2+a2)y = tanax12. Solve + 4 = 2 by the method of variation of parameters

13. Solve: (x2D2 +3xD +5)y = x cos(logx)

14. Solve: (x2D2 - xD +4)y = x2 sin(logx)

15. Solve (x2D2-3xD+4)y = x2 cos(logx)

16. Solve: (x2D2 - 2xD -4)y = x2 + 2logx



SCE 1 Dept of S&H

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 = 2

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

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

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

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

UNIT – III LAPLACE TRANSFORM

PART-A1. Find L

2. Is the linearity property applicable to L ? Reason out.3. Find the Laplace transform of4. Find the Laplace transform of 25. State the conditions under which Laplace transform of ( ) exists6. Find L[ ]7. State the first shifting theorem on Laplace transforms.

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

cos − , >

9. Find the Laplace transform of unit step function.10. Find the inverse Laplace transform of11. Find [ ]12. Find the inverse Laplace transform of

13. Find the inverse Laplace transform of



SCE 1 Dept of S&H

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 theorem19. State and prove initial value theorem.20. State and prove final value theorem21. Verify the initial and final value theorems for f(t) = 322. Verify initial value theorem for 1 + ( + ).

PART-B

1. Find the Laplace transform of 32. Find L

3. Find( )( )

4. Find

5. Find the Laplace transform of ∫ .

6. Solve by using Laplace transform + 9 = 2 given that (0) = 1, = −1.

7. Solve the differential equation + = 2 with y (0) = 0 and (0) = 0 by usingLaplace transforms method.

8. Using Laplace Transform solve the differential equation − 3 − 4 = 2with (0) = 1 = (0)

9. Using Laplace Transform solve the differential equation − 3 + 2 = 4with (0) = −3 , (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) usingLaplace transforms.

13. Find the inverse Laplace transform of( )( )

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( )( )



SCE 1 Dept of S&H

19. Find the Laplace transform of square wave function defined by( ) = 1 0 < <

−1 < < 2 with period 2

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

( ) = , 0 ≤ ≤2 − , ≤ ≤ 2 ( + 2 ) = ( ).

21. Find the Laplace transform of ( ) = , 0 < < 10 ,1 < < 2 ( + 2) = ( ) > 0.

22. Find the Laplace transform of23. Find the Laplace transform of the periodic function

( ) = , 0 ≤ ≤2 − , 0 < ≤ 2 ( + 2 ) = ( )

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

25. Find the Laplace transform of ( ) = ∈ ,0 ≤ ≤−∈ ,0 < ≤ 2 ( + 2 ) = ( ) for all

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

( ) =0 ≤ ≤

2−

2≤ ≤

( + ) = ( )

27. 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=2z3. Prove that a bilinear transformation has atmost two fixed points4. State the basic difference between the limit of a function of a real variable and that of a

complex variable5. Show that an analytic function with constant imaginary part is constant6. Find the invariant points of the transformation w=7. Show that u=2x-x3+3xy2 is harmonic8. Verify whether the function u=x3-3xy2+3x2-3y2+1 is harmonic9. Find the constants a,b,c if f(z)=x + ay + i(bx+cy) is analytic10. Find the invariant points of the transformation w =



SCE 1 Dept of S&H

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

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

PART-B1. 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 theupper 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 = 0,1,∞ =, 1, −

7. Show that the image of the hyperbola x2-y2=1 under the transformation =is the lemniscates r2=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 u10. Show that the map = maps the totality of circles and straight lines as circles or

straight lines11. Prove that the transformation = maps the family of circles and straight lines

into the family of circles or straight lines12. 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. Findalso 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 = 3z17. Find the bilinear transformation that maps the points z = 0,− ,−1 =

, 1,0

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



SCE 1 Dept of S&H

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 andv=constant are orthogonal

22. Find the image of the circle | − 1| 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=y2). Also find the conjugateharmonic function v.

25. Prove that u = x2-y2 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, ∞ ofthe w-plane. Also show that the transformation maps interior of the unit circle of thez-plane onto upper half of the w-plane

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

UNIT-V COMPLEX INTEGRATION

PART-A

1. Evaluate ∫ ( )( ), C is the circle | | =

2. If f(z) = − 2[1 + ( − 1) + ( − 1) + ], ( ) = 1

3. Define Singular point4. 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 origin12. Expand f(z)=sinz in a Taylor series about Z=

13. Evaluate ∫ | | = 2

14. Evaluate ∫ , where C is | | =



SCE 1 Dept of S&H

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 | | = 3 2.⁄

19. State Cauchy’s integral theorem

20. Use Cauchy’s integral formula to evaluate ∫ , where C is the circle with centre 1and 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 region4. Evaluate ∫ .

5. Evaluate the integral ∫

6. Evaluate the integral ∫ , a>b>0

7. Evaluate ∫ ( )( )where C is | − 2| = by using Cauchy’s integral formula.

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.



SCE 1 Dept of S&H

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

in the region2 < | | < 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


in the region1 < | + 1| < 3.

22. Evaluate ∫ , (0 < < 1)

23. Prove that ∫ ( )( )∞∞ = , > > 0


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) | − 1| < 1 (ii)1 < | | < 2


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 � �2 2 1 cosxD D y e x�� .

2. Solve the equation 2 0x y xy ycc c� � .

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

� �2x cy z 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 transformation2 6

7zw

z�

�

.

7. Evaluate tan c³ where C is 2z .

8. Find the Taylor series for ( ) sinf z z about4

z S .

9. Find the Laplace transform of1 cos t

t�

.

10. Find the inverse Laplace transform of 1cot ks

� § ·¨ ¸© ¹

.


PART B – (5 x 16 = 80 marks)

11. (a) (i) Solve the equation � �2 24 cos 2D y x x� .

(ii) Solve the equation � �2 2 tanD a y ax� by the method of variation of

parameters.

Or

(b) (i) Solve the equation � � � �2 2 3 5 cos logx D xD y x x� � .

(ii) Solve sin , cosdx dyy t x tdt dt

� � given that 2x and 0y at 0t .

12. (a) (i) Find the angle between the normals to the surface 3 2 4xy z at the points

� �1, 1, 2� � and � �4,1, 1� .

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

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

XY plane.

Or

(b) (i) Find the directional derivative of 22xy zI � at the point � �1, 1, 3� in the direction

of 2 2i j k� � .

(ii) Verify Gauss divergence theorem for 2 2 2F x i y j z k � � where S is the surface of

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

13. (a) (i) Find the analytic function ( )f z P iQ � , if sin 2

cosh 2 cos 2xP Q

y x�

�.

(ii) Find the bilinear transformation which maps the points 0, , 1z i � � into

,1,0w i respectively.

Or

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

2 22 2 ( ) 4 ( )f z f z

x y§ ·w w c� ¨ ¸w w© ¹

.


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

.

Show the regions graphically.

14. (a) (i) Evaluate � � � �21 2c

zdzz z� �³ where c is the circle

122

z � using Cauchy’s integral

formula.

(ii) Evaluate � �2

20

, 0 11 2 sin

d xx x

S TT

� �� ³ , using contour integration.

Or

(b) (i) Find the Laurent’s series of �

� �

2

21( )

5 6zf z

z z valid in the region � �2 3z .

(ii) Evaluate � � � �2

2 2 2 2

x dxx a x b

f

�f � �³ , using contour integration, where 0a b! ! .

15. (a) (i) Find the Laplace transform of 2 cos 3tte t� .

(ii) Find the inverse Laplace transform of� � � �2

11 4s s� �

.

(iii) Solve the equation 9 cos 2 , (0) 1y y t ycc � and 12

y S§ · �¨ ¸© ¹

using Laplace

transform. Or

(b) (i) Find the Laplace transform of , in 0

( )2 , in 2t t a

f ta t a t a

d d ® � d d¯

and ( 2 ) ( )f t a f t� .

(ii) Find the Laplace transform of 4

0

sin 3 t

te t t dt� ³ .


www.rejin

paul.com

www.rejin

paul.com

www.rejin

paul.com


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

Regulations 2008

Second Semester

Common to all branches

MA2161 – Mathematics II

Time: Three Hours Maximum: 100 Marks

Answer ALL Questions

PART A – (10 x 2 = 20 Marks)

1. Transform the equation 2x y xy xcc c� into a linear differential equation with constant

coefficients.

2. Find the particular integral of � �2 1 sinD y x� .

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

4. State Gauss divergence theorem.

5. Verify whether the function 3 2 2 23 3 3 1u x xy x y � � � � is harmonic.

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

7. What is the value of the integral 23 7 1

1c

z z dzz

§ ·� �¨ ¸�© ¹³ where C is

12

z ?

8. If 21( ) 2 1 ( 1) ( 1) ...1

f z z zz� ª º � � � � � �¬ ¼�

, find the residue of ( )f z at 1z .

9. Find the Laplace transform of unit step function.

10. Find ^ `1 1cot ( )L s� � .

PART B – (5 x 16 = 80 marks)

11. (a) (i) Solve the equation � �2 4 3 sinxD D y e x�� .

(ii) Solve the equation � �2 1 sinD y x x� by the method of variation of

parameters.


OR

(b) (i) Solve � �2 2 22 4 2logx D xD y x x� � � .

(ii) Solve

22 3 2 ,

3 2 0.

tdx x y edtdy x ydt

� �

� �

12. (a) (i) Prove that � � � � � �3 2 26 3 3F xy z i x z j xz y k � � � � � is irrotational vector and

find the scalar potential such that F M � .

(ii) Verity Green’s theorem for � � � �2 23 8 4 6C

x y dx y xy dy� � �³ where C is the

boundary of the region defined by 2 2, x y y x .

OR

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

� �3 22 2f x yz i x yj k � � � over the cube bounded by 0, 0, 0x y z 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 .

(ii) Find the bilinear transformation which maps the points 0,1,z f into

,1,w i i � respectively.

OR

(b) (i) find the image of the hyperbola 2 2 1x y� under the transformation 1wz

.

(ii) Prove that the transformation1

zwz

�

maps the upper half of z - plane on to the

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

14. (a) (i) Find the Laurent’s series of 7 2( )

( 1)( 2)zf z

z z z�

� �

in 1 1 3z� � � .


(ii) Using Cauchy’s integral formula, evaluate4 3

( 1)( 2)C

z dzz z z

�� ³ , Where ‘C ’ is the circle

32

z .

OR

(b) (i) Evaluate 2

4 22

10 9x x dx

x x

f

�f

� �� ³ using contour integration.

(ii) Evaluate2

0 2 cosdS T

T�³ using contour integration.

15. (a) (i) Apply convolution theorem to evaluate � �

122 2

sLs a

�ª º« »« »�¬ ¼

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

, 0( )

2 , 2t t

f tt t

SS S S

d d ® � d d¯

and ( 2 ) ( )f t f tS� .

OR

(b) (i) Verify initial and final value theorems for the function ( ) 1 (sin cos )tf t e t t� � � .

(ii) Using Laplace transform solve the differential equation 3 4 2 ty y y e�cc c� � with

(0) 1 (0)y yc .


S.NO CONTENTS PAGE NO UNIT-1 VECTOR …...MA6251 MATHEMATICS-II S.NO CONTENTS PAGE NO UNIT-1 VECTOR...

Documents

Transcript of S.NO CONTENTS PAGE NO UNIT-1 VECTOR …...MA6251 MATHEMATICS-II S.NO CONTENTS PAGE NO UNIT-1 VECTOR...