CALCULUS AND DIFFERENTIAL EQUATIONS - mpbou

B.Sc. First Year

Mathematics, Paper - II

CALCULUS AND DIFFERENTIALEQUATIONS

MADHYA PRADESH BHOJ (OPEN) UNIVERSITY - BHOPAL

COURSE WRITERS

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V K Khanna, Formerly Associate Professor, Department of Mathematics, Kirori Mal College, University of DelhiS K Bhambri, Formerly Associate Professor, Department of Mathematics, Kirori Mal College, University of DelhiUnit: (3.0-3.2)

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SYLLABI-BOOK MAPPING TABLECalculus and Differential Equations

UNIT-I: Successive Differentiation and AsymptotesSuccessive Differentiation, Leibnitz Theorem, Maclaurin and TaylorSeries Expansions, Asymptotes.

UNIT-2: CurvatureCurvature, Tests for Concavity and Convexity, Points of Inflection,Multiple Points, Tracing of Curves in Cartesian and PolarCoordinates.

UNIT-3: Integration of Transcendental Functions, Reduction,Quadrature and RectificationIntegration of Transcendental Functions, Definite Integrals,Reduction Formulae, Quadrature, Rectification.

UNIT-4: Differential EquationsLinear Differential Equations and Equations Reducible to the LinearForm, Exact Differential Equations, First Order Higher DegreeEquations Solvable for x, y and p, Clairaut's Equation and SingularSolutions, Geometrical Meaning of a Differential Equation,Orthogonal Trajectories.

Unit-5: Linear Differential EquationsLinear Differential Equations with Constant Coefficients,Homogeneous Linear Ordinary Differential Equations, LinearDifferential Equation of Second Order, Transformation of Equationsby Changing the Dependent Variable / the Independent Variable,Method of Variation of Parameters.

Unit-1: Successive Differentiation andAsymptotes(Pages 3-46)

Unit-2: Curvature(Pages 47-104)

Unit-3: Integration of TranscendentalFunctions, Reduction, Quadrature

and Rectification(Pages 105-178)

Unit-4: Differential Equations(Pages 179-224)

Unit-5: Linear Differential Equations andMethod of Variation of Parameters

(Pages 225-258)

INTRODUCTION

UNIT 1 SUCCESSIVE DIFFERENTIATION AND ASYMPTOTES 3-46

1.0 Introduction1.1 Objectives1.2 Successive Differentiation1.3 Leibnitz Theorem1.4 Maclaurin and Taylor Series Expansions

1.4.1 Taylor’s Infinite Series1.4.2 Maclaurin’s Infinite Series1.4.3 Application of Taylor’s Infinite Series to Expand f(x + h)

1.5 Asymptotes1.5.1 Asymptotes Parallel to Axes of Co-Ordinates1.5.2 Oblique Asymptotes1.5.3 Total Number of Asymptotes of a Curve1.5.4 Miscellaneous Methods of Finding Asymptotes1.5.5 Intersection of a Curve and Its Asymptotes1.5.6 Asymptotes of Polar Curves

1.6 Answers to ‘Check Your Progress’1.7 Summary1.8 Key Terms1.9 Self-Assessment Questions and Exercises

1.10 Further Reading

UNIT 2 CURVATURE 47-104

2.0 Introduction2.1 Objectives2.2 Curvature

2.2.1 Radius of Curvature for Cartesian Curves2.2.2 Radius of Curvature for Polar and Pedal Equations2.2.3 Radius of Curvature at the Origin2.2.4 Centre of Curvature2.2.5 Chord of Curvature

2.3 Tests for Concavity and Convexity2.4 Points of Inflection2.5 Multiple Points2.6 Tracing of Curves in Cartesian and Polar Coordinates2.7 Answers to ‘Check Your Progress’2.8 Summary2.9 Key Terms

2.10 Self-Assessment Questions and Exercises2.11 Further Reading

UNIT 3 INTEGRATION OF TRANSCENDENTAL FUNCTIONS,REDUCTION, QUADRATURE AND RECTIFICATION 105-178

3.0 Introduction3.1 Objectives3.2 Integration and Definite Integrals

CONTENTS

3.3 Reduction Formula3.3.1 Walli’s Formula

3.3.2 Reduction Formulae for tann x dx and cotn x dx

3.3.3 Gamma Function3.4 Integration of Transcendental Functions3.5 Quadrature3.6 Rectification3.7 Answers to ‘Check Your Progress’3.8 Summary3.9 Key Terms


UNIT 4 DIFFERENTIAL EQUATIONS 179-224

4.0 Introduction4.1 Objectives4.2 Linear Differential Equations

4.2.1 Geometrical Meaning of a Differential Equation4.2.2 Solving Linear Differential Equations

4.3 Equations Reducible to the Linear Form4.4 Exact Differential Equations4.5 First Order Higher Degree Equations Solvable for x, y and p4.6 Clairaut’s Equation4.7 Singular Solutions4.8 Orthogonal Trajectories4.9 Answers to ‘Check Your Progress’

4.10 Summary4.11 Key Terms4.12 Self-Assessment Questions and Exercises4.13 Further Reading

UNIT 5 LINEAR DIFFERENTIAL EQUATIONS ANDMETHOD OF VARIATION OF PARAMETERS 225-258

5.0 Introduction5.1 Objectives5.2 Linear Differential Equations with Constant Coefficients5.3 Homogeneous Linear Ordinary Differential Equations5.4 Linear Differential Equation of Second Order5.5 Method of Variation of Parameters5.6 Answers to ‘Check Your Progress’5.7 Summary5.8 Key Terms5.9 Self-Assessment Questions and Exercises


Introduction

NOTES

Self - LearningMaterial 1

INTRODUCTION

Mathematics is the most important subject for achieving excellence in any field ofScience and Engineering. Calculus, originally called infinitesimal calculus or ‘thecalculus of infinitesimals’, is the mathematical study of continuous change, in thesame way that geometry is the study of shape and algebra is the study ofgeneralizations of arithmetic operations. Calculus is that branch of mathematicsthat focusses on limits, functions, derivatives, integrals and infinite series. Calculushas two major branches, differential calculus and integral calculus. The differentialcalculus studies instantaneous rates of change, and the slopes of curves, whileintegral calculus concerns accumulation of quantities, and areas under or betweencurves. These two branches are related to each other by the fundamental theoremof calculus, and they make use of the fundamental notions of convergence of infinitesequences and infinite series to a well-defined limit.

Infinitesimal calculus was developed independently in the late 17th centuryby Isaac Newton and Gottfried Wilhelm Leibniz. Today, calculus has widespreaduses in science, engineering, and economics. In mathematics, calculus denotescourses of elementary mathematical analysis, which are mainly devoted to thestudy of functions and limits, propositional calculus, Ricci calculus, calculus ofvariations, lambda calculus, and process calculus. Knowledge of calculus is,therefore, significant in mathematical analysis.

In mathematics, a differential equation is an equation that relates one ormore functions and their derivatives. Differential equations first came into existencewith the invention of calculus by Newton and Leibniz. He solved these examplesand others using infinite series and discussed about the non-uniqueness of solutions.Jacob Bernoulli proposed the Bernoulli differential equation in 1695.

Differential equations can be divided into several types. Apart from describingthe properties of the equation itself, these classes of differential equations can helpinform the choice of approach to a solution. Commonly used distinctions includewhether the equation is ordinary or partial differential equations, linear or non-linear differential equations, and homogeneous or heterogeneous differentialequations. In applications, the functions generally represent physical quantities,the derivatives represent their rates of change, and the differential equation definesa relationship between the two. Because such relations are extremely common,differential equations play a prominent role in many disciplines including engineering,physics, economics, and biology.

This book, Calculus and Differential Equations, is designed to be acomprehensive and easily accessible book covering the basic concepts of calculusand differential equations. It will help readers to understand the basics of successivedifferentiation and asymptotes, Leibnitz theorem, Maclaurin and Taylor series,curvature, tests for concavity and convexity, points of inflection, multiple points,tracing of curves, integration of transcendental functions, reduction, quadratureand rectification, differential equations, linear differential equations, exact differentialequations, Clairaut’s equation and singular solutions, geometrical meaning of a

Introduction

NOTES

Self - Learning2 Material

differential equation, orthogonal trajectories, linear differential equations withconstant coefficients, homogeneous linear ordinary differential equations, lineardifferential equation of second order, transformation of equations by changing thedependent variable and the independent variable, and method of variation ofparameters. The book is divided into five units that follow the self-instruction modewith each unit beginning with an Introduction to the unit, followed by an outline ofthe Objectives. The detailed content is then presented in a simple but structuredmanner interspersed with Check Your Progress Questions to test the student’sunderstanding of the topic. A Summary along with a list of Key Terms and a set ofSelf-Assessment Questions and Exercises is also provided at the end of each unitfor understanding, revision and recapitulation. The topics are logically organizedand explained with related mathematical theorems, analysis and formulations toprovide a background for logical thinking and analysis with good knowledge ofcalculus. The examples have been carefully designed so that the students cangradually build up their knowledge and understanding.

NOTES

Successive Differentiationand Asymptotes


UNIT 1 SUCCESSIVEDIFFERENTIATION ANDASYMPTOTES

Structure

1.0 Introduction1.1 Objectives1.2 Successive Differentiation1.3 Leibnitz Theorem1.4 Maclaurin and Taylor Series Expansions

1.4.1 Taylor’s Infinite Series1.4.2 Maclaurin’s Infinite Series1.4.3 Application of Taylor’s Infinite Series to Expand f(x + h)

1.5 Asymptotes1.5.1 Asymptotes Parallel to Axes of Co-Ordinates1.5.2 Oblique Asymptotes1.5.3 Total Number of Asymptotes of a Curve1.5.4 Miscellaneous Methods of Finding Asymptotes1.5.5 Intersection of a Curve and Its Asymptotes1.5.6 Asymptotes of Polar Curves

1.6 Answers to ‘Check Your Progress’1.7 Summary1.8 Key Terms1.9 Self-Assessment Questions and Exercises


1.0 INTRODUCTION

Successive differentiation is the process of differentiating a given functionsuccessively n times and the results of such differentiation are called successivederivatives. The higher order differential coefficients are of utmost importance inscientific and engineering applications.

We know that the derivative f of a differentiable function f is a function andis called the derived function of f. The concept of differentiation was motivated bysome physical concepts (like the velocity of a moving particle) and also bygeometrical notions (like the slope of a tangent to a curve). The second and higherorder derivatives are also similarly motivated by some physical considerations(like the acceleration) and some geometrical ideas (like the curvature of a curve).We shall also expand some functions in terms of infinite series by using thesetheorems.

Leibnitz Theorem is basically the Leibnitz Rule defined for derivative of theanti-derivative. As per the rule, the derivative on nth order of the product of twofunctions can be expressed with the help of a formula. Therefore, the LeibnitzTheorem gives us a formula for finding the nth derivatives of a product of twofunctions.


NOTES


In analytic geometry, an asymptote of a curve is a line such that the distancebetween the curve and the line approaches zero as one or both of the x or ycoordinates tends to infinity. Principally, an asymptote is a line that a curveapproaches, as it heads towards infinity. There are three types horizontal, verticaland oblique. This involves taking limits as orx y . Before discussingthe concept of asymptotes, first we shall study the branches of a curve and meaningof P (where P is a point on an infinite branch).

If in an equation, y has two or more values for every value of x, then we cansuppose that we are given two or more distinct functions. Generally, the curvescorresponding to these functions are regarded as different branches of one curve, and

not as different curves. For example, if 2 2 2 2 2, thenx y a y a x and thus

2 2 2 2andy a x y a x are the two branches of the curve x2 + y2 = a2

(circle), which is called the upper half and lower half of the circle. We know thatthe circle x2 + y2 = a2 is bounded by a square whose sides are ,x a y a .Therefore, we say that the two branches of the circle are finite. Now, consider the

hyperbola 2 2

2 21

x y

a b . Solving for y, we get 2 2b

y x aa

and

2 2by x a

a . Here, and thus x y . Therefore, we say that the

two branches of the hyperbola are infinite.

If any point P(x, y) on an infinite branch of a curve P moves along thebranch of the curve, then it is said to tend to along the curve. In this case one atleast of x and y or – .

In this unit, you will study about successive differentiation, Leibnitz Theorem,Maclaurin and Taylor series expansions, and asymptotes.

1.1 OBJECTIVES

After going through this unit, you will be able to:

Understand successive differentiation

Calculate higher order derivatives of a given function f

Find the nth order derivatives of products of functions using Leibnitz theorem

Expand the functions in the form of Maclaurin and Taylor infinite series

Define asymptotes and obtain their equations parallel to the axes

Understand the oblique asymptotes and generate its equation

Use different methods for finding asymptotes

Prove the concept of intersection of a curve and its asymptotes

Learn the asymptotes of polar curves

NOTES



1.2 SUCCESSIVE DIFFERENTIATION

If y be a function of x, then dy

dx is said to be the first differential coefficient or first

derivative of y with respect to x. Since, dy

dx is a function of x, therefore, it can be

further differentiated. The differential coefficient of dy

dx is called the second

differential coefficient or second derivative of y with respect to x and is denoted

by2

2

d y

dx. Similarly, the differential coefficient of

2

2

d y

dx is called the third differential

coefficient or third derivative of y with respect to x and is written as3

3

d y

dx and so

on.

In general, the nth differential coefficient of y or nth derivative of y with

respect to x is denoted byn

n

d y

dx.

This process of finding differential coefficients of a given function is knownas successive differentiation and these coefficients are called successive differentialcoefficients of y.

Note: The successive differential coefficients of y can be denoted by any one ofthe following ways:

(i)dy

dx,

2

2

d y

dx,

3

3

d y

dx, …

n

n

d y

dx

(ii) 1 2 3, , , ..., ny y y y

(iii) ', '', ''', ..., ny y y y

(iv) ( )'( ), ''( ), '''( ), ..., ( )nf x f x f x f x

(v) 2 3, , , ..., nDy D y D y D y

The ‘nth’ Differential Coefficient of Some Functions

Differential coefficient of a function is what is also called as derivative, themultiplicative factor or coefficient of the differential.

nth Differential Coefficient of xm

Let y = xm

By the successive differentiation of the given function with respect to x, wehave

y1 = mxm – 1

y2 = m(m – 1)x m – 2


NOTES


y3 = m(m – 1)(m – 2) x m – 3

…………….

..……………

yn = m(m – 1)(m – 2) … (m – n + 1) xm – n

If m is a positive integer, then

( 1)( 2) ....( 1)

( 1)( 2) ...( 1)( ) ...3.2.1

( )( 1) ...3.2.1

m nn

m n

y m m m m n x

m m m m n m nx

m n m n

!

!m nm

xm n

Corollary: If n = m, then !

!m m

m

my x

m m

= m!

nth Differential Coefficient of (ax + b)m

Let y = (ax + b)m

By the successive differentiation of the given function with respect to x, wehave

y1 = m(ax + b)m – 1 a

y2 = m(m – 1)(ax + b)m – 2 a2

y3 = m(m – 1)(m – 2)(ax + b)m – 3 a3

………………

……………..

yn = m(m – 1)(m – 2) … (m – n + 1)(ax + b)m – n an

If m is a positive integer, then

( 1)( 2)....( 1)( )

( 1)( 2)...( 1)( ) ...3.2.1( )

( )( 1)...3.2.1

m n nn

m n n

y m m m m n ax b a

m m m m n m nax b a

m n m n

!

( )!

m n nmax b a

m n

Corollary: If n = m, then !

( )!

m m mm

my ax b a

m m

= m!am

And if m < n, then yn = 0

nth Differential Coefficient of 1

,b

xax b a

Let1

yax b

NOTES



1 2

( 1)

( )

ay

ax b

2

2 3

( 1)( 2)

( )

ay

ax b

3

3 4

( 1)( 2)( 3)

( )

ay

ax b

……………………

……………………

1

( 1)( 2)( 3) ... ( )

( )

n

n n

n ay

ax b

1

( 1)

( )

n n

n

n a

ax b

Hence,

1

( 1)

( )

n n

n n

n ay

ax b

nth Differential Coefficient of log (ax + b)

Let y = log (ax + b)

1

ay

ax b

2

2 2

( 1)

( )

a

yax b

3

3 3

( 1)( 2)

( )

a

yax b

……………….

……………….

( 1)( 2)( 3) ... [ ( 1)].

( )

n

n n

n ay

ax b

1( 1) 1.2.3.... ( 1).

( )

n n

n

n a

ax b

1( 1) ( 1)

( )

n n

n

n a

ax b

Hence,11 1

( ) ( )

( )

n n

n n

n ay

ax b

Corollary 1: If y = log (x + a), then 1( 1) ( 1)

( )

n

n n

ny

x a

Corollary 2: If y = log x, then 1( 1) ( 1)n

n n

ny

x

nth Differential Coefficient of amx

Let y = amx

y1 = mlog a. amx

y2 = (mlog a)2. amx


NOTES


y3 = (mlog a)3. amx

………………

………………

yn = (mlog a)n. amx

Corollary: If y = mxe , then n mxny m e

nth Differential Coefficient of sin (ax + b)

Let y = sin (ax + b)

1 cos( ) sin2

y a ax b a ax b

sin cos2

2 2

2 cos sin2 2 2

y a ax b a ax b

2 sin

2a ax b

3

3 cos2

y a ax b

3 sin

2a ax b

……………………………………………………

Thus, sin2

nn

ny a ax b

nth Differential Coefficient of cos (ax + b)

Let y = cos (ax + b)

1 sin ( ) cos2

y a ax b a ax b

cos sin

2

2 2

2 sin cos2 2 2

y a ax b a ax b

2 cos

2a ax b

3

3 sin2

y a ax b

3 cos

2a ax b

…………………………...…………………………..

Thus, cos2

nn

ny a ax b

nth Differential Coefficient of eax sin bx

Let y = eax sin bx

y1 = eax (a sin bx + b cos bx)

Put a = r cos f and b = r sin f, so that

r2 = a2 + b2 and f = tan–1 b/a

NOTES



y1 = r eax (cos f sin bx + sin f cos bx) = r eax sin (bx + f)

y2 = r eax [a sin(bx + f) + b cos (bx + f)] = r2 eax sin (bx + 2f)

y3 = r3 eax sin (bx + 3f)

……………………….

…………………………

yn = rn eax sin (bx + nf)

Thus yn = (a2 + b2)n/2 eax sin (bx + n tan–1b/a)

Corollary: If y = eax sin (bx + c), then

yn = (a2 + b2)n/2 eax sin (bx + c + n tan–1b/a)

nth Differential Coefficient of eax cos bx

Let y=eax cos bx

y1=eax (a cos bx – b sin bx)

Put a=r cos f and b = r sin f, so that

r2=a2 + b2 and f = tan–1 b/a y

1=r eax (cos f cos bx – sin f sin bx) = r eax cos (bx + f)

y2=r eax [a cos(bx + f) – b sin (bx + f)]=r eax [r cos f cos(bx + f) – r sin f sin (bx + f)]=r2 eax cos (bx + 2f)

y3=r3 eax cos (bx + 3f)……………………….………………………..

yn=rn eax cos (bx + nf)

Thus yn=(a2 + b2)n/2 eax cos (bx + n tan–1b/a)

Corollary: If y = eax cos (bx + c), then

yn = (a2 + b2)n/2 eax cos (bx + c + n tan–1b/a)

Example 1.1: If p2 = a2 cos2 + b2 sin2, prove that2 2 2

2 3

d p a bp

d p

.

Solution: Given, 2 2 2 2 2cos sinp a b

Differentiating with respect to , we get

2 22 2 cos sin sin cosdp

p a bd

= (b2 – a2) sin 2q … (i)Again differentiating, we get

222 2

22 2 2( )cos 2

d p dpp b a

d d

22

2 22

( ) cos 2d p dp

p b ad d


NOTES


Multiplying by p2, we get22

3 2 2 22

( ) cos 2d p dp

p p b a pd d

2 2 2 2

2 2 2 2 ( ) sin 2( )(cos sin )

4

b ap b a

23 2 2 2 2 2 2 2 2 2 4 4 2 2 2

2( cos sin ) ( cos sin ) ( 2 )sin cos

d pp p b a p a b a b a b

d

2 2 2 2 2 2 2 4 4 4 2 2 2( cos sin )( cos sin ) ( 2 )sin cosa b b a p a b a b 2 2 4 4 2 4(cos sin sin cos )a b p 2 2 2 2 4(cos sina b p 2 2 4a b p

2

4 3 2 22

d pp p a b

d

2 2 2

2 3

d p a bp

d p

‘nth’ Derivative by the Use of Partial Fractions

Fractional expressions whose numerators and denominators are both rationalalgebraic expressions are differentiated n times by splitting them into partial fractions.

To split a given rational fraction into its partial fractions, the degree of thenumerator of the function must be less than the degree of the denominator. In caseif it is not so, then divide the numerator by the denominator and resolve thedenominator into simple factors, linear or quadratic as possible.

Example 1.2: If y = 3

2,

1

x

x prove that for x = 0, n > 1

ny = 0, if n is even

= – (n!), if n is odd.

Solution: Let y = 3

2 1

x

x = 2 1

xx

x

y = 1 1 1

2 1 1x

x x

[By Partial fractions]

Differentiating n time with respect to x, we get

ny = 1 1

1 ( 1) ! ( 1) !

2 ( – 1) ( 1)

n n

n n

n n

x x

= 1 1

( 1) ! 1 1

2 ( 1) ( 1)

n

n n

n

x x

When x = 0, ny = 1

( 1) ! 11

2 ( 1)

n

n

n

Case 1. If n is even, n + 1 is odd

ny = ![ 1 1]

2

n = 0

NOTES



Case 2. If n is odd, n + 1 is even

ny = –![1 1]

2

n = – n!.

Check Your Progress

1. If x = 2 cos t – cos 2t and y = 2 sin t – sin 2t; find 2

2

d y

dx when t =

2

.

2. If y = sin -1( sin )m x , then find the value of 2 22 1(1 – ) –x y xy + m y .

3. Find the nth derivative of 2 3sin cosx x .

4. Find the nth differential coefficient of 2

1

1 5 6 x x.

5. Find the nth derivative of 1cot x .

1.3 LEIBNITZ THEOREM

This theorem is used to find the nth differential coefficient of the product of twofunctions.

Statement: It states that if u and v are any two functions of x possessing derivativesupto nth order, then

Dn(uv) = (Dn u) v + nC1 Dn–1 u D v + nC

2 Dn–2 u D2 v + …. + nC

r Dn–r uDrv +

…. + u Dn v

Proof: We shall prove this theorem by mathematical induction.

By actual differentiation, we haveD (uv) = (Du) v + uD v = (Du)v + nC

1 uDv, where n = 1

Differentiating againD2(uv) = (D2u)v + D(u)D(v) + D(u)D(v) + uD2v

= (D2u)v + 2D(u)D(v) + uD2(v)= D2(u)v + 2C

1 D(u)D(v) + 2C

2 uD2(v)

Thus, the theorem is true for n = 1, 2.

Let us assume that theorem is true for a particular value of n, say m.

i.e., Dm(uv) = Dm(u)v + mC1 Dm – 1 uDv + mC

2 Dm – 2 uD2v + … + mC

r

Dm – r uDrv + … + uDmv

Again differentiating this result, we get

Dm+ 1(uv) = Dm + 1(u)v + DmuDv + mC1{DmuDv + Dm – 1 uD2v}

+ mC2{Dm– 1uD2v + Dm – 2 uD3v} + …

+ mCr{Dm– r + 1 uDrv + Dm – r uD r + 1v}

+ … … … + {DuDmv + uDm + 1 v}

By rearranging the above terms, we get

Dm+ 1(uv)= (Dm + 1u)v + (1 + mC1)DmuDv + (mC

1 + mC

2)Dm– 1uD2v + …

+ (mCr–1

+ mC r )Dm – r + 1 uDrv + … + uDm + 1v


NOTES


As we know that mCr–1

+ mC r = m + 1C

r

Putting r = 1, mC0 + mC

1 = m + 1C

1;

Putting r = 2, mC1 + mC

2 = m + 1C

2 etc.

1 1 1 1 1 2 1 1 11 2( ) ( ) ..... ..... .m m m m m m m m r r mD uv D u v C D uDv C D uD v C D uD v u D v

1 1 1 1 1 2 1 1 1( ) ( ) ..... ..... .m m m m m m m m r r mrD uv D u v C D uDv C D uD v C D uD v u D v

This shows that if the theorem is true for n = m, it is also true for n = m + 1.

But the theorem is true for n = 1; therefore by the principle of mathematicalinduction, the theorem is true for every positive integral value of n.

Note: In applying this theorem we take u as that function whose nth derivative isknown and v as that function which is of the form xn.

Example 1.3: If y = a cos (log x) + b sin (log x), then show that

(a) 22 1x y xy y = 0

(b) 2 22 1(2 1) ( 1)n n nx y n xy n y = 0

Solution: We have y = a cos (log x) + b sin (log x)

1y =1 1

sin (log ) cos (log ) .a x b xx x

1xy = sin (log ) cos (log )a x b x

Differentiating again, we have

2 1xy y =1 1

cos(log ) . sin (log ) .a x b xx x

22 1x y xy = cos (log ) sin (log )a x b x = – y

22 1x y xy y =0 …(i)

This proves the first part.Again differentiating Equation (i) n times by Leibnitz’s Theorem, we have

22 1 1

( 1)2 . 2 .

2!n n n n n nn n

x y ny x y y x n y y

= 0

2 22 1(2 1) ( 1)n n nx y n xy n y = 0

Example 1.4: If y = 2 2[log { (1 )}]x x , prove that

2 22 1(1 ) (2 1)n n nx y n xy n y = 0.

NOTES



Solution: Given, y = 2 2[log { (1 )}]x x ,

y1= 2

2 2

1 22 log (1 ) . . 1

(1 ) 2 (1 )

xx x

x x x

= 22

2 2

1 12 log (1 ) . .

(1 ) 1

x xx x

x x x

1y = 2

2

12 log { (1 )}.

(1 )x x

x

Cross-multiplying and squaring both sides, we get

2

2 2 21(1 ) 4 log( 1 4x y x x y

Differentiating both the sides with respect to x, we get2 2

1 2 1(1 ) . 2 2x y y xy = 14y

22 1(1 )x y xy = 2 [Dividing by 2y

1]

Differentiating n times by Leibnitz’s theorem, we get

22 1 1(1 ) 2 ( 1)n n n n nx y nxy n n y xy ny = 0

2 22 1(1 ) (2 1)n n nx y n xy n y = 0

Calculation of ‘nth’ Derivative at x = 0

Sometimes it is difficult to calculate the nth derivative of a function in general. Inthat case its value can be calculated at x = 0 by the application of Leibnitz’sTheorem.

Working Rule:

(i) Let the given function be y.

(ii) Find dy

dx and then, take L.C.M, if possible.

(iii) If square roots are present, square both sides and then try to get y in R.H.S.

(iv) Find the equation in y2, y

1, y by differentiating both sides of the equation

obtained in Step (iii).

(v) Differentiate the equation so obtained n times, by Leibnitz’s Theorem.

(vi) Put x = 0 in equation obtained in all the steps above from Step (i) toStep (v).

(vii) Put n = 1, 2, 3, 4 in the last equation of Step (vi).

(viii) Discuss the cases when n is even or odd.


NOTES


Example 1.5: If y = 2[ (1 )]mx x , find 0( )ny .

Solution: We have y= 2[ (1 )]mx x …(i)

1y = 2 1

2

2[ (1 )] 1

2 (1 )

m xm x x

x

= 2

2[ (1 )]

(1 )

mmx x

x

1y =2(1 )

my

x…(ii)

2 21(1 )x y = 2 2m y

Differentiating again, we have

2 21 2 1(1 ) 2 2x y y xy = 2

12m yy

i.e., 22 1(1 )x y xy = 2m y

… (iii)

Differentiating Equation (iii) n times by Leibnitz’s Theorem, we have

22 1 1

( 1)(1 ) (2 ) 2

1.2n n n n nn n

x y ny x y xy ny

= 2nm y

2 2 22 1(1 ) (2 1) ( )n n nx y n xy n m y = 0 …(iv)

Putting x = 0 in Equations (i), (ii), (iii) and (iv), we get

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

2 0( )ny = 2 20( ) ( )nm n y . …(v)

Case 1: When n is odd. Putting n = 1, 3, 5,… in Equation (v), we have

1 0( )y = m.

3 0( )y = 2 21 0( 1 ) ( )m y = 2 2( 1 )m m ,

5 0( )y = 2 23( 3 ) ( )m y

= 2 2 2 2( 3 ) ( 1 ) ,m m m … … ….

0( )ny = 2 2[ ( 2) ]m n … 2 2 2 2( 3 ) ( 1 ) .m m m

Case 2: When n is even. Putting n = 2, 4, 6,… in Equation (v), we have

2 0( )y = 2m .

4 0( )y = 2 22 0( 2 ) ( )m y = 2 2 2( 2 )m m ,

6 0( )y = 2 24 0( 4 ) ( )m y

= 2 2 2 2 2( 4 ) ( 2 ) ,m m m … … ….

0( )ny = 2 2[ ( 2) ]m n … 2 2 2 2 2( 4 ) ( 2 )m m m

NOTES



Example 1.6: If y = 1 2(sin )- x , prove that

(a) 22 1(1 ) 2x y xy = 0

(b) 2 22 1(1 ) (2 1)n n nx y n xy n y = 0

Deduce that 2

0lim

n

x n

y

y

= 2n and find (0)ny .

Solution: Given, y = 1 2(sin )x

1y = 1

2

12 (sin ) .

1x

x

… (i)

211 x y = 12sin x

Squaring both sides, we have

2 21(1 )x y = 1 24 (sin )x = 4y

Differentiating both sides with respect to x, we have2 2

1 2 1(1 ) 2 2x y y xy = 14y

22 1(1 ) 2x y xy = 0 [Dividing both sides by 12y ] …(ii)

Which proves (a).

Differentiating 22 1(1 ) 2x y xy = 0 n-times by Leibnitz’s Theorem, we get

22 1 1

( 1)(1 ) ( 2 ) ( 2) [ .1]

2!n n n n nn n

y x ny x y y x ny

= 0 …(iii)

2 22 1(1 ) (2 1)n n nx y n xy n y = 0

2 1 2(2 1)

1n n

xy n y

x

= 2

21nn y

x

When x 0,2

0lim n

x n

y

y

= 2

20lim

1x

n

x = 2n

From Equation (i), when x = 0, 1 (0)y = 0

From Equation (ii), when x = 0, 2 (0)y = 2

From Equation (iii), when x = 0, 2 (0)ny = 2nn y (0)

From Equation (iv), when n = 1, 3 (0)y = 211 (0)y = 0

n = 2, 4 (0)y = 222 (0)y = 22 . 2

n = 3, 5 (0)y = 233 (0)y = 0

n = 4, 6 (0)y = 244 (0)y = 2 24 . 2 . 2

If n is odd, ny (0) = 0

and if n is even, ny (0) = 2 2 2 22 . 2 . 4 . 6 ... ( 2)n .


NOTES


Check Your Progress

6. Find the nth derivative of 3 axx e .

7. If y = 2 sinx x , then find the value of yn.

8. If y = 1sin ,x find ( )n xy = 0.

9. If u = tan-1x, prove that (1 + x)2u2 + 2xu

1 = 0,

22 1(1 ) 2( 1) ( 1) 0n n nx u n xu n n u

and hence determine the values of all the derivatives of n with respect to x,when x = 0.

1.4 MACLAURIN AND TAYLOR SERIESEXPANSIONS

A set of terms which are connected by positive or negative signs and are arrangedaccording to some fixed definite law are called a series. There are two types ofseries, finite series and infinite series according to the number of terms (finite orinfinite) it contains.

For example, 1 + 4 + 7 + 10 + 13 is a finite series containing five terms

whereas 1 1 1

1 ...2 4 6

is an infinite series.

If nS = 1 2 3 ... na a a a , then (S )n is called the sequence of partialsums of the series. If { }na is a sequence, then 1 2 3 ... ....na a a a is called

an infinite series which is written as 1

nn

a

.

1.4.1 Taylor’s Infinite Series

Theorem 1.1: Let f(x) be a function such that:

(a) It has continuous derivatives of all order in the open interval (a, a + h) and

(b) Taylor’s remainder, Rn = ( )

!

nnh

f a hn

, 0 < < 1 tends to 0 as n

then f (a + h) = 2

( ) ( ) ( ) ... ( ) ...2! !

nnh h

f a hf a f a f an

…(1)

Proof: Let f(x) be a function, which possesses continuous derivates of all order inthe open interval (a, a + h). Then, for every integer n, however large, there is acorresponding Taylor’s development with Lagrange’s form of remainder, i.e.,

f (a + h)=2 1

1( ) ( ) ( ) ... ( ) R2! ( 1)!

nn

nh h

f a hf a f a f an

,

NOTES



where Rn = ( ),!

nnh

f a hn

0 < < 1

We can write the above expression as f (a + h) = S Rn n ,

whereSn = 2 1

1( ) ( ) ( ) ... ( )2! ( 1)!

nnh h

f a hf a f a f an

.

Suppose, Rn tends to 0 as n . Then, we have

lim Snn =

2

( ) ( ) ( ) ...2!

hf a hf a f a

And hence f (a + h) = 2

( ) ( ) ( ) ...2!

hf a hf a f a

Hence proved.

This theorem expands f (a + h) to an infinite series of ascending integral powers ofh and the infinite series thus obtained is called Taylor’s infinite series.

Expressing Taylor’s Infinite Series in Different Forms

(1) On putting a = x in Equation (1), we get

f (x + h) = 2

( ) ( ) ( ) ... ( ) ...2! !

nnh h

f x hf x f x f xn

(2) On putting a + h = b or h = b – a in Equation (1), we getf(b)

=2( ) ( )

( ) ( ) ( ) ( ) ... ( ) ...2! !

nnb a b a

f a b a f a f a f an

(3) On putting a + h = x or h = x – a, we get

f(x) =2( ) ( )

( ) ( ) ( ) ( ) ... ( ) ...2! !

nnx a x a

f a x a f a f a f an

Here, we get the expansion of f(x) to an infinite series of ascending integral powersof (x – a).

1.4.2 Maclaurin’s Infinite Series

Let f(x) be a function such that

(a) It has continuous derivatives of all orders in the open interval (0, x), and

(b) Maclaurin’s remainder Rn = ( )!

nnx

f xn

0 as n

then f (x) = 2

(0) (0) (0) .... (0) ...2! !

nnx x

f xf f fn

.


NOTES


This is called Maclaurin’s infinite series.

We can obtain the above expression by putting a = 0, and h = x in Taylor’s infiniteseries.

Note: Let the function f (x) be denoted by y, then the above expansion can bewritten as:

y = 2

1 2(0) . (0) (0) ... (0) ...2! !

n

nx x

y x y y yn

where 1 2(0), (0), (0) ,..., (0)ny y y y are the values of 1 2, , ,..., ny y y y respectivelyat x = 0.

Example 1.7: Expand sin ( 1)xe upto and including the term of 4x .

Solution: We know that xe = 2 3 4

1 ...2! 3! 4!

x x xx

1xe =2 3 4

...2! 3! 4!

x x xx = t (say)

sin ( 1)xe = sin t = 3 5

...3! 5!

t tt

=

32 3 4 2 3 41... ... ...

2 6 24 6 2 6 24

x x x x x xx x

=2 3 4 2 3

3 21... 3 ... ...

2 6 24 6 2 6

x x x x xx x x

= 2 41 5...

2 24x x x

Example 1.8: Expand sin x and cos x in powers of x and hence find cos 18º uptofour decimal places.

Solution: (a) Let f(x) = sin x f(0) = 0

Now ( )f x = cos x (0)f = 1

( )f x = – sinx (0)f = 0

( )f x = cos x (0)f = –1

( )ivf x = sin x (0)ivf = 0

( )vf x = cos x (0)vf = 1

………………….

By Maclaurin’s expansion, we have

f (x) = 2 3

(0) (0) (0) (0) ...2! 3!

x xf xf f f

NOTES



sin x = 2 3 4 5

0 .1 .0 ( 1) .0 .1 ...2! 3! 4! 5!

x x x xx

= 3 5

...3! 5!

x xx

(b) Let f(x) = cos x f(0) = 1

Now ( )f x = – sin x (0)f = 0

( )f x = – cos x (0)f = –1

( )f x = sin x (0)f = 0

( )ivf x = cos x (0)ivf = 1

( )vf x = – sin x (0)vf = 0

( )vif x = – cos x (0)vif = –1

cos x = 2 3 4 5 6

1 . 0 ( 1) . 0 .1 . 0 . ( 1) ...2! 3! 4! 5! 6!

x x x x xx

= 2 4 6

1 ...2! 4! 6!

x x x

Let x = 18º = 10

= 0.314

cos 18º = 2 4 6

1 1 11 ...

2! 10 4! 10 6! 10

= 2 4 61 1 11 (0.314) (0.314) (0.314) ...

2! 4! 6!

= 1 – .04929 + .00040 + … = .9511 (upto four decimal places)

1.4.3 Application of Taylor’s Infinite Series to Expandf(x + h)

1. Let f (x + h) be the given function.

2. Obtain f(x) by putting h = 0.

3. Differentiate f(x) a number of times and obtain ( ), ( ), ( ), ...f x f x f x , etc.

4. Substitute the values of ( ), ( ), ( ), ...f x f x f x in

f (x + h) = 2 3

( ) ( ) ( ) ( ) ...2! 3!

h hf x hf x f x f x

Example 1.9: If f(x) = 3 22 5 11,x x x find the value of 9

10f

with the help

of Taylor’s series for f(x + h).


NOTES


Solution: f (x) = 3 22 5 11x x x

Now by Taylor’s Theorem, we have

f(x + h) = 2 3

( ) ( ) ( ) ( ) ...2! 3!

h hf x hf x f x f x ...(i)

Putting x = 1 and h = 1

10 in Equation (i), we get

11

10f

= 2 3

1 1 1 1 1(1) (1) (1) (1) ...

10 2! 10 3! 10f f f f

…(ii)

Here, f(x) = 3 22 5 11x x x f(1) = 9

Now, ( )f x = 23 4 5x x (1)f = 2

( )f x = 6x + 4 (1)f = 10

( )f x = 6 (1)f = 6

Putting these values in Equation (ii), we have

9

10f

= 1 1 1

9 (2) (10) (6)10 200 6000

= 9 0.2 0.05 0.001 = 8.849

Example 1.10: Expand log sin x in powers of (x – 2).

Solution: Let f(x) = log sin x

log sin x = f(x) = f(2 + (x – 2)]

= f(2 + h) (where, h = x – 2)

By Taylor’s Theorem for x = 2, we have

f(2 + h) = 2 3

(2) '(2) ''(2) '''(2) ...2

h hf hf f f

… (i)

Now, f(x) = log sin x f(2) = log sin 2

f ‘(x) = cos

cotsin

xx

x , f (2) = cot 2

f (x) = – cosec2 x, f (2) = – cosec2 2 f (x) = – 2cosec x(– cosec x cot x) = 2 cosec2 x cot x f(2) = 2cosec2 2 cot 2 …………………

From Equation (i), we have

2 3

2 2( 2) ( 2)logsin logsin 2 ( 2)cot 2 cosec 2 2cosec 2cot 2 ...

2 3

x xx x

NOTES



Solutions Using Differential Equations

Example 1.11: Show that1

2 2 2 2 2 2 2 2 2sin 3 4 5( 1) ( 2 ) ( 1)( 3 )

1 ...2

a x a x a a a a a a ae ax x x x

and hence deduce that 2 31 sin sin sine

Solution: Let y = 1sina xe … (i)

1sin

1 2 2

.

1 1

a xe a ayy

x x

… (ii)

By squaring and cross multiplying,

(1 – x2)y12 = a2y2

Differentiating again, we get

(1 – x2)2y1y

2 – 2xy

12= a2 2yy

1 (1 – x2)y

2 – xy

1 – a2 y = 0 … (iii)

Differentiating the Equation (iii), n times by Leibnitz’s Theorem, we have

yn+2

(1 – x2) + nC1y

n + 1(–2x) + nC

2y

n (–2) –[y

n + 1x + nC

1y

n.1]= a2y

n

(1 – x2)yn+2

– (2n + 1)xyn+1

– (n2 + a2)yn = 0

Putting x = 0, we have yn+2

(0) = (n2 + a2)yn(0) … (iv)

From Equations (i), (ii) and (iii), we have

y(0) = 1, y1(0) = a, y

2(0) = a2,

Now, putting n = 1, 2, 3, …, in Equation (iv), we get

y3(0) = a(a2 +12),

y4(0) = a2(a2 + 22),

y5(0) = a(a2 +12) (a2 +32) …………………………..

Thus, by Maclaurin’s Theorem,

2 3 4 5

1 2 3 4 5(0) (0) (0) (0) (0) (0) ....2

x x x xy y xy y y y y

1

2 2 2 2 2 2 2 2 2 2 2sin 3 4 5( 1 ) ( 2 ) ( 1 )( 3 )

1 ...2

a x a x a a a a a a ae ax x x x

2 2 2 2 2 2 2 2 2 2 2

sin 3 4 5( 1 ) ( 2 ) ( 1 )( 3 )1 ...

a x a a a a a a ae ax x x x

To get the second part, let a = 1 and x = sin q,

2 3 41 2 51 sin sin sin sin ...

2e

Solutions Using Differentiation and Integration Method

Example 1.12: Expand 2log 1x + x =

3 51 1.3. . ...

2 3 2.4 5

x xx


NOTES


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

y1 = 2 2

1 21

1 2 1

x

x x x

= 2

1

1 x = (1 +x2)–1/2

Expanding by binomial theorem, we get

2 41

1 1.31 ...

2 2.4y x x

Integrating both sides w.r.t x within the limits 0 to x, we get

y = 3 51 1.3

. . ...2 3 2.4 5

x xx

2log 1x + x = 3 51 1.3

. . ...2 3 2.4 5

x xx

Check Your Progress

10. Expand tan x in power of 4

x

upto first four terms.

1.5 ASYMPTOTES

A straight line at a finite distance from the origin, is said to be an asymptote to aninfinite branch of a curve, if the perpendicular distance of any point P on thatbranch from the straight line tends to be zero as P tends to infinity along the branchof the curve.

We can also define asymptotes as a straight line which is at a finite distancefrom the origin and cuts a curve in two points at an infinite distance from the origin.

For example, the co-ordinate axes are asymptotes of the curve y =1

x. The

graph of the curve y = 1

x is given in the Figure 1.1

Y

O X

P

Fig. 1.1 Graph of the Curve

NOTES



Asymptotes are usually classified as the following:

Vertical Asymptotes

Horizontal Asymptotes

Oblique Asymptotes

Vertical and Horizontal Asymptotes

Vertical Asymptotes: The asymptotes parallel to y-axis are called verticalasymptotes.

A line x = a is a vertical asymptote of a curve if either

lim ( )x a

f x

or lim ( )x a

f x

.

Horizontal Asymptotes: The asymptotes parallel to x-axis are calledhorizontal asymptotes.

A line y = b is a horizontal asymptote of a curve y = f(x) if either

lim ( )x

f x = b or lim ( )

x

f x b .

1.5.1 Asymptotes Parallel to Axes of Co-Ordinates

An asymptote of a curve is a line such that the distance between the curve and theline approaches zero as they tend to infinity.

Asymptotes Parallel to y-Axis of the Curve y = f(x)

Let the line x = c … (2)

Parallel to y-axis be an asymptote of the curve y = f(x).

Now, it is required to determine the value of c (Refer Figure 1.2).

Y

O X

M

N A

c

P(x, y)

Figure 3.2 Fig. 1.2 Asymptotes Parallel to y-Axis

Let PM = be the distance of a point P(x, y) on the curve from the linex = c.

= x c


NOTES


(Perpendicular distance of a point (x1, y

1) from the line,

ax + by + c = 0 is 1 1

2 2

ax by c

a b

)

Then, by the definition of an asymptote, if line Equation (2) is an asymptote of thecurve, then 0 as P .

As

P , = x c 0 or x c.

P(x, y) is tend to infinity, y co-ordinate must tend to (+ or )

limy

x c

, i.e., x c as y … (3)

Thus, to find the asymptotes parallel to y-axis, we find from the given equation,the definite values c

1, c

2, … to which x tends, as y + or . Then

x = c1, x = c

2,… are the asymptotes of the curve parallel to y-axis.

Corollary: Similarly, to find asymptotes parallel to x-axis, we find the definitevalues d

1, d

2, … from the equation of the curve to which y tends as x + or

. Then, the asymptotes parallel to x-axis are y = d1, y = d

2, and so on.

Asymptotes Parallel to the Axes for an Algebraic Curve

Asymptotes Parallel to the y-Axis

Let the equation of the curve be,

1 2 2 1 1 2 2 1

0 1 2 1 1 2 1

2 3 3 22 3 1

( ... ) ( ... )

( ... ) 0

n n n n n n n n nn n n n

n n n nn n

a y a y x a y x a yx a x b y b y x b yx b x

c y c y x c yx c x

X O

Y

(0, 2)

y = 1

Figure 3.3

Fig. 1.3 Asymptotes Parallel to the Axis for an Algebraic Curve

It can be arranged in descending powers of y as follows:

1 2 20 1 1 2 2 2( ) ( ) .... 0n n na y a x b y a x b x c y

1 20 2) ) ) .... ) 0

n n nny x y x y x x … (4)

Where, 1), ),..., ) nx x x are polynomials in x.

NOTES



Dividing Equation (4) by ny , we have

0 22) ) ) .... ) 0

nn

x x x xy y y ...(5)

Let x = c be an asymptote of Equation (4) parallel to y-axis, then lim

y

x c .

Therefore, Equation (3) gives 0(c) = 0, so that c is a root of the equation

0(x) = 0.

Working Rule: Asymptotes parallel to the axis of y can be obtained byequating to zero the coefficient of the highest degree term of y in the equation ofthe curve.

If the coefficient of the highest powers of y is a constant or if its linearfactors are all imaginary, there will be no asymptote parallel to y-axis.

Asymptotes Parallel to the x-Axis

Working Rule: Asymptotes parallel to the axis of x can be obtained by equatingto zero the coefficient of the highest degree term of x in the equation of the curve.

If the coefficient of the highest powers of x is a constant or if its linearfactors are all imaginary, there will be no asymptote parallel to x-axis.

Example 1.13: Find all the asymptotes, parallel to the axes of the given curve2 2

2 21

a b

x y .

Solution: The given curve is 2 2

2 21

a b

x y

2 2 2 2 2 2 0x y b x a y … (i)

As Equation (i) is of fourth degree, it cannot have more than four asymptotes.

In order to find out the asymptotes parallel to x -axis, equating the coefficientof highest power of x, i.e., x2 in (i) to zero

2 2y b = 0 y = ± bi

Which gives the imaginary asymptotes.

Again equating to zero, the coefficient of highest power of y, i.e., y2 in (i),the asymptotes parallel to the y-axis are given by x2 – a2 = 0, x = + a whichgives two real asymptotes parallel to y-axis.

Hence, the only asymptotes are x = + a.

1.5.2 Oblique Asymptotes

An asymptote which is neither parallel to x-axis nor parallel to y-axis is called anoblique asymptote of the curve (Refer Figure 1.4).


NOTES


Y

O X

M P(x, y)

Figure 3.4

Fig. 1.4 Oblique Asymptote

Theorem 1.2: If y = mx + c is an asymptote to the curve f(x, y) = 0, prove that

lim and lim ( )x x

ym c y mx

x

Proof: Let y = mx + c, be an oblique asymptote to the curve f(x, y) = 0,

where m and c are both finite.

Let P(x, y) be any point on the infinite branch of the curve f(x, y) = 0. If PM = p,be the perpendicular distance of the point P from the line y = mx + c. Then

2PM

1

y mx cp

m

… (6)

In this case, the abscissa x tends to infinity as P(x, y) along the infinitebranch of the curve.

Now, from Equation (6), 21y mx c p m

On dividing both sides by x , we get

21y c p

m mx x x

Since, y = mx + c is an asymptote, it follows from the definition that p 0 asx .

2lim lim 1x x

y c pm m m

x x x

Hence, limx

ym

x

… (7)

Again from Equation (6), 21y mx c p m

2

0

lim ( ) lim ( 1 )x x

p

y mx c p m

Thus, lim ( )x

y mx c

… (8)

NOTES



Thus, if y = mx + c is an oblique asymptote to any curve f(x, y) = 0, then

lim and lim ( )x x

ym c y mx

x

.

Oblique Asymptotes of the General Algebraic Curves

Let the general algebraic equation of the curve of nth degree be

1 2 2 1 2 3 2 10 1 2 1 2 3

2 3 4 2 21 2 3

( ... ) ( ... )

( ... ) ... ( ) 0

n n n n n n n nn n

n n n nn

a y a y x a y x a x b y b y x b y x b x

c y c y x c y x c x ax by c

The above equation can be put into the form

1 21 2 0... 0n n n

n n n

y y y yx x x

x x x x

… (9)

where n

y

x

represents an expression of nth degree in y

x.

Let y = mx + c … (10)

be an asymptote of the curve, where m and c are finite,

where, lim and lim ( )x x

ym c y mx

x

Dividing both sides of Equation (9) by xn, we get

1 2 02

1 1 1.... 0n n n n

y y y y

x x x x x x x

Taking limit x , limx

ym

x

on both sides, we get

( ) 0n m … (11)

which is in general an nth degree polynomial equation in m. The real values of mobtained from Equation (11) give slopes of the asymptotes. If m = 0 is a root ofEquation (11), then it corresponds to an asymptote parallel to x-axis.

Let y mx = p, so that p c as x . lim ( )x

y mx c

Now, y = mx + p

Dividing both sides by x ,we get y p

mx x

On substituting the value of y

x in Equation (9), we get

1 21 2 .... 0n n n

n n n

p p px m x m x m

x x x

… (12)


NOTES


By Taylor’s theorem, we know that2

( ) ( ) '( ) ''( ) ....2

hf x h f x hf x f x

On expanding the Equation (12) by Taylor’s theorem, we get

2 21

1 1 12 2

22

1 1( ) ' ( ) . '' ( ) ... ( ) ' ( ) . '' ( ) ...

2 2

( ) ... .... 0

n nn n n n n n

nn

p p p px m m m x m m m

x x x x

x m

Arranging the above equation in descending powers of x, we get

2

1 21 1 2( ) ' ( ) ( ) '' ( ) ' ( ) ( ) ... 0

2n n n

n n n n n n

px m x p m m x m p m m

… (13)

From Equation (13) the asymptotes n (m) = 0

2

1 21 1 2' ( ) ( ) '' ( ) ' ( ) ( ) ... 0

2n n

n n n n n

px p m m x m p m m

Dividing both the sides by xn 1, we have

2

1 1 2

1' ( ) ( ) '' ( ) ' ( ) ( ) ... 0

2n n n n n

pp m m m p m m

x

Taking limits as x , so that p c, we getc

n (m) +

n – 1 (m) = 0 … (14)

If n(m) 0, which will be the case if

n(m) = 0 has no repeated roots, then

we get

1( )

' ( )n

n

mc

m

… (15)

Thus, if m1, m

2, m

3, … are the distinct roots of

n(m) = 0 and c

1, c

2, c

3, … are

the corresponding values of c determined from Equation (15), then the asymptotesare given by

1 1 2 2 3 3, , ,...y m x c y m x c y m x c

When n(m) = 0 for m = m

1 (say) and

n - 1(m) 0, then there is no any asymptote

in this case.

Case of Two Parallel Asymptotes

Let n (m) = 0 and

n – 1

(m) = 0 for m = m1 (say). Since m

1 is a repeated root

of n(m) = 0, thus for m = m

1, there are two parallel asymptotes. Substituting

these values of n (m) and

n

1 (m) in Equation (13), we get

22

1 2'' ( ) ' ( ) ( ) ... 02

nn n n

px m p m m

NOTES



Dividing both sides by xn – 2, we get

2

1 2'' ( ) ' ( ) ( ) ... 02 n n n

pm p m m

Taking limits as x , so that p c, we get

2

1 2'' ( ) ' ( ) ( ) 02 n n n

cm c m m

The above equation, being a quadratic in c, gives two values of c, say c1

and c2 corresponding to two values of m (i.e., m = m

1).

Thus, there are two parallel asymptotes y = m1x + c

1 and y = m

1x + c

2

corresponding to m = m1.

Working Rule to Find Oblique Asymptote of an Algebraic Curveof nth Degree

1. Find n (m). This can be obtained by putting x = 1, y = m in nth degree

terms (highest degree terms) of the given equation of the curve. Similarly,find out all the lower degree terms by putting x = 1, y = m .

2. Equate n (m) = 0 and solve for m. Then its roots m

1, m

2, … are the slope

of the asymptotes.

3. For all the distinct roots of n (m) = 0, find c from the equation 1( )

' ( )n

n

mc

m

corresponding to each values of m. Then the required asymptotes arey = m

1x + c

1, y = m

2x + c

2, …

4. If n(m) = 0 for some values of m and

n

1 0 then there will be no

asymptote corresponding to that values of m.

5. If n(m) = 0 and

n

1(m) = 0 for some values of m, then find the values

of c from the quadratic equation2

1 2'' ( ) ' ( ) ( ) 02 n n n

cm c m m

which

gives two values of c. Thus, there are two parallel asymptotes correspondingto this value of m.

6. If n(m) = 0,

n

1(m) = 0 and

n

2(m) = 0 for some values of m, then

the values of c are determined from the equation3 2

1 2 3''' ( ) '' ( ) ' ( ) ( ) 03 2n n n n

c cm m c m m

which gives three values

of c. Thus, there are three parallel asymptotes corresponding to this valueof m.

Note: Before finding the oblique asymptotes of a curve, first find the asymptotesparallel to the axes if any.


NOTES


1.5.3 Total Number of Asymptotes of a Curve

As the equation n (m) = 0 is a polynomial of degree n, we can get n value of m.

In general one value of m gives one asymptote. Hence, a curve of degree n cannothave more than n asymptotes (real or imaginary) or we can say that the totalnumber of asymptotes of a curve cannot exceed the degree of a curve.

Example 1.14: Find the asymptote to the curve

2x3 – x2y – 2xy2 + y3 + 2x2 – 7xy + 3y2 + 2x + 2y + 1 = 0

Solution: The given curve is

2x3 – x2y – 2xy2 + y3 + 2x2 – 7xy + 3y2 + 2x + 2y + 1 = 0 ... (i)

Since, the co-efficients of x3 and y3 are constants, therefore there are noasymptotes parallel to the co-ordinate axes.

To find the oblique asymptotes, substitute x = 1, y = m in the third, secondand first degree terms, one by one in Equation (i), we get

3(m) = 2 – m – 2m2 + m3 …(ii)

2(m) = 2 – 7m + 3m2 …(iii)

1(m) = 2 + 2m

Now, by differentiating Equation (ii) with respect to x, we get

3(m) = –1 – 4m + 3m2

Slopes of asymptotes are obtained by equating 3(m) = 0

2 32 2 0m m m

m2(m – 2) – (m – 2) = 0

(m – 2)(m2 – 1) = 0, which gives m = 2, 1, –1

Now, c is given by c = 2

22

3

( ) 3 7 2

( ) 3 4 –1

m m m

m m m

Substituting m = 2, c = 12 14 2

012 8 1

For m = 1, c = 3 7 2 2

13 4 1 2

For m = –1, c = 3 7 2

23 4 1

By substituting the values in y mx c , the asymptotes are y = 2x, y = x– 1 and y = –x – 2.

1.5.4 Miscellaneous Methods of Finding Asymptotes

The asymptotes can also be determined by the following methods:

First Method: Let us consider a non-repeated factor y – mx, of the highestdegree terms of the equation of the curve.

NOTES



If Pn – 1

is an homogeneous polynomial of degree (n – 1) only and Fn – 1

is apolynomial of degree n – 1 in x and y, then the equation of the curve can bewritten in the form

1 1( ) 0n ny mx P F … (16)

Here m is a root of n(m) = 0. Thus, y = mx + c will be an asymptote of the

curve Equation (16), where

lim( )x

c y mx

From Equation (16), 1

1

lim( ) lim n

x xn

Fc y mx

P

One asymptote of the curve Equation (16) is

1

1

lim n

xn

Fy mx

P

Put lim

x

ym

x

Corollary: If ax + by + c be a factor of the polynomial of degree n, thenthe equation of the curve of nth degree is written as,

1 1( ) 0n nax by c P F … (17)

where both Pn – 1

and Fn – 1

are polynomials of degree n – 1.

One of the asymptote of Equation (17) is

1

1

lim n

xn

Fax by c

P

Put limx

y a

x b

Second Method: Let the equation of the curve is of the formF

n + F

n – 2 = 0, so that every non-repeated linear factor of F

n equated to

zero is an asymptote of the curve.

Suppose ax + by + c is a factor of Fn. Then, the equation of the curve can

be put in the form

1 2( ) 0n nax by c P F

As in first method one asymptote is 2

1

lim n

xn

Fax by c

P

By dividing the numerator and denominator of the above equation by xn– 1,

we see that 1

x is a factor of the numerator..

When x , then 1

x 0

Hence, in general, the line ax + by + c = 0 is an asymptote.


NOTES


But if (ax + by + c)2 is a factor of Fn , then there will be two parallel

asymptotes given by

2 2

2

lim n

xn

Fax by c

P

.

Third Method: Let (y – mx)2 is a factor of the highest degree terms of theequation of the curve.

So, the equation of the curve is written as

22 1( ) 0n ny mx P F … (18)

We see that here no asymptote exists corresponding to the repeated factor(y – mx).

If (y – mx) is also a factor of terms of (n – 1)th degree of the equation of thecurve, then the equation of the curve is written as

22 2 2( ) ( ) 0n n ny mx P y mx F G

where Pn – 2

, Fn – 2

and Gn – 2

are the polynomial of degree n – 2, then theasymptote parallel to y – mx = 0 are given by

2 2 2

2 2

( ) ( ) lim lim 0n n

x xn n

F Gy mx y mx

P P

1.5.5 Intersection of a Curve and Its Asymptotes

Theorem 1.3: Any asymptote of an algebraic curve of nth degree cuts the curvein (n – 2) points.

Proof: Let y = mx + c … (19)

be an asymptote of the curve given by

1 21 2 ... 0n n n

n n n

y y yx x x

x x x

… (20)

Now, eliminating y from Equations (19) and (20), the abscissae of the pointsof intersection are given by

1 21 2 ... 0n n n

n n n

c c cx m x m x m

x x x

Expanding each term by Taylor’s theorem and arranging in descendingpowers of x, we get

1 21 2( ) '( ) ( )] ''( ) '( ) ( ) ... 0

2n n n

n n n n n n

cx m x c m m x m c m m

… (21)

Since, Equation (19) is an asymptote of (20), the coefficients of xn andxn–1 are both zero.

m and c are given by the equations

n(m) = 0 and c

n(m) +

n–1(m) = 0

NOTES



Putting these values in Equation (21), we get

2 32''( ) '( ) ( ) [...] ... 0

2n n

n n n

cx m c m m x

which is of degree (n – 2) and this gives (n – 2) values of x.

Hence an asymptote of a curve of nth degree cuts the curve in (n – 2) points.

Cor. I: If a curve of nth degree has n asymptotes, then the point of intersection ofa curve and its asymptotes are n(n – 2).

Cor. II: Let the joint equation of n asymptotes be given by

Fn = 0 … (22)

and the equation of the curve be F

n + F

n – 2 = 0 … (23)

The Equations (22) and (23) hold simultaneously at the points of intersection ofthe curve and its asymptotes. Hence for such points, we have

Fn – 2

= 0

Thus, Fn – 2

= 0 is a curve on which lie all the points of intersection of thecurve and its asymptotes.

Example 1.15: Show that the point of intersection of the curve

3 2 2 3 2 22 2 4 4 14 6 4 6 1y x y xy x xy y x y = 0

and its asymptotes lie on the straight line 8x + 2y + 1 = 0

Solution: The equation of the given curve is

3 2 2 3 2 22 2 4 4 14 6 4 6 1y x y xy x xy y x y = 0 …(i)

The curve has no asymptotes parallel to x-axis or y-axis.

( the co-efficients of 3x and 3y , the highest degree terms in x and y are constants)

To obtain the oblique asymptote, substitute x = 1, y = m in the two highest (i.e.,third and second) degree terms

( )n m = 3 ( )m = 3 22 2 4 4m m m

and 1( )n m = 2 ( )m = 214 6 4m m

The slopes of the asymptotes are given by ( )n m = 0 3 ( ) 0m

3 22 2 4 4m m m = 0

2 22 ( 1) 4 ( 1)m m m = 0

2(2 4) ( 1)m m = 0

m = 2, 1, –1


NOTES


Also ( )n m = 3 ( )m = 26 2 8m m

c = 1 ( )

( )n

n

m

m

=

2

3

( )

( )

m

m

= 2

2

( 14 6 4)

6 2 8

m m

m m

= 2

2

7 3 2

3 4 1

m m

m m

When m = 2, c = 14 12 2

12 8 1

= 0,

When m = 1, c = 7 3 2

3 4 1

= 2

2 = – 1.

When m = – 1,

c = 7 3 2

3 4 1

= – 2

Putting these values of m and c in y = mx + c, the asymptotes arey = 2x + 0, y = x – 1, and y = –x –2

2x – y = 0, x – y – 1 = 0 and x + y + 2 = 0.Thus, the three asymptotes will cut the curve again in 3(3 – 2) = 3 points.The joint equation of the asymptotes is

(2 ) ( 1) ( 2)x y x y x y = 0

3 2 2 3 2 22 2 7 3 2 2 4y x y xy x xy y x y x = 0

Multiplying by 23 2 2 3 2 22 2 4 4 14 6 4 4 8y x y xy x xy y x y x = 0.

Now the equation of the curve is3 2 2 3 2 22 2 4 4 14 6 4 6 1y x y xy x xy y x y = 0

which can be re-written as3 2 2 3 2 2[2 2 4 4 14 6 4 4 8 ] [2 8 1]y x y xy x xy y x y x y x = 0

which is of the form 2F Fn n = 0

Hence, the points of intersection lie on the curve

2Fn = 0 or 2 8 1y x = 0 or 8 2 1x y = 0

which is the required straight line.

1.5.6 Asymptotes of Polar Curves

Theorem 1.4: If the equation of the polar curve be r = f(), then the equation ofthe asymptote is given by

P sin( )r

NOTES



where 1P lim

du

du r

and is a root of the equation obtained by putting

u = 0 (Refer Figure 1.5).

O

L

T’

X

P

– /2

T

Fig. 1.5 Asymptotes of Polar Curves

Proof: Let P(r, ) be any point which recedes to the infinity along the curve. Then

remains finite. Also, let r , then 1 (say). So, by putting 1

ur

, we get

u 0 and 1

And let the asymptote to the curve be

cos( )p r …(24)

where p and are finite quantities such that the Equation (24) lies at a finitedistance from the pole.

By definition of asymptote, we have

= p – r cos ( ) ; 0 as r …(25)

where = PL is the length of perpendicular from P(r, ).

Dividing Equation (25) by r, we get

cos( )p

r r

Let r , so that 0 as 1 (say)

1cos( ) = 0

1

1

…(26)

Again, if we put r , 0 and 1 in Equation (25), we get

1θ θ0 lim cos( )p r

1θ θ

cos( )limp

u

, where

1u

r

Applying L’Hospital rule, 1θ θ

sin( )lim

θ

pdu

d

1θ θ

θlim

d

du …(27)


NOTES


By putting the values of p and from Equations (27) and (26), respectively inEquation (24), we get

1

1 1θ θ

θ πlim cos θ θ sin θ θ

2

dr r

du

11

θ θ

θlim sin θ θ

dr

du

This is the required equation of the asymptote in polar co-ordinates.

Working Rule for Finding Asymptotes of Polar Curves

1. Put r = 1

uin the given equation.

2. If the given equation has trigonometric ratios, convert them to sin andcos . Find the limit of as u 0. Let

1,

2, ……

n be the limits

obtained.

3. Find p = 0

lim –i

u

d

du

for each of the i, obtained in step 2.

4. The corresponding asymptotes are given by the equation p = r sin (i – ).

Example 1.16: Find the asymptotes of the curve r = 2

1+2cos

a

.

Solution: Putting, u = 1

,r

the equation of the curve becomes

2au = 1 + 2 cos i

When u 0, then cos1 =

1 2– cos

2 3

1 = 2n ±

2;

3

n = 0, ±1, ±2, ±3….

Differentiating Equation (i) with respect to , we get

2du

ad

= –2 sin d

du

= –a cosec

p = 1 1

lim – lim cosecd

adu

= a cosec

2 2 22 cosec

3 3 3

an a

The equation of asymptotes is p = r sin (1 – ).

2

3

a =

2 2sin 2 – sin –

3 3r n r

2

3

a= r sin

2

3

Hence, the asymptotes are 2 2

sin .33

ar

NOTES



Example 1.17: Find the asymptotes of

rn fn() + rn – 1 f

n – 1 () + …. + f

0() = 0

Solution: Changing r into 1

,u

the given equation becomes

fn() + uf

n – 1() + u2f

n – 2() + … + unf

0() = 0 …(i)

Now u 0 gives fn() 0

Let 1 be any root of f

n() = 0

Differentiating Equation (i) with respect to u, we get

fn() + f

n – 1()

du

d + uf

n-1() + (terms containing power of u) = 0

Taking u = 0, we get

fn() + f

n – 1()

du

d= 0

du

d=

– 1

( )–

( )n

n

f

f

p =1

limd

du

=

– 1 1

1

( )

( )n

n

f

f

Thus, the equation of asymptote is p = r sin (1 – )

–1 1

1

( )

( )

n

n

f

f

= r sin [1 – ]

where 1 is any root of equation f

n () = 0.

Check Your Progress

11. Find the asymptotes parallel coordinate axes for the curve

x2y2 – a2(x2 + y2) – a3(x + y) + a4 = 0.

12. Find the asymptote of the following curve x3 + 3x2y – 4y2 – x + y + 3 = 0.

13. Find all the asymptotes of the curve 2 2 2 2 2 2( ) ( 4 )y x a x x a

14. Find the equation of the quartic curve which has x = 0, y = 0, y = x and y= –x for asymptotes and which passed through (a, b) and which cuts itsasymptotes again in eight points that lie on a circle whose centre is originand radius a.

15. Find the equation of the cubic which has the same asymptotes as the curve3 2 2 36 11 6 1 0x x y xy y x y which touches the axis of y at the

origin and passes through the point (3, 2).

16. Show that the eight points of intersection of the curve4 2 2 4 2 25 4 1x x y y x y x y = 0.

and its asymptotes lie on a rectangular hyperbola.

17. Find the asymptotes of the curve r = a.

18. Find the asymptotes of the curve r cos = a sin .


NOTES


1.6 ANSWERS TO ‘CHECK YOUR PROGRESS’

1.3

2

2. 0

3.1 5 3

cos cos 5 cos 38 2 16 2 16 2

n nn n nx x x

4.1 1

1 1

2 3( 1) ! .

(2 1) (3 1)

n nn

n nn

x x

5. ( 1) ( 1)!sin sin ,n nn ny y where y = 1cot x

6. 3 2 1 2 3[ 3 3 ( 1) ( 1) ( 2) ]ax n n n ne a x nx a n n xa n n n a

7. 2 2 sin 2 cos .2 2

n nx n n x nx x

8. 2 2 2 2( 2) . ( 4) .... 5 . 3 .1.n n

9. (- 1)(n - 1)/2 (n - 1) !

10.2 3

81 2 2 ...

4 4 3 4x x x

11. x = ± a

12. x – y = 0, x + 2y – 1 = 0 and x + 2y + 1 = 0

13. y = + x, x = + a

14.2 2

2 2 2 2 2( )( ) .( )

a a bxy x y x y a

b

= 0

15. 3 2 2 36 11 6 0x x y xy y x

16. The given curve is,

4 2 2 4 2 25 4 1x x y y x y x y = 0 ... (i)

4 2 2 4 2 2( 5 4 ) ( ) ( ) 1x x y y x y x y = 0

4 ( )m = 2 41 5 4m m , 4' ( )m = 310 16m m and 3( )m = 0

Equating ( )n m to zero, we get

4 24 5 1m m = 0

2 2( 1)(4 1)m m = 0

m = 1

1,2

NOTES



We know that

c = 1( )

' ( )n

n

m

m

=

3

4

( )

' ( )

m

m

= 0

[ 3( )m = 0]

Hence the asymptotes are y = ± x and y = 1

2x

y x = 0, y x = 0, 2y x = 0 and 2y x = 0.

The combined equation of the asymptotes is

( )( )(2 )(2 )y x y x y x y x = 0

4 2 2 45 4x x y y = 0 …(ii)

Subtracting Equation (ii) from Equation (i), we get the points of intersection

for the curve and its asymptotes lie on 2 2 1x y x y = 0, which

represents a rectangular hyperbola, and the sum of coefficients of 2x and

2y is zero.

The asymptotes cut the curve in n(n – 2), i.e., 4(4 – 2) 8 points.

17. r sin = a

18. r cos = a

1.7 SUMMARY

If any function can be differentiated upto nth terms, then the differentialcoefficients are called successive differential coefficients and this process offinding the differential coefficients is called successive differentiation.

The fractional expressions whose numerators and denominators are bothrational algebraic expressions are differentiated n times by splitting theminto partial fractions.

If the denominator of algebraic fraction cannot be resolved into real linearfactors then solve it by splitting it into imaginary factors.

If u and v are any two functions of x possessing derivatives upto nth order,then Dn(uv) = (Dn u) v + nC

1 Dn–1 u D v + nC

2 Dn–2 u D2 v + …. +

nCr Dn–r uDrv + …. + u Dn v. This is called Leibnitz’s theorem.

The nth derivative of a function can also be calculated at x = 0 by theapplication of Leibnitz’s theorem.

Taylor’s infinite series: Let f(x) be a function such that(a) it has continuous derivatives of all order in the open interval (a, a + h)

and


NOTES


(b) Taylor’s remainder, Rn = ( )

!

nnh

f a hn

, 0 < < 1 tends to 0 as

n

then f (a + h) = 2

( ) ( ) ( ) ... ( ) ...2! !

nnh h

f a hf a f a f an

.

Taylor’s infinite series can be expressed in different forms.

(a) f (x + h) = 2

( ) ( ) ( ) ... ( ) ...2! !

nnh h

f x hf x f x f xn

,

where a = x

(b) f(b) = 2( ) ( )

( ) ( ) ( ) ( ) ... ( ) ...2! !

nnb a b a

f a b a f a f a f an

,

where a + h = b or h = b – a

(c) f(x) = 2( ) ( )

( ) ( ) ( ) ( ) ... ( ) ...2! !

nnx a x a

f a x a f a f a f an

, where

a + h = x or h = x – a . This is the expansion of f(x) to an infiniteseries of ascending integral powers of (x – a).

Maclaurin’s infinite series: Let f(x) be a function such that(a) it has continuous derivatives of all orders in the open interval (0, x) and

(b) Maclaurin’s remainder Rn = ( )!

nnx

f xn

0 as n

then f (x) = 2

(0) (0) (0) .... (0) ...2! !

nnx x

f xf f fn

.

Asymptotes are classified as vertical asymptotes, horizontal asymptotesand oblique asymptotes.

A line x = a is a vertical asymptote of a curve of either lim ( )x a

f x

or

lim ( )x a

f x

.

A line y = b is a horizontal asymptote of a curve y = f(x) if either lim ( )x

f x

or lim ( )isx

f x b

.

If y = mx + c is an oblique asymptote to any curve f(x, y) = 0, then

lim and lim ( )x x

ym c y mx

x

.

The general algebraic equation of the curve of nth degree be

1 21 2 0... 0n n n

n n n

y y y yx x x

x x x x

where n

y

x

represents an expression of nth degree in y

x.

The general nth degree polynomial equation in m is ( ) 0n m .

NOTES



The value of c corresponding to a value of m is given by the equation1( )

' ( )n

n

mc

m

.

If n (m) = 0 and

n – 1

(m) = 0, then there are two parallel asymptotes. Inthis case the value of c is given by the equation

2

1 2'' ( ) ' ( ) ( ) 02 n n n

cm c m m

.

The total number of asymptotes of a curve cannot exceed the degree of acurve.

If y = mx + c be an asymptote of the curve1 2

1 2 ... 0n n nn n n

y y yx x x

x x x

then an asymptote of a curve of nth degree cuts the curve in (n – 2)points.

Then equation of degree (n –2) is given by

2 32''( ) '( ) ( ) [...] ......... 0

2n n

n n n

cx m c m m x

If the joint equation of n asymptotes is Fn = 0 and the equation of the curve

is Fn + F

n – 2 = 0, then F

n – 2 = 0 is a curve on which lies all the points of

intersection of the curve and its asymptotes.

If the equation of the polar curve be r = f(), then the equation of the

asymptote is given by p = sin( )r , where 1P lim

du

du r

and is

a root of the equation obtained by putting u = 0.

1.8 KEY TERMS

Successive differentiation: If y = f (x) be a function of x and it can befurther differentiated with respect to x upto nth terms, then the differentialcoefficients are called successive differential coefficients of y and this processof finding the differential coefficients of a function is called successivedifferentiation.

Leibnitz theorem: In leibnitz theorem, If u and v are any two functions ofx possessing derivatives upto nth order, then Dn(uv) = (Dn u) v + nC

1 Dn–1

u D v + nC2 Dn–2 u D2 v + …. + nC

r Dn–r uDrv + …. + u Dn v

Infinite series: Infinite series are a set of terms, which are connected bypositive or negative signs and are arranged according to some fixed definitelaw.

Asymptotes: A straight line at a finite distance from the origin, is said to bean asymptote of an infinite branch of a curve, if the perpendicular distanceof any point P on that branch from the straight line tends to zero as P tendsto infinity along the branch of the curve.


NOTES


Vertical asymptotes: The asymptotes parallel to y-axis are called verticalasymptotes.

Horizontal asymptotes: The asymptotes parallel to x-axis are calledhorizontal asymptotes.

Oblique asymptotes: An asymptote which is neither parallel to x-axis norparallel to y-axis is called an oblique asymptote of the curve.

1.9 SELF-ASSESSMENT QUESTIONS ANDEXERCISES

Short-Answer Questions

1. If y = 21 x , prove that 4

4

d y

dx =

2

2 7 / 2

12 3

(1 )

x

x

.

2. If y = log x

x, show that

2

2

d y

dx = 3

2 log 3x

x

.

3. If x = ( sin )a , y = (1 cos )a , find 2

2

d y

dx at =

2

.

4. If y = 2

1 12

1 1 2tan tan

1

x x

x x

; prove that 2

2

d y

dx = 2 2

5

(1 )

x

x

.

5. If y = mx mxae be ; prove that 22y m y = 0.

6. If y = 1 2(tan )x , prove that 2 2 22 1( 1) 2 ( 1)x y x x y = 2.

7. If x = sin t, y = sin pt, prove that 22 1(1 )x y xy + 2p y .

8. If y = log (1 cos )x , prove that 3 1 2y y y = 0

9. If x = ( ) nta bt e , show that 2

22

2d x dx

n n xdtdt

= 0

10. If 1/ 2 1/ 2x y = 1/ 2a , find the value of 2

2

d y

dx for x = a.

11. If y = 2A [ 1 ]nx x , prove that 2 22 1( 1)x y xy n y = 0.

12. Find the asymptotes, parallel to coordinate axes of the following curves:(i) y = tan–1 x

(ii) 2 xy e

(iii) y = xe1/x 1(iv) y = sec x(v) y = cosec x

NOTES



(vi) x2y2 = a2(x2 + y2)(vii) y = x(x – 2)(x – 3)

(viii) y = 2 –1

x

x

13. Find the asymptotes of the following curves:(i) x4 y4 + xy = 0(ii) x3 + 4x2y + 5xy2 + 2y3 + 2x2 + 4xy + 2y2 x 9y + 1 = 0(iii) x3 – 6x2y + 11xy2 – 6y3 + x + y + 1 = 0(iv) 4x3 – 3xy2 – y3 + 2x2 – xy y2 1 = 0(v) x3 – 2y3 + xy (2x – 1) + y(x – 1) + 1 = 0 (vi) y2 (x – 2a) = x3 –

a3

(vii) (x2 – y2) 2 – 4x2 + x = 0(viii) (x – 1) (x – 2) (x + y) + x2 + x + 1 = 0(ix) (x2 y2) (2x + 3y) (3x + 2y) + 2x2 + 3y2 – 6 = 0(x) (x + y + 1)2 (x2 + y2 – xy) + 3xy – 7x2 – 2y2 – 7x + 8 = 0

14. Prove that the asymptotes of the curve x2y2 = a2(x2 + y2) are the sides of asquare.

15. Find the asymptotes of the curve x2y – xy2 + xy + y2 + x – y = 0 and showthat they cut the curve in three points on the straight line x + y = 0.

16. Find the equation of the curve on which lie the points of intersection of thecurve (x2 – 4y2) (x2 – y2) + 2x3 – 2xy2 + 2x2 – 2xy – 2x – 2y + 1 = 0 andits asymptotes.

Long-Answer Questions

1. Find the nth derivatives of the following :

(i) 2

1

5 6x x (ii) 4

( 1) ( 2)

x

x x

(iii) 2

1

6 7 3

x

x x

(iv)

2

3( 1) ( 1)

x

x x

(v) 2 2 2 2

1

( ) ( )x a x b

2. Find the nth derivatives of the following :

(i)1 1

tan1

x

x

(ii) 4 4

1

x a (iii) 2

1

1x x

[Hint. Let = 1 2 1

cot3

x ]

3. Find the nth derivatives of the following:

(i) 2 2[ 2 ( 1)]axe a x nax n n (ii) logxe x

(iii) n xx e (iv) 3 sinx ax

(v) 2 cosaxx e bx


NOTES


4. Find the nth derivative of

(i)2

2 22

(1 )d y dy

x x a ydxdx

= 0 (ii) 22 1x y xy y = 0.

5. If y = lognx x , show that 1ny = !n

x.

6. If y = 1tan x , prove that

(i) 22 1(1 ) 2x y xy = 0

(ii) 21 1(1 ) 2 ( 1)n n nx y nxy n n y = 0

7. If y = 1sinm xe prove that

(i) 22 1(1 )x y xy = 2m y

(ii) 2 2 22 1(1 ) (2 1) ( )n n nx y n xy n m y = 0.

8. If y = 1cosm xe , prove that 2 2 2

2 1(1 ) (2 1) ( )n n nx y n xy n m y = 0.

9. If y = 2( 1)nx , prove that 22 1( 1) 2 ( 1)n n nx y xy n n y = 0.

10. If 1cos

y

b

= logn

x

n

; prove that 2 22 1(2 1) 2n n nx y n xy n y = 0.

11. If y =

1

2

sin

1

x

x

, prove that 2 22 1(1 ) (2 3) ( 1)n n nx y n xy n y = 0.

12. If y = 1 2(sin ) ,h x prove that 2 22 1(1 ) (2 1)n n nx y n xy n y = 0.

Also find (0)ny .

13. Show by means of Maclaurin’s infinite series expansion that:

(i) 1tan (1 )x = 2 3

...4 2 4 12

x x x

(ii)2 4

log (1 ) log 2 ...2 8 192

xe

x x xe

(iii) esinx = 2 43

1 ...1.2 1.2.3.4

x xx

(iv)2 32 4

1 ...cos 2! 3!

xe x xx

x

(v) log cos x = 2 4 62 16

...2! 4! 6!

x x x

14. Prove that:

(i)2 3 42 7

log(1 tan ) ...2 3 12

x x xx x

NOTES



(ii)2

1 1 1tan

x

x

= 3 5 71

...2 3 5 7

x x xx

(iii)2

14

2sin

1

x

x

= 6 10

22 ...3 5

x xx

(iv) 2log (1 )x x = 2 3 4 5 6 7 82 2

...2 3 4 5 6 7 8

x x x x x x xx

15. Expand the following by Maclaurin’s infinite series expansion in the powersof x as far as the term containing x3

(i) secxe x

(ii) log(1 )xe x

(iii) log [1 + log (1 + x)](iv) log [1 – log (1 – x)]

16. Expand sinaxe bx as an infinite series of ascending powers of x. Give also

the (n + 1) term of the series.

17. Expand 2log (1 sin )x in powers of x as far as terms containing 4x .

18. Find the approximate value of 26 to three decimal places by Taylor’ssexpansion.

19. Expand the following:

(i) xa in powers of (x – a)

(ii) log x in powers of (x – k)

(iii) 24 7 5x x in powers of (x + 2)

(iv) 1tan x in the powers of4

x

20. Prove that

22 1( ) '( ) . ''( ) ...

1 1 1 2

x x xf f x f x f x

x x x

21. Find the asymptotes of the curve xy(x2 – y2) + x2 + y2 – a2 = 0. Show thatthe eight points of intersection of the curve with its asymptotes lie on a circlewhose centre is at the origin.

22. Find all the asymptotes of curve (x + y + 1) (y + 2x + 2) (y + 3x + 3)(y – x) + x2 + y2 – 8x = 0 and show that the eight points of intersection ofthe curve and its asymptotes lie on a circle. Find also the centre and theradius of the circle.

23. Find the equation of the cubic which has the same asymptotes as the curve

x3 – 6x2y – 11xy2 – 6y3 + x + y + 4 = 0 and which passes through (0, 0),(2, 0) and (0, 2).


NOTES


24. Find the asymptotes of the following curves:(i) r2 = a2

(ii) r = 1

a

(iii) r = 1 cos

a

(iv) r = 2

1 2cos

a

(v) r(1 – e) = a(vi) r( + ) = ae(vii) r log = a(viii) r cos = a cos 2(ix) r cos = a cos2(x) r cos = a sin2 (xi) r = a cosec + b cot (xii) r = a + b cot n(xiii) r2 = a2(sec2 + cosec2 )

(xiv) r = 3 3

3 sin cos

sin cos

a

(xv) rn sin n = an

1.10 FURTHER READING

Prasad, Gorakh. 2016. Differential Calculus. Allahabad: Pothishala Private Ltd.

Prasad, Gorakh. 2015. Integral Calculus. Allahabad: Pothishala Private Ltd.

Murray, Daniel A. 1967. Introductory Course in Differential Equations. India:Orient Longman.

Coddington, Earl A. 1987. An Introduction to Ordinary Differential Equations.New Delhi: Prentice Hall of India.

Coddington, Earl A. and N. Levinson. 1972. Theory of Ordinary DifferentialEquations. New Delhi: Tata Mc Graw-Hill.

Boyce, W.E. and Richard C. DiPrima. 1986. Elementary Differential Equationsand Boundary Value Problems. New York: John Wiley and Sons, Inc.

Gelfand, J.M. and S.V. Fomin. 1963. Calculus of Variations. New Jersey: PrenticeHall.

Ross, S. L. 1984. Differential Equations, 3rd Edition. New York: John Wileyand Sons.

Somasundaram, D. 2002. Ordinary Differential Equations. Chennai: NarosaPublishing House.

NOTES

Curvature


UNIT 2 CURVATURE

Structure

2.0 Introduction2.1 Objectives2.2 Curvature

2.2.1 Radius of Curvature for Cartesian Curves2.2.2 Radius of Curvature for Polar and Pedal Equations2.2.3 Radius of Curvature at the Origin2.2.4 Centre of Curvature2.2.5 Chord of Curvature

2.3 Tests for Concavity and Convexity2.4 Points of Inflection2.5 Multiple Points2.6 Tracing of Curves in Cartesian and Polar Coordinates2.7 Answers to ‘Check Your Progress’2.8 Summary2.9 Key Terms


2.0 INTRODUCTION

In mathematics, curvature is any of several strongly related concepts in geometry.Spontaneously, the curvature is the amount by which a curve deviates from beinga straight line, or a surface deviates from being a plane. For curves, the canonicalexample is that of a circle, which has a curvature equal to the reciprocal of itsradius. Smaller circles bend more sharply, and hence have higher curvature.

The numerical measures of the sharpness of the bending of a curve at apoint is called the curvature of the curve at that point. Here, curvature of PA isgreater than that of PB. If we roughly regard PA and PB as arcs of circles then wesee that radius of PA is less than the radius of PB. Thus we can say that radius ofcircle would be small when curvature is large and vice-versa.

Curve tracing or curve sketching means how to draw the rough sketch ofthe given curve. Principally, curve tracing helps in finding the approximate shape ofa curve without plotting a large number of points. Mathematically, the term curvetracing includes techniques that can be used to produce a rough idea of overallshape of a plane curve given its equation without computing the large numbers ofpoints required for a detailed plot. Many properties of the curve can be determinedeasily by knowing its graph. The theory of curves method is typically used to tracea curve whose equation is given in Cartesian, polar or parametric equations.

In this unit, you will study about curvature, tests for concavity and convexity,points of inflection, multiple points, tracing of curves in Cartesian and polarcoordinates.

Curvature

NOTES


2.1 OBJECTIVES


Define curvature and obtain the radius of curvature for intrinsic and Cartesiancurves

Derive the equation of radius of curvature for polar and pedal equations

Use different methods to find the radius of curvature at the origin

Define the centre of curvature and the meaning of evolute and involute

Understand the chord of curvature and generate its equation for Cartesianco-ordinates

Explain the test of concavity as well as convexity

Define point of inflexion

Define singular point and multiple points

Define double points and its different types like node and conjugate point

Determine the nature of origin in case it is a double point

Explain the condition for the existence of double point on the curve

Trace some curves whose equations are given in Cartesian, parametric orpolar forms

2.2 CURVATURE

The curvature is the amount by which a curve deviates from being a straight line,or a surface deviates from being a plane. For curves, the canonical example is thatof a circle, which has a curvature equal to the reciprocal of its radius. Smallercircles bend more sharply, and hence have higher curvature.

The curvature at a point of a differentiable curve is the curvature of itsosculating circle, i.e., the circle that best approximates the curve near this point.The curvature of a straight line is zero. The curvature of a curve at a point isnormally a scalar quantity such that it is expressed by a single real number.

The curvature of a differentiable curve was originally defined throughosculating circles. A French mathematician, engineer, and physicist Augustin-LouisCauchy showed that the center of curvature is the intersection point of two infinitelyclose normal lines to the curve.

X O

Y

P

R

Q N

C

A s

+

s

Fig. 2.1 Curvature

NOTES

Curvature


In Figure 2.1, Let P, Q be two neighbouring points on a curve AB. Let arcAP = s and arc AQ = s + ds so that the length of the arc PQ = ds, A being the fixedpoint on the curve, from where arc is measured. Let the tangents at P and Q makeangles y and y + dy, respectively with a fixed line say x-axis. Then, angle dythrough which the tangent turns as its point of contact travels along the arc PQ iscalled the total curvature of arc PQ.

The ratio δψ

δsrepresents the average rate of change in the angle y per unit of

arc length along the curve. It is called the average curvature of arc PQ.

The limiting value of the average curvature when Q P is called the curvatureof the curve at the point P.

In general, the ratioδψ

δs approaches a limit

ψd

ds as s 0.

Thus, the curvature at a point P = 0

lim limQ P s

d

s s ds

.

The curvature of a curve C at a point (x, y) on C is usually denoted by the

Greek letter (kappa). It is given by the equation κ =d

ds

where s is the arc

length measured along the curve and is the angle made by tangent line to C at(x, y) with positive x-axis.

The reciprocal of the curvature of the curve at P, is called the radius of

curvature of the curve at P and is usually denoted by . Thus, 1

ρκ

ds

d

If PC is normal at P and PC = , then C is called the centre of curvature ofthe curve at P.

The circle with centre C and radius PC = is called the circle of curvatureof the curve at P.

The length of the chord drawn through P, intercepted by the circle of curvatureat P, is called a chord of curvature.

Note: A straight line does not bend at all (because ψd

dsis zero as is constant).

Therefore, the curvature of a straight line is zero (Refer Figure 2.2).

P T

Q

O

r

s

Fig. 2.2 Chord of Curvature

Curvature

NOTES


Curvature of a Circle

Let O and r be the centre and radius of a circle, respectively. Let P and Q be twopoints on the circle.

So that, arc PQ = s and the tangent at Q make an angle with the tangentat P. Then, POQ = . Therefore, s (= arc PQ ) = r .

Differentiating with respect to , we get ds

d = r

Curvature = ψd

ds=

1

r(constant)

Thus, the curvature at every point of the circle is equal to the reciprocal ofits radius and therefore, it is constant.

Note: Circle is the only curve of constant curvature.

Radius of Curvature for Intrinsic Curves

X O

Y

P

R

Q N

C

A s

+

s

Fig. 2.3 Radius of Curvature for Intrinsic Curves

Let P, Q be two neighbouring points on a curve AB. Let the lengths of arcAP = s and arc AQ = s + s.

Therefore, the length of the arc PQ = s.

Let angles made by the tangents at P and Q with x-axis be be + s,respectively. Also let the normals at P and Q intersect at N. Join P and Q.

PNQ =

From the triangle PQN, by sine-rule, we get

chord PQPN

sin PQN sin

chord PQ

PN .sin PQNsin

chord PQ

. . .sin PQNsin

s

s

If be the radius of curvature, then we have

lim PNQ P

NOTES

Curvature


0

chord PQlim . . .sin PQN

sins

s

s

[ s 0, 0, 0

chord PQlim 1s s

, PQN

2

and 1

sin

]

ds

d

The angle between the tangents to the curve at P and Q, i.e., is calledthe angle of contingence of the arc PQ. The relation between s and for a curveis called its intrinsic equation.

Thus, we can say that at a point P

arc PQlim

Angle of Contigence of arc PQQ P.

2.2.1 Radius of Curvature for Cartesian Curves

Analytical measure of the radius of curvature in Cartesian co-ordinates include thegeneral conic, cissoid and conchoid.

Radius of Curvature for Explicit Equation y = f(x)

Equation of the curve is y = f(x).

Let be the angle which the tangent at any point P(x, y) on the curvemakes with the x-axis.

Then sin , cos tandy dydx

ds ds dx ...(1)

Differentiating both sides with respect to x, we get,

2

22

(sec )d y d

dx dx

2(1 tan ) .

d ds

ds dx

2

1 .dy d ds

dx ds dx

[using Equation (1)]

But 1d

dx

121 2

2 2and = sec (1 tan ) = 1+dyds

dx dx

Using these,

12 22 2

2

11 1

d y dy dy

dx dx dx

Hence,

32 2

2

2

1dy

dx

d y

dx

The above formula can also be written as

32 2

1

2

1 y

y

Curvature

NOTES


Note: This formula is not appropriate when the tangent is parallel to y-axis,

i.e., when dy

dx is infinite.

Since, the value of radius of curvature depends only on the curve and noton the axes, therefore in such cases we can interchange the axes of x and y and

we get

32 2

2

2

1dx

dy

d xdy

Radius of Curvature for Parametric Equations x = f(t), y = (t)

The equations of curve are given by x = f(t), y = (t),

Then /dy dy dx

dx dt dt , i.e.,

'

'

dy y

dx x

where xand y denotes the differential coefficients of x and y with respect to t,respectively.

Also,2

2

'

'

d y d y

dx dx x

'

.'

d y dt

dt x dx 2

' '' ' '' 1.

' '

x y y x

x x

1dt

dx x

3

' '' ' ''

'

x y y x

x

Putting the values of dy

dx and

2

2

d y

dx in the equation

32 2

2

2

1dy

dx

d y

dx

,we get

32 2

3

'1

'

' '' ' ''

'

y

x

x y y x

x

2 2 3/ 2( ' ' )

' '' ' ''

x y

x y y x

Radius of Curvature for Implicit Equation f(x, y) = 0

Equation of the curve is f(x) = 0

At a point where, fy 0, we get

x

y

fdy

dx f

and

2 22

2 3

2xx x y xy yy xy

y

f f f f f f fd y

dx f

NOTES

Curvature


Since,

32 2

2

2

1dy

dx

d y

dx

32 2

3

2 2

1 .

2

xy

y

xx y x y xy yy x

ff

f

f f f f f f f

32 2 2

2 22

x y

xx y x y xy yy x

f f

f f f f f f f

Radius of Curvature when x and y are Function of s

We know that, cosdx

ds … (2)

and sindy

ds … (3)

By differentiating Equation (2) with respect to s, we get

2

2sin

d d x

ds ds

… (4)

2

2

1.

dy d x

ds ds

2

2

dy

dsd x

ds

Now, differentiating Equation (3) with respect to s, we have2

2cos

d d y

ds ds

… (5)

2

2

1.

dx d y

ds ds

2

2

dx

dsd y

ds

By squaring and adding Equations (4) and (5), we get2 22 2 2

2 2

d d x d y

ds ds ds

2 22 2

2 2 2

1 d x d y

ds ds

….(6)

Again,1

sin .dy dy d dy

ds d ds d

….(7)

And1

cos .dx dx d dx

ds d ds d

….(8)

By squaring and adding Equation (7) and (8), we get2 2

2 dx dy

d d

Example 2.1: Prove that for the curve 3 3cos , sinx a y a or the curve

2 / 3 2 / 3 2 / 3x y a at 3cos , sin3 ; 3 sin cosa a a .

Curvature

NOTES


Solution: The equation of the curve is 3 3cos , sinx a y a …(i)

2' 3 cos .sindx

x ad

2' 3 sin .cosdx

y ad

Also, 2 2'' 3 cos 2sin cos ,x a

2 2'' 3 sin 2cos sin ,y a

3/ 2'2 '2

' '' '' '

x y

x y x y

3/ 2 3/ 22 2 2 2 2

2 2 2 2 2 2 2

9 sin cos cos sin

9 sin cos 2cos sin 2sin cos

a

a

2 2 2 2 2 2 29 sin cos 2cos sin 2sin cosa 3 sin cosa (numerically)

Example 2.2: If 1 2, be radii of curvature at the extremities of a pair of semi-conjugate diameters of an ellipse, prove that

2 / 3 2 / 3 2 / 3 2 21 2 .ab a b

Solution: Let the ellipse be 2 2

2 21

x y

a b

Its parametric equations are cos ,x a siny b

' sin , ' cosx a y b ; cos , sinx a y b

3/ 22 2 2 2

3 32 2

sin coscos , sin

sin cos

a ba b

ab ab

3/ 22 2 2 2sin cosa b

ab

… (i)

Let CP and CD be two semi-conjugate diameters. If the co-ordinates of P

are cos , sina b , then co-ordinates of D are cos , sin2 2

a b

If 1 is the radius of curvature at P, then

3/ 22 2 2 2

1

sin cosa b

ab

Changing ,2

we get

NOTES

Curvature


2 , the radius of curvature at 3/ 22 2 2 2cos sina b

Dab

2 2 2 2 2 2 2 22 / 3 2 / 3

1 2 2 /3 2 / 3

sin cos cos sina b a b

ab ab

2 / 3 2 / 3 2 / 3 2 2 2 2 2 21 2 sin cos cos sinab a b 2 2a b

2.2.2 Radius of Curvature for Polar and Pedal Equations

The radius of curvature of a curve at a point is the radius of the osculating circle atthat point.

Polar Equations

X O

T

P(r, )

r

L

p

Figure 4.5

Fig. 2.4 Polar Equations

The radius of curvature for the curve r = f() or f(r, ) = 0 is given by

3

2 2 21

2 22 12

r r

r rr r

.

Let the tangent at any point P(r,) to the curve make an angle with the initial lineOP, then

θ θtan , sin and cos

rd rd dr

dr ds ds , … (9)

Again, if the tangent at any point P(r,) to the curve make an angle with theinitial line OX, then

From the Figure 2.4,

= +

Differentiating with respect to s, we get

d d d

ds ds ds

.

d d d

ds d ds

1

d d

ds d

… (10)

Now,1

tand r r

rdrdr rd

Curvature

NOTES


Again differentiating with respect to , we get

22 1 1 2 1 2

2 21 1

sec .r r rr r rrd

d r r

2 2 2

1 2 1 2 1 22 2 2 2 2

1 1 21 2

1

sec (1 tan1

r rr r rr r rrd

d r r rr

r

1

tanr

r

21 2

2 21

r rr

r r

… (11)

Also,1

sincosec

dr

ds

2 2 21

1 1 1.

1 cot

d

ds r r r

… (12)

From Equations (10), (11) and (12), we get

2 2 21 2 2 1

2 2 32 22 21 21

1

211

r rr r rr rd

ds r rr r r r

32 2 2

1

2 22 12

r rds

d r rr r

Where 2

1 2anddr d r

r rd d

Corollary: If the equation of the curve is in the form u = f() where 1

ur

, then

2 2 3/ 21

32

( )ρ=

( )

u u

u u u

.

Proof: Given,1

ur

1

ru

Differentiating it with respect to , we get

11 2 2

1

θ

udur

du u , where 1 θ

duu

d

and2 2

2 1 1 1 22 4 3

.2 2u u u uu u uur

u u

, where

2

2 2θ

d uu

d

32 21

2 42 2 3/ 21

2 2 2 22 1 1 2 1

2 4 4

1

( )ρ =

2 2 21

u

u ur r

r rr r u uu u

u u u

2 2 3/ 2

13

2

( )ρ=

( )

u u

u u u

NOTES

Curvature


Radius of Curvature for Pedal Equations

Theorem 2.1: If be the radius of curvature for the curve

p = f(r), thendr

rdp

.

Proof: Let the tangent at any point to the curve make an angle with the initialline OX.

Then, = +

d d d

ds ds ds

... (13)

Also, p = r sin …(14)

Differentiating Equation (14) with respect to r, we get

cos . sindp d

rdr dr

. .dr d d

r rds dr ds

sin ,cosd dr

rds ds

d d

rds ds

dr

ds

[Using Equation (13)]

1

.dp

rdr

drr

dp

Rules to Transform Polar Equation to Pedal Equation

1. Use the relation tand

rdr

and find .

2. Put this value of in sinp r

3. Eliminate to get the equation in Pedal form.

Example 2.3: Find the radius of curvature for the parabola 2

1 cos .a

r and

show that the square of the radius of curvature varies as the cube of the focaldistance.

Solution: Given, the equation of parabola is 2

1 cosa

r …(i)

Putting 1

ur

in (i), we get 2 1 cosau …(ii)

Differentiating (ii) with respect to we have 12 sinau …(iii)

Curvature

NOTES


Differentiating (iii) with respect to , we have 22 cosau

3/ 22 21

3 22

u u

u u u

3/ 22 2

2

3

1 cos sin

2 4

1 cos 1 cos cos

2 2 2

a a

a a

3 / 2

33 / 2

3 / 23

4

12 1 cos

4 2 281 1 cos1 cos

16

aa ra

a

2 34r

a

In the given equation of the parabola, i.e., 2

1 cosa

r , the pole is at the

focus. Therefore, the focal distance of the point on the parabola is r. Hence, thesquare of the radius of curvature varies as the cube of the focal distance.

Example 2.4: Find the radius of curvature for the curve cos .n nr a n

Solution: The equation of the curve is cos .n nr a n …(i)

log log logcosn r n a n

Differentiating, 1. sin

cos

n drn n

r d n

cot

dr n

dr

Also, tan cotd

r ndr

tan tan2

n

2n

…(ii)

Now, sin sin2

p r r n

cos .

n

n

rr n r

a

Pedal equation of the given curve is 1n

n

rp

a

…(iii)

1.

1 1

n n

n n

dr a ar r

dp n r n r

NOTES

Curvature


Radius of Curvature for the Polar Tangential Equations p = f()

Y

A

O X

L

P(x, y)

B

p

Figure 4.6 Fig. 2.5 Radius of Curvature for Polar Tangential Equation

A polar tangential or tangential polar equation is a relation between p and withusual notations.

From the Figure 2.5, we have 2

...(15)

The equation of the line AB in normal form is

cos sinX Y p ...(16)

From Equations (15) and (16), cos sin2 2

p X Y

sin cosX Y

Since P (x, y) lies on this line

sin cosp x y ...(17)

Differentiating both sides with respect to , we get

cos sin . sin cos .dp dx dy

x yd d d

cos sin sin . . cos . .dydx ds ds

x yds d ds d

cos sin sin .cos . cos sin .x y

cos sindx dy

andds dx

cos sinx y

Curvature

NOTES


Again differentiating with respect to , we have2

2sin cos . cos sin

d p dx dyx y

d d d

sin cos cos . sin . .dx ds dy ds

x yds d ds d

2 2cos . sin .p p

2

2

d pp

d

Example 2.5: If be the angle which the radius vector of the curve r = f()

makes with the tangent, prove that sin 1ρ θ

r d

d

, where is the radius of

curvature. Apply this result to show that

2

a for the circle r = a cos .

Solution: We know that = + Differentiatng with respect to s

.d d d d d d

ds ds ds ds d ds

1

d d

ds d

1 sin

1ρ θ

d

r d

sind

rds

sin 1ρ θ

r d

d

… (i)

Now, equation of the circle is r = a cos

Differentiatng with respect to , we get

sindr

ad

tan = cos

cot tansin 2

d ar

dr a

12

d

d

From (i), sin 1ρ θ

r d

d

=

cosec

1θ

rdd

NOTES

Curvature


cos .cosec

2cosθsecθ =

1+1 2 2

aa a

cos sec

2 2

a a

2.2.3 Radius of Curvature at the Origin

The following methods are used to find the radius of curvature at the origin.

Method of Direct Substitution

In this method, we put x = 0 and y = 0 directly in the the formula,

32 2

1

2

1 y

y

.

We can also find first the values of dy

dx and

2

2

d y

dx at (0, 0) and then substitute

these values in this formula.

Method of Expansion

When the x-axis or the y-axis is not the tangent at the origin, we use the method ofexpansion.

Let the equation of the curve be y = f(x)

Since it passes through the origin (0, 0)

f(0) = 0.

By Maclaurin’s Expansion, we have

2 3

(0) '(0) ''(0) '''(0) ...2

x xy f xf f f

2 3

'(0) ''(0) '''(0) ...2

x xy xf f f

…. (18)

If the equation of the curve is such that y can be expanded in powers of x bytrignometrical or algebraic methods, then we have

2 31 1...

2y px qx rx

…. (19)

On comparing Equations (18) and (19), we get

p = (0)f = (y1)

0 ; q = (0)f = (y

2)

0, etc.

Now differentiating Equation (19) with respect to x, 21

1...

2y p qx rx

Differentiating again 2 ...y q rx At the origin, y

1 = p and y

2 = q

i.e., (y1)

0 = p and (y

2)

0 = q

at the origin

3 3

2 2 2 21 0

2 0

1 1y p

y q

Curvature

NOTES


Newton’s Method

Theorem 2.2: If the curve passes through the origin and axis of x is the tangent atthe origin, then

(at the origin) =

2

00

lim2x

y

x

y

Proof: Since, x-axis is the tangent at the origin, then

f(0) = 0, (0)f = 0 and (0)f = (0,0)

dy

dx

Hence, by Maclaurin’s expansion,2 3

(0) '(0) ''(0) '''(0) ...2

x xy f xf f f

2 3

0 0 ...,above2 3

x xx q r

2 3

...2

x xq r

2

2...termsof

y xq r x

x r

At the origin x 0, y 0 so that 200

2limxy

yq

x

Hence

32 2

1

2 2

1 1y

y y

1 0y

1

q

2

00

lim2x

y

x

y

Note: Similarly, if the curve passes through the origin and y-axis is the tangent atthe origin, then the radius of curvature at the origin is given by

2

00

lim2x

y

y

x

.

Curvature at the Origin when Polar Equation of the Curve is Given

If the initial line, i.e., 0 is a tangent to the curve r f , then

(at the pole)

2 2 2

0 00 0

coslim lim

2 2 sinxy r

x r

y r

(Changing to polar co-ordinates)

NOTES

Curvature


2

0 00 0

lim . .cos lim2 sin 2

r r

r r

0lim 1

sin

By L’ Hospital’s rule,

00

1lim

2r

dr

d

Corollary: Tangents at the origin.

Let the equation of the curve which passes through the origin be

2 21 1 1 .... 0ax by a x b xy c y …(20)

Let 2 3

....2! 3!

x xy px q r be its Maclaurin’s expansion.

Putting this value of y in Equation (20), we get22 2

21 1... .... ... 0,

2! 2!

x xax b px q a x b x px q

an identity..

Equating the co-efficients of x on both sides, we get

0 , 0a

a bp p bb

Since, ,a

pb

where ' 0p f is the slope of the tangent at the origin,

thus the equation of the tangent at the origin is

0 0a

y xb

0ax by

which are the lowest degree terms of the equation of the curve equated to zero.

Similarly, if the terms of the first degree in x and y are wanting in theequation of the curve, then the tangents at the origin are given

by 2 21 1 1 0a x b xy c y which are again the lowest degree terms of the equation

equated to zero and so on.

Note: Newton’s method for finding the radius of curvature shall be applied onlywhen either x-axis, i.e., 0y or y-axis, i.e., 0x is a tangent to the curve at theorigin.

Example 2.6: Find the radius of curvature at the origin for the curve

4 3 2 3 2 22 4 6 3 2 4 0x x y xy y x xy y x .

Solution: Since, the equation does not contain any constant term, therefore, thecurve passes through the origin, and the tangent at the origin is 0x ,

i.e., y-axis is a tangent at the origin. Thus, by Newton’s method,

(at the origin) =

2

00

lim2x

y

y

x

Curvature

NOTES


Dividing both sides of the given equation by 2x, we get

2 2 23 2 3

2 6. . 2 02 2 2 2

y y yx x y y x y

x x

Taking limits when 0, 0x y and

2

00

lim2x

y

y

x

, we get

2 0 2 .

2.2.4 Centre of Curvature

In Figure 2.6, point C is the centre of curvature and the circle with centreC and radius is called the circle of curvature of the curve at P.

Let C(, ) be the centre of curvature corresponding to any point P(x, y)on the curve y = f(x).

Then CP = .

Draw CN and PM perpendiculars on the axis of x and then draw PLperpendicular on CN.

Let the tangent PT at P to the curve make an angle with x-axis.

PCL = 90 – CPL

= 90 – (90 – LPT)

X O

Y

P (x, y)

N

L

C

M

T X

Y

(, ) D

Figure 4.7

Fig. 2.6 Centre of Curvature

=LPT = PTX = PCL=

Also tan = 1

dyy

dx

1

2 21 1

1sin ,cos

1 1

y

y y

And

32 2

1

2

1 y

y

NOTES

Curvature


Now, = ON = OM – NM = OM – PL = x – CP sin = x – sin [from LPC]

=

32 2

1 11

22 21

1.

1

y yx

yy

…(21)

= CN = CL +LN = CL + PM

= y + CP cos = y + cos =

32 2

1

122 2

1

1 1.

1

yy

yy

or 21

2

11y y

y …(22)

Corollary 1: Putting ρ ,sin ψ ,cosψψ

ds dy dx

d ds ds , we get

ξ .ds dy dy

x xd ds d

and .

ds dx dxy y

d ds d

Corollary 2: The equation of the circle of curvature at a given point of the curve.

Let be the radius of curvature and (, ) the centre of curvature at the givenpoint. Then, the equation of the circle of curvature is given by(x – )2 + (y – )2 = 2.

Evolute and Involute

The locus of the centre of curvature of the given curve is called the evolute of thecurve and the curve itself is called the involute of its evolute.

Example 2.7: Find the coordinates of the centre of curvature at any point(x, y) of a parabola y2 = 4ax. Also find the equation of the evolute of the parabola.

Solution: Equations of the parabola is y2 = 4ax

1 12

2 4 ora

yy a yy

and2

2 12 3

2 4.

a ay y

y y

If (, ) is the centre of curvature, then

2

221

1 22

3

41

ξ 14

a a

y yyx y x

ayy

[ y2 = 4ax]

= 3x + 2a … (i)

And

2

22

1 22

3

41

1η 1

4

a

yy y y

ayy

3 / 22x

a … (ii)

The centre of curvature at (x, y) is 3 / 22

3 2 ,x

x aa

Curvature

NOTES


To find the evolute, we have to eliminate x from Equations (i) and (ii).

Now, 33

2 4 4 ξ 2η

3

x a

a a

2 327 η 4(ξ 2 )a a

The locus of (, ), i.e., the evolute of the parabola is

27ay2 = 4(x 2a)3

2.2.5 Chord of Curvature

Any chord drawn through P intercepted by the circle of curvature at P, is knownas the chord of curvature.

Let PQ be any chord through P making an angle with CP, the radius ofcurvature at P.

Join DQ, where PD is the diameter, thus PD = 2

From PQD, PQD = π

2and DPQ

PQ = 2 cos

Cartesian Coordinates

Chords of curvature can be parallel to the axes (Refer Figure 2.7).

Let the tangent at P make an angle with the x-axis. Then, chord of curvature PAparallel to x-axis makes angle 90 – with CP, the radius of curvature at P andthe chord of curvature PB, parallel to y-axis, make angle with CP.

X T O

P(x,y) A

D

C

90–

Cy

Cx

B

Y

Fig. 2.7 Chords of Curvature Parallel to the Axis

Cx = Length of the chord of curvature PA, parallel to x-axis

= PD cos (90 – ) = 2 sin

=

2 3/ 21 1

22 1

(1 )2 .

1

y y

y y

1

1

21

tan

sin1

y

y

y

= 2

1 1

2

(1 )2

y y

y

NOTES

Curvature


Cy = Length of the chord of curvature PB, parallel to y-axis

= PD cos = 2 cos

=

2 3/ 21

22 1

(1 ) 12 .

1

y

y y

=2

1

2

(1 )2

y

y

Hence, chord PA (parallel to x-axis) = 2 sin =2

1 1

2

(1 )2

y y

y

and chord PB (parallel to y-axis) = 2 cos =2

1

2

(1 )2

y

y

Example 2.8: If Cx and C

y be the chords of curvature parallel to the axis of x and

y respectively at any point of the curve y = aex/a, prove that 2 2

1 1 1

2x y xC C aC .

Solution: Equation of the curve is y = aex/a

/1 2 1 2

1 1. , .x a yy

y ae y ya a a a

tan = y1

sin = 1

2 21

y

a y

cos = 2 2

1

a

a y

Cx = 2 sin =

32

1 11

2 221

2. 1

1

y y

y y

3 221

2

1P

y

y

211

2

2. 1y

yy

2 2

22. 1

y a y

a y a

2 2

22

a ya

a

2 22 a y

a

Cy = 2 cosψ

32 2

1

12 22

1

1 12.

1

y

yy

2

1

2

2. 1 y

y

2

2 22

2

2. 12.

ya ya

y ya

2 2 2 2

2 2 2 2 2 2 2 2 2 2 2

1 1

4( ) 4( ) 4( )x y

a y a y

C C a y a y a y

Curvature

NOTES


2 2

1

4( )a y

1

2 xaC

2 22( )x

a yC

a

Check Your Progress

1. Find the radius of curvature at any point of the catenary coshx

y cc

.

2. Show that for the curve 8 , ρ 4 12

ys ay a

a .

3. Find at any point of the cycloid x = a ( + sin ), y = a (1 cos ).

4. Find the radius of curvature of x y a , at the point where the liney = x cuts it.

5. Find the radius of curvature for the hyperbola 2 2

2 2 22

a br a b

p .

6. Find the radius of curvature of the curve r = f().

7. Prove that for curve with usual notations

2 2

2

θ θρ

d d r dr r

ds dsds

.

8. Find the radius of curvature at the origin of the curve2 26 2 0x y x y

9. Show that the chord of curvature parallel to y-axis for the curve y = a log

sec x

a

is of constant length.

2.3 TESTS FOR CONCAVITY AND CONVEXITY

X

Y

Fig. 2.8 Concavity Curve

Let f be any function continuous on a closed interval [a, b] and differentiable onan open interval (a, b). Then,

The graph of f is concave upwards or convex downwards on [a, b] if,

throughout (a, b), it lies above the tangent lines of f, i.e., if f is increasingin the interval.

NOTES

Curvature


The graph of f is concave downwards or convex upwards on [a, b] if,

throughout (a, b), it lies below the tangent lines of f, i.e., if f is decreasingin the interval.

Conditions for the Concavity or Convexity of the Curve in theUpward Direction

Y Y = f (x)

A

B C

(x,y)

0 XN

B [x+h, f(x+h)]

N

C

Y

(x,y)A

B

C

0 N X

B[x+h, r (x+h)]

C

N

y = f(x)

Fig. 2.9 (a) Fig. 2.9 (b)

Let B[x + h, f(x + h)] be a point close to the point A(x, y) of the curve.Draw a line BN perpendicular to the x-axis. Suppose the tangent at A meet theordinate in C. As, C lies on the ordinate BN, hence the abscissa of C is the sameas that of B (Refer Figure 2.9(a) and 2.9(b)).

Let the ordinate of C be ty and that of B be cy .

Then, cy f x h … (23)

The equation of the tangent to the curve y f x at A(x, y) is

Y (X )dy

y xdx

Y = ( ) ( )(X )f x f x x … (24)

Since point C lies on the tangent at A, hence the co-ordinates of

C(x + h, ty ) must satisfy Equation (24).

( ) ( )ty f x hf x … (25)

We can see from the Figure 2.9 (a) and (b), that the curve is concave or convex

upwards at A(x, y) depending upon whether 0c ty y or 0c ty y ,respectively.

So, by subtracting Equation (25) from Equation (23), we get

( ) ( ) ( )c ty y f x h f x hf x

= 2

( ) ( ) ( ) ( ) ( )2!

hf x hf x f x h f x hf x

= 2

( )2!

hf x h … (26)

Let ( )f x be a continuous function of x, hence for very small value of h, the sign

of ( )f x h is the same as that of ( )f x . Therefore, the sign of c ty y depends

Curvature

NOTES


upon ( )f x because2

2!

his always positive. Thus, the curve is concave or convex

in upwards direction depending upon whether c ty y > 0 or

c ty y < 0, i.e., 2

20

d y

dx or

2

20

d y

dx .

Similarly, the curve is concave or convex in the positive direction of x-axis dependingupon whether

2

20

d x

dy or

2

20

d x

dy .

Note: The above proof can also be generalized for ( )nf x . In this case, the curve

is concave or convex in the upward direction depending upon whether

( ) 0nf x or ( ) 0nf x , n being even.

2.4 POINTS OF INFLECTION

A point of inflexion is a point on a curve at which the curve changes from beingconcave upwards to concave downwards or vice-versa.

If P(x, y) is a point of inflexion on the curve, then on one side of P, the curveis concave and on the other side it is convex. At the point of inflexion, the tangentcrosses the curve at that point.

Test for Point of Inflexion

If , ( )c f c is a point of inflexion on the graph of f, then either ( ) 0f c or f does

not exist at c. The converse may not necessarily be true.

Thus, a point is said to be a point of inflexion, if 2

20

d y

dx at this point and

2

2

d y

dxchanges sign while passing through this point, i.e.,

3

30

d y

dx at this point.

Steps for finding point of inflexion

(i) Find all the numbers at which ( ) 0f x

(ii) Use the test of concavity

(iii) There is an inflexion point, if the concavity changes; otherwise, there is noinflexion point.

NOTES

Curvature


Example 2.9: Find the points of inflexion of the curve 6 5( 2) ( 3) .y x x

Solution: The given curve is 6 52 3 .y x x ...(i)

6 4 5 52 .5 3 6 2 3

dyx x x x

dx

5 42 3 5 2 6 3x x x x 5 4

2 3 11 28x x x

2

5 4 5 3 4 4

22 3 .11 2 11 28 .4 3 3 11 28 .5 2

d yx x x x x x x x

dx

= 4 32 3 11 2 3 4 2 11 28 5 3 ( 11 28x x x x x x x x

4 3 22 3 110 560 710x x x x

4 3 210 2 3 11 56 71x x x x

4 3 28 3 28 310 2 3

11 11x x x x

At the points of inflexion2

20

d y

dx

28 32,3, .

11x

At 2

22,

d yx

dx does not change sign as the corresponding factor 4

2x

remains +ve whether 2 or 2.x x

2x does not correspond to the point of inflexion.

Also, 2

2

d y

dx changes sign at 3x and

28 3

11x

.

Thus, these correspondences to the abscissa of the points of inflexion of thegiven curve.

Check Your Progress

10. Show that the curve y = log x is convex upwards everywhere.

11. Determine the point of inflexion on the curve

(2 sin ), (2 cos )x a y a

2.5 MULTIPLE POINTS

Singular point is a point on a curve at which the curve shows an extraordinary orunusual behaviour. Now, in this unit, we shall study about two particular types ofsingular points: multiple points and points of inflexion.

Curvature

NOTES


If more than one branch of a curve passes through a point on the curve,then this point is said to be a multiple point of the curve.

If two branches of the curve pass through a point on the curve, then thispoint is called a double point.

There are two tangents, one to each branch can be drawn at the doublepoints of the curve. The tangents drawn may be real and distinct, real and coincidentor imaginary.

Types of Double Points

Depending upon whether the tangents at the double points are distinct, coincidentor imaginary, these double points are classified into the following three categories:

Node: If the two tangents at the double point are real and distinct, then thedouble point is known as a node. (Refer Figure 2.10)

Cusp: If the two tangents at the double point are real and coincident, thenthe double point is known as a cusp. (Refer Figure 2.11)

Conjugate or Isolated Point: If there are no real points of the curve in theneighbourhood of the point, then this point P is called a conjugate or anisolated point of the curve. (Refer Figure 2.12)

Fig. 2.10 Node

Remark 1: Normally the tangents at the isolated points are imaginary. Insome cases, they are real but then the two branches of the curve are imaginary.

P

Fig. 2.11 Cusp Fig. 2.12 Conjugation Isolated Point

Remark 2: In case the curve passes through the origin, if you equate thelowest degree terms of x and y to zero, you will get the equations of thetangents to the curve at the origin.

Rules to Determine the Nature of the Origin, if it is a Double Point

(1) (i) In case the two tangents at the origin are imaginary, the conjugate or anisolated point of the curve is origin.

NOTES

Curvature


(ii) In case the two tangents at the origin are real and distinct, the originmay be a node or a conjugate point depending upon the two branchesthrough the origin real or imaginary.

(iii) In case the two tangents at the origin are real and coincident, the originmay be a cusp or a conjugate point depending upon the two branchesthrough the origin real or imaginary.

(2) When you have doubt about the nature of the double points as in (ii) and(iii) above, use the following methods for testing the nature of the curve atthe double point.

(i) If a cuspidal tangent at the origin is given by x = 0, then neglect cubesand higher powers of x and solve the equation to get the value of x. Ifx is real for small enough values of y, then the branches of the curvethrough the origin are real, otherwise they are imaginary.

(ii) If a cuspidal tangent at the origin is given by y = 0, then neglect cubesand higher powers of y and solve the equation to get the value of y. Ify is real for small enough values of y, then the branches of the curvethrough the origin are real, otherwise they are imaginary.

Note: While neglecting cubes and higher powers of x or y, keep in mindthat the reduced equation of the curve may not coincide with that of thejoint equation of the two tangents at the origin.

Example 2.10: Show that the point (1, 2) is a node on the curve y(x – 1)2

= x(y – 2)2 and also determine the equations of the tangents to the curve at thispoint.

Solution: Shift the origin to the point (1, 2) by putting x – 1 = X and y – 2 = Y, weget

(Y + 2)X2 = (X + 1)Y2 … (i)

So, the tangents at this new origin are

2X2 = Y2 Þ Y = 2X

Thus, the new origin is either a conjugate point or a node.

Now, equation (i) can also be rewritten as,

(X + 1)Y2 – X2Y – 2X2 = 0

By solving it for Y, we get

Y = 2 4 2X X 8(X +1)X

2(X + 1)

The quantity given in above equation will have positive sign for small valueof X. Hence, y is real.

Thus, the two branches of the curve which pass through the origin are realand so the new origin, i.e., the point (1, 2) is a node.

Curvature

NOTES


Now, the tangents at the new origin are

Y = 2X and Y = 2X

y – 2 = 2( 1)x and y – 2 = 2( 1)x

y = 2 2 2x and y = 2 2 2x

Condition for the Existence of a Double Point on a Curve

Suppose the equation of the curve is given by f (x, y) = 0 … (27)

By differentiating it with respect to x, we get 0f f dy

x y dx

… (28)

Equation (28) is a first degree equation in dy

dxand hence gives only one

value at any point P(x, y) on the curve. We know that there are two tangents to

the curve at double point. Hence, dy

dx should have two values which satisfy Equation

(28) But this is possible only if,

f

x

= 0 and f

y

= 0 … (29)

On solving Equation (29) for x and y simultaneously, we get the possibledouble points. As P(x, y) which lies on the curve f(x,y), therefore only those valueof x and y are to be considered as double point that satisfy Equation (27) also.

We now see the nature of the double point that is obtained from the above.

By differentiating Equation (28), x, we get 0f f dyd d

dx x dx y dx

2 2 2 2 2

2 2 20

f f dy f d y dy f f dy

y x dx y dx x y dxx dx y

Using f

y

= 0 from Equation (27) in the above equation, we get

22 2 2

2 22 0

f f dy f dy

x y dx dxx y

2 2f f

x y y x

22 2 2

2 22 0

f dy f dy f

dx x y dxy x

… (30)

Equation (30) is a quadratic equation in dy

dxand hence gives two values

dy

dx,

which provides the slopes of the tangents at the double point of the curve. Thesetangents may be real and distinct, real and coincident or imaginary dependingupon whether the discriminant (D) of Equation (30) is positive, zero or negative.

NOTES

Curvature


i.e.,

22 2 2

2 2.

f f f

x y y x

>, = or < 0, respectively … (31)

Now, if(i) D > 0, double point is a node or a conjugate point(ii) D = 0, double point is a cusp or conjugate point(iii) D < 0, double point is a conjugate point.

Example 2.11: Find the position and nature of the double points on the curve2 2( 2) ( 1)y x x

Solution: Let 2 2, 2 1 0f x y x x y …(i)

Then 22 2 2 1

fx x x

x

and 2f

yy

For the double points 0f

x

and 0

f

y

2 2 2 2 0x x x

x = 2 and x = 4

3And y = 0

Thus, the possible double points are (2, 0) and4

,03

. But only (2, 0) satisfies

Equation (i). Therefore (2, 0) is the only double point.

Shifting the origin at (2, 0), Equation (i) transforms into

Y2 = X2(X + 1) …(ii)

The tangents at the new origin are Y2 = X2 i.e., Y = X

Thus the new origin is either a node or a conjugate point,

Solving Equation (ii) for Y, we get Y X X 1 which gives Y real forsmall values of X. Thus, the branches of the curve near the new origin are real.Therefore, the new origin or the point (2, 0) on Equation (i) is a node.

Check Your Progress

12. Find the nature of the origin for the curve 4 2 4 2 2( ).a y x x a

13. Find the position and nature of the double points of the curve

3 2 2 4 3 0x x y x y .

Curvature

NOTES


2.6 TRACING OF CURVES IN CARTESIANAND POLAR COORDINATES

The procedure of tracing a curve in Cartesian coordinates is as follows (ReferFigure 2.13 (a) (b) (c) (d) and (e).

1. Symmetry. Apply the following rules to ascertain whether the given curveis symmetrical about any line or not:

(i) Symmetry about the x-Axis: If the equation of a curve remainsunaltered when y is changed to – y, i.e., if all the powers of y in thegiven equation are even, the curve is symmetrical about the x-axis.For example, since in the parabola y2 = 4ax, power of y is even,hence it is symmetrical about the x-axis.

(ii) Symmetry about the y-Axis: If the equation of a curve remainsunaltered when x is changed to – x, i.e., if all the powers of x are evenin the given equation, the curve is symmetrical about the y-axis. Forexample, since the curve x2 = 4ay has the even power of x, hence it issymmetrical about the y- axis.

(iii) Symmetry about both Axes: If all the powers of both x and y areeven in the given equation, the curve is symmetrical about both the

axes. For example, since in the ellipse 2 2

2 21

x y

a b , powers of both

x and y are even, thus it is symmetrical about both the axes.(iv) Symmetry in Opposite Quadrants: If the equation of the curve

remains unchanged when x and y are changed to –x and –y,respectively, the curve is symmetrical in opposite quadrants. Forexample, the hyperbola xy = c2 and the curve y = x3 are symmetricalin opposite quadrants.

(v) Symmetry about the Line y = x: If the equation of the curve isunchanged when x and y are interchanged (i.e., x is changed to y andy to x), the curve is symmetrical about the line y = x. For example, thecurve x3 + y3 = 3axy is symmetrical about the line y = x.

Y

O

(x, –y)

y = 4ax2

(x, y)

X

x = 4ay2

(–x, y) (x, y)

O X

Y

y = x3

Y

O X

(–x, y)–

(x, y)

(a) (b) (c)

NOTES

Curvature


Y

OX

(y, x)

(x, y)

Y

(x, y)

(–y, –x)(–2, 2)

XOx + Y + 4x – 4y + 1 = 02 2

(d) (e)

Fig. 2.13 Tracing a Curve in Cartesian Co-ordinates

(vi) Symmetry about the Line y = –x: If the equation of the curve isunchanged when x is changed to –y and y is changed to –x, the curveis symmetrical about the line y = –x. For example, the curve x2 + y2 +4x – 4y + 1 = 0 is symmetrical about the line y = –x.Note: The curve which has symmetry about both the axes also hassymmetry in opposite quadrants but the converse is not true. For

example, in ellipse 2 2

2 21

x y

a b (symmetrical about both the axes), the

equation of the curve remains unchanged when x and y are changedto –x and –y, respectively. Thus, it is symmetrical in opposite quadrants

also. But the curve 1

y xx

(symmetrical in opposite quadrants) does

not have even powers of x or y. Thus, it is not symmetrical about boththe axes.

2. Origin: Find whether the curve passes through the origin or not.(i) It will pass through the origin if the equation of the curve does not

have any constant term. In this case, find the equation of the tangentsat the origin by equating the lowest degree terms (present in the equationof curve) to zero.

(ii) If the origin is a double point, then find its nature. The origin is a nodeif the tangents are real and distinct; cusp if the tangents are real andcoincident; and conjugate point if the tangents are imaginary.

3. Asymptotes: To find the direction in which the curve extends to infinity,find the following asymptotes:

(i) Asymptotes parallel to x-Axis or y-Axis: Find the asymptotesparallel to x-axis or y-axis by equating the co-efficients of highestpowers of x or y (in the equation of the curve) to zero.

(ii) Oblique Asymptotes: Put x = 1 and y = m in the highest degreeterms of x and y to find f

n(m). Then, equate f

n(m) to zero and solve to

get the values of m, which gives the slopes of the asymptotes. Putx = 1 and y = m in the next lower degree terms of x and y to findf

n–1(m) and similarly find f

n–2(m) and obtain the values of m.

Curvature

NOTES


If m has real and distinct values, then find c using 1( )

( )

n

n

mc

m

and

if m has real and equal values, then find c using2

1 2( ) ( ) ( ) 02! n n n

cm c m m

.

Finally, find the oblique asymptotes by putting the values of m and c inthe equation y = mx + c.

4. Point of Intersection with the Axes(i) Find the points where the curve meets the co-ordinate axes. If the

curve meets the x-axis, then on putting y = 0, we get the real values ofx. Similarly, examine for y-axis where x = 0.

(ii) Shift the origin to all such points and find the tangents at all such points(new origin) by equating to zero the lowest degree terms. If any oneof them is a double point, find its nature.

(iii) Find the points of intersection of the curve with the lines y = ± x if thecurve is symmetrical about these lines.

5. Region: Find the region to which curve is bounded in the quadrants usingthe following steps:

(i) Solve the equation of the curve for y. Find the corresponding valuesof x for which y is real. The curve lies between these obtained valuesof x.

(ii) Ignore the values of x and y for which L.H.S. and R.H.S. of theequation has opposite sign.

(iii) Put x = 0. Examine how y varies when x increases and tends to ,keeping in mind those values of x for which y = 0 or y .

(iv) Examine how y varies when x decreases from 0 and tends to –. Ifthe curve is symmetrical about y-axis, take only the positive values ofx and trace the curve for negative values of x by symmetry.

6. Special Points:(i) Find the points on the curve where:

(a) 0dy

dx , i.e., the points where the tangents are parallel to the x-

axis.

(b)dy

dx , i.e., the points where the tangents are parallel to the

y-axis.

(c)dy

dx is positive, i.e., the points where the curve is increasing or

(d)dy

dx is negative, i.e., the points where the curve is decreasing.

(ii) Find the points on the curve where:

NOTES

Curvature


(a)2

2

d y

dx is positive, i.e., the points where the curve is concave

upward

(b)2

2

d y

dx is negative, i.e., the points where the curve is concave

downward(iii) Find points of inflexion on the curve by equating

2 3

2 3to zero where 0

d y d x

dx dy

Some Elementary Curves

Some curves which are useful in drawing the approximate shape of the curvesnear the origin when either x-axis or y-axis is a tangent at the origin as given inFigure 2.14(a), (b), (c), (d), (e), (f), (g) and (h).

These figure display the cubic parabola.

Figure 2.15(a), (b), (c) and (d) show semi-cubic parabola.

y = 4ax2

Y(x, y)

AxisX

S (a, 0)

(a)Y΄

X΄

S (–a, 0)

X΄Axis

Y

y = –4ax2

X

(b)Y΄

O O

x = 4ay2

Y Axis

S (0, a)

OX

(c)

X´ X´ X

Y

x = –4ay2

AxisYÝ´

(d)

O

S (0, – a)

y = – x3

Y

O

Y´

X´ X

Y

XO

X´

Y´

x = y3

(e) (f)

Curvature

NOTES


Y

Y´

XX´X´

Y

x = – y3

O

Y´

X

(g) (h)

Fig. 2.14 Cubical Parabola

x = y32

Y

Y´

XÓ X O XX´

Y

Y´

y = – x32

(a) (b)

Y

XO

x2 = – y

3

X´

Y´

Y

XOX´

Y´

Fig. 2.15 Semi-Cubical Parabola

Example 2.12: Trace the curve y2(2a – x) = x3.

Solution: The given equation of the curve is y2(2a – x) = x3 …(i)I. Symmetry: Since (i) contains even powers of y, the curve is

symmetrical about x-axis.II. Origin: The curve passes through the origin. Equating the lowest

degree terms to zero, the tangents at the origin are given by y2 = 0 ory = 0, i.e., the two tangents are real and coincident. Thus, origin is acusp and y = 0, i.e., x-axis is the tangent.

III. Asymptotes: Equating to zero the coefficient of y2 (the highest degreeterm in y), the asymptote parallel to y-axis is x – 2a = 0 or x = 2a.There is no other asymptote of the curve.

IV. Point of Intersection with Co-Ordinate Axes: The curve meetsx-axis and y-axis at the origin (0,0) only.

NOTES

Curvature


V. Region: From (i) , we have 3

2

2

xy

a x

or

2

xy x

a x

We see that the y is real for x < 2a or 0 < x < 2a, i.e., the whole curvelies between lines x = 0 and x = 2a.

VI. Special Points: Differentiating 3

2

2

xy

a x

w.r.t x,

2 3

2

3(2 ) ( 1)2

(2 )

dy a x x xy

dx a x

0dy

dx when 2 33(2 ) ( 1)a x x x = 0 which gives x = 0, 3a.

When x = 3a, y is imaginary. So, reject x = 3a. Thus, the tangent is parallelto x-axis at x = 0.

Example 2.13: Trace the curve 2 / 3 2 / 3

1x y

a b

Solution: The given equation of the curve is 2 / 3 2 / 3

1x y

a b

…(i)

I. Symmetry: Equation (i) can be written as

2 21/3 1/ 3

1x y

a b

.

Clearly, the equation contains even power of both x and y and hence thecurve is symmetrical about both the axes.

II. Origin: The curve does not pass through the origin.

III. Asymptotes: There is no asymptote of the curve since it is a closed curve.

IV. Points of Intersection with Co-Ordinate Axis: The curve meets x-axis

where y = 0, i.e., which gives 2

1x

a

21/ 3

obtain by cubing both sides of 1x

a

or .x a . Similarly, it

meets y-axis where x = 0, which gives .y b

V. Region: From Equation (i), we have2 / 3 2 / 3

1y x

b a

1/ 3 2 / 3

1y x

b a

3/ 22 / 3 2 / 3a xy

b a

3/ 22 / 3 2 / 3b

y a xa

We see that y is real only when | |x a . Thus, the whole curve lies betweenthe lines .x a . Similarly, expressing the equation in terms of x, we see

Curvature

NOTES


that the whole curve lies between the lines .y b Thus, the whole curve

lies between the rectangle formed by the lines x a and y b .

Y

X´

Y´

XO

(0, b)

(–a, 0) (a, 0)

(0, –b)

VI. Special Points: Differentiating Equation (i) with respect to x,

1/ 3 1/ 32 1 2 1

03 3

x y dy

a a b b dx

2 / 3 1/ 3

2 /3 1/ 3.

dy b y

dx a x

Since 0dy

dx , thus 0y

2 / 3

1x

x aa

Thus, x-axis is tangent to the curve at (a, 0) and (–a, 0). As the curve issymmetrical about x-axis, thus x-axis is a cuspidal tangent at both these points.Similarly, y-axis is a cuspidal tangent at the points (0, b) and (0, –b). Also, on

finding 2

2

d y

dx we can observe that when both x and y are positive,

2

20

d y

dx and

hence the curve is concave upwards in the Quadrant I.

Since, the curve is symmetrical about both the axes, thus the shape of thecurve is as shown in the above Figure.

Example 2.14: Trace the curve 1

y xx

.

Solution: The given equation of the curve is 1

y xx

or 2 1xy x …(i)

I. Symmetry: Since Equation (i) remains unchanged on replacing x by –xand y by –y, thus the curve is symmetrical in opposite quadrants.

NOTES

Curvature


II. Origin: The curve does not pass through the origin.

OXX´

Y´

(–1, –2)

dy

dx

dy

dx

= 0

= 0

(1, 2)

Yx = 0

y = x

III. Asymptotes: Equation (i) is the equation of second degree in x and y andthe term y2 is absent. Since, the co-efficient of y is x, thus x = 0, i.e., y-axisis an asymptote. The expansion of the equation in the form

2

A B...y mx c

x x gives asymptote as y = mx + c. Thus, the

asymptote of 1

y xx

is y x . Also if the ordinates of the curve and the

asymptotes are cy and ay , then1

c ay yx

. When x > 0, 1

c ay yx

is

positive and when x < 0, it is negative.

IV. Points of Intersection with Co-Ordinate Axes: The curve does not meetany of the axes.

V. Region: Since, when x > 0, then y > 0 and when x < 0, then y < 0, thus,the curve lies in I and II quadrants. From step III, it is clear that the curvelies above the asymptote y = x in I quadrant and it lies below the asymptotein III quadrant.

VI. Special Points: When x = –1 and x = 1, the maximum and minimum

values of y are –2 and 2. Also, 0dy

dx at (1, 2) and (–1, –2).

Thus, the shape of the curve is as shown the above Figure.

Tracing of Parametric Curve

Let us learn to trace a curve whose equations are of the form x = f(t) and y = (t),where t is the parameter.

Method 1. We try to eliminate the parameter t if it can be easily eliminated.The equation, thus, obtained is the corresponding Cartesian equation of the curve.These curves can then be easily traced.

For example,

(i) Ellipse: x = a cos t, y = b sin t or 2

2 2

(1 ) 2,

1 1

a t btx y

t t

cos , sinx y

t ta b or

2

2 2

(1 ) 2,

1 1

x t y t

a t b t

Curvature

NOTES


Eliminating t by squaring and adding, we get2 2

2 21

x y

a b , which is an ellipse.

(ii) Circle: x = a cos t, y = a sin t or 2

2 2

(1 ) 2,

1 1

a t atx y

t t

Eliminating t by squaring and adding, we get

x2 + y2 = a2, which is a circle.

(iii) Cissoid: 3

2 sinsin ,

cos

tx a t y a

t

2sinx a t 3 3 6sinx a t

Or 2sinx a t 21 cosx a t 2cos 1x

ta

3sin

cos

ty a

t

62 2

2

sin

cos

ty a

t

3

2

1

x

ayx

a

3

2 xy

a x

,

which is called Cissoid.

(iv) Astroid: 3 3cos , sinx a t y a t

2 2 2 2

2 23 3 3 3cos , sinx a t y a t

Eliminating t by adding, we get2 2 2

3 3 3x y a , which is called Astroid.

(v) Hypocycloid: 3 3cos , sinx a t y b t

3 3cos , sinx y

t ta b

2 2

3 32 2cos , sin

x yt t

a b

Eliminating t by squaring and adding, we get2 2

3 31

x y

a b

, which is called Hypocycloid.

(vi) Simple Curve: 2 31,

3x t y t t

2

3 2 2 21 11

3 3y t t y t t

Eliminating t by putting t2 = x, we get2

2 13

xy x

9y2 = x(x – 3)2, which is a form of a simple curve.

NOTES

Curvature


(vii) Folium of Descartes’: 2

3 3

3 3,

1 1

at atx y

t t

Dividing, we get y

tx

Putting this value of t in 3

3

1

atx

t

, we get

x3 + y3 = 3axy, which is called Folium of Descartes’.

(viii) Lemniscate of Bernoulli: 3 3

4 4

( ) ( ),

1 1

a t t a t tx y

t t

Eliminating t, we get (x2 + y2)2 = a2(x2 – y2), which is called Lemniscate ofBernoulli.

Method 2. When the parameter t cannot be easily eliminated, then we adopt thefollowing steps to trace the curve:

I. Symmetry:

(i) Symmetry about x-Axis: The curve is symmetrical about x-axis if byreplacing t by –t or t by – t, y changes to –y and x remains unchanged.For example, the parabola x = at2, y = 2at is symmetrical about x-axis.

(ii) Symmetry about y-Axis: The curve is symmetrical about y-axis if byreplacing t by –t or t by – t, x changes to –x and y remains unchanged.For example, the ellipse x = a cos t, y = b sin t is symmetrical abouty-axis.

(iii) Symmetry in Opposite Quadrants: The curve is symmetrical in oppositequadrants if by replacing t by –t, x changes to –x and y changes to –y.

For example, the rectangular hyperbola x = ct, y = c

t is symmetrical in

opposite quadrants.

II. Origin: Find whether the curve passes through the origin. The curve passesthrough the origin if the same real value of t satisfies both f(t) = 0 and (t) = 0.

III. Asymptotes:

(i) Asymptotes Parallel to Co-Ordinate Axes: Obtain the asymptotesparallel to x-axis by finding the values b

1, b

2, … to which y tends as x

or x – . We get y = b1, y = b

2 as asymptotes. Similarly, obtain the

asymptotes parallel to y-axis by finding the values a1, a

2, … to which x

tends as y or y – . We get x = a1, x = a

2 as asymptotes.

(ii) Oblique Asymptotes: y = mx + c is an oblique asymptote if for somevalue of t (say t t

0), both x and y tends to and

0 0

lim and lim( )t t t t

ym y mx c

x .

Curvature

NOTES


IV. Points of Intersection with Co-Ordinate Axes: The curve meets the x-axis where y = (t) = 0 and it meets the y-axis where x = f(t) = 0.

V. Region: For some real values of t, find the greatest and the least values of xand y if possible. This gives the lines parallel to the axes of the co-ordinate, betweenwhich the curve lies or does not lie.

VI. Special Points: Find dy

dt and

dx

dt. Divide them to get

dy

dx, which is the slope

of the tangent at the point [f(t), (t)]. Then obtain the points where dy

dx = 0 or ,

i.e., the points where the tangent is parallel to x-axis or y-axis. If required, also

find 2

2

d y

dx and discuss the concavity or convexity and points of the inflexion on the

curve.

Example 2.15: Trace the curve 2 31,

3x t y t t .

Solution: The given equation of the curve is 2 31,

3x t y t t . Eliminating t from

these equations, we get 229 3y x x . … (i)

I. Symmetry: Since Equation (i) contains only even powers of y, thus thecurve is symmetrical about x-axis.

II. Origin: The curve passes through the origin and the tangent at the origin isgiven by x = 0, i.e., y-axis is the tangent at the origin.

III. Asymptotes: There is no asymptote of the curve.

IV. Points of Intersection with the Co-Ordinate Axes: The curve meetsy-axis only at (0, 0). It meets x-axis at x = 0 and x = 3. When the origin isshifted at (3, 0), the equation transforms into 9Y2 = (X + 3) X2. So, the

tangents at the new origin are 9Y2 = 3X2 or 1

Y X3

. Thus, the new

origin at the point (3, 0) is a node.

V. Region: Since, for 0,x y is imaginary, thus no part of the curve lies to

the left of the y-axis. Also, for 3,x y is real. For larger values of 2 3,x y x ,

i.e, the curve is inclined towards y-axis.

Thus, the shape of the curve is as shown in the above Figure .

NOTES

Curvature


Example 2.16: Trace the curve sin , 1 cosx a y a , – .

Solution: The given equation of the curve is

sin , 1 cosx a y a …(i)

I. Symmetry: Since in Equation (i), when is changed into –, x changesinto –x and y remains unchanged, thus, the curve is symmetrical about y-axis.

II. Origin: Since, does not have any common value for which x and y bothare zero, thus the curve does not pass through the origin.


IV. Points of Intersection with Co-Ordinate Axes: The curve meets x-axiswhere y = 0, which gives 1 cos 0 and – and thecorresponding values of x thus obtained are a and –a . The curve alsomeets y-axis where x = 0, which gives = 0 and the corresponding value ofy thus obtained is 2a.

V. Region: Since, | cos | 1,0 2y a , i.e., the whole of the curve liesbetween y = 0 and y = 2a.

VI. Special Points: Differentiating Equation (i), we have 1 cosdx

ad

and

sindy

ad

x = –aπ

(–a , 0)π

0 = –π OX´

Y´

X

Yθ = 0 x = aπ

y = 2aB(0, 2a)

A0 = –π

(a , 0)π

Now, sin

1 cos

dy a

dx a

2

2sin cos2 2 tan

22cos2

Since 0dy

dx , thus tan 0

2

This gives = 0. So, the tangent is parallel to x-axis where = 0, whichgives x = 0, y = 2a, i.e, at (0, 2a).

Now,dy

dx or

This gives the corresponding points ,0a and ,0a .

As2

2 42

1 1sec . sec

2 2 2 2

d y d

dx dx

,

1

1 cos

d

dx a

Curvature

NOTES


2

20

d y

dx for all values of and hence the curve is concave downwards.

When or , the point traces similar curves.

The shape of the curve is as shown in the above Figure .

θ = 0

Y

B y = 2a

X´

Y´

O 0 – πA´ A

(a , 2a)π

θ = 2π(2ap , 0)

x = 2ap

X X´ X

A (aπ, 2a)

y =2a

θ = –π

A´ , 2a) = (–aπ

x = –ax θ = 0O x = aπ

θ = π

Y´

Y

Note: The above equation is the equation of the cycloid. A cycloid is a curve whichis described by a fixed point in the plane of the circle (generating circle) when thecircle rolls without sliding along some straight line (directix). There are three othertypes of equations of the cycloid, which can be traced on the similar lines.

Curve Tracing in Polar Coordinates

The procedure of tracing a curve in polar co-ordinates is as follows:

I. Symmetry: Apply the following rules to ascertain whether the given curveis symmetrical about any line or not:

(i) Symmetry About the Initial Line, = 0 or x-Axis: If the equationof the curve remains unchanged when is changed to – or the pair (r,) to the pair (–r, – ), the curve is symmetrical about the initial line.For example, the curve r cos = a sin2 is symmetrical about the x-axis.

(ii) Symmetry About the Ray = 2

or y-Axis: If the equation of the

curve remains unchanged when is changed to – or the pair (r,) to the pair (–r, – ), the curve is symmetrical about the y-axis. Forexample, the curve r = a is symmetrical about y-axis (seeFigure 2.16).

NOTES

Curvature


(a) (b)

(c)

Fig. 2.16 Curve Tracing in Polar Coodinates

(iii) Symmetry About the Pole: If the equation of the curve is unchangedwhen r is changed – r or the pair (–r, ) to the pair (r, + ), thecurve is said to be symmetrical about the pole. For example, thecurve r2 = a2 cos 2 is symmetrical about the pole.

II. Origin or Pole: Find whether the curve passes through the pole or not byputting r = 0 in the equation. If we get =

1,

2,

3, … where

1,

2,

3, …

are real numbers, then the curve passes through the pole otherwise thecurve does not passes through the pole. In case the curve passes throughthe pole, the tangents at the pole are given by =

1, =

2, =

3, …

III. Asymptotes: When r as 1, there is an asymptote. If the equation

f() = 0 has the root , then the asymptotes of the curve 1

( )fr is given

by r sin( – ) = 1

( )f .

IV. Point of Intersection: Find the corresponding values of r for some suitablevalues of , particularly for those values of for which the curve issymmetrical.

V. Region. Solve the given equation for r and examine how r varies when increases from 0 or

1 to and when it decreases from 0 or

1 to –.

Find the regions in which the curve does not lie. This can be done in the followingmanner.

(i) If r is imaginary, when < < , then there is no part of the curve liesbetween the rays = and = .

Curvature

NOTES


(ii) If a be the greatest numerical value of r, the curve lies entirely within thecircle of radius a. Also, if the least numerical value of r be b, the curve liesoutside the circle of radius b.

Note: When r is a periodic function of , the negative values of need not betaken into account. We may consider values from = 0 to those values of ,where the values begin to repeat.

VI. Value of :

(i) Find the angle between the tangent and the radius vector, at P(r, )

using tand

rdr

.

(ii) Find the points where = 0 or 2

. When

dr

dt is positive, the value of

increases and as a result r increases. Similarly, when dr

dt is negative, the

value of decreases and as a result r decreases.

Example 2.17: Trace the curve r2 = a2 cos 2.

Solution: The given equation of the curve is r2 = a2 cos 2 …(i)

I. Symmetry: Since Equation (i) remains unchanged when is replaced by –, the curve is symmetrical about x-axis. The curve is also symmetrical about

the line = 2

since it remains unchanged when is changed to – . Again,

Equation (i) remains unchanged when r is changed to – r. Thus, the curve isalso symmetrical about the pole.

II. Origin or Pole: When r = 0, cos 2 = 0 2 = 2

=

4

, which is

a real value.

Thus, the curve passes through the pole and = 4

are the tangents at the

pole when = 0, r2 = a2 or r = + a. Thus, the curve meets the initial line( = 0) in the points (+ a, 0).

III. Value of : Differentiating Equation (i), we get 22 2 sin 2dr

r ad

2

2 2

cos 2tan cot 2 tan 2

sin 2 sin 2 2

d r r ar

dr adr ad r

22

when = 0,

2

.

Thus, at the points (+a, 0), the tangent are perpendicular to the initial line.

NOTES

Curvature


IV. Asymptotes: Curve has no asymptotes because there is no finite value of for which r .

V. Special Points and Region: From Equation (i), we have cos2r a

sin 2

cos2

dr a

d

dr

d is negative for 0 < <

4

, r decreases in this range.

Again becausedr

d is positive for

3

4

, thus r decreases in this range

when = 0, r = a. As increases from 0 to4

, r decreases from a to 0. For

4

< <

4

, r is imaginary. Thus, no part of the curve lies between the lines

= 4

and =

4

. Again as increases from

4

to , r remains positive

and increases from 0 to a. Thus, we trace the part of the curve below theinitial line. The part of the curve below the initial line can be traced bysymmetry.

Example 2.18: Trace the curve sin3 .r a

Solution: The given equation of the curve is sin 3r a …(i)

I. Symmetry: Since, sin3 sin 3 3 sin3 , thus the symmetry

is about the line 2

. In other words, the equation remains unchanged

when is changed to 2 and thus the curve is symmetrical is about .

II. Pole: On putting 0r , we get sin3 0 .

3 n where In

3 0,3 ,3 2 ,3 3 ,...

2 3 4 5

0, , , , ,3 3 3 3 3

which are tangents at the origin.


IV. Region: As | sin 3 | 1 , thus ,r a i.e., the whole curve lies within a circle ofradius a.

When 0 3 or 03

, 0,r

When 2

3 3

, 0r etc.

Curvature

NOTES


The maximum value of r is given by sin3 1 , i.e., when 5 9

3 , ,2 2 2

or

5 9, ,

6 6 6

and the minimum value of r is given by sin 1 , i.e., when

3 7 113 , ,

2 2 2

or

7 11, ,

2 6 6

V. Value of : Differentiating Equation (i),

3 cos3dr

ad

1

tan tan 33

dr

dr

2

when 3

2

, i.e.,

3 5 7, , , ,...

6 6 6 6

Thus, the tangent is perpendicular to the radius vector at the points where

5 7 9 11, , , , , 0 2

6 2 6 6 6 6

VI. Point of Intersection: We make a table for 3

2 2

or

3 93

2 2

since the curve is symmetrical about the line 2

(or y-axis)

6

12

7

12

8

12

9

12

10

12

11

12

12

12

13

12

14

12

15

12

16

12

17

12

18

12

r –a – 0.7a 0 0.7a a 0.7a 0 –0.7a –a –0.7a 0 0.7a a

Since, sin 3r a is a periodic function of period 2

3

, thus the same curve

is obtained after a period of 2

3

.

NOTES

Curvature


Example 2.19: Trace the curve 1 secr a .

Solution: The equation of the curve is 1 cos

cos

ar

…(i)

I. Symmetry: Since Equation (i) remains unchanged on replacing by – ,thus, the curve is symmetrical about x-axis.

II. Pole or Origin: On putting r = 0, we get cos 1 , which gives .Thus, is the tangent to the curve at the origin.

III. Asymptote: Equation (i) is the equation of fourth degree in x and y in which

both 4y and 3y are absent. The co-efficient of 2y is 2x a . Thus, the

asymptote to the curve is x – a = 0 or x = a.

IV. Region: From (i) we have cos 1 cosr a . Converting it into Cartesian

co-ordinates by putting cosx

r , we get

.x

x a ar

ax

x ar

2 2 2 2x a r a x 2 2 2 2 2x a x y a x

2 22 2 2 2x a y a x x x a 2 2 3 2x a y x a x

32

2

2x a xy

x a

We see that for both 0x and x > 2a, y is imaginary. Thus, the whole ofthe curve lies between the lines x = 0 and x = 2a. Also the origin is a cuspand y = 0, i.e., x-axis is cuspidal tangent.

V. Intersection with the Axes of Co-Ordinates: The curve meets x-axiswhen y = 0, which gives x = 0 and x = 2a. When the origin is shifted at

2 ,0a , the equation transforms into 2 22X Y X 2 Xa a .

So, the tangent at the new origin is X = 0 and the line x = 2a is a tangent at(2a, 0).

Note: We can change the equations from the polar to the Cartesian co-ordinatesor from Cartesian to polar co-ordinates so that we can trace the curve easily.

Curvature

NOTES


Check Your Progress

14. Trace the curve 2 2 2 2 2 2y a x x a x or 2 2 2 2 2 2x x y a x y .

15. Trace the curve2 2 2 2 2 2( ) ( ) 0y x y a x y or 2 2 2 2 2 2( ) ( )x a y y a y .

16. Trace the curve 4 4 24x y a xy .

17. Trace the curve 3 3 23x y ax .

18. Trace the curve r = a cos 2 .

19. Trace the curve r = a(1 – sin ).


1.2y

c

2. The equation of the curve is 28 8s ay s ay

Differentiating with respect to s, we have

2s = 8a.dy

ds= 8a sin sin

dy

ds

s = 4a sin

Also, 4 cosds

ad

24 1 sina 2

2 2

84 1 4 1

16 16

s aya a

a a 4 1

2

ya

a

3. 4 cos2

a

4.2

a

5.2 2

3

a b

p (Numerically)

6.

cosec

1θ

rdd

7. We have cosdr

ds

2

2sin . sin .

d r d d

ds ds ds

sind d

ds ds

NOTES

Curvature


22

2

1 1.

d r d d d dr r r

ds ds ds ds ds

2 2

2

1.

d d d rr r

ds ds ds

2 2

2

d d d rr r

ds ds ds

8.1

2 5 (Numerically)

9. We have y = a log sec x

a

y1 = tan

x

a = tan or =

x

a(tan = y

1)

and y2 =

1

asec2

x

a

We know that the chord of curvature parallel to y-axis

= 2 cos

= 2 2 3/ 2

1

2

(1 ).cos

y

y

3/ 22

2

2 1 tan.cos

1sec

xxa

x aa a

= 2a, which is constant

10. Given, logy x

The above equation is defined only when x > 0.

11

yx

, x > 0 and 2 2

1y

x

Since 2

20

d y

dx for all real value x > 0, hence the given curve is convex

upwards everywhere.

11.2 3 3

4 ,3 2 2

ax a n y

12. Origin is a conjugate point.

13. (–1, 2) is node.

14.

X

Curvature

NOTES


15.

X

Y

O

(0, a)

y = x y = –x

(0, –a)

16.

17.

18.

NOTES

Curvature


19. π2

π2

32π

θ = 32π

2a,

θ =

Ф =

Y

θ = 0θ π = (a, )π O

B

XXÁ´ A(a, 0)

Y´

2.8 SUMMARY

The curvature is the amount by which a curve deviates from being a straightline, or a surface deviates from being a plane. For curves, the canonicalexample is that of a circle, which has a curvature equal to the reciprocal ofits radius.

Smaller circles bend more sharply, and hence have higher curvature.

The curvature at a point of a differentiable curve is the curvature of itsosculating circle, i.e., the circle that best approximates the curve near thispoint.

The curvature of a straight line is zero.

The curvature of a curve at a point is normally a scalar quantity such that itis expressed by a single real number.

The curvature of a differentiable curve was originally defined throughosculating circles.

A French mathematician, engineer, and physicist Augustin-Louis Cauchyshowed that the center of curvature is the intersection point of two infinitelyclose normal lines to the curve.

The numerical measures of the sharpness of the bending of a curve at apoint is called the curvature of the curve at that point.

The reciprocal of the curvature of the curve at P, is called the radius of

curvature of the curve at P and is usually denoted by . Thus 1

ρκ

ds

d

.

The curvature at every point of the circle is equal to the reciprocal of its

radius and therefore it is constant. Curvature = ψd

ds=

1

r(constant).

The explicit equation of the radius of curvature at any point of the curve

y = f(x) is given by

32 2

2

2

1

ρ

dy

dx

d y

dx

or

3

2 21

2

1 y

y

Curvature

NOTES


The radius of curvature defined by parametric equations x = f(t), y = (t) is

given by 2 2 3/ 2( ' ' )

ρ' '' ' ''

x y

x y y x

.

The implicit equation of the radius of curvature at any point of the curve

f(x,y) = 0 is given by

32 2 2

2 2ρ

2

x y

xx y x y xy yy x

f f

f f f f f f f

.

Radius of Curvature when x and y are function of s is given by the equation2 2

2ρdx dy

d d

.

The graph of f is concave downwards or convex upwards on [a, b] if,

throughout (a, b), it lies below the tangent lines of f, i.e., if f is decreasingin the interval.

The curve is concave or convex in upwards direction depends upon

whether c ty y > 0 or c ty y < 0, i.e., 2

20

d y

dx or

2

20

d y

dx

A point of inflexion is a point on a curve at which the curve changes frombeing concave upwards to concave downwards or vice-versa.

Singular point is a point on a curve at which the curve shows an extraordinaryor unusual behaviour.

If more than one branch of a curve passes through a point on the curve,then this point is said to be multiple point of the curve.

If two branches of the curve pass through a point on the curve, then thispoint is called a double point.

If the two tangents at the double point are real and distinct, then the doublepoint is known as a node.

If there are no real points of the curve in the neighbourhood of the point,then this point P is called a conjugate or an isolated point of the curve.

The graph of f is concave upwards or convex downwards on [a, b] if,

throughout (a, b), it lies above the tangent lines of f, i.e., if f is increasingin the interval.

If the equation of a curve remains unaltered when y is changed to – y, i.e.,if all the powers of y in the given equation are even, the curve is symmetricalabout the x-axis.

If the equation of a curve remains unaltered when x is changed to – x, i.e.,if all the powers of x are even in the given equation, the curve is symmetricalabout the y-axis.

NOTES

Curvature


If all the powers of both x and y are even in the given equation, the curve issymmetrical about both the axes.

If the equation of the curve remains unchanged when x and y are changedto –x and –y, respectively, the curve is symmetrical in opposite quadrants.

If the equation of the curve is unchanged when x and y are interchanged(i.e., x is changed to y and y to x), the curve is symmetrical about the liney = x.

If the equation of the curve is unchanged when x is changed to –y and y ischanged to –x, the curve is symmetrical about the line y = –x.

A cycloid is a curve which is described by a fixed point in the plane of thecircle (generating circle) when the circle rolls without sliding along some straightline. There are three other types of equations of the cycloid, which can betraced on the similar lines.

2.9 KEY TERMS

Curvature: It is the numerical measure of the sharpness of the bending of acurve at a point.

Radius of curvature: It is the reciprocal of the curvature of the curve at P,and is usually denoted by .

Double point: If two branches of the curve pass through a point on thecurve, then this point is called a double point.

Inflexion point: It is a point on a curve at which the curve changes frombeing concave upwards to concave downwards or vice-versa.

Curve tracing: It means to find the approximate shape of a curve withoutplotting a large number of points.

Cycloid: It is a curve which is described by a fixed point in the plane of thecircle (generating circle) when the circle rolls without sliding along somestraight line.



1. Find the radius of curvature at ,s for each of the following curves:

(i) log tan sec tan secs a a (ii) 28 sin6

s a

2. Find the radius of curvature at the given point for the following curves:

(i) 2xy c at the point (x, y) (ii) 4sin sin 2y x x at 2

x

Curvature

NOTES


3. Show that the radius of curvature at any point of the astroide3 3cos , sinx a y a is equal to three times the length of the perpendicular

from the origin to the tangent.

4. The tangents at two points P, Q on the cycloid

sin ; 1 cosx a y a are at right angles. Show that if 1 2, are

the radii of curvature at these points then 2 2 21 2 16 .a

5. Prove that for an ellipse 2 2

2 21

x y

a b

2 2

3

a b

p , where p is the length of the

perpendicular from the centre upon the tangent at the point P(x, y).

6. Show that the radius of curvature at any point P(x,y) of the curve2 / 3 2 / 3 2 / 3x y a is given by 3 27 .axy

7. Show that for the cardioid 1 cosr a , the radius of curvature

4

cos3 2

a and

2 8

9

a

r

8. Find the radius of curvature for the following curves at (r, ):

(i) cosr a n (ii) 2 2cos2r a (iii) sinm mr a m

9. Find the radius of curvature for the following curves at (p, r):

(i) the cardioid 2 32ap r (ii)4

22 2

rp

r a

10. Find the radius of curvature for the following curves at ,p :

(i) the circle 1 sinp a (ii) 1 1 sin1

m m

m m mp a

m

11. Show that the radius of curvature for the curve 2 2

1cosr a a

a r

is

2 2r a .

12. Show that the radius of curvature of the curve r2 = a2 cos 2 at the point

where the tangent is parallel to x-axis is 2

3

a.

13. Find the radius of curvature at the origin for the following curves:

(i) 3 2 22 6 0x y x y

(ii) 2 22y x x xy y

NOTES

Curvature


(iii) 2 2 31 2 1 2 3 1 .... 0a x a y b x b xy b y c x

(iv) 2 2 3 4 53 4 5 0y xy x x x y y

(v) 2 2 3a y x x

14. Show that for the hyperbola 2 2

2 21,

x y

a b whose parametric equations are

sec , tanx a y b , the centre of curvature at any point ‘ ’ is

2 2 2 23 3sec , tan

a b a b

a b

and the equation of its evolute is

2 / 32 / 3 2 /3 2 2ax by a b .

15. Show that the chord of curvature parallel to y-axis for the catenary

coshx

y cc

is double the ordinate of the point and the chord of curvature

parallel to x-axis is 2

sinhx

cc

.


1. Find the nature of the origin on the curves given below:

(i) 3 3 2y x ax

(ii) 4 4 4 0x y xy

(iii) 4 2 2 2 2 0x ax y axy a y

(iv) 2 2 2 2 34a y a x x

2. Determine the position and nature of double points on the curves givenbelow:

(i) 2 2( 2) ( 1)x y y

(ii) 2 6( ) 0y x x

(iii) 4 3 3 2 2 2 2 44 2 4 3 0x ax ay a x a y a

(iv) 2 3( 6) ( 2) 9y y x x

(v) 3 2 3 2 24 4 1 0x xy y x y x y

(vi) 2 2 siny x x

3. Find the equations of the tangents at the multiple point of the

curve 2 2( 2) ( 1)x y y .

Curvature

NOTES


4. Prove that the curve 2 tanx

y bxa

has a conjugate point or a node at the

origin according as a and b have unlike or like signs.

5. Discuss the convexity or concavity in the upward direction of the curve

(sin cos )xy e x x for0 2x .

6. Determine the intervals of x for which the curve 2 4 5 xy x x e is

(i) Concave Upwards(ii) Concave Downwards

7. Find the points of inflexion for the following curve:

(i) 3(log )x y

(ii) sinx

y ca

(iii) 2 354 ( 5) ( 10)y x x

(iv) ( 1)( 2)( 3)x y y y

8. Show that the abscissa of the points of inflexion of the curve 2 ( )y f x ,

satisfy the equation

2( ) 2 ( ) ( )f x f x f x .

9. Show that the points of inflexion of the curve 2 2( ) ( )y x a x b lies on

the line 3x + a = 4b.

10. Show that the abscissa of the points of inflexion of the curve2 2

cos , sin is .a b

x a b y a ba

11. Trace the following curves in Cartesian co-ordinates:

(i) 9ay2 = x(x – 3a2)

(ii) a2y2 = x2(a2 – x2)

(iii) (a2 + x2)y = a2x

(iv) x2 = y2(x + a)3

(v) y2 = (x – 2)2(x – 5)

(vi) xy2 = 4a2 (2a – x)

(vii) x2 = (y – 1)(y – 2)(y – 3)

(viii)2

2

1

1

xy

x

NOTES

Curvature


(ix) y(x2 + 4a)2 = 8a3

(x) (x2 – a2)(y2 – b2) = a2b2

(xi) y2(a + x) = x2(a – x) or x(x2 + y2) = a(x2 – y2)

(xii) x3 + y3 = 3axy

(xiii) x2y2 = a2 (y2 – x2)

(xiv) x2/3 + y2/3 = a2/3

(xv) y = x3 – x

(xvi) 3

1xy

x

12. Trace the following curves in parametric form:

(i) x = a( – sin ), y = a(1 – cos )

(ii)2

3 3

3 3,

1 1

at atx y

t t

(iii)3 3

4 4

( ) ( ),

1 1

a t t a t tx y

t t

(iv)2 2

2 2

(1 ) (1 ),

1 1

a t at tx y

t t

(v) 2cos log tan , sin2 2

a tx a t y a t

(vi) x = a[cos – log( 1+ cos )], y = a sin

13. Trace the following curves in polar co-ordinates.

(i) r = a(1 + sin )

(ii) r = a(1 – cos )

(iii) r = 3(1 + cos )

(iv) r = a

(v) r = a sin 2(vi) r = a cos 3

(vii)2

21

ar

(viii) r2 cos = a2 sin 3

(ix) r2 cos 2 = a2 or x2 – y2 = a2

(x) x5 + y5 = 5a2x2y [Hint: change equation into polar co-ordinates].




Curvature

NOTES









NOTES

Integration of TranscendentalFunctions, Reduction,

Quadrature and Rectification


UNIT 3 INTEGRATION OFTRANSCENDENTALFUNCTIONS, REDUCTION,QUADRATURE ANDRECTIFICATION

Structure

3.0 Introduction3.1 Objectives3.2 Integration and Definite Integrals3.3 Reduction Formula

3.3.1 Walli’s Formula


3.3.3 Gamma Function3.4 Integration of Transcendental Functions3.5 Quadrature3.6 Rectification3.7 Answers to ‘Check Your Progress’3.8 Summary3.9 Key Terms


3.0 INTRODUCTION

In mathematics, an integral assigns numbers to functions in a way that can describedisplacement, area, volume, and other concepts that arise by combining infinitesimaldata. Integration is one of the two main operations of calculus, with its inverseoperation, differentiation, being the other.

A transcendental number is a number that is not the root of any integerpolynomial. Similarly, a transcendental function is a function that cannot be writtenusing roots and the arithmetic found in polynomials. Integration can be used to findareas, volumes, central points, and the area under the curve of a function. Integrationmeans summation. Integral of a function y = f(x) under the limit a to b gives us thearea enclosed by the curve y = f(x) and the two coordinates x = a, and y = b.

Sometimes we have the integrand which is neither immediately integrablenor reducible by substitution to any of the standard forms. But sometimes it ispossible to express the given integral in terms of another integral of similar butsimpler form which can be integrated. This is done by the method of integration byparts or by connecting the integrals. Once the relation between the given integral

Integration of TranscendentalFunctions, Reduction,Quadrature and Rectification

NOTES


and the simpler integral of the same type is established, the integration of the givenintegral can be completed by repeated application of the already established relation.

A reduction formula for a given integral is an integral which is of the sametype as the given integral but of a lower degree or order. The reduction formula isused when the given integral cannot be evaluated otherwise. The repeated applicationof the reduction formula helps us to evaluate the given integral.

Integration by reduction formula in integral calculus is a technique orprocedure of integration, in the form of a recurrence relation. It is used when anexpression containing an integer parameter, usually in the form of powers ofelementary functions, or products of transcendental functions and polynomials ofarbitrary degree, cannot be integrated directly.

In mathematics, quadrature is a historical term which means the process ofdetermining area. This term is still used nowadays in the context of differentialequations, where ‘solving an equation by quadrature’ means expressing its solutionin terms of integrals. The term rectification is s used to refer to the determination ofthe length of a curve. Rectification also refers to the operation which converts themidpoints of the edges of a regular polyhedron to the vertices of the related rectifiedpolyhedron.

In this unit, you will study about the integration of transcendental functions,definite integrals, reduction formulae, quadrature, and rectification.

3.1 OBJECTIVES


Define what integration is

Understand the various methods of integration

Explain about the definite integrals

Discuss about the integration of transcendental functions

Describe the significance of reduction formulae

Elaborate on the quadrature process

Understand the importance of rectification methods

3.2 INTEGRATION AND DEFINITEINTEGRALS

In mathematics, an integral assigns numbers to functions in a way that can describedisplacement, area, volume, and other concepts that arise by combining infinitesimaldata. Integration is one of the two main operations of calculus, with its inverseoperation, differentiation, being the other.

Sometimes we have the integrand which is neither immediately integrablenor reducible by substitution to any of the standard forms. But sometimes it ispossible to express the given integral in terms of another integral of similar but

NOTES




simpler form which can be integrated. This is done by the method of integration byparts or by connecting the integrals. Once the relation between the given integraland the simpler integral of the same type is established, the integration of the givenintegral can be completed by repeated application of the already established relation.

Integration can be used to find areas, volumes, central points, and the areaunder the curve of a function. Integration means summation. Integral of a functiony = f(x) under the limit a to b gives us the area enclosed by the curve y = f(x) andthe two coordinates x = a, and y = b.

To give a precise shape to the definition of integration, we observe. If g (x)is a function of x such that,

d

dx g (x) = f (x)

then we define integral of f (x) with respect to x, to be the function g (x). This isput in the notational form as,

( )f x dx = g (x)

The function f (x) is called the Integrand. Presence of dx is there just toremind us that integration is being done with respect to x.

For example, since d

dxsin x = cos x

cos x dx = sin x

We get many such results as a direct consequence of the definition of integra-tion, and can treat them as ‘formulas’. A list of such standard results are given:

(1) 1 dx = x because d

dx(x) = 1

(2) nx dx = 1

1

nx

n, (n – 1) because

1

1

nd x

dx n = xn, n – 1

(3)1

dxx

= log x because d

dx(log x) =

1

x

(4) x xe dx e because d

dx (ex) = ex

(5) sin x dx = – cos x because d

dx (–cos x) = sin x

(6) cos x dx = sin x because d

dx (sin x) = cos x

(7) 2sec xdx = tan x because d

dx (tan x) = sec2 x

(8) 2cosec x dx = – cot x because d

dx(– cot x) = cosec2x

(9) sec tanx x dx = sec x because d

dx(sec x) = sec x tan x


NOTES


(10) cosec cotx x dx = – cosec x becaused

dx(–cosec x) = cosec x cot x

(11)2

1

1 x dx = sin–1 x because

d

dx(sin–1 x) =

2

1

1 x

(12)2

1

1 x dx = tan–1 x because

d

dx(tan–1 x) =

2

1

1 x

(13)2

1

1x x dx = sec–1 x because

d

dx(sec–1 x) =

2

1

1x x

(14)1

ax bdx =

log ( )ax b

a because

d

dx log ( )ax b

a = 1

ax b

(15) ( ) nax b dx = 1( )

1

nax b

n .

1

a, (n – 1)

because d

dx

1( )

( 1)

nax b

a n = (ax + b)n,

n – 1

(16) xa dx = log

xa

abecause

d

dxax = ax log a

One might wonder at this stage that since,

d

dx (sin x + 4) = cos x

then by definition, why cos x dx is not (sin x + 4)? In fact, there is nothing very

sacred about number 4 and it could very well have been any constant. This sug-gests perhaps a small alteration in the definition.

We now define integration as:

If,d

dxg(x) = f (x)

Then, ( )f x dx = g(x) + c

Where c is some constant, called the constant of integration. Obviously, c canhave any value and thus integral of a function is not unique! But we could say onething here, that any two integrals of the same function differ by a constant.

Since c could also have the value zero, g(x) is one of the values of ( )f x dx . By

convention, we will not write the constant of integration (although it is there), and

thus ( )f x dx = g(x), and our definition stands.

The above is also referred to as Indefinite Integral (indefinite, because weare not really giving a definite value to the integral by not writing the constant ofintegration). We will give the definition of a definite integral also.

NOTES




Some Properties of Integration

(i) Differentiation and integration cancel each other.

The result is clear by the definition of integration.

Let d

dx g(x) = f (x)

then ( )f x dx = g(x) [by definition]

d

dx( )f x dx =

d

dx [g(x)] = f (x)

which proves the result.

(ii) For any constant a, ( ) a f x dx = ( )a f x dx

Since d

dx ( ( )a f x dx) = a

d

dx ( ) f x dx

= a f (x)

By definition, ( ) a f x dx = a ( ) f x dx

(iii)For any two functions f (x) and g(x),

[ ( )f x ± g(x)] dx = ( ) f x dx ± ( ) g x dx

As ( ) ( )d

f x dx g x dxdx

= d

dx ( )f x dx

d

dx( )g x dx

= f (x) ± g(x)It follows by definition that

[ ( ) ( )]f x g x dx = ( ) ( )f x dx g x dx This result could be extended to a finite number of functions, i.e., in general,

1 2[ ( ) ( )f x f x ± . . . ± f n (x)]dx = 1 2( ) ( )f x dx f x dx ± ... ±

( )nf x dx

Let us now solve some problems to illustrate the use of these results.

Example 3.1: Find 2(2 3)x dx.

Solution: We have2(2 3)x dx = 2(4 9 –12 )x x dx

= 24 9 – 12 `x dx dx x dx

= 24 9 – 12x dx dx xdx

= 3

43

x + 9x – 12

2

x

= 4

3 x3 – 6x2 + 9x


NOTES


Example 3.2: Find 1/ 3(2 1)x dx.

Solution: We have 1/ 3(2 1)x dx = 1/ 3 1(2 1) 1

1 213

x

= 3

8(2x + 1)4/3

Example 3.3: Solve3

1

x

x dx.

Solution: By division, we note3

1

x

x= (x2 – x + 1) –

1

1x

Thus3

1

x

x dx = 2( 1)x x dx –

1

1x dx

= 2 1

1x dx xdx dx dx

x

= 3 2

3 2

x x + x – log (x + 1)

Example 3.4: Find 1 cos 2+ x dx .Solution: We observe

1 cos 2x dx = 22 cos 2 cosx dx x dx = 2 sin x

Example 3.5: Evaluate 1x

x dx

Solution: 1x

x dx = 1

xdxx

dx

= 1/ 2 1

11

2

x – 1/ 2 1

11

2

x

= 3/ 22

3x – 2 x

Example 3.6: Evaluate

(i) 3 2ax bx cx d

x dx (ii) 3sin xdx

Solution: We have

(i)3 2ax bx cx d

x dx = 2 d

ax bx cx

dx

= 2 1a x dx b x dx c dx d dx

x

= a 3 2

3 2

x bx + cx + d log x

NOTES




(ii) We have

3sin x dx = 1

(3 sin sin 3 )4

x x dx

= 3 1sin sin 3

4 4x dx x dx

= 3 cos 3cos 9

4 12

xx

= 1(cos 3 9 cos )

12x x

Note: Note that as 1 d

a dxcos ax = – sin ax

sin ax dx = –cos ax

a

Similarly, cos ax dx = sin ax

aand bxe dx =

bxe

b.

Definite IntegralSuppose f (x) is a function such that,

( )f x dx = g(x)

The definite integral ( )b

af x dx is defined by

( )b

af x dx = ( ) b

ag x = g(b) – g(a)

where a and b are two real numbers, and are called respectively, the lower andthe upper limits of the integral.

Example 3.7: Evaluate 20

cos x dx

.

Solution: We know that cos x dx = sin x

Thus 20

cos x dx

= 20

sin x

= sin 2

– sin 0

= 1 – 0 = 1

Example 3.8: Evaluate 1b

adx .

Solution: We know that 1dx x . Thus 1b

adx = b

ax = b – a.

Note: It might be noticed that in case of a definite integral, there is no constant ofintegration. Suppose

( )f x dx = g(x) + c

then ( ( ) )b

af x dx = ( ) b

ag x c

= [g(b) + c] – [g(a) + c]

= g(b) – g(a)So when we talk of a definite integral, we note it has a unique definite value.


NOTES


Methods of Substitution

In this method, we express the given integral ( )f x dx in terms of another integral

in which the independent variable x is changed to another variable t through somesuitable relation x = (t).

Let I = ( )f x dx

dI

dx= f (x)

dI

dt=

dI

dx.

dx

dt = f (x) dx

dt

Thus I = ( )dx

f xdt . dt = [φ ( )]f t (t) dt

Note that we replace dx by '(t) dt, which we get from the relation

dx

dt = (t) by assuming that dx and dt can be separated.

In fact this is done only for convenience.

The following example will make the process clear.

Example 3.9: Integrate x(x2 + 1)3.

Solution: Put x2 + 1 = t 2xdx

dt = 1

Thus 2xdx = dt

2 3( 1)x x dx = 31

2t dt = 1

23t dt =

41

2 4

t =

4

8

t =

2 4( 1)

8

x

Example 3.10: Find tan θe sec2 d

Solution: Put tan = t, then sec2 d = dt

Thus tane sec2 d = te dt = et = etan

Example 3.11: Integrate 2

cos

1 sin

x

x.

Solution: Put sin x = t, then cos x dx = dt

Thus 2

cos

1 sin

x

x dx =

21

dt

t = tan–1 t = tan–1 (sin x).

Example 3.12: Integrate 2

6

4

1

x

x.

Solution: Put x3 = t then 3x2 dx = dt

NOTES




Thus 2

6

4

1

x

x dx =

4

3 21

dt

t =

4

3 sin–1 t =

4

3 sin–1 (x3).

To Evaluate ( )( ) f ' x

f x dx, where f(x) is the Derivative of f(x)

Put f (x) = t, then f (x)dx = dt

Thus '( )

( )

f x

f x dx =

dt

t = log t = log f (x)

To Evaluate [ ( )]nf x f(x)dx, n –1

Put f (x) = t, then f (x)dx = dt

Thus [ ( ) ( )nf x f x dx nt dt = 1

1

nt

n =

1[ ( )]

1

nf x

n

To Evaluate ' ( + )f ax b dx

Put ax + b = t, then adx = dt

( )f ax b dx = 1

( )dt

f ta a

( )f t dt = ( )f t

a =

( )f ax b

a

Example 3.13: Evaluate (i) tan xdx (ii) sec xdx

Solution: (i) tan xdx = sin

cos

xdx

x =

sec tan

sec

x xdx

x = log sec x

(ii) sec xdx = sec (sec tan )

sec tan

x x xdx

x x= log (sec x + tan x)

Example 3.14: Find 2 1x x dx .

Solution: We have

2 1x x dx = 1

2 21(2 ) ( 1)

2x x dx

=

11

2 21 ( 1)12 12

xz

[see 5.8]

= 1

2(x2 + 1)3/2

Example 3.15: Evaluate (i) cos x cos 2x dx

(ii) sin 4x cos 2x dx


NOTES


Solution: (i) cos x cos 2xdx = 1

2 cos cos 22

x x dx

= 1

(cos 3 cos )2

x x dx

= 1 1

cos 3 cos2 2

x dx x dx = 1 sin 3 1

2 3 2

x sin x

(ii) We have

sin 4 x cos 2 x = 1

22 sin 4x cos 2x

= 1

2[sin (4x + 2x) + sin (4x – 2x)]

= 1

2[sin 6x + sin 2x]

Thus

sin 4x cos 2 xdx = 1

2(sin 6 sin 2 )x x dx

= 1

2 sin 6 xdx +

1

2sin 2xdx

= cos 6

12

x – cos 2

4

x

= –1

4cos 6

cos 23

xx

Six Important Integrals

We will now evaluate the following six integrals:

(i)2 2

1

–a x dx (ii)

2 2

1

a xdx (iii)

2 2

1

x adx

(iv) 2 2a x dx (v) 2 2a x dx (vi) 2 2x a dx

(i) To evaluate 2 2

1

a xdx

Put x = a sin , then dx = a cos d

Thus

2 2

1

a x dx=

2 2 2

cos θ θ

sin θ

a d

a a =

cos θ

cos θ

a

ad

= 1. dθ =

= sin–1 x

a

(ii) To evaluate 2 2

1

a x dx

NOTES




Put x = a sin h , then dx = a cos h d

Thus

2 2

1

a x dx =

2 2 2

cos θ θ

sin θ

a h d

a a h = cos θ

cos θ

a h

a h d

as cos h2 – sin h2 = 1

= θd = = sin h–1 x

a

(iii)To evaluate 2 2

1

x a dx

Put x = a cos h then dx = a sin h d.

Thus

2 2

1

x a dx =

2 2 2

sin θ θ

sin θ

a h d

a h a =

sin

sin

a hd

a h = d

= = cos h–1 x

a

(iv)To evaluate 2 2a x dx

Put x = a sin , then dx = a cos dThus

2 2a x dx = 2 2 2sin θa a . a cos d = 2a cos2 d

= a2 1 cos 2θθ

2d

= 2 sin 2θ

θ2 2

a

= 2

2

a ( + sin cos )

= 2

2θ sin θ 1 sin θ2

a

= 2

2

a 21

2sin 1

x x x

a a a

and hence

2 2a x dx 2 2

2

xa x +

2

2

a sin–1 x

a


NOTES


(v) To evaluate 2 2a x dx

Put x = a sin h , then dx = a cos h d

Thus

2 2a x dx = 2 2 2sin θa x h . a cos h d

= 2a cos h2 d

= a2 (cos 2θ 1)

2

h d

(As 2 cos h2 = 1 + cos h 2

= 2

2

a sin 2θθ

2

h

= 2

2

a sin cosh h

(As sin h 2 = 2 sin h cos h

= 2

2sin θ 1 sin θ θ2

ah h

= 2 2

12

1 sin2

a x x xh

a aa

and hence

2 2a x dx = 2

x 2 2a x + 2

2

a sin h–1 x

a

(vi) To evaluate 2 2x a dx

Put x = a cos h , then dx = a sin h d

Thus

2 2x a dx = 2 2 2cos θa h a a sin h d

= a2 2sin θ θh d

= a2 (cos 2θ 1)

θ2

hd

= 2

2

a sin 2θθ

2

h

= 2

2

a (sin h cos h – )

= 2

2

a [ 2cos θ 1h . cos h – ]

NOTES




= 2

2

a 21

21. cos /

x xh x a

aa

and hence

2 2x a dx = 2

x 2 2x a – 2

2

a cos h –1 x/a.

Example 3.16: Solve 2

1

1x x dx.

Solution: We have

I = 2

1

1x x dx =

22

1

1 3

2 2x

dx

Put x + 12

= t, then dx = dt

Thus

I = 2

2 3

2

dt

t

= sin h–1 3/ 2

t

(By the second integral evaluated above)

= sin h–1 1/ 2

3/ 2

x = sin h–1 2 1

3

x

The above result could, of course, be written directly without actually making

the substitution x + 12 = t by taking x as x + 1

2 in the formula.

Example 3.17: Evaluate 23 2x x dx.

Solution: 23 2x x dx = 2 23

3 3

xx dx

= 223

3 3

xx dx

= 22 1 1

33 36 6

x dx

= 2 2

5 13

6 6x dx


NOTES


=

2

2 21

51 15 1 66 63 sin

2 6 6 2 5 / 6

x xx

= 1

12(6x – 1) 23 2x x

25

723 sin–1 6 1

5

x .

Integration by Parts

We will now learn a formula which will help us in finding the integral of a productof two functions.

We know that if u and v are two functions of x

then ( )d

uvdx

= dv

udx

+ du

vdx

dvu

dx=

d

dx (uv) – v

du

dx

Integrating both sides with respect to x , we get

dvu

dx dx =

d

dx(uv) dx –

duv

dx dx

ordv

udx dx = uv – .

duv

dx dx

Put u = f (x), dv

dx = g(x), then v = ( )g x dx

The above reduces to

( ) ( )f x g x dx = f (x) ( )g x dx – [ ' ( ) ( ) ]f x g x dx dx

where f (x) denotes the derivative of f (x). This is the required formula. In words,integral of the product of two functions

= First function × Integral of the second – Integral of

(Differential of first × Integral of the second function).

It is clear from the formula that it is helpful only when we know (or can easilyevaluate) integral of at least one of the two given functions. The following ex-amples will illustrate how to apply this rule.

Example 3.18: Find 2 xx e dx .

Solution: Taking x2 as the first function and ex as the second function, we note

2 xx e dx = x2ex – (2 ) xx e dx= x2ex – 2 xx e dx

= x2ex – 2 x xxe e dx (Integrating by parts again)

= x2ex – 2 [ ]x xxe e

Note: If we had taken ex as the first function and x2 as the second function, we would nothave got the answer.

NOTES




To Evaluate the Integrals

Evaluate

(i) xe [f (x) + f (x)] dx

(ii) I1 = axe sin (bx + c) dx

(iii) I2 = axe cos (bx + c) dx

(i) Consider xe f (x) dx

Integration by parts yields

xe f (x) dx = f (x)ex – '( ) xf x e dx

xe f (x) dx + xe f (x) dx = f (x) ex

i.e., xe [f (x) + f (x)] dx = f (x) ex

(ii) Using integration by parts, we find

I1 = axe sin (bx + c) dx

= axe

a sin (bx + c) –

axe

a. cos (bx + c). b dx

= axe

a sin (bx + c) –

b

aI2

Similarly,

I2 = axe cos (bx + c) dx

= axe

a cos (bx + c) –

axe

a sin (bx + c) bdx

= axe

a cos (bx + c) +

b

aI1.

and thus

I1 = axe

a sin (bx + c) –

b

a 1cosaxe b

bx c Ia a

2

121

bI

a=

axe

asin (bx + c) – 2

b

a eax cos (bx + c)

2 2

12

a bI

a

= eax 2

sin( ) cos ( )a bx c b bx c

a

I1 = axe sin (bx + c) dx = axe2 2

sin ( ) cos ( )a bx c b bx c

a b


NOTES


Similiarly,

I2 = eax2 2

cos( ) sin ( )a bx c b bx c

a b

The above two integrals could be put into another form by the substitution

a = r cos , b = r sin

I1 = eax 2 2

cosθ. sin ( ) sin θ cos ( )r bx c r bx c

a b

= eax2 2

sin ( θ)r bx c

a b

axe sin (bx + c) dx = eax 1

2 2

sin ( tan / )bx c b a

a b

(as r2 = a2 + b2 and

tan = b

a)

Similarly,

(iii) axe cos (bx + c) dx = eax1

2 2

cos ( tan / )bx c b a

a b

Example 3.19: Find xe [sin x + cos x] dx.

Solution: Since d

dx (sin x) = cos x

xe (sin x + cos x) dx = ex sin x

Example 3.20: Find 21

xxedx

x

Solution: We have 21

xxedx

x= xe 2

1 1

1 1x x dx

= ex.1

1x (as

d

dx

1

1x

= – 2

1

( 1)x)

Example 3.21: Evaluate 2xe cos (3x + 4) dx.

Solution: According to the formula,

2xe cos (3x + 4) dx = e2x 1 3cos(3 4) tan

213

x

NOTES




Some More Examples

We give below a few more examples, where both substitution and integration byparts are used.

Example 3.22: Find 1tan

2 3 / 2(1 )

m xe

x

dx

Solution: Put tan–1 x = t, then x = tan t

Also 2

1

1dx

x = dt and 1 + x2 = 1 + tan2 t = sec2 t

Thus

1tan

2 3 / 2(1 )

m xe

x

dx = sec

mtedt

t = cosmte t dt

= emt

1

2

1cos tan

1

tm

m

where t = tan–1 x

Example 3.23: Evaluate 2 1sinx x dx .

Solution: We have, by integration by parts,

2 1sinx xdx = (sin–1 x) 3

3

x –

3

2

1.

3 1

xdx

x

= 3

3

xsin–1 x – 1

3

3

21

xdx

x. . . .(1)

To evaluate 3

21

xdx

x, put 21 x = t

1 – x2 = t2

–2x dx = 2t dt

x dx = – t dt

Also x2 = 1 – t2

Thus3

21

x dx

x= –

2(1 )t tdt

t = 2( 1)t dt =

3

3

t – t

= 2 3/ 2

2(1 )1

3

xx

Hence the required value is [from Equation (1)]

2 1sinx xdx = 3

3

x sin–1 x – 1

3

2 3/ 22(1 )

13

xx


NOTES


Some More Methods

If the integrand consists of even powers of x only, then the substitution x2 = t, ishelpful while resolving into partial fractions.

Note: The substitution is not to be made in the integral.

Example 3.24: Evaluate 2

2 2( 1)(3 1)

x dx

x x

Solution: Put x2 = t in 2

2 2( 1)(3 1)

x

x x

Then( 1)(3 1)

t

t t

1 3 1

A B

t t

t A (3t + 1) + B (t + 1)

Putting t = – 1 and – 13

, we get

A = 12

, B = – 12

( 1)(3 1)

t

t t=

1

2( 1)t –

1

2(3 1)t

Thus2

2 2( 1)(3 1)

x

x x=

2

1

2( 1)x –

2

1

2(3 1)x

2

2 2( 1)(3 1)

x

x x dx = 1

2

2 1

dx

x –

212 3 1

dx

x

= 12

tan–1 x – 2

16 1

3

dx

x

= 12

tan–1 x – 1 1.

6 1/ 3 tan–1

1/ 3

x

= 12

tan–1 x – 2 3

x tan–1 ( 3)x

We give two more examples that would help us in solving many other prob-lems.

Example 3.25: Solve 2

4

1

1

xdx

x

Solution: We have I = 2

4

1

1

xdx

x =

2

2 2

1 1/

1/

xdx

x x

Put x – 1

x= t

Then 2

11

x dx = dt

NOTES




Also x2 + 2

1

x – 2 = t2

So I = 2 2

dt

t =

2

t tan–1 2

t = 1

2 tan–1 1 1

2x

x

Substitution before Resolving into Partial Fractions

The integration process is sometimes greatly simplified by a substitution as is seenin the following examples:

Example 3.26: Solve 4( 1)

dx

x x

Solution: Put x4 = t, then 4x3 dx = dt

Thus4( 1)

dx

x xdt =

3

4 4( 1)

x dx

x xdt = 1

4

( 1)

dt

t t

Now1

( 1)t t =

1

1t –

1

t

and hence the given integral

= 1 114 1t t

dt = 14

[log (t – 1) – log t]

= 14 log 1t

t

= 14 log

4

4

1x

x

Example 3.27: Solve π/4

0

tan x dx

Solution: Put tan x = t, then tan x = t2

and sec2 x dx = 2 t dt

dx = 2

2

1 tan

t

x dt =

4

2

1

t dt

t

Also when x = 0, t = tan 0 = 0

when x = π

4, t = π

tan 14

Hence the given integral becomes

1

40

.2

1

t t dt

t= 2

1 2

40 1

t

t

= 12 2

12

0

2 2 1 112 log 2 tan48 22 1

t t t

tt t


NOTES


By integrating,

= 1 12 2 2 2 2 22 log tan 0 log 1 tan

8 4 8 42 2

= 2 2 2 22 log

8 4 22 2

= 2

4 log

2 2 2

42 2 = 1

2 2 log

2 2 1

2 2 2 2

Integrals of the Type cos + sin

dx

a + b x c x, b2 + c2 0

The substitution tan 2

x = t, converts every rational function of sin x and cos x into

a rational function of t and we can then evaluate the integral by using the previousmethods.

Example 3.28: Evaluate (i) π/2

20 4 5 cos

dx

x(ii)

π

05 3 cos

dx

x

Solution: (i) Put tan 2

x = t, then 12

sec2 2

x dx = dt

dx = 2

2

1 tan / 2

dt

x =

2

2

1

dt

t

Also as cos x = 2

2

1 tan / 2

1 tan / 2

x

x =

2

2

1

1

t

t

the given integral reduces to

1

220

2

2

5 5(1 ) 4

1

dt

tt

t

Note, when = 0

tan 0 0

When π / 2

tan π/4 1

x

t

x

t

= 21

2 20 4 4 5 5

dt

t t = 2

1

20 9

dt

t

= 2. 1

0

1 3log

2 3 3

t

t

= 13

log 13

3 1

3 1 log

3

3

= 13

log 2

NOTES




(ii) Put tan 2

x = t, then 12

sec2 2

x dx = dt

dx = 2

2

1 tan / 2

dt

x =

2

2

1

dt

t

Also as cos x = 2

2

1 tan / 2

1 tan / 2

x

x =

2

2

1

1

t

t

the given integral reduces to

220

2

2

(1 )(1 ) 5 3

1

dt

tt

t

Note, when 0, tan 0 0

πWhen π, tan

2

x t

x t

= 2 20

25 5 3 3

dt

t t = 2

0

22 8

dt

t = 2

0 4

dt

t

= 112

0

tan2

t = 12 tan–1 – 1

2 tan–1 0

= π π0

2 2

Example 3.29: Evaluate 23 2 cos

dx

x.

Solution: Put tan x = t, then sec2 x dx = dt

Now23 2 cos

dx

x=

2

2

sec

3 sec 2

x dx

x =

23(1 ) 2

dt

t

= 23 5

dt

t =

213 5

3

dt

t

= 13 × 3

5 tan–1 3

5

t

= 1

15tan–1 3

tan5

x

Integration of cos sincos sin

a x + b xc x + d x

, (a2 + b2) (c2 + d2) 0

We determine two constants and such that,

a cos x + b sin x = (–c sin x + d cos x) + (c cos x + d sin x)

where – c sin x + d cos x = d

dx(c cos x + d sin x)


NOTES


Comparing coefficients of cos x and sin x, we get

a = d + c

b = – c + d

= 2 2

ad bc

d c, =

2 2

ac bd

d c

Hence

cos sin

cos sin

a x b x

c x d x dx =

sin cosμ 1.

cos sin

c x d xdx dx

c x d x

= log (c cos x + d sin x) + x

Integration of cos sincos sin

a x + b x + cd x+ e x + f

In this case we determine three constants , , , such that a cos x+b sin x+c= (d cos x + e sin x + f) + (–d sin x + e cos x) + v and proceed as in the earliercase.

Example 3.30: Find 4 sin + 2 cos + 3

2 sin + cos + 3

x x

x x dx

Solution: We determine , , such that,

4 sin x + 2 cos x + 3 = (2 sin x + cos x + 3) + (2 cos x – sin x) + Comparing coefficients of sin x, cos x and the constant terms, we get

4 = 2 – ,2 = + 2 3 = 3 +

= 2, = 0, = –3

Thus

4 sin x + 2 cos x + 3 = 2(2 sin x + cos x + 3) + 0 (2 cos x – 1) – 3

4 sin 2 cos 3

2 sin cos 3

x x

x x

dx= 2

11. 3

2 sin cos 3dx

x x dx

= 2x – 31

2 sin cos 3x x dx

Example 3.31: To solve 1

2 sin cos 3x xdx

Solution: We put tan 2

x = t, then dx = 2

2

1

dt

t

and the integral

= 2

22 2

2

2.2 1(1 ) 3

1 1

dt

t tt

t t

NOTES




= 2 2

2

4 1 3 3

dt

t t t

= 2 2 2

dt

t t

= 2( 1) 1

dt

t

= tan–1 (t + 1) = tan–1 1 tan2

x

Hence the required result is

2x – 3 tan–1 1 tan2

x

Check Your Progress

1. Define the terms integration and integrand.

2. Integrate the following functions with respect to x:

(a) (i)1x

(ii) 3/ 2

1

x(iii) (1 )x x (iv)

3

2

ax bx

(b) (i) e2x (ii) e2x – 3 (iii) 1xe

(c) Integrate the following functions:

(i)1

2

sin

1

x

x

(ii) cos (log )x

x

(iii) 1x xe e

3. Integrate the following functions:

(a) (i)2

1

2 2x x (ii) 2

1

1 12 9x x (iii)2 2

1

2 / 2a x

(b) (i) 22 3 4x x (ii) 22 3x (iii) 29 4x

3.3 REDUCTION FORMULA

A formula which connects an integral with another integral, having the same formas the given integral but of lower degree or order and easier to integrate is calleda reduction formula.

A reduction formula is usually obtained by the method of integration byparts and is used to integrate by successive reduction.


NOTES


Reduction Formulae for sinn x dx And cosn x dx

Let In = sinn x dx.

For applying the method of integration by parts, we write

sinnx = sinn1x.sin x

In

= sinn1 x sin x dx

= sinn1 x ( cos x) 1sin ( cos )ndx x dx

dx

= sinn1 x.cos x + 2( 1)sin .cos .cosnn x x x dx= 1 2 2sin .cos ( 1) sin .(1 sin )n nx x n x x dx = 1 2sin .cos ( 1) sin ( 1) sinn n nx x n xdx n x dx = 1 2sin cos ( 1) sin ( 1)In n

nx x n xdx n By transposing the last term to the left, we get

1 2I sin cos ( 1) sinn nnn x x n xdx

Hence, 1

2sin .cos ( 1)sin sin

nn nx x n

xdx xdxn n

…(1)

which is the required reduction formula.

If In = sinn x dx, then I

n

2 = sinn2 x dx

Equation (1) can also be written as

1

2

sin cos ( 1)I I

n

n n

x x n

n n

Similarly, we can get the reduction formula for cosn xdx as

1

2cos sin ( 1)cos cos

nn nx x n

xdx xdxn n

… (2)


3.3.1 Walli’s Formula

To evaluate / 2

0sinn xdx

and/ 2

0cosn xdx

where ‘n’ is a natural number..

From the reduction formula of sinn xdx , we have

1

2sin .cos ( 1)sin sin

nn nx x n

xdx xdxn n

NOTES




/ 21/ 2 / 2 2

0 00

sin .cos ( 1)sin sin

nn nx x n

xdx xdxn n

/ 2 2

0

( 1)sinnn

xdxn

By writing n 2, n 4, …, etc., for n in it, we get

/ 2 / 22 4

0 0

3sin sin

2n nn

xdx xdxn

/ 2 / 24 6

0 0

5sin sin

4n nn

xdx xdxn

, etc.

/ 2 / 2 6

0 0

1 3 5sin . . sin ...

2 4n nn n n

xdx xdxn n n

… (3)

Case I: When n is an Even Positive Integer

Since, in Equation (3) the power is being reduced by two, at each step of reduction,finally we shall get an integrand in which power of sin x is 0.

/ 2 / 2 0

0 0

1 3 5 1sin . . ..... sin

2 4 2n n n n

xdx xdxn n n

/ 2

0

1 3 5 1. . .....

2 4 2

n n nx

n n n

Hence,

/ 2

0

( 1) ( 3) ( 5) 1sin . . .... .

2 4 2 2n n n n

dxn n n

… (A)

Case II: When n is an Odd Positive Integer

In this case, we shall get an integrand in which power of sin x is 1.

/ 2 / 2

0 0

1 3 5 2sin . . ..... sin

2 4 3n n n n

xdx xdxn n n

/ 2

0

1 3 5 2. . ..... cos

2 4 3

n n nx

n n n

Hence, / 2

0

( 1) ( 3) ( 5) 2sin . . ....

2 4 3n n n n

dxn n n

… (B)

The above formulae (A) and (B) are called Walli’s formula.

Similarly, we can get the Walli’s formula for / 2

0cosn dx

as

/ 2

0

( 1) ( 3) ( 5) 1cos . . .... .

2 4 2 2n n n n

dxn n n

… (C)


NOTES


when n is even.

And, / 2

0

( 1) ( 3) ( 5) 2cos . . ....

2 4 3n n n n

dxn n n

… (D)

when n is odd.

Example 3.32: Evaluate / 2 6

0sin d

Solution: / 2 6

0sin d

=

5.3.1 5.

6.4.2 2 32

Example 3.33: Evaluate/ 4 3/ 2

0(cos 2 ) cos d

.

Solution:/ 4 3/ 2

0(cos 2 ) cos d

=

/ 4 2 3/ 2

0(1 2sin ) cos d

Now putting 2 sin sin t so that 1

cos cot2

d t dt

and when = 0, then t = 0, when = 4

, then sin t = 2 sin 1

4

2

t

3/ 2/ 4 / 4 2 3/ 2

0 0

1(cos 2 ) cos (1 sin ) cos

2d t t dt

/ 2 4

0

1cos

2t dt

1 3 1. . .4 2 22

16 2


Reduction formula for tann xdx , where ‘n’ is a positive integer

tann xdx = tann 2 x. tan2x dx

= tann 2 x. (sec2x 1) dx

= tann 2 x. sec2x dx tann 2 x dx

= 1 21tan tan

1n nx xdx

n

Therefore, If In = tann xdx ,

then, In = 11

tan1

n xn

In–2


NOTES




Similarly cotn x dx = 11cot

1n x

n

In–2

Reduction Formulae for secn x dx and cosecn x dx

Let In = secn xdx .

For apply the method of integration by parts, we write

secnx = secn 2x.sec2 x

In

= secn 2 x sec2 x dx

Integration by parts, we get

secn 2 x tan x (n 2) 2tan sec tannx x x dx= 2 2 2sec tan ( 2) sec tann nx x n x x dx = 2 2 2sec tan ( 2) sec (sec 1)n nx x n x x dx = 2 2sec tan ( 2) sec secn n nx x n xdx xdx = 2

2sec tan ( 2)( )nn nx x n I I

22{1 ( 2)} sec tan ( 2)n

n nn I x x n I

2

2

sec tan ( 2)

1 ( 1)

n

n n

x x nI I

n n

Hence, 2

2sec tan ( 2)sec sec

1 ( 1)

nn nx x n

xdx xdxn n

Which is the required reduction formula.

Similarly, we get the reduction formula for cosec n x as

22cosec cot ( 2)

cosec cosec1 ( 1)

nn nx x n

xdx xdxn n

Reduction Formulae for sinm x cosn x dx

This integral can be connected with any one of the following six integrals dependingupon the sign of m and n in the integral to be connected:

(i) 2sin cosm nx xdx(ii) 2sin cosm nx xdx(iii) 2 2sin cosm nx xdx (iv) 2 2sin cosm nx xdx


NOTES


(v) 2sin cosm nx xdx(vi) 2sin cosm nx xdx

We shall take these cases one by one.

Case I: When Both m and n are Positive Integers.

Here, ,I sin cosm nm n x x dx

1(sin cos ) cosm nx x x dx

1

1 1 2sin 1cos sin cos sin

1 1

mn m nx n

x x x x dxm m

(integration by parts)

1

1 2 2sin 1cos sin cos (1 cos )

1 1

mn m nx n

x x x x dxm m

1

1, 2 ,

sin 1cos I

1 1

mn

m n m n

x nx I

m m

1

1, , 2

1 sin 1I 1 cos

1 1 1

mn

m n m n

n x nx I

m m m

1 1

2,

sin cos 1I sin cos

m nm n

m n

x x nx x dx

m n m n

… (4)

Successive applications of this reduction formula will reduce the integral tothe form

sin or sin cosm mx dx x x dx according as n is an even or odd positive integer.

Similarly, we can reduce the power of sin x instead of that of cos x by theformula

1 12

,

sin cos 1I sin cos

m nm n

m n

x x mx x dx

m n m n

…(5)

Case II: When Either m or n is a Negative Integer.

In this case, let m be a negative integer.

sin cosm nx x dx = 1(sin cos ) cosm nx x x dx1 1 2 21 1

sin cos sin cos1 1

m n m nnx x x x dx

m m

… (6)

which enables us to bring the power of sin x nearer to zero. This formula isuseful when m is a negative integer and n is a positive integer.

NOTES




Similarly, when m is a positive integer and n is a negative integer, we use

sin cosm nx x dx = 1 1

2 2sin cos 1sin cos

1 1

m nm nx x m

x xdxn n

…(7)

Case III: In Case I, if we Put n + 2 in Place of n

2sin cosm nx x dx = 1 1sin cos 1

sin cos2 2

m nm nx x n

x x dxm n m n

On transposition and simplification, we get


2sin cos 2sin cos

1 1

m nm nx x m n

x x dxn n

…(8)

Similarly, we have


2sin cos 2sin cos

1 1

m nm nx x m n

x x dxm m

...(9)

Note: The Case III is useful when either m or n or both are negative.

Method to connect sin cosm nx x dx with any integral of the above type.

Step I. Take P = sin + 1 x cos + 1 x

where is smaller of the indices of sin x and is smaller of the indices ofcos x in the two integral to be connected.

Step II. Differentiate P with respect to x to getPd

dx. Express it in terms of integrands

of the two integrals to be connected.

Step III. Integrate both sides with respect to x and solve to get the value of thegiven integral.

3.3.3 Gamma Function

The Gamma function (n) can be evaluated by the following relations:

(1) = 1; (n + 1) = n(n); 1

2

Example 3.34: Evaluate 4 2 2 5/ 2

0( )

ax a x dx .

Solution: Let x = a sin so that dx = a cos d

Now x = a sin gives = 0 when x = 0 and = /2 when x = a.

/ 24 2 2 5/ 2 4 4 2 2 5/ 2

0 0( ) sin ( cos ) cos

ax a x dx a a a d


NOTES


/ 210 4 6 10

0

5 72 2

sin cos12

22

a d a

10

3 1 5 3 1. . . .

2 2 2 2 2.2.5.4.3.2.1

a

10

10.512 512

aa

Example 3.35: Evaluate/ 2

0sin cosm nx x dx

where m and n are positive integers

and hence find / 2 5 4

0sin cosx x dx

.

Solution:/ 2

m, 0I sin cosm n

n x xdx

is connected with

/ 2 2

m 2, 0I sin cosm n

n x xdx

Let P = 1 1sin cosm nx x

Differentiating with respect to x, we have

2 1 1P1 sin .cos .cos sin . 1 cos sinm n m nd

m x x x x n x xdx

2 21 sin .cos 1 sin .cosm n m nm x x n x x

2 21 sin .cos 1 sin 1 sin .cosm n m nm x x x n x x

21 sin .cos 1 sin cos 1 sin .cosm n m n m nm x x m x x n x x

21 sin .cos sin .cosm n m nm x x m n x x

Integrating both sides, we get

P = 21 sin cos sin .cosm n m nm x x dx m n x x dx

1 1

2sin .cos 1sin cos sin cos

m nm n m nx x m

x x dx x x dxm n m n

Inserting the limits between 0 to2

, we

have

/ 21 1/ 2 / 2 2

0 00

sin cos 1sin cos sin cos

m nm n m nx x m

x x dx x x dxm n m n

m, 2,

1I In m n

m

m n

...(i)

Changing m to m – 2

m 2, 4,

3I I

2n m n

m

m n

...(ii)

NOTES




From Equations (i) and (ii), we get

m, 4,

1 3I . .I

2n m n

m m

m n m n

Case I: When m is a +ve Odd Integer.

m, 1,

1 3 2I . .... .I

2 3n n

m m

m n m n n

But/ 21

/ 2

1, 00

cosI sin cos

1

nn

n

xx x dx

n

1 1

01 1n n

m,

1 3 2 1I . .... .

2 3 1n

m m

m n m n n n

Case II: When Both m and n are +ve Even Integer.

m, 0,

1 3 1I . .... I

2 2n n

m m

m n m n n

But/ 2

0, 0I cosn

n x dx

=1 3 1

.... .2 2 2

n n

n n

[ n is even]

m,

1 3 1 1 3 1I . .... . . .... .

2 2 2 2 2n

m m n n

m n m n n n n

Case III: When Both m is a +ve Even Integer but n is a +ve Odd Integer.

m, 0,

1 3 1I . .... .I

2 2n n

m m

m n m n n

But / 2

0, 0I cosn

n x dx

1 3 2

....2 3

n n

n n

[ n is odd]

m,

1 3 1 1 3 2I . .... . . ....

2 2 2 3n

m m n n

m n m n n n n

The value of given integral can be written by a simple rule as

/ 2

0

1 go on diminishing by 2 1 go on diminishing by2sin cos

go on diminishing by 2 2m n

m nx x dx

m n

/ 2 5 4

0

4.2.3.1 8sin cos

9.7.5.3 315x x dx

[We do not write2

as both m and n are not even].


NOTES


Check Your Progress

4. Evaluate

(i)/ 8 3

0cos 4

xdx (ii) 2 7

0sin

4d

(iii)

31

0 21

xdx

x

5. Evaluate4 1

sin2

d

.

6. Evaluate 3tan x dx .

7. If/ 3

0I tann

n x dx

, find 2( 1) n nn I I .

8. Find the value of 2 40 (1 )

dx

x

.

3.4 INTEGRATION OF TRANSCENDENTALFUNCTIONS

Let ,I sin sinmm n x nx dx

Integrating by parts, taking sin nx as second function,

1,

cosI sin (sin cos ) cosm m

m n

nx mx x x nx dx

n n

and integrating by taking cos nx as second function,

,

1 2 1

1I sin cos

sin 1(sin cos ) ( 1)sin cos cos sin sin sin

mm n

m m m

x nxn

m nxx x m x x x x x nx dx

n n n

1 2 2

1sin cos

1 1sin sin cos ( 1)sin (1 sin ) sin sin

m

m m m

x nxn

mnx x x m x x x nx dx

n n n

2

12, ,2 2 2

1 ( 1)sin cos sin sin cos I Im m

m n m n

m m m mx nx nx x x

n n n n

2

1, 2,2 2 2

1 ( 1)I 1 sin cos sin sin cos Im m

m n m n

m m m mx nx nx x x

nn n n

1

22 2 2 2

sin cos sin cos sin ( 1)sin sin sin sin

m mm mn x nx m x x nx m m

x nxdx x nxdxm n m n

NOTES




Similarly,

12

2 2 2 2

sin sin sin cos cos ( 1)sin cos sin cos

m nm mn x nx m x x nx m m


Reduction Formulae for cosmx sin nx dx and cosmx cos nx dx

Proof: Let ,I cos sinmm n x nx dx ; now integrating by parts, we have

1,

1I cos cos cos sin cosm m

m n

mx nx x x nx dx

n n

Now sin (n 1)x = sin nx cos x cos nx sin x

cos nx sin x = sin nx cos x sin(n 1)x

1

,

1I cos cos cos {sin cos sin( 1) }m m

m n

mx nx x nx x n x dx

n n

11cos cos cos sin cos sin( 1)m m mm m

x nx x nxdx x n xdxn n n

, 1, 1

1cos cosm

m n m n

mx nx I I

n n

, 1, 1

1I 1 cos cos Im

m n m n

m mx nx

n n n

Hence, 1cos coscos sin cos sin( 1)

mm mx nx m

x nx dx x n xdxm n m n

.

Similarly, we can prove that,

1cos sincos cos cos cos( 1)

mm mx nx m


.

Example 3.36: If/ 2

, 0I cos sin ;m

m n x nx dx

prove that

, 1, 1

1I Im n m n

m

m n m n

.

Hence, evaluate 5,3I .

Solution: / 2

, 0I cos sinm

m n x nx dx

/ 2

/ 2 1

00

cos coscos . cos sin .m mnx nx

x m x x dxn n

[Integrating by parts]

/ 2/ 2 1

0 0

1cos cos cos sin cosm mm

x nx x x nxdxn n


NOTES


/ 2 1

0

10 1 cos sin cosmm

x x nx dxn n

/ 2 1

0

1cos sin cosmm

x x nx dxn n

...(i)

Now, sin 1 sin cos cos sinn x nx x nx x

sin cos sin cos sin 1x nx nx x n x

From Equation (i), we get

/ 2 1

, 0

1I cos sin cos sin 1m

m n

mx nx x n x dx

n n

/ 2 / 2 1

0 0

1cos sin cos sin 1m mm m

x nxdx x n xdxn n n

, 1, 1

1I Im n m n

m m

n n n

m, m 1, 1

11 I In n

m m

n n n

m, m 1, 1

1I In n

m

m n m n

... (ii)

(ii) Putting m = 5, n = 3 in Equation (ii), we have

5,3 4,2

1 5I I

8 8 … (iii)

Putting m = 4, n = 2 in Equation (ii), we have

4,2 3,1 3,1

1 4 1 2I I I

6 6 6 3 ... (iv)

Putting m = 3, n = 1 in Equation (ii), we have

/ 2 2

3,1 2,0 0

1 3 1 3I I cos sin 0

4 4 4 4x dx

/ 2 / 2

00

1 3 1 30

4 4 4 4dx c

1 3 1

4 4 4c c

From Equation (iv), 4,2

1 2 1 1I .

6 3 4 3

From Equation (iii), 5,3

1 5 1 8 1I .

8 8 3 24 3

Reduction Formulae for xnsin mx dx and xn cos mx dx

1cos cossinn n nmx mx

x mxdx x nx dxm m

(Integrating by parts)

1coscosn nmx n

x x mxdxm m

1 21 sin sin

cos ( 1)n n nn mx mxx mx x n x dx

m m m m

NOTES




1

21 1cos sin sin

nn nn x n

x mx mx x mxdxm m m m

1 2

2 2

1 ( 1)sin cos sin sinn n n nn n n

x mxdx x mx x mx x mxdxm m m

which is the required reduction formula of the given integral. Its repeatedapplication reduces it to either x sin mx or sin mx according as n is even or odd.

Similarly,

12

2 2

sin cos ( 1)cos cos

n nn nx mx nx mx n n

x mxdx x mxdxm m m

.

Reduction Formulae for x sinnx dx

Let I sinnn x xdx 1( sin ).sinnx x xdx 1 1 2( sin ).( cos ) 1.sin ( 1)sin cos . ( cos )n n nx x x x n x x x x dx

1 1 2 2cos sin sin .cos ( 1) sin cosn n nx x x x xdx n x x xdx

1 2 21

cos sin sin ( 1) sin .(1 sin )n n nx x x x n x x x dxn

1

2

1cos sin sin ( 1)I ( 1)In n

n nx x x x n nn

Hence by transposing the last term, we get

12

1cos sin sin ( 1)n n

n nnI x x x x n In

122

1 1sin cos sin sin ( 1)n n n

nx xdx x x x x n In n

Similarly, 1

22

1 1cos sin cos cos ( 1)n n n

nx xdx x x x x n In n

Example 3.37: If / 2

0sinn

nu x xdx

, n > 1, then prove that

1

2( 1)2

n

n nu n n u n

. Hence evaluate u5.

Solution: Here, / 2

0sinn

nu x xdx

/ 2 1

0 0( cos ) ( cos )n nx x nx x dx


/ 2 1

00 cosnn x xdx


NOTES


/ 2 / 21 2

00(sin ) ( 1) (sin )n nn x x n n x x dx

1/ 2 2

0sin ( 1) (sin )

2 2

nnn n x x dx

1

2( 1)2

n

n nu n n n u

1

2( 1)2

n

n nu n n u n

… (i)

Putting n = 5, 3 respectively in Equation (i), we get4

5 35.4. 52

u u

...(ii)

2

3 13.2. 32

u u

...(iii)

Also, / 2

1 0sinu x xdx

/ 2/ 2

0 0cos 1. cosx x x dx

/ 2 / 2

000 cos sin 1 0 1x dx x

...(iv)

From Equations (iii) and (iv),2

3

36

4u

...(v)

From Equations (ii) and (v), we have4 2 4

25

3 55 20. 6 15 120.

2 4 16u

Reduction Formula for xn eax dx

1ax ax

n ax n ne ex e dx x nx dx

a a = 1

axn n axe n

x x e dxa a

Thus, if 1I , then I In ax

n axn n n

x e nx e

a a which is the required reduction formula.

Reduction Formulae for eaxcosn bx dx and eax sinn bx dx

11cos cos (cos sin )ax n n ax ax nnb

e bx dx bx e e bx bx dxa a


1 2 2 11 1cos cos sin ( 1)cos sin cos cosax n n ax ax n nnb b

e bx bx bx e e n bx bx bx bx dxa a a a

1 2 21 1cos cos sin ( 1)cos (1 cos ) cosax n ax n ax n nnb b

e bx e bx bx e n bx bx bx dxa a a a

NOTES




2 2 21 2

2 2 2

1cos cos sin ( 1) cos cosax n ax n ax n a x nnb b n b

e bx e bx bx n n e bxdx e bx dxa a a a

Transposing the last integral to the left, we have

2 2 21 2

2 2 2

11 cos cos cos sin ( 1) cosax n ax n ax n ax nn b nb b

e bx dx e bx e bx bx n n e bx dxa a a a

1 2

22 2 2 2 2 2

cos ( cos sin ) ( 1)cos cos

ax nax n ax ne bx a bx nb bx n n b

e bx dx e bx dxa n b a n b

Similarly

1 22

2 2 2 2 2 2

sin ( sin cos ) ( 1)sin sin



Reduction Formulae for xn eaxcos bx dx and xneax sin bx dx

1cos cos sinax ax axb

e bx dx e bx e bx dxa a

1 1

cos sin cosax ax axb be bx e bx e bx dx

a a a a

2

2 2

11 cos cos sinax ax axb b

e bx dx e bx e bxa a a

2 2cos cos sin

axax e

e bx dx a bx b bxa b

= 2 2 11

cos( ), ( ), tanax be bx r a b

r a .

Applying the above formula and integrating by parts, we have

1cos cos( ) cos( )n

n ax ax n axx nx e bx dx e bx x e bx dx

r r

Similarly,

1sin sin( ) sin( )n

n ax ax n axx nx e bx dx e bx x e bx dx

r r

Example 3.38: Integrate by parts twice or otherwise, obtain a reduction formulafor

0I sin ,x m

m e x dx where m > 2, in the form 2

2(1 )I ( 1)Im mm m m and

hence evaluate I4.

Solution: Here,0

I sin ,x mm e x dx

= 1

0 0sin ( sin cos ( )m x m xx e m x x e dx

, [Integrating by parts]


NOTES


= 0 + 1 2 2 1

0 0sin cos ( 1)sin cos sin sin ( )m x m m xm x xe m x x x x e dx

= 2 2

0 0{0} ( 1) sin (1 sin ) sinm x m xm m x x e dx xe dx

= 2

0 0 0( 1) sin ( 1) sin sinm x m x m xm m xe dx m xe dx xe dx

= 2[( 1)I ( 1)I I ]m m mm m m

= 2( 1)I { ( 1) }Im mm m m m m

2[1 ( 1) ]I ( 1)Im mm m m m m

22[1 ]I ( 1)Im mm m m .

This gives the reduction formula.

To evaluate 2nd part, 22

( 1)I I

1m m

m m

m

…..(i)

Putting m = 4, 2, successively in Equation (i), we get

4 2

4.3I I

(1 16)

0 4

4.3 2.1 24I I

17 5 85

0

0

24sin

85xe x dx

0

24 24

85 85xe

Reduction Formula for xm (log x)n dx

Integrating by parts on taking xm as second function, we have1 1 1(log )

(log ) (log ) .1 1

m n mm n n x n x x

x x dx x dxm x m

1

1(log )(log )

( 1) 1

m nm nx x n

x x dxm m

This is the required reduction formula.

Reduction Formula for(log )

m

n

xdx

x

1 (log )

(log )

m nm

n

x xdx x dx

x x

1 1 1(log ) (log )

( 1)1 1

m n nmx x x

m x dxn n

1

1 1

1

( 1)(log ) 1 (log )

m m

n n

x m xdx

n x n x

This is the required reduction formula.

NOTES




Example 3.39: If 1

0I log

mnm x x dx (n 0 and m is positive integer); prove

that 1I I1m m

m

n

and hence evaluate 3I .

Solution: 1

0I log

mnm x x dx

1 111

00

log1log .

1 1

mn nm xx x

x m dxn n x

[Integrating by

parts]

1 1

00 0 . log

1mnm

x x dxn

m 1I I1m

m

n

… (i)

Putting 3,2,1m successively in Equation (i), we get

3 2

3I I

1n

… (ii)

2 1

2I I

1n

... (iii)

1

1 0 0

1 1I I

1 1nx dx

n n

11

0

1

1 1

nx

n n

2

1 1 10

1 1 1n n n

1 2

1I

1n

..(iv)

From Equations (ii), (iii) and (iv), 3 2 4

3 2 1 6I

1 1 ( 1) 1n n n n

Reduction Formula forcos

cos

nxdx

x

Here the given integral is connected with the integral cos 2

cos

n xdx

x

in order to

get the reduction formula.

Now, cos 2 cos cos 2cos

cos cos cos

n x nx n xnx

x x x

2cos cos

2cos 1cos

nx x xn x

x

By taking the integral of both sides, we get

cos 2cos2 cos 1

cos cos

n xnxdx dx n x dx

x x


NOTES


sin 1 cos 2cos

2.cos 1 cos

n x n xnxdx dx

x n x


Note: The reduction formulae forsin

sin

nxdx

x can also be obtained by the above

method.

Reduction Formula for nxdx

x

sin

cos

Consider, sin 2 sin sin 2sin

cos cos cos

n x nx n xnx

x x x

sin 2 2sin 1 cossin

cos cos cos

n x n x xnx

x x x

sin 2sin

2sin 1cos cos

n xnxn x

x x

Integrating both sides with respect to x, we get

sin 2 cos 1sin2 sin 1 2.

cos cos 1

n x n xnxdx dx n xdx

x x n

2cos 1 sin 2sin

cos 1 cos

n x n xnxdx dx

x n x


Example 3.40: If cos cosec ,nu n d prove that

2

2cos 1

1n n

nu u

n

.

Solution: Consider cos 2 cos cos 2cos

sin sin sin

n n nn

2sin 1 sin

sin

n

cos 2cos

2sin( 1)sin sin

nnn

cos 2cos2 sin 1

sin sin

nnd d n d

cos 2 cos 1cos

2sin sin 1

n nnd d

n

i.e.,

2

2cos 1

1n n

nu u

n

NOTES




Reduction Formula for ( )m n px a bx dx pm nx a bx dx can be connected with any one of the following six integrals:

(i) 1pm nx a bx dx

(ii) 1pm nx a bx dx

(iii) pm n nx a bx dx (iv) 1pm n nx a bx dx

(v) pm n nx a bx dx (vi) 1pm n nx a bx dx

We cannot connect the given integral to that in which the indices of

both na bx and mx are increased or decreased.

Method to connect p

m nx a bx dx with any integral of the above type

Step 1: Let 11P nx a bx , where is smaller of the two indices of x and

is smaller of the two indices of na bx .

Step 2: After defining P, differentiate P with respect to x to get Pd

dx. Express it in

terms of integrands of the two integrals to be connected.

Step 3: Integrate both sides with respect to x and solve to get the value of thegiven integral.

Example 3.41: Prove that

1 122 2 2 2 2 2 2

2 3

2 12 1n n n

ndx x dx

a na x a n a x a x

Solution: Let us connect 2 2 n

dx

a x with 12 2 n

dx

a x

Now

0 2 2

2 2

n

n

dxx a x dx

a x

and

10 2 2

12 2

n

n

dxx a x dx

a x

i.e.,

12 2

12 2P

n

n

xx a x

a x

...(i)

12 2 2 2P1 .2

n nda x x n a x x

dx

2

12 2 2 2

12 1

n n

xn

a x a x


NOTES


2 2 2

12 2 2 2

12 1

n n

a x an

a x a x

2

1 12 2 2 2 2 2

2 1 2 11n n n

n a n

a x a x a x

2

12 2 2 2

2 1P 3 2n n

a nd n

dx a x a x


2

12 2 2 2

1 1P 3 2 2 1

n nn dx a n dx

a x a x

2

12 2 2 2

1 12 1 P (3 2 )

n na n dx n dx

a x a x

… (ii)

Using Equations (i) in (ii) and dividing by 22 1a n , we get

1 122 2 2 2 2 2 2

2 31 1

2 12 1n n n

nxdx dx

a na x a n a x a x

=

1 2 12 2 2 2 2

2 3 1

2 12 1n n

nxdx

a na n a x a x

Check Your Progress

9. If logn

nu x dx , find the value of 1n nu n u .

10. Connect pm nx a bx dx with 1pm nx a bx dx

11. If 1/ 2I ( ) ,nn x a x dx prove that

3 / 21(2 3)I 2 I 2 ( )n

n nn an x a x

12. If 0

I sin(2 1)nn x p xdx

, prove that

1

22 2

( 1)I I ( 1)

2(2 1) (2 1)

np

n n

n n n

p p

, n and p being positive

integers.

NOTES




3.5 QUADRATURE

The process of determining the area of a plane region is known as quadrature.

Area for Cartesian Curves

Theorem 3.1: The area of the region bounded by the curve y f x , the x-axis

and the two ordinates x a and x b is given by the definite integral

A = b

af x dx

where f(x) 0, on the interval [a, b]

Proof: Let f x be a continuous function of x in the domain (a, b). Suppose y

increases continuously as x increases from a to b and let AC, BD be the ordinates

x = a and x = b, respectively. Choose any points P (x, y) and Q ,x x y y on

the curve and let PM and QN be their ordinates, respectively.

Draw PS QS and PR QN.

Then,OM , ON , MP , NQx x x y y y

MN x and QR y

If the areas CMPA and CNQA be denoted by S and S S , respectively..

Then, S S S S

= Area CNQA – Area CMPA = Area MNQP

Clearly, Area of rectangle MNQS > Area MNQP > Area of rectangleMNRP

y x S y y x

S

y y yx

... (10)

When Q P, 0x and 0y

Sdy f x

dx

Sd f x dx ...(11)

Integrating with respect to x between the limits a and b, we get

SS S

b b x b x b

x aa a x a

df x dx dx d

dx

= (value of S at x = b) – (value of S at x = a)

= Area CDBA – 0

Hence, required area CDBAb

af x dx .


NOTES


Similarly, we can find the area of the region bounded by the curve x f y ,

the axis of y and the two abscissae y = c and y = d is given by A = .d

cf y dy

Example 3.42: Find the area between the curve 2 2 2 2 2x y a y x and its

asymptote.

Solution: The equation of the curve is given by 2 2 2 2 2x y a y x ... (i)

To trace the given curve, we observe:1. The curve is symmetrical about both the axes.2. The curve passes through the origin and the tangents at the origin are

given by 2 2 0y x or ,y x , which are real and distinct and so

the origin is a node.3. The curve meets both x-axis and y-axis at the origin only.

Y

x a = x – a =

y x =

XO

y– x

=

4. The curve has no other asymptotes except parallel to y-axis whichare given by 2 2 0a x or x a .

5. The given curve can be rewritten as

2 2 2 2

2

2 2

a x a xy

a x a xa x

... (ii)

L.H.S of Equation (ii) is always positive, thus its R.H.S must also be positive.

If x a or x a , than 2y becomes negative. Thus, the curve does not lie beyond

the lines .x a .

Also, when x increase from 0 to a, y increase from 0 to .

x varies from 0 to a in first quadrant and the curve is symmetrical aboutboth the axes.

Required area between the curve and its asymptotes

= 4 [Area of the curve in the first quadrant and its asymptote]

04

aydx 0 2 2

4a ax

dxa x

Put sinx a , so that cosdx a d

NOTES




Now for x = 0, 0 and for ,2

x a

Required area = / 2

0

. sin4 . cos

cos

a aa d

a

= / 2/ 22 2

004 sin 4 cosa d a

= 2 24 0 1 4a a

Area between Two Curves

Let f and g be the two functions (Refer Figure 3.1) which are continuous on the

interval [a,b]. If f and g both are positive and f x g x for all values of x in

interval [a, b]. Then, the area between two graphs is given by

A = b b

a af x dx g x dx

b

af x g x dx

Y

d

c

O

g f

X

Fig. 3.1 Area of Boundary Curves

I II

b

ay y dx

Note: If the boundary curves are functions of y, then the required area is given by

the integral d

cf y g y dy .

Corollary: Now, we derive a formula for the regions where neither f nor g remainspositive and also neither function remains greater than the other (Refer Figure3.2).

Y

O

f

a g

g

f

f

gb X

Fig. 3.2 Derivation of Formula for Regions


NOTES


Y

O

F

a

G

G

F

F

G

Xb1 ba1

Fig. 3.3 Raising the Region by a fixed Number of Units

Raise the entire region by a fixed number of units so that both f and gbecome positive (Refer Figure 3.3).

Now, the new boundaries become of the form

F(x) = f (x) + K

G(x) = g(x) + K

So, area of the first part is:

1 1 1

F Ga a a

a a ax x dx f x g x dx f x g x dx

Area of the second part is:

1 1

1 1

G Fb b

a ax x dx g x f x dx 1

1

b

af x g x dx

Area of the third part is:

1 1

G Fb b

b bx x dx g x f x dx

1

b

bf x g x dx

Thus, the total area:

1 1

1 1

a b b

a a bf x g x dx f x g x dx f x g x dx

Required area between the curves = b

af x g x dx

Example 3.43: Find the area common to the parabola 2 24 and 4 .y ax x ay

Solution: The equations of the given parabolas are 2 4y ax ... (i)

and 2 4x ay ... (ii)

x ay2 = 4 Y

y ax2 = 4

A(4 , 4 )a a

XO

NOTES




Solve the Equations (i) and (ii) to get the values of x and y

4

24

16

xax

a 3 364 0x x a 0, 4x x a

and 0at 0 and 4 at 4y x y a x a

Thus, the points of intersection of two parabolas are (0, 0) and (4a, 4a).

Required area (common to both the parabolas)

4

0

a

upper lowery y dx 2

4

04

4

a xax dx

a

[From Equations (i) and (ii)]4

3/ 2 3

0

4 1

3 12

a

ax x

a

3/ 2 34 14 4 0

3 12

aa a

a

2 2 232 16 16

3 3 3

a a a

Area for Parametric Curves

If the functions , have continuous derivatives, then the area bounded by the

curves , ,x t y t the x-axis and the ordinates of the points where

,t a t b is

b

a

dxy dt

dt

The area bounded by the curves , ,x t y t the y-axis and the

abscissae at the point when ,t c t d isd

c

dyx dt

dt .

Example 3.44: Find the area of the curve 3 3cos , sinx a t y b t .

Solution: The equations of the given curve are 3 3cos , sinx a t y b t

Y

B

O A X

t = 0

t =

The curve is symmetrical about both the axes.

Also, 3cosx a t x a

and 3siny b t y b


NOTES


So the curve lies within the rectangle bounded by x a and y b .

Also in the first quadrant t varies from 0 to 2

Required Area = 4 [area OAB]

/ 2

04

dxy dt

dt

/ 2 3 2

04 sin 3 cos sinb t a t t dt

/ 2 4 2

012 sin cosab t tdt

3.1.1 312 . .

6.4.2 2 8ab ab

(Numerically).

Area for Polar Curves (Sectorial Area)

Theorem 3.2: Let f() be a continuous function for every value of in the domain(,), then the area of the section bounded by the curve r = f() and the two radii

vectors = , = ( < ) is equal to 21

2r d

.

Proof: As shown in the Figure 3.4, let OC and OD be the two radii vectors =

, = (<), respectively. Let P(r,) and Q( , )r r be twoneighbouring points on the curve. By taking O as the centre, draw two circulararcs PR and QS with radii OP and OQ, respectively. Then PR = r and QS = (r+ r) .

D

RQ

PS

C

A

r

O

Fig. 3.4 Determining the Area of Polar Curves

Sectorial area21 1

OPR .2 2

r r r

and sectorial area 1

OSQ ( )( )2

r r r r

21

( )2

r r .

NOTES




Let A and A + A denote the area of sectors OCP and OCQ respectively,then A = Area OCQ Area OCP

= Area OPQ

Also, the area OPQ lies between the areas OSQ and OPR

2 21 1( )

2 2r r A r

2 2( )

2 2

r r A r

… (12)

Since, f() is a continuous function so, as 0 and r 0

2A

2

d r

d

21

2r d dA .

On integrating both sides between the limits = , = , we get

21 A

[A]2

dr d d

d

= (Value of A when = ) (Value of A when = )

= Area of OCD

Hence, sectorial area OCD = 21

2r d

.

Note: The following points should be remembered while doing the problems forpolar curves:

1. It is easy to transform a cartesian equation to polar form than to solvefor y in some cases.

2. To obtain the limits of integration for a loop, find two successive valuesof for which r = 0.

3. If the curve is symmetrical about the x-axis only, evaluate the integralfrom 0 to and multiply the result by 2.

4. If the curve is symmetrical about both the axes, evaluate the integral

from 0 to2

and multiply the result by 4.

5. If n is odd, the curve cosr a n or sinr a n have n equal loopsand if n is even the curve have 2n equal loops.

Example 3.45: Find the area of one loop of the curve r = a sin 3.


NOTES


Solution: The equation of the given curve isY

XO

r = a sin 3

By comparing it with r = a sin n, we get n = 3, which is odd.

So, the curve has 3 equal loops.

Put r = 0, so 3 = 0, i.e., = 0, 3

Thus, there is a loop between = 0 and = 1

3

Hence, the area of the loop lying in the positive quadrant

= / 3 2

0

1

2r d

/ 3 2 2

0

1sin

2a d

= 2

0

1 cos 6

2 2

ad

/ 32

0

sin 6

4 6

a

2 2

4 3 12

a a

and Total Area of the Curve = 3 Area of One Loop

= 32 2

12 4

a a .

Area between Two Polar Curves

The area bounded by the curves ,r f r g and the radii vectors

, is 2 21 2

1

2r r d

where 1r and 2r are the radii for the outer and

inner curves, respectively.B

D

A

C

XO

Fig. 3.5 Area Between Two Curve

Let the curves AB and CD are two curves (Refer Figure 3.5) denoted

by r f and r g , respectively

NOTES




OCA and ODB are the radii vectors for and .

So, Area CABD = Area OAB – Area OCD

2 21 1

2 2f d g d

2 21

2f g d

2 21 2

1

2r r d

Example 3.46: Find the area common to the circle r = a and the cardioid

1 cosr a .

Solution: The equations of the given curves are r a ... (i)

and 1 cosr a ... (ii)

Solve Equations (i) and (ii) to get the points of intersection

1 cosa a cos 0 2

The two curves intersect at the points where2

.

Required Area = 2 [Area OMNPQO] = 2 [Area OMNPO + Area OPQO] ... (iii)

Y

PN

Q

O ML X

Now, Area OMNPO/ 2 2

0

1

2r d

for r a

2

/ 2 / 22

00

1

2 2

aa d

2 2

2 2 4

a a

... (iv)

Area OPQO 2

/ 2

1

2r d

for 1 cosr a

22

/ 2

11 cos

2a d

2 2

/ 2

11 2cos cos

2a d


NOTES


2

/ 2

1 cos21 2cos

2 2

ad

2

/ 2

3 12cos cos2

2 2 2

ad

2

/ 2

3 12sin sin 2

2 2 4

a

2 23 3

2 3 82 2 4 8

a a ... (v)

Using the values from Equations (iv) and (v) in (iii), we get

Required Area 2 2

2 52 3 8 2

4 8 4

a aa

Check Your Progress

13. Trace the curve y2 (2a - x) = x3 and find the area between the curve and itsasymptote.

14. Find the area of the ellipse2 2

2 21

x y

a b .

15. Find the area enclosed by the curve

2 2 2 34 and ( 4 ) 8x ay y x a a

16. Find the area bounded by the cardioide 1 cosr a .

17. Find the area of the loop of the curve 5 5 2 25 0x y ax y

18. Find the area of the region included between the cardioids 1 cosr a

and 1 cosr a .

3.6 RECTIFICATION

Rectification is the process of finding out the length of an arc of a curve between twodistinct points (Refer Figure 3.6). Suppose f is a function, which is continuous on the

closed interval [x0, x

n]. Then an arc from 0 0A ,x y to point B ,n nx y has the length

L if L has the property: for 0 , there exists 0 , such that for every partition

0 1 2 1A P ,P ,P ,...,P ,P Bn n it is true that if , then 1P P Li i .

NOTES




YP2

P1 Pn – 1B = Pn

A = P0

P3 P4

XO

Fig. 3.6 Rectification

An arc is known as rectifiable if it has length.

Length of Cartesian Curves

Theorem 3.3: Let the Cartesian curve is given by y f x , where f x exist

and is continuous on the interval ,a b . Then, the length s of the arc is given by

2

1b

a

dys dx

dx

Y B

PS

A

x a = x b = X

Fig. 3.7 Length of the Cartesian Curve

Proof: Let us consider a curve y f x with two points on it as A and B, as

shown in Figure 3.7. Let the abscissa of points A and B be a and b. Also let s bethe length of the curve from a fixed point A to a variable point P(x, y) on the curve.

From differential calculus, we have 2

1ds dy

dx dx

On integrating both sides with respect to x from x a to x b , we get

2

1b

b

aa

ds dydx dx

dx dx

2

1b

x b x a a

dys s dx

dx


NOTES


Since, the length is being measured from fixed point A, so s for x = a iszero, thus

(Arc AB) – (0) = 2

1b

a

dydx

dx

Thus, arc AB = 2

1b

a

dydx

dx

Note: The length of the curve x f y where f y exists and is continuous on

the interval [c, d] is 2

1d

c

dxdy

dy

.

Working Rule to Determine the Length of an Arc of Cartesian Curve

y f x

1. If the end points of the arc are given, then the limits of the integration areknown and there is no need to trace the curve. However, if the end pointsare not given, trace the curve roughly and find the limits of integration.

2. Express y in terms of x, and then find dy

dx.

3. Find the length of the required arc using2

1b

a

dydx

dx , where a and b

are the limits of integration obtained in the above step.

Example 3.47: Find the length of the arc of the parabola 2 4y ax from the

vertex to an extremity of the latus rectum.

Solution: Let the vertex be O and an extremity of the latus rectum of the givenparabola be L.

The equation of the parabola is 2 4y ax

YL(a, 2a)

XSO

2

4

yx

a

On differentiating with respect to y, we get

2

4 2

dx y y

dy a a

NOTES




As y varies from 0 to 2a for the length of the arc from O (0, 0) to L ,2a a ,

Required length of the arc 2

2

01

a dxdy

dy

2

2

01

2

a ydy

a

2 2 2

0

12

2

ay a dy

a

22 22

2 2

0

41 44 log

2 2 2 2

a

y y ay ay a

a a

2 22 2 21 2 2 4 4

4 4 2 log 02 2 2

a a a aa a a

a a

2 log 1 2a a 2 log 2 1a

Example 3.48: Show that the length of the curve 2 2 2 2 28x a x a y is 2a .

Y

( , 0)aX

AO

(– , 0)a

Solution: The given equation of the curve is

2 2 2 2 28a y x a x ...(i)

To find the limits of integration, we need to trace the curve roughly.1. The curve has symmetry about both x-axis and y-axis.2. The curve passes through the origin and the tangents at the origin are

given by1

2 2y x , which are real and distinct. Thus, the origin is a

node.

3. If 0, then 0,y x a . Thus, the curve meets the x-axis at (0, 0), (a,0) and (–a, 0).

If 0, then 0x y . Thus, the curve meets the y-axis at the origin only..

4. There is no asymptote to the curve.

5 In the curve 2 2 2 2 28a y x a x , L.H.S. is always positive which

means R.H.S. must also be positive, i.e, 2 2 0a x 2 2x a


NOTES


It is clear that –a x a and hence the whole curve lies between the linesx a and x a .

The curve has two symmetrical loops as shown in the above Figure.

From Equation (i), 2 2

2

1

8y x a x

a

Differentiating, we get

2 2

2 2 2

1 2

8 2

dy xx a x

dx a a x

2 2

2 2 2

1 2

8

a x

a a x

Required Length = 4 (Arc OA)2

04 1

a dydx

dx

22 2

2 2 20

24 1

8

a a xdx

a a x

4 2 2 4 4 2 2

2 2 20

8 8 4 44

8

a a a x a x a xdx

a a x

4 2 2 4

2 22 0

4 9 12 4

8

a a a x xdx

a xa

2 2

0 2 2

4 3 2

2 2

a a xdx

a a x

2 2 2

0 2 2

22 a a x adx

a a x

2

2 2

0 2 2

22

a aa x dx

a a x

2 2 2

1 2 1

0

22 sin sin

2 2

a

a x a x xx a

a a a

2 2 2 1

0

22 sin

ax

x a x aa a

2 1 1 22 22 sin 1 sin 0 .2 . 2

2a a a

a a

NOTES




Length of Parametric Curves

Y

O

t t = 1

A

SP

t t = 2

B

X

Fig. 3.8 Length of Parametric Curve

Theorem 3.4: Let the parametric curve is given by x f t and y g t , where

f t and g t exists and are continuous on the interval 1 2,t t , as shown in

Figure 3.8. Then, the length s of the arc between the points where 1t t to 2t t isgiven by

2 22

1

t

t

dx dys dt

dt dt

Proof: Let AB represents the portion of the given parametric curve

x f t and y g t , for 1 2t t t .

If the length of the arc AB, measured from the fixed point A to a variablepoint P(t) on the given curve is taken as s, then from differential calculus, we have

2 2ds dx dy

dt dt dt

On integrating both sides with respect to t from 1t t to 2t t , we get

2 22 2

1 1

t t

t t

ds dx dydt dt

dt dt dt

2 2

2

2 1 1

t

x t x t t

dx dys s dt

dt dt

Since the length is being measured from a fixed point A, so s for x = t1 is

zero,

(Arc AB) – (0) = 2 2

2

1

t

t

dx dydt

dt dt

Thus, arc AB = 2 2

2

1

t

t

dx dydt

dt dt


NOTES


Example 3.49: Find the length of the arc in the first quadrant of the curve2 / 3 2 / 3

1.x y

a b

Solution: The given equation of the curve is 2 / 3 2 / 3

1.x y

a b

Y

B

AX

O

Its parametric equations are 3 3cos , sinx a y b . …(i)

It is clear from the parametric equations that x cannot be numerically greaterthan a and y cannot be numerically greater than b.

To find the limits of integration, we need to trace the curve roughly.1. The curve has symmetry about both x-axis and y-axis.2. There is no value of for which both x and y are zero. Thus, the

curve does not pass through the origin.

3. The curve meets the x-axis at ,0a and the y –axis at (0, b)

4. If = 0, then x = a, y = 0 and if 2

, then x = 0, y = b. So, when

varies from 0 to2

in the first quadrant, x decreases from a to 0

and y increases from 0 to b.

Differentiating Equation (i), with respect to , we get 23 cos sindx

ad

And 23 sin cosdy

bd

Required length of the arc in the first quadrant

2 2

/ 2

0

dx dyd

d d

/ 2 2 2 2 2 2 2

09sin cos cos sina b d

/ 2 1/ 22 2 2 2

03 cos sin sin cosa b d

NOTES




2 2 2 2 2

2 2Put cos sin andsin cos

zdza b z d

b a

.

Also, when 0, and when ,2

z a z b

Length of the arc in the first quadrant 2 23 .

b

a

zdzz

b a

2

2 2

3 b

az dz

b a

3

2 2

3

3

b

a

z

b a

3 3

2 2

b a

b a

2 2 2 2b a b ba a a ab b

b a b a a b

Length of Polar Curves

Theorem 3.5: Let the polar curve is given by r f , where f and f are

continuous on a interval , . Then the length s of arc of the curve between the

points for which to is given by 2

2 drs r d

d

Proof: The given equation of the curve is r f .

We know that cosx r , siny r …(13)


sin cosdx dr

rd d

…(14)

and cos sindy dr

rd d

…(15)

Also, we know that the length s of the parametric curve from to isgiven by

2 2

dx dys d

d d

On substituting the values from Equations (14) and (15), we get

s 2 2

sin cos cos sindr dr

r r dd d

On solving, we get 2

2 drs r d

d

which is the required equation for finding out the length of an arc of the polarcurve.


NOTES


Theorem 3.6: Let the equation of the curve is given by f r . Then the length

s of arc between two points with radii vectors 1r and 2r is given by

2

2

11

r

r

ds r dr

dr

.

Proof: We know that the length s of arc of the curve between the points for which

to is 2

2 drs r d

d

2

2 2

1

r

r

dr dr dr

d dr

2

2 2

11

r

r

dr dr

dr

22

11

r

r

dr dr

dr

Theorem 3.7: Let the equation of the curve is given by p = f(r). Then the lengths of the arc of the curve between the points r = a and r = b is given by

2 2

b

a

rs dr

r p

.

Proof: The length s of the arc of the curve is2

1b

a

ds r dr

dr

Let tand

rdr

21 tanb

as dr sec

b

as dr …(16)

As sinp r sinp

r

2 2sec

r

r p

On substituting the value of sec in Equation (16), we get 2 2

b

a

rs dr

r p

Example 3.50: Find the entire length of the cardioid 1 cos ,r a and show

that the arc of the upper half is bisected by .3

Solution: The given equation of the curve is 1 cosr a .


XO

sindr

ad

NOTES




Length of whole arc = 2Length of upper half2

2

02

drr d

d

22 2 2

02 1 cos sina a d

0

2 2 1 cosa d

00

4 cos 4 2sin2 2

a d a

8 1 0 8a a

Since the complete length of cardioid is 8a, thus the length of the upper halfis 4a.

The length s of arc from 0 to3

is

2/ 3 2

0

drs r d

d

/3

02 cos

2a d

/3

0

2 2sin2

a

14 0 2

2a a

1

2 ×Length of the upper half arc

Thus, the length of the upper half arc is bisected at .3

Check Your Progress

19. Find the length of the arc 2 2 2 0x y ax in the first quadrant.

20. Find the length of the loop of the curve 223 .ay x x a

21. Find the length of the arc of the curve x = e sin , y = e sin from

0 to .2

22. Find the whole length of the asteroid 2 / 3 2 / 3 2 / 3.x y a 23. Find the whole length of the curve 1 cosr a and show that the arc of

the upper half of the curve is bisected by 2

3

.

24. Find the length of arc of the equiangular spiral cotr ae between the points

for which the radii vectors are 1r and 2 .r


NOTES



1. To give a precise shape to the definition of integration, we observe. If g (x)is a function of x such that,

d

dx g (x) = f (x)

then we define integral of f (x) with respect to x, to be the function g (x).This is put in the notational form as,

( )f x dx = g (x)


2. (a) (i) 2 x (ii) 2

x

(iii)3 / 2 5 / 22 2

3 5

x x(iv)

4 2

8 4

ax bx

(b) (i)2

2

xe(ii)

2 3

2

xe(iii) xe

(c) (i) 1 21 (sin )2

x (ii) sin (log x) (iii) tan–1 ex

3. (a) (i) sin h–1 (x + 1) (ii) 13 sin h–1

3 2

5

x

(iii) 12 sin2

x

a

(b) (i)1

8(4x – 3) 22 3 4x x +

412

32sin–1

4 3

41

x

(ii) 2

x 2 3

2 33

x sin–1 3

2

x

(iii) 2 19 29 4 sin

2 4 3

x xx

4. (i) 1

6(ii)

64

35(iii)

2

3

5.22

cosec cot cot3

NOTES




6.2tan

log sec2

xx

7. 1

3n

8.5

32

9. logn

x x

10.

11

1 1

pm npm n

x a bx pnax a bx dx

m pn m pn

11. Here, connect 3/ 2( )nx a x dx with 1 1/ 2( )nx a x dx .

Let 1 1 1/ 2 1 3 / 2P ( ) ( )n nx a x x a x

1 3 / 2 1/ 2P 3( ) ( )

2n nd

nx a x x a xdx

1 1/ 2 1/ 23( ) ( ) ( )

2n nnx a x a x x a x

1 1/ 2 1/ 2 1/ 23( ) ( ) ( )

2n n nnax a x nx a x x a x

1 1/ 2 1/ 21( ) (2 3) ( )

2n nnax a x n x a x .

This, on integration and transposition, gives

1/ 2 1 1/ 2(2 3) ( ) 2 ( ) 2n nn x a x dx an x a x ax P 3/ 2

1(2 3) 2 2 ( )nn nn I anI x a x

12. We have 0

I sin(2 1)nn x p xdx

/ 2

1

00

cos(2 1)cos(2 1)

(2 1) (2 1)

nnx p x n

x p xdxp p

/ 21 2

2 20 0

( 1)sin(2 1) sin(2 1)

(2 1) (2 1)n nn n n

x p x x p xdxp p

1

22 2

( 1)I ( 1) I

2(2 1) (2 1)

np

n n

n n n

p p

1

22 2

( 1)I I ( 1)

2(2 1) (2 1)

np

n n

n n n

p p


NOTES


13. 3a2

14. ab

15.22

(3 2)3

a

16. 23

2a

17.25

2

a

18. 2

3 82

a

19. a

20.4

3

a

21. / 22 1e

22. 6a

23. 8a

24. 2 1 secr r

3.8 SUMMARY

An integral assigns numbers to functions in a way that can describedisplacement, area, volume, and other concepts that arise by combininginfinitesimal data.

Integration can be used to find areas, volumes, central points, and the areaunder the curve of a function.

Integration means summation. Integral of a function y = f(x) under the limita to b gives us the area enclosed by the curve y = f(x) and the twocoordinates x = a, and y = b.

To give a precise shape to the definition of integration, we observe, If g (x)is a function of x such that,

d

dx g (x) = f (x)

then we define integral of f (x) with respect to x, to be the function g (x).This is put in the notational form as,

( )f x dx = g (x)


NOTES




Define integration as:

If,d

dxg(x) = f (x)

then, ( )f x dx = g(x) + c

where c is some constant, called the constant of integration. Obviously. The c can have any value and thus integral of a function is not unique! But

we could say one thing here, that any two integrals of the same functiondiffer by a constant.

Differentiation and integration cancel each other.

For any constant a, ( ) a f x dx = ( )a f x dx

For any two functions f (x) and g(x),

[ ( )f x ± g(x)] dx = ( ) f x dx ± ( ) g x dx

The definite integral ( )b

af x dx is defined by

( )b

af x dx = ( ) b

ag x = g(b) – g(a)

where a and b are two real numbers, and are called respectively, the lowerand the upper limits of the integral.

If u and v are two functions of x

then ( )d

uvdx

= dv

udx

+ du

vdx

dvu

dx =

d

dx (uv) – v

du

dx

Integral of the product of two functions

= First function × Integral of the second – Integral of

(Differential of first × Integral of the second function).

If the integrand consists of even powers of x only, then the substitutionx2 = t, is helpful while resolving into partial fractions.

The substitution tan 2

x = t, converts every rational function of sin x and

cos x into a rational function of t and we can then evaluate the integral.

Reduction Formulae for sinn x dx and cosn x dx

12sin .cos ( 1)

sin sinn

n nx x nxdx xdx

n n

1

2cos sin ( 1)cos cos

nn nx x n

xdx xdxn n


NOTES


Walli’s FormulaWhen n is an even positive integer

/ 2

0

( 1) ( 3) ( 5) 1sin . . .... .

2 4 2 2n n n n

dxn n n

When n is an odd positive integer

/ 2

0

( 1) ( 3) ( 5) 2sin . . ....

2 4 3n n n n

dxn n n

When n is an even positive integer

/ 2

0

( 1) ( 3) ( 5) 1cos . . .... .

2 4 2 2n n n n

dxn n n

When n is an odd positive integer

/ 2

0

( 1) ( 3) ( 5) 2cos . . ....

2 4 3n n n n

dxn n n

Reduction Formulae for tann x dx and cotn x dx

1 21tan tan tan

1n n nxdx x xdx

n

1 21

cot cot cot1

n n nxdx x xdxn

Reduction Formulae for secn x dx and cosecn x dx2

2sec tan ( 2)sec sec

1 ( 1)

nn nx x n

xdx xdxn n

22cosec cot ( 2)

cosec cosec1 ( 1)

nn nx x n

xdx xdxn n

Reduction Formulae for sinm x cosn x dx

sin cosm nx x dx1 1

2sin cos 1sin cos

m nm nx x n

x x dxm n m n

Reduction Formulae for sin sin and sin cosm mx nx dx x nx dx .

12

2 2 2 2

sin cos sin cos sin ( 1)sin sin sin sin

m mm mn x nx m x x nx m m


12

2 2 2 2

sin sin sin cos cos ( 1)sin cos sin cos

m nm mn x nx m x x nx m m


.

Reduction formulae for cos sin , cos cosm mx nx dx x nx dx .

1cos coscos sin cos sin( 1)

mm mx nx m


.

1cos sincos cos cos cos( 1)

mm mx nx m


.

NOTES




Reduction formulae for sinnx mxdx and cosnx mxdx .

1 22 2

1 ( 1)sin cos sin sinn n n nn n n

x mxdx x mx x mx x mxdxm m m

12

2 2

sin cos ( 1)cos cos

n nn nx mx nx mx n n

x mxdx x mxdxm m m

.

Reduction formula for sinnx xdx1

22

1 1 ( 1)sin cos sin sin In n n

n

nx xdx x x x x

n nn

Reduction formula for n axx e dx

1I , then I In ax

n axn n n

x e nx e

a a Reduction formulae for cosax ne bx dx and sinax ne bx dx

1 22

2 2 2 2 2 2

cos ( cos sin ) ( 1)cos cos



1 22

2 2 2 2 2 2

sin ( sin cos ) ( 1)sin sin



Reduction formulae for cosn axx e bx dx1cos cos( ) cos( )

nn ax ax n axx n

x e bx dx e bx x e bx dxr r

Reduction formula for (log )m nx x dx

(log )m nx x dx 1

1(log )(log )

( 1) 1

m nm nx x n

x x dxm m

Reduction formula for (log )

m

n

xdx

x

(log )

m

n

xdx

x1

1 1

1

( 1)(log ) 1 (log )

m m

n n

x m xdx

n x n x

Reduction formula for cos

cosnx

dxx

sin 1 cos 2cos2.

cos 1 cos

n x n xnxdx dx

x n x

The process of determining the area of a plane region is known as quadrature.

The area under a given curve can be calculated when the equation of thecurve is given in Cartesian, parametric or polar form.

The area of the region bounded by the curve y f x , the x-axis and the

two ordinates x a and x b is given by the definite integral

A = b

af x dx

where f(x) 0, on the interval [a, b].


NOTES


The area of the region bounded by the curve x f y , the axis of y and

the two abscissae y = c and y = d is

A = d

cf y dy .

If f and g are both continuous on the interval [a,b] and are positive and if

f x g x for all x in [a, b], then the area between the graphs is given by

A I II .b

ay y dx

where Iy f x is the upper curve and IIy g x is the lower curve.

If the functions , have continuous derivatives, then the area bounded by

the curves , ,x t y t the x-axis and the ordinates of the points

where ,t a t b is b

a

dxy dt

dt .

The area bounded by the curves , ,x t y t the y-axis and the

abscissae at the point when ,t c t d isd

c

dyx dt

dt .

If f() be a continuous function for every value of in the domain (,),then the area bounded by the curve r = f() and the radii vectors = ,

= ( < ) is equal to 21

2r d

.

The area bounded by the curves ,r f r g and the radii vectors

, is 2 21 2

1

2r r d

, where 1r and 2r are the radius for the

outer and inner curves, respectively.

The Cartesian curve is y f x , where f x exists and is continuous on

the interval ,a b . Then the length s of the arc is given by

2

1b

a

dys dx

dx .

The parametric curve is given by x f t and y g t , where

f x and g x exist and are continuous on the interval 1 2,t t . Then, the

length s of the arc between the points where 1t t to 2t t is given by2 2

2

1

t

t

dx dys dt

dt dt .

NOTES




The polar curve is given by r f , where f and f are continuous

on an interval , . Then, the length s of arc of the curve between the

points for which to is given by

22 dr

s r dd

.

The equation of the curve is given by f r . Then, the length s of arcbetween two points with radii vectors 1r and 2r is given by

22

11

r

r

ds r dr

dr

.

The equation of the curve is given by p = f(r). Then, the length s of the arcof the curve between the points r = a and r = b is given by

2 2

b

a

rs dr

r p

.

3.9 KEY TERMS

Integration: The process of computing an integral; the inverse ofdifferentiation.

Integrand: A function to be integrated.

Definite integral: Definite integral is an integral with two real numberswhich are the upper and lower limits of the integral.

Reduction formula: It is a formula which connects an integral with anotherintegral, having the same form as the given integral but of lower degree ororder and is easier to integrate.

Successive reduction: It is a method in which you obtain the value of thegiven integral in terms of another integral in which the integral is of lowerdegree.

Rectification: It is the process of finding out the length of an arc of a curvebetween two distinct points.

Quadrature: It is the process of determining the area of a plane region.



1. What is integration? Explain with the help of an example.

2. Write any five formulas obtained as a result of direct consequence of inte-gration.


NOTES


3. What is the significance of dx in integration?

4. What do you mean by integration by parts?

5. How is the integration of algebraic functions done?

6. What do you mean by integration of irrational functions?

7. Evaluate (i) 0 2

na x

dxax x

(ii)9 / 2

2

0 2

a xdx

a x (iii) 4

20

1 cossin .

1 cos

xx dx

x

8. If / 2

0I cosn

n x x dx

, prove that 2 2

1 1I In n

n

n n

.

9. If/ 4

0I tan ,n

n d

where n is a positive integer; prove that

1 1(I I ) 1n nn

10. Prove that / 4 3

0

1sec 2 log(1 2)

2xdx

.

11. Prove that

(i) / 2 5 6

0

8sin cos

693x x dx

(ii)

/ 2 6 8110

5sin cos

2 2 2d

12. Show that2

2 20 0

( 1)sin sin sin sinm mm m

x nxdx x nxdxm n

.

13. If0

( , ) cos cosmf m n x nxdx

, show

that 2 2

( 1)( , ) ( 2; )

m mf m n f m n

m n

14. Obtain a reduction formula for cosnx xdx and hence evaluate 3 cosx xdx .

15. Find the length of the arc of the curve 1

log1

x

x

ey

e

from 1x to 2x .

16. Find the length of the curve 2 33 2ay x from 0,0 to ,x y .

17. Find the whole length of the loop of the curve 2 23ay x a x .

18. Find the length of the boundary of the region enclosed by the curve y = x2

+ 1 and the lines y = x, x = 0 and y = 2.

19. Find the length of an arc of parabola 2 4x ay :

(i) from the vertex to an extremity of the latus rectum.

(ii) cut off by latus rectum.

20. Find the length of the complete cycloid given by

sin , 1 cosx a y a .

NOTES




21. Rectify the cycloid sin , 1 cosx a y a .

22. Show that the length of an arc of the curve 33sin sinx a and3cosy a , measured from 0,a to any point

,x y is 3sin cos

2

a .

23. Show that the length of the arc of the curve sin 2 1 cos 2x a

and cos2 1 cos2y a measured from the origin to any point (x, y) is

4sin 3

3

a .

24. Show that the length of the cycloid ( sin ), (1 cos )x a y a

measured from the vertex to any point ,x y is 8ay and hence find the

whole length of the curve.


1. Explain the properties of integration with the help of examples.

2. Define six important integrals and their results.

3. Describe the methods of integration.

4. What is a definite integral? Also define the properties of definite integrals.

5. Integrate the following functions with respect to x:

(a) (i)2

1–x

x(ii)

1

1 1x x (iii)1

1 1x x

[Hint. Rationalize]

(b) (i) (sin x – cos x)2 (ii) sin2 x (iii) cos3 x

(iv)1

1 sin x (v) sin (4x + 3) (vi)sin 2

sin

x

x

(c) (i) 2

1

4x (ii) 2

1

1

x

x (iii)4 2

2

1

2( 1)

x x

x(iv) 3

12x

x

(v)

23 2x x

x

6. Evaluate the following integrals:

(a)π /2

0

sin x dx (b)π/4

2

0

sin x dx (c)3

2

1dx

x (d)3

2

2

( 1)x dx

(e)1

20

1

1 x dx (f) 2b

ax dx


NOTES


(g) (i) sin x

x(ii) tan θ sec2 (iii) ecos x sin x

(h) (i) sec x log (sec x + tan x) (ii) tan x (iii)log x

x(iv) tan2 x

(i) (i) sin6 x cos x (ii) cos6 x sin x (iii)2

4 1

x

x

e

e

(j) (i)2

61

x

x(ii)

1

1

x

x(iii) 2

2 3

3 7

x

x x

[Hint. Put x = cos ]

7. Evaluate 2 2

1

1x x dx.

8. Integrate the following:

(a) (i) cot x (ii) cosec x (iii) 1

logx x (iv) 1

(1 log )x x

(b) (i) 2

2 3

3 7

x

x x(ii)

5

61

x

x(iii)

cos

sin

x

a b x (b 0)

(c) (i) tan3x sec2x (ii) 2 2x a x (iii) 1 4

2

(sin )

1

x

x

(iv) x4(1 + x5) 12

(d) (i) 1 tan

1 tan

x

x (ii)1

x x (iii) sec cosec

log tan

x x

x

(e) (i) 1 sinxe x (ii) (ex + e–x) (ex – e–x)

(iii)ex(a + bex)n, (b 0)

(f) (i) cos3x (ii) sin3x (iii) sin 2x cos 3 x

(iv) 1 sin x (v) 1 cos x

(g) (i) 2

2 1

x

x(ii)

3

8 1

x

x (iii) 3

2 2

1

a x

(h) (i) 23 2x x (ii) 23 5x (iii) 3 8 16x x

(iv) cos x 24 sin x [Put x4 = t]

9. Show that 2

2

2 3

1

x x

xdx =

5

22 11 sinx h x .

NOTES




10. Evaluate the integrals:

(i)1

logxe x dxx (ii) 2[tan sec ]xe x x dx

(iii)

2

2

1

1x x

e dxx

(iv) 2

1 sin cos

cosx x x

e dxx

11. Find a reduction formula for 0mx

n

edx n

x .

12. Prove that (i) 2 20cos ( 0)ax a

e bx dx aa b

(ii) 2 20sin ( 0)ax b

e bx dx aa b

.

13. Obtain a reduction formula for lognmx x dx and hence show that

1

1 !log ,

1

nnm

n

nx x dx

m

where 0m and n is a positive integer..

14. If sin

sinn

nxu dx

x , prove that 2

2sin( 1)

1n n

n xu u

n

Hence evaluate/ 2

0

sin 7

sin

xdx

x

.

15. Prove that

2 2 2

0

2 1 2 1 .....3.12 .

2 1 .....2.1

a m m m mx ax x dx a

m m m

, where m is

positive integer.

16. Obtain a reduction formula for / 22 2 nx a dx and hence evaluate

5/ 22 2x a dx .

17. If 1

, 0I 1

nmm n x x dx , prove that 1m n m, m, 1I In nn

Hence find the value of 1 34

01x x dx

18. Show that 1 12 2 2

2 3

2 11 2 1 1 1n n n

dx x n dx

nx n x x

.

19. If m,2

I1

m

n n

xdx

x

then prove that

11 2m, m 2, 12 1 I 1 1 I

nmn nn x x m


NOTES


20. Find the whole area of the curve 2

2 2

1 2, .

1 1

t atx a y

t t

21. Show that the area of a loop of the curve 3 cos3 sin3r is 3

.

22. Show that the area contained between the circle r = a and the curvecos5r a is equal to three-fourths of the area of the circle.

23. Find the area inside the circle sinr and outside the cardioid

1 cosr .

24. Find the area outside the circle 2 cosr a and inside the cardioid

1 cosr a .











NOTES

Differential Equations


UNIT 4 DIFFERENTIAL EQUATIONS

Structure

4.0 Introduction4.1 Objectives4.2 Linear Differential Equations

4.2.1 Geometrical Meaning of a Differential Equation4.2.2 Solving Linear Differential Equations

4.3 Equations Reducible to the Linear Form4.4 Exact Differential Equations4.5 First Order Higher Degree Equations Solvable for x, y and p4.6 Clairaut’s Equation4.7 Singular Solutions4.8 Orthogonal Trajectories4.9 Answers to ‘Check Your Progress’

4.10 Summary4.11 Key Terms4.12 Self-Assessment Questions and Exercises4.13 Further Reading

4.0 INTRODUCTION

In mathematics, a Differential Equation (DE) is defined as an equation of the formthat interconnects certain function with its derivatives, where usually the functionrepresents the physical quantity while the derivatives denote their rates of changeand the relationship between the two is defined by the equation. Fundamentally,the ‘solutions of differential equations’ are functions which precisely ‘represent therelationship or correlation between a continuously varying or fluctuating quantityand its rate of change’. A Differential Equation (DE) comprises of one or moreexpressions including derivatives of one dependent variable ‘y’ with reference toanother independent variable ‘x’. An ordinary differential equation is a differentialequation that includes a function of a single variable and some of its derivatives.Principally, a differential equation is an equation for a function that relates thevalues of the function to the values of its derivatives. Therefore, a differential equationis an equation between specified derivative on an unknown function, its values andknown quantities and functions or a Differential Equation (DE) is an equation thatcomprises of a function and its derivatives. Differential equations are categorizedas Partial Differential Equations (PDE) or Ordinary Differential Equations (ODE)in accordance with whether or not they hold partial derivatives, while the order ofa differential equation is defined on the basis of the highest order derivative thatoccurs in the equation.


NOTES


In mathematics, a linear differential equation is a differential equation that isdefined by a linear polynomial in the unknown function and its derivatives, that isan equation of the form,

0 1 2 0nna x y a x y a x y a x y b x

where a0(x), ..., a

n(x) and b(x) are arbitrary differentiable functions that do not

essential to be linear, and y, …, y (n) are considered as the successive derivativesof an unknown function y of the variable x. A linear differential equation may alsobe a Linear Partial Differential Equation (Linear PDE), if the unknown functiondepends on several variables, and the derivatives that appear in the equation arepartial derivatives. If a linear differential equation or a system of linear equationsare such that the associated homogeneous equations have constant coefficientsthen these may be solved by means of quadrature mathematics, i.e., the solutionsmay be expressed in terms of integrals. This is also true for a linear equation oforder one with non-constant coefficients.

By a differential equation, you mean an equation which involves derivatives of adependent variable with respect to one or more independent variables. Some

examples of differential equations are 6dy

xdx

, 2

25 8 0

d y dyy

dx dx and

2 22

2 2

z zx y

x y

. Symbolically, a differential equation can be written

as2

2, , , ,...,

n

n

dy d y d yf x y

dx dx dx

. There are two types of differential equations: ordinary

differential equations and partial differential equations. Ordinary differentialequations involve only one independent variable, while the partial differentialequations involve the partial derivatives with respect to two or more independentvariables. In this unit, you will learn about only ordinary differential equations,including the necessary conditions for an ordinary differential equation to be exact.

The first order differential equations are used to formulate the problem offinding the other parameter family of curves, which is orthogonal to the givenparameter family of curves.

In this unit, you will study about the linear differential equations and equationsreducible to the linear form, exact differential equations, first order higher degreeequations solvable for x, y and p, Clairaut’s equation and singular solutions,geometrical meaning of a differential equation, and orthogonal trajectories.

4.1 OBJECTIVES


Give the geometrical meaning of differential equations

Know the necessary and sufficient condition required for an ordinarydifferential equation to be exact, i.e., exact differential equations

NOTES



Find the solution of a given differential equation

Elaborate on the linear differential equations and equations reducible to thelinear form

Find integrating factor of an equation by inspection method and using variousrules

Define differential equation of first order and higher degree

Solve the first order higher degree differential equations solvable for x, yand p

Understand Clairaut’s equation and singular solutions

Find the singular solution of a first order higher degree differential equation

Define p-discriminant and c-discriminant forms

Understand the meaning of trajectory and orthogonal trajectories

Derive and apply differential equations of orthogonal trajectories in Cartesianand polar co-ordinates

4.2 LINEAR DIFFERENTIAL EQUATIONS

In mathematics, a Differential Equation (DE) is defined as an equation of theform that interconnects certain function with its derivatives, where usually the functionrepresents the physical quantity while the derivatives denote their rates of changeand the relationship between the two is defined by the equation. Fundamentally,the ‘solutions of differential equations’ are functions which precisely ‘representthe relationship or correlation between a continuously varying or fluctuating quantityand its rate of change’.

A Differential Equation (DE) comprises of one or more expressions includingderivatives of one dependent variable ‘y’ with reference to another independentvariable ‘x’, such as

2dy

xdx

An ordinary differential equation is a differential equation that includes afunction of a single variable and some of its derivatives, such as

22 3 5dy

x xdx

Principally, a differential equation is an equation for a function that relatesthe values of the function to the values of its derivatives. Therefore, a differentialequation is an equation between specified derivative of an unknown function, itsvalues and known quantities and functions or a Differential Equation (DE) is anequation that comprises of a function and its derivatives. Differential equations arecategorized as Partial Differential Equations (PDE) or Ordinary DifferentialEquations (ODE) in accordance with whether or not they hold partial derivatives,


NOTES


while the order of a differential equation is defined on the basis of the highest orderderivative that occurs in the equation.

In mathematics, a linear differential equation is a differential equationthat is defined by a linear polynomial in the unknown function and its derivatives,that is an equation of the form,

0 1 2 0 nna x y a x y a x y a x y b x

where a0(x), ..., a

n(x) and b(x) are arbitrary differentiable functions that do not

essential to be linear, and y, …, y (n) are considered as the successive derivativesof an unknown function y of the variable x.

This is an Ordinary Differential Equation (ODE). A linear differential equationmay also be a Linear Partial Differential Equation (Linear PDE), if the unknownfunction depends on several variables, and the derivatives that appear in the equationare partial derivatives.

If a linear differential equation or a system of linear equations are such thatthe associated homogeneous equations have constant coefficients then these maybe solved by means of quadrature mathematics, i.e., the solutions may be expressedin terms of integrals. This is also true for a linear equation of order one with non-constant coefficients.

4.2.1 Geometrical Meaning of a Differential Equation

Let a differential equation be ( , )dy

f x ydx

. Then the slope of the tangent to the

curve at the point (x, y) is given bydy

dx.

A1

A2

A3

A4

A5

Y

X0

Fig. 4.1 Geometrical Differential Equation

As shown in Figure 4.1, take a point 1 1 1A ( , )x y in the xy-plane. If m1 is the

slope of the tangent to the curve at the point (x1, y

1), then at this point

dy

dxis equal

to m1. Suppose the point moves in the direction of m

1 from (x

1, y

1) to A

2(x

2, y

2)

for an infinitesimal distance and the slope of the tangent at this new point is givenby m

2. Again, the point moves from A

2(x

2, y

2) for an infinitesimal distance in the

direction of m2 to the point A

3(x

3, y

3) and the slope of the tangent at this new point

is given by m3.

NOTES



Choosing the successive points A1, A

2, A

3, … near to the one another

makes the broken curve approximate to a smooth curve ( )y x which is

associated with the initial point 1 1 1A ( , )x y . The slope of the tangent to the curve at

a point and the co-ordinate of that point always satisfy the equation ( , )dy

f x ydx

.

On choosing different initial points, different curves with the same property

will be obtained. Thus, we can say that the differential equation ( , )dy

f x ydx

represents a family of curves such that through each point of the xy-plane, therepasses one curve of the family.

4.2.2 Solving Linear Differential Equations

The linear differential equation is a differential equation in which the dependentvariable and all its derivatives appear only in the first degree and are not multipliedtogether.

A linear differential equation of order n is of the form

1 2

0 1 21 2P P P ....... P X

n n n

nn n n

d y d y d yy

dx dx dx

… (1)

where, P0, P

1, P

2,….., P

n and X are either the functions of x or constants.

When in Equation (1), P0, P

1, P

2,…, P

n are all constants and not the function

of x while X is the some function of x, the equation is called a linear differentialequation with constant co-efficients of nth order.

Differential Operator ‘D’

Sometimes, it is easier to write a linear differential equation in the simple form by

replacing the part d

dx of the symbol

dy

dxwith D. This D or

d

dxis regarded as

differential operator.

So, Equation (1) can be written in terms of differential operator D as

1 20 1 2P D P D P D ....... P Xn n n

n y

Or f(D)y = X where f(D) = 1 20 1 2P D P D P D ....... Pn n n

n .

Solution of Linear Differential Equations

Theorem 4.1: If y = y1, y = y

2, …, y = y

n be the linear independent solutions of

1 20 1 2D D D ....... 0n n n

na a a a y , then 1 1 2 2 .... n ny c y c y c y

where c1, c

2, c

3, ...,c

n are arbitrary constants, is the general or complete solution

of the differential Equation (1).

Proof: Given, 1 20 1 2D D D ....... 0n n n

na y a y a y a y … (2)


NOTES


As y = y1, y = y

2,…, y = y

n are solutions of Equation (2),

1 20 1 1 1 2 1 1D D D ....... 0n n n

na y a y a y a y 1 2

0 2 1 2 2 2 2D D D ....... 0n n nna y a y a y a y

………………………………………………… …………………………………………………… ……………………………………………………

1 20 1 2D D D ....... 0n n n

n n n n na y a y a y a y

1 1 2 2 .... n ny c y c y c y

… (3)

Now, by putting 1 1 2 2 .... n ny c y c y c y in Equation (2), we get

10 1 1 2 2 1 1 1 2 2

1 1 2 2

D ... D ...

.................................... ... 0

n nn n n n

n n n

a c y c y c y a c y c y c y

a c y c y c y

1 11 0 1 1 1 1 2 0 2 1 2 2

10 1

D D ..... D D .....

.................................... D D ..... 0

n n n nn n

n nn n n n n

c a y a y a y c a y a y a y

c a y a y a y

Substituting the values from Equation (3) in the above equation, we get

c1.0 + c

2.0 + ... + c

n.0 = 0 0 = 0

Thus, 1 1 2 2 .... n ny c y c y c y satisfies the Equation (2), which istherefore its solution.

Auxiliary Equation

Consider 1 20 1 2D D D ....... 0n n n

na y a y a y a y … (4)

where, a0, a

1, a

2, ...,a

n are all constants.

Now, let one of the solution of Equation (4) be mxy e .

Then, D mxy me , 2 2D mxy m e ,…, Dn n mxy m e

By putting all these values including mxy e in Equation (4), we get

1 20 1 2 ....... 0n n n mx

na m a m a m a e

Since, 0mxe for any m, therefore,

1 20 1 2 ....... 0n n n

na m a m a m a … (5)

This equation in m is known as auxiliary equation of the given differentialEquation (4).

If m is a solution of Equation (5), then emx will be a solution of Equation (4).To solve the Equation (4), write its auxiliary equation and then solve it first.

NOTES



Note: The values of D obtained from

1 20 1 2(D) D D D ....... 0n n n

nf a a a a … (6)

is same as the values of m obtained from Equation (5). Hence, Equation (6), ingeneral can be regarded as the auxiliary equation which is obtained by equatingthe symbolic coefficient of y in Equation (4) to zero. Thus, in practice we do notreplace D by m to form an auxiliary equation.

The roots of the auxiliary equation can be as follows:

Real and Different

Real and Repeated

Complex

Case I: When all the roots of Equation (5) are real and different.

Let 1 2 3, , , ....... nm m m m be the n different and real roots of Equation (5).Then

31 2, , ,...,m x m xm x m x ne e e e will be the n distinct solutions of Equation (4).

Thus, the general solution of Equation (4) is

31 21 2 3 .....m x m xm x m x n

ny c e c e c e c e … (7)

where c1, c

2,…, c

n are arbitrary constants.

Case II: When all the roots of Equation (5) are real, two of them are repeatedand all others are different.

Let 1m and 2m be the two equal roots of the auxiliary equation.

Then, Equation (7) becomes 311 2 3 .....m x m xm x n

ny c c e c e c e

which has only (n – 1) arbitrary constants and so it is no longer a generalsolution.

Consider a differential equation of second order

1 1D D 0m m y …(8)

whose auxiliary equation 21D 0m has equal roots.

Now, putting 1D m y = v in the Equation (8), we get

1D 0m v 1dv

m vdx

1dv

m dxv

On integrating both sides, we get

1log logv m x c

1m xv ce 11D m xm y ce 1

1m xdy

m y cedx


NOTES


This is a linear equation of first order having I.F. as 1 1m dx m xe e

Its solution is 1 1. .m x m x mxy e ce e dx c

1. m xy e cdx c 1. m xy e cx c 1. m xy cx c e 1

1 2m xy c c x e

Thus, the general or complete solution of equation (4) is

311 2 3 .....m x m xm x n

ny c c x e c e c e

Similarly, we can prove that if r roots of auxiliary equation are equal, then

the general solution would be 2 1 11 2 3 ..... ..... m xm xr n

r ny c c x c x c x e c e

Case III: When two roots of Equation (5) are complex and others are real anddifferent.

Let a pair of complex roots of the auxiliary Equation (5) be i and other

real and different roots be 3 4 5, , , ....... nm m m m . Then, the complete solution ofEquation (4) is

( ) ( ) 31 2 3 .....m x m xi x i x n

ny c e c e c e c e

31 2 3 .....m x m xx i x i x n

ne c e c e c e c e

31 2 3(cos sin ) (cos sin ) .....m x m xx n

ne c x i x c x i x c e c e

cos sinie i

31 2 1 2 3( )cos ( ) sin .....m x m xx n

ne c c x i c c x c e c e

33A cos Bsin .....m x m xx n

ne x x c e c e

Corollary: If the complex roots pair i occur twice, the above equation reducesto

31 2 3 4 3( )cos ( )sin ...m x m xx n

ny e c c x x c c x x c e c e

Example 4.1: Solve the differential equation 4

44

0d y

a ydx

.

Solution: The given differential equation is4

44

0d y

a ydx

.

Symbolic form of the equation is 4 4D 0a y

Its auxiliary equation is 4 4D 0a

22 2 2 2D 2D 0a a

NOTES



222 2D 2D 0a a

2 2 2 2D 2D D 2D 0a a a a

D ,2 2 2 2

a a a ai i

Since, there are two pairs of complex roots, thus the general solution is

2 21 2 3 4cos sin cos sin

2 2 2 2

a ax xa a a a

y e c x c x e c x c x

2 21 2 3 4cos cos

2 2

a ax xa a

y e c x c e c x c

Example 4.2: If 1 2, are real and distinct roots of the auxiliary equation2

1 2 0a a and 1 21 2,x xy e y e , are its solution, then prove that

1 1 2 2y c y c y is a solution of 2

1 220

d y dya a y

dx dx . Is this solution general?

Solution: The given auxiliary equation is 21 2 0a a

21 2D D 0a a ...(i)

[On replacing by D]

Since the real and distinct roots of the above equation are 1 and 2 .

11

xy e and 22

xy e are two independent solution of

21 2D D 0a a y

21 2D D 0y a y a y ...(ii)

2

21 1 1 2 1

2 1 2 2 2

D D 0

D D 0

y a y a y

y a y a y

...(iii)

On putting 1 1 2 2y c y c y in Equation (ii), we have

21 1 2 2 1 1 1 2 2 2 1 1 2 2D D 0c y c y a c y c y a c y c y

2 21 1 1 1 2 1 2 2 1 2 2 2D D D D 0c y a y a y c y a y a y

1 20 0 0c c which is true. [Using Equation (iii)]

1 1 2 2y c y c y is the solution of Equation (ii) or2

1 220

d y dya a y

dx dx .


NOTES


As the solution has two arbitrary constants and the order of the given equationis also 2, thus this solution is general.

Complementary Function and Particular Integral

Theorem 4.2: If the complete solution of the equation

1 20 1 2P D P D P D ....... P 0n n n

n y … (9)

is y = Y and the particular solution (containing no arbitrary constants) of the equation

1 20 1 2P D P D P D ....... P Xn n n

n y … (10)

where X is a function of x is y = u, then y = Y + u will be the complete solution ofthe Equation (10).

Proof: By putting y = Y + u in Equation (10), we get

1 20 1 2P D P D P D ....... P Y Xn n n

n u

1 20 1 2P D Y P D Y P D Y ....... P Y Xn n n

nu u u u

1 2 1 20 1 2 0 1 2P D P D P D ....... P Y + P D P D P D ....... P Xn n n n n n

n n u ...(11)

As y = Y is the solution of Equation (9), we have

1 20 1 2P D P D P D ....... P Yn n n

n = 0 … (12)

Also, since y = u is the solution of Equation (10), we have

1 20 1 2P D P D P D ....... P Xn n n

n u … (13)

By putting the values from Equations (12) and (13) into equation (11),we get

0 + X = X

X = X, which is true.

Thus, y = Y + u is the complete or general solution of Equation (10).

The expression Y is called Complementary Function (C.F.) and u is knownas Particular Integral (P.I.).

Thus, Complete Solution = Complementary Function + Particular Integral

Inverse Operator

The operator 1

(D)f is known as the inverse operator of the operator (D)f , if

1X

(D)y

f gives the solution of the equation (D) Xf y .

NOTES



If 1

X(D)

yf

does not contain any arbitrary constant, then it is called the

particular solution of the equation (D) Xf y . Thus, 1

X(D)f , a function of x,

free from constants, when operated on f(D) gives X.

Some Theorems for Finding Particular Integrals

1

X(D)f

is the particular integral of (D) Xf y .

1X .X

Dax axe e dx

a

, no arbitrary constant being added.

1.

!D

nax ax

n

xe e

na

The particular integral of (D) Xf y is

1 1 2 21 2A . .X A .X ....... A .Xx xx x x x n n

ne e dx e e dx e e dx .

Some Special Cases in Particular Integral

Case I: Prove that 1 1

. .(D) ( )

ax axe ef f a

provided that f(a) 0, where a is a

constant.

Case of Failure

If f(a) = 0, then 1 1

. .(D) (D)

D

ax axe x edf f

d

Example 4.3: Solve the differential equation2

32

6 25 104 xd y dyy e

dx dx .


32

6 25 104 xd y dyy e

dx dx .

Symbolic form of the equation is 2 3D 6D 25 104 xy e

Its auxiliary equation is 2D 6D 25 0

6 36 100

D2

6 64 6 8

D 3 42 2

ii

C.F. 31 2cos 4 sin 4xe c x c x


NOTES


And P.I. 3 32

1 1.104 .104

D 6D 25 9 18 25x xe e

[ D = 3]

3

31042

52

xxe

e

Thus, the complete solution is C.F. + P.I.y

3 31 2cos 4 sin 4 2x xy e c x c x e


216 sec 4

d yy x

dx

Solution: The given differential equation is 2

216 sec 4

d yy x

dx

Symbolic form of the equation is 2D 16 sec4y x

Its auxiliary equation is 2D 16 0

D = 4i

C.F. 1 2cos4 sin 4c x c x

And P.I. 2

1sec 4

D 16x

1 1 1

sec 48 D 4 D 4

xi i i

4 4 4 41sec 4 sec 4

8i x ix ix ixe e xdx e e x dx

i

4 411 tan 4 1 tan 4

8ix ixe i x dx e i x dx

i

4 4 4 41 1. log sec 4

8 4ix ix ix ixe e x i e e x

i

1 1

sin 4 cos 4 log sec 44 4

x x x x

1 1sin 4 cos 4 log cos 4

4 16x x x x

Thus, the required solution is

1 2

1 1cos4 sin 4 sin 4 cos 4 log cos4

4 16y c x c x x x x x

Case II: Prove that 2 2

1 1sin sin

Dax ax

f f a

if 2 0.f a

NOTES



Case of Failure: If 2 0f a , then

(i) 2

2

1 1sin . sin

D DD

ax x axdf f

d

(ii) 2

2

1 1cos . cos

D DD

ax x axdf f

d


2sin 2

d y dyy x

dxdx .


2sin 2

d y dyy x

dxdx

Symbolic form of the equation is 2D D 1 sin 2y x

Its auxiliary equation is 2D D 1 0

1 1 4 1 3

D=2 2 2

i

C.F. / 2

1 2

3 3cos sin

2 2xe c x c x

And P.I. 2

1sin 2

D D 1x

2

1sin 2

2 D 1x

2 2D 2

1

sin 2D 3

x

2

D 3 D 3sin 2 sin 2

D 3 D 3 D 9x x

2

D 3sin 2

2 9x

2 2D 2

1D 3 sin 2

13x

1Dsin 2 3sin 2

13x x

1

sin 2 3sin 213

dx x

dx

12cos2 3sin 2

13x x


/ 21 2

3 3 1cos sin 2cos 2 3sin 2

2 2 13xy e c x c x x x


NOTES


Case III: To calculate1

(D)mx

f , where m is a positive integer..

To calculate the P.I.1

(D)mx

f , follow the steps given below:

Step 1: Take out the lowest degree term from f(D) such that the remaining

factor will be of the form 1 (D) .

Step 2: Take 1 (D) in the numerator with negative index and expand it in

ascending powers of D by Binomial Theorem up to the terms Dm.

Step 3: Operate on xm by each term of the expansion.

Note: Do not write the terms of the expansion which contains powers of D greaterthan m because Dm+1 when multiplied by xm will be zero.


22

4 4 cos2xd y dyy x e x

dx dx


22

4 4 cos2xd y dyy x e x

dx dx

Symbolic form of the equation is 2 2D 4D 4 cos2xy x e x


2D 2 0 D 2,2

C.F. 21 2

xc c x e

And P.I. 22

1cos2

D 4D 4xx e x

22 2 2

1 1 1cos2

D 4D 4 D 4D 4 D 4D 4xx e x

22 22

1 1 1cos2

1 4.1 4 2 4D 4D4 1 D

4

xx e x

12

21 D 1 11 D . cos 2

4 4 4 Dxx e x

22 221 D D 1

1 D D ... cos24 4 4 4

xx e x dx

NOTES



2

2 21 D 1 sin 21 D D ... .

4 4 4 2x x

x e

2 21 3 1

1 D D ... .sin 24 4 8

xx e x

2 2 2 21 3 1D D .sin 2

4 4 8xx x x e x

21 3 1

2 sin 24 2 8

xx x e x


2 21 2

1 3 12 sin 2 .

4 2 8x xy c c x e x x e x

Case IV: 1 1

( )( ) ( )

ax axe V e Vf D f D a

, where V is a function of x.


2 32

2 . .cos 2x xd yy x e e x

dx .


2 32

2 . .cos 2x xd yy x e e x

dx

Symbolic form of the equation is 2 2 3D 2 cos2x xy x e e x

Its auxiliary equation is 2D 2 0

D 2 0 2i i

C.F. 01 2 1 2cos 2 sin 2 cos 2 sin 2xe c x c x c x c x

And P.I. 2 32

1cos2

D 2x xx e e x

3 22 2

1 1cos 2

D 2 D 2x xe x e x

3 2

2 2

1 1. . cos 2

D 3 2 D 1 2x xe x e x

3 22 2

1 1. . cos 2

D 6D 11 D 2D 3x xe x e x

3 2

2

1 1. cos 2

4 2D 36 D11 1 D

11 11

x xe x e x

123 21 6 D 1

1 D . cos211 11 11 2D 1

x xe x e x


NOTES


22 23 21 6 D 6 D 2D 1

1 D D ..... . . cos211 11 11 11 11 2D 1 2D 1

x xe x e x

2

3 2 22

1 6 D 36 2D 11 D D ..... . cos 2

11 11 11 121 4D 1x xe x e x

3 2 2 2D 11 6 251 D D ..... . .cos 2

11 11 121 4 4 1x xe x e x

3 2 2 2 21 6 25D D 2D 1 cos2

11 11 121 17

xx e

e x x x x

3 21 6 25.2 .2 2D cos2 cos2

11 11 121 17

xx e

e x x x x

3 21 12 50 1.2 2sin 2 cos2

11 11 121 17

xx e

e x x x x

3 21 12 50 14sin 2 cos2

11 11 121 17x xe x x e x x

3 21 12 50 14sin 2 cos2

11 11 121 17x xe x x e x x

Thus, the required solution is:

3 21 2

1 12 50 1cos 2 sin 2 4sin 2 cos2

11 11 121 17x xy c x c x e x x e x x

Case V: To calculate 1

. VD

xf

, where V is any function of x.

Or prove that 1 1 1

. V . .V .VD D D D

dx x

f f d f


22 sin .xd y dy

y xe xdx dx


22 sin .xd y dy

y xe xdx dx

Symbolic form of the given equation is 2D 2D 1 sinxy xe x


2D 1 0 D 1,1

C.F. = 1 2xc c x e

And P.I. 2

1sin

D 2D 1xxe x

NOTES



2

1sin

D 1xxe x

2

1. sin

D 1 1xe x x

2

1. sinD

xe x x

2 2

1 1. sin sinD D D

x de x x x

d

2 3

1 2. sin .sinD D

xe x x x

2

1 2 1. sin .sin

1 D Dxe x x x

2 1

sin . sinD 1

xe x x x

2

sin sinD

xe x x x

sin 2 sine x x xdx sin 2 cos sin 2cosx xe x x x e x x x

Thus, the required solution is 1 2 sin 2cos .x xy c c x e e x x x

Check Your Progress

1. Solve the differential equation (2D3 – 7D2 + 7D – 2)y = 0.

2. Solve the differential equation4 3 2

4 3 24 8 8 4 0

d y d y d y dyy

dxdx dx dx .

3. Solve the differential equation 2

22

secd y

y xdx

.

4. Solve the differential equation .3 2

23 2

5 7 3 coshxd y d y dyy e x

dxdx dx


2sin sin 2

d yy x x

dx .

6. Solve 2 2D 2 cos D cosn n x a nt , given that x = 0, 0dx

dt

when t = 0.

7. Solve3 2

3 22 3 cos

d y d yy x

dx dx .

8. Solve the differential equation 4 3 2 2D D D y x a bx .


NOTES


Check Your Progress

9. Solve the differential equation4 3 2

2 24 3 2

2 3 3 4sinxd y d y d yx e x

dx dx dx .


2 32

2 xd y dyy x e

dx dx .

11. Solve the differential equation 3D 2D 4 cosxy e x .


24 sin

d yy x x

dx .

13. Find the particular integral of 4 2 2D 2D 1 cosy x x .

4.3 EQUATIONS REDUCIBLE TO THE LINEARFORM

An equation of the form

1

10 1 11

P ( ) P ( ) ... P ( ) P ( )n n

n nn nn n

d y d y dya bx a bx a bx y f x

dx dx dx

…(14)

where a, b, P0, P

1, P

2, …, P

n are constants and f(x) is either a constant or

a function of x.

This equation can be reduced to a homogeneous linear equation.

Taking a + bx = u (say)

Differentiating with respect to x, we get du

bdx

Now, we have

.dy dy du

dx du dx .

dyb

du

2

2

d y d dy

dx dx dx

.dyd

bdx du

2

2. .d y du

bdxdu

2

2

2

d yb

du

……………………………….n n

nn n

d y d yb

dx dx

NOTES



Substituting these values in Equation (14), we get1

1 10 1 11

P P ... P Pn n

n n n nn nn n

d y d y dy u ab u b u bu y f

dz bdz dz

…(15)

which is a homogeneous linear equation.

Now, putting (a + bx) = ez z = log(a + bx) and Dd

dz in the Equation

(15), we get

10 1 1P D(D 1)...(D 1) P D(D 1)...(D 2) ... P D Pn n

n nb n b n b y

ze af

b

…(16)

which is a linear equation with constant coefficients and can be solved for y interms of z by the standard methods.

Suppose the solution of Equation (16) is y = F(z). Then, the solution ofEquation (14) can be obtained as y = f[log (a + bx)] by putting z = log(a + bx).

Example 4.9: Solve the linear differential

equation 2

2

21 1 2 3 2 4

d y dyx x x x

dx dx .

Solution: Putting 1 zx e log 1z x and = Dd

dz in the given equation,

we get

D D 1 D 2 1 2 2z zy e e 2D 2 1 2 2z zy e e

Then the A.E. is 2D 0 D 0,0

C.F. 01 2 1 2. zc c z e c c z

And P.I. 22 2

1 12 1 2 2 4 6 2

D Dz z z ze e e e

214 6 2

Dz ze e dz

1mean integration with respect to

Dz

21 4

6 2D 2

zze

e z

212 6 2

Dz ze e z

22 6 2z ze e z dz

2 2

2. 6 2.2 2

zze z

e 2 26z ze e z


NOTES


Thus, the solution of the given equation is y = C.F. + P.I.

2 21 2 6z zy c c z e e z

22

1 2 log 1 1 6 1 log 1y c c x x x x

1 log 1ze x z x

221 2 .log 1 8 log 1y c c x x x x

[Constant term 7 can be considered to be included in 1c ]

Check Your Progress

14. Which equation form can be reduced to the homogeneous linear equation?

4.4 EXACT DIFFERENTIAL EQUATIONS

A differential equation which can be obtained from its solution ,f x y c directly

by differentiation without any further multiplication, elimination is called an exactdifferential equation.

We can define the exact differential equations in another way also.

The differential equation M N 0dx dy , where M and N are the functions

of x and y is called exact if there exist a function F ,x y such that M N Fdx dy d

(i.e., M N dx dy is an exact differential of a function of x and y) and F(x, y) = cis its solution.

The following differential equations are exact:

(i) 0xdy ydx because xdy + ydx = d(xy) and xy = c

(ii) 20

xdy ydx

x

because 2

xdy ydx yd

x x

and y

cx

(iii) sin cos cos sin 0x ydy x ydx because sin x cosydy+cos x sin ydx=d

(sin x sin y) and sin sinx y = c

Theorem 4.3: The necessary and sufficient condition for the ordinary differential

equation M N 0dx dy to be exact is M N

y x

.

Proof: Necessary Condition

Let the given equation M N 0dx dy be exact.

Then by definition there exists a function F ,x y such that

M N Fdx dy d …(17)

NOTES



Also, from differential calculus, we know that

F FFd dx dy

x y

…(18)

Comparing Equations (17) and (18), we get

FM

x

and

FN

y

2M F F

y y x y x

and

2N F F

x x y x y

As,2 2F F

y x x y

, thusM N

y x

which is the required condition for the equation M N 0dx dy to be exact.

Sufficient Condition

Let M N

y x

. Then, we have to show that M N 0dx dy is an exact differential

equation.

Let M ,dx x y …(19)

where M ,x y is integrated with respect to x taking y as a constant such that

Mdxx x

Mx

…(20)

2M

y y x y x

2N

x y x

M N

y x

Also,2 2

y x x y

2N

x x y x y

Integrating with respect to x, taking y as a constant, we get

N g yy

…(21)

where g y is a function of y.

Using Equations (20) and (21), we get

M Ndx dy dx dy g y dyx y

d d g y dy

d g y dy

which is an exact differential.


NOTES


Solving an Exact Differential Equation

If M N 0dx dy is an exact differential equation, then from the above section,we have

0d g y dy

Integrating, we get

g y dy c , where c is an arbitrary constant …(22)

From Equation (19), we have

constant

M

y

dx Putting the value of in Equation (21), we get

constant

M

y

dx g y dy c

constant

M (Terms of N not conatining )

y

dx x dy c which is the required solution.

Example 4.10: Solve the differential

equation 4 2 4 2 32 2 4 sin 0x xy y dx x y xy y dy .

Solution: The given equation is

4 2 4 2 32 2 4 sin 0x xy y dx x y xy y dy …(i)

We know that M N 0dx dy …(ii)

Comparing Equations (i) and (ii), we get 4 2 4M 2x xy y and2 3N 2 4 sinx y xy y

Now,3M

4 4xy yy

and 3N4 4xy y

x

Since M N

y x

, thus the Equation (i) is exact and its solution is given by

constant

M Terms of N not containing

y

dx x dy c 4 2 4

constant

2 sin

y

x xy y dx y dy c

5 22 42 . . cos

5 2

x xy y x y c

52 2 4 cos

5

xx y xy y c

which is the required solution.

NOTES



Integrating Factors

If the left side of the equation M N 0dx dy is not an exact but it can be made

exact on multiplying it by a function ,x y , then ,x y is called the IntegratingFactor (IF).

Theorem 4.4: The number of integrating factors of the differentialequation M N 0dx dy is infinite.

Proof: The given differential equation is M N 0dx dy … (23)

Let ,x y be its integrating factor. Then, M N 0dx dy …(24)

is an exact differential equation.

Suppose Equation (24) is an exact differential of some function F(x, y) (say).

M N Fdx dy d

Multiplying by any function of F, say F , we get

F M N F Fdx dy d …(25)

But the right side of Equation (25) is an exact differential.

F M Ndx dy is also an exact differential.

Thus, F is also an integrating factor of differential Equation (23).

As F is an arbitrary function of F, thus the equation has infinite number of

integrating factors.

Sometimes the integrating factor of the equation M N 0dx dy , which isnot an exact can be found easily just by inspection after recognising certain groupterms in the given equation. Table 4.1 shows the list of some integrating factors ofcertain group of terms, which are parts of an exact differential equation.

Table 4.1 List of Some Integrating Factors

Group of Terms Integrating Factors Exact Differential xdy ydx xdy ydx xdy ydx xdy ydx xdy ydx

2

1

x

2

1

y

1

xy

2 2

1

x y

2 2

1

x y

2

xdy ydx yd

x x

2

xdy ydx xd

y y

1 1log

xdy ydx ydy dx d

xy y x x

1 12 2

tan tanxdy ydx y x

d dx y x y

2 2

1log

2

xdy ydx x yd

x y x y

xdy ydx

1

nxy

1 1log

xdy ydxdy dx d xy

xy y x

, if n =1

1

1

1n n

xdy ydxd

xy n xy

, if n ? – 1

xdx ydy

2 2

1n

x y 2 2

2 2

1log

2

xdx ydyd x y

x y

, if n =1

12 2 2 2

1

2 1n n

xdx ydyd

x y n x y

, if n ? – 1


NOTES


Example 4.11: Solve the differential equation 2 2 2 2 0x y x dx y dy .

Solution: The given equation is 2 2 2 2 0x y x dx y dy …(i)


Comparing Equations (i) and (ii), we get 2 2M 2x y x and N 2y

Now,M

2yy

and

N0

x

SinceM N

y x

, thus the differential Equation (i) is not exact.

By inspection, we take xe as integrating factor of Equation (i).

Multiplying Equation (i) by xe , we get

2 2 2 2 0x xe x y x dx ye dy …(iii)

Comparing Equations (iii) and (ii), we get

2 2M 2xe x y x and N 2 xye

Now, M

2 xyey

and

N2 xye

x

Since M N

y x

, thus Equation (iii) is exact and its solution is given by

constant


y

dx x dy c 2 2

constant

2 0x

y

e x y x dx c 2 2

constant constant

2x x

y y

e x y dx xe dx c 2 2

constant constant

( ) 2 2x x x

y y

x y e xe dx xe dx c [Integrating by parts]

2 2( ) xx y e c which is the required solution.

Rules for Finding the Integrating Factors

Sometimes, it is not easy to find integrating factor of a differential equation usinginspection method. In that case, we have to search for a rule which help us indetermining the integrating factor for a given differential equation so it becomesexact differential equation. Here, we are going to learn some rules for determiningthe integrating factor for a given differential equation.

NOTES



Rule 1: If M N 0x y and the equation M N 0dx dy is homogeneous, then

1

M Nx y is its integrating factor..

Example 4.12: Solve 2 2 2 32 3 0x y xy dx x y x dy .

Solution: The given equation is 2 2 2 32 3 0x y xy dx x y x dy …(i)


Comparing Equations (i) and (ii), we get 2 2M 2x y xy and 2 3N 3x y x

Now, 2M4x xy

y

and 2N

6 3xy xx

SinceM N

y x

, thus the differential Equation (i) is not exact. Here Equation (i) is

homogeneous equation.

Also, 2 2 2 3M N 2 3x y x y xy x x y x y = 2 2x y ‘“ 0

2 2

1 1

M Nx y x y

is an integrating factor of Equation (i).

Now, multiplying Equation (i) by 2 2

1

x y , we get

2

1 2 30

xdx dy

y x y y

…(iii)

which is an exact differential equation and its solution is given by

constant


y

dx x dy c

constant

1 2 3

y

dx dy cy x y

2log 3logx

x y cy

3

2log

x yc

y x which is the required solution.

Rule 2: If M N 0x y and the equation M N 0dx dy is of the form

0f xy ydx g xy xdy , then 1

M Nx y is its integrating factor..

Rule 3: If

M N

N

y xf x

is a function of x alone, then the integrating factor of

the equation M N 0dx dy is f x dxe .


NOTES


Rule 4: If

M N

M

y xf y

is a function of y alone in the equation M N 0dx dy ,

then f y dy

e is an integrating factor .

Rule 5: If the equation M N 0dx dy can be written as

0x y mydx nxdy x y pydx qxdy

where , , , , , , andm n p q are all constants, then its integrating factor is a bx y .

Check Your Progress

15. Solve the differential equation 2 2 22 ( ) 0a xy y dx x y dy .


2 2

( )a xdy ydxxdx ydy

x y

.

17. Find the integrating factor of the equation by inspection and hence solve3 33 0ydx xdy x y dy .

18. Find the integrating factor of the equation by inspection and hence solve 2 2 2 0x y dx xydy .

19. Solve the differential equation 2 3 3 0x ydx x y dy .

20. Solve the differential equation 2 2 2 21 1 0x y xy ydx x y xy xdy .

21. Solve the differential equation 2 2 2 2 0x y x dx ydy .

22. Solve the differential equation 3 2 2 42 0xy y dx x y x y dy .

23. Solve the differential equation 2 2 32 2 0y x y dx x xy dy .

4.5 FIRST ORDER HIGHER DEGREEEQUATIONS SOLVABLE FOR X, Y AND P

The differential equation of first order and nth degree is generally represented by1 2 3

1 2 3 1A A A A A 0n n n nn np p p p p

where p denotes dy

dxand 1 2 3 1A , A , A , , A , An n are functions of x and y.

The above equation can also be written as , , 0f x y p … (26)

The differential Equation (26) can be classified into three categories: Equationssolvable for p, i.e., of the form ,p f x y , Equations solvable for y, i.e., of the

form ,y f x p and Equations solvable for x, i.e., of the form ,x f y p .

NOTES



Equation Solvable for p

Let the equation 1 2 31 2 3 1A A A A A 0n n n n

n np p p p p … (27)

be solvable for ‘p’ and can be written in form

1 2 3F , F , F , ... F , 0np x y p x y p x y p x y … (28)

1 2F , 0, F , 0,p x y p x y

3F , 0,p x y ... , F , 0np x y

1 2 3F , , F , , F , , , F ,n

dy dy dy dyx y x y x y x y

dx dx dx dx

which are the equations of first order and first degree.

Let the solutions of these equations be

1 1 2 2 3 3, , 0, , , 0, , , 0,f x y c f x y c f x y c , , , 0n nf x y c

where, 1 2 3, , , , nc c c c are arbitrary constants.

Then, the most general solution of the differential Equation (27) is

1 1 2 2 3 3, , , , , , , , 0n nf x y c f x y c f x y c f x y c

As the Equation (27) is of first order thus, its general solution can contain only onearbitrary constant. Hence, without the loss of generality we can take

1 2 3 nc c c c c (say)

Thus, the general solution of the Equation (27) can be written as

1 2 3, , , , , , , , 0nf x y c f x y c f x y c f x y c


2 2 22 2 0dy dy

x xy y xdx dx

Solution: The given equation is 2

2 2 22 2 0dy dy

x xy y xdx dx

… (i)

By putting dy

pdx

in Equation (i), we get 2 2 2 22 2 0x p xyp y x … (ii)

Solving Equation (ii), we get

2 2 2 2 2

2

2 4 4 2

2

xy x y x y xp

x

2 2 2 2 4

2

2 4 8 4

2

xy x y x y x

x

4 2 2

2

2 4 4

2

xy x x y

x


NOTES


2 2 2

2

2 4

2

xy x x y

x

2 2 2 2

2

2 2

2

xy x x y y x y

x x

2 2y x ydy

dx x

…(iii)

which is a homogeneous equation in x and y.Let y vxDifferentiating both sides with respect to x, we get

dy dvv x

dx dx

By putting the values of y and dy

dxin Equation (iii), we get

2 2 2dv vx x v x

v xdx x

21dv

v x v vdx

21dv

x vdx

Separating the variables, we get

21

dv dx

xv


21

dv dxc

xv

1sin log logv x c 1sin logv xc 1sin log ,y

cxx


Equation Solvable for y

A differential Equation is said to be solvable y, if it can be expressed as

,y f x p … (29)

Differentiating both sides with respect to x, we get

F , ,dy dp

p x pdx dx

… (30)

The differential Equation (ii) is in two variables p and x.

Let the solution of the Equation (ii) be

, , 0x p c … (31)

So, by eliminating p from Equation (29) and Equation (31), we get thesolution of differential Equation (29). If it is not easier to eliminate p, then putEquations (29) and (31) in the form

1 2, , ,x p c y p c

NOTES



In this case these two equations together constitute the solution of Equation (29).

Note: When the Equation (30) can be written as 1 2( , ). , ,dp

f x p f x pdx

= 0, neglect

the factor which does not contain dp

dx, i.e., 1( , )f x p since if 1( , )f x p = 0, then

elimination of p from this equation and Equation (29) does not give a generalsolution.

Example 4.14: Solve the differential equation 3 log .y x p

Solution: The given equation is 3 logy x p … (i)


13

dy dp

dx p dx

13

dpp

p dx

dyp

dx

3

dpdx

p p

1 1 1

3 3dx dp

p p

(By Partial Fraction)


1

1log 3 log log

3x p p c

2

33 log log

px c

p

2

33 log .

px c

p

Taking exponent of both sides, we get

32

3.x p

e cp

33 xpce

p

33

1 xcep

3

3

1 xp

ce

Substituting this value of p in Equation (i), we get 3

33 log

1 xy x

ce


Equation Solvable for x

A differential equation is said to be solvable for x, if x can be expressed as

,x f y p … (32)

Differentiating both sides with respect to y, we get 1

F , ,dx dp

y pdy p dy

… (33)

The differential Equation (33) is in two variables y and p.

Let the solution of the Equation (33) be , , 0y p c … (34)

Then, by eliminating p from Equations (32) and (34), we get the solution ofdifferential Equation (32). If it is not easier to eliminate p, then express Equation(32) and (34) in the form

1 2, , ,x p c y p c


NOTES


In such case, the above two equations taken together constitute the solution ofEquation (32).

Note: When the Equation (33) is written as 1 2( , ). , ,dp

f y p f y pdy

= 0, then neglect

the factor which does not contain dp

dy , i.e., 1( , )f y p , since if 1( , )f y p = 0, then

eliminating p from this equation Equation (33) doe not give a general solution.

Example 4.15: Solve the differential equation 2x y p

Solution: The given equation is 2x y p … (i)

Differentiating both sides with respect to y, we get

1 2dx dp

pdy dy

11 2

dpp

p dy 21 2

dpp p

dy

22

1

pdy dp

p


22

1

pdy dp c

p

1

2 11

y p dp cp

2

2 log 12

py c p p

2 2 2log 1y c p p p

Substituting the value of y in Equation (i), we get

2 2 log 1x c p p which is the required solution.

Check Your Progress

24. Solve the differential equation 2 2 2 2 0xyp x xy y p x xy .

25. Solve the differential equation4 2

2 2

2 2

21 0

y y yy p p

xx x

.

26. Solve the differential equation y = p tan p + log cos p.

27. Solve the differential equation y – 2px = f(xp2).

28. Solve the differential equation 2tan

1

pp x

p

.

NOTES



4.6 CLAIRAUT’S EQUATION

The differential equation of the form y px f p is termed as Clairaut’ss

Equation.

Solution of Clairaut’s Equation

The given equation is y px f p … (35)


dy dp dpp x f p

dx dx dx

dp dpp p x f p

dx dx 0

dpx f p

dx

Neglecting x f p , we get 0dp

dx

Integrating both sides, we get p = cSubstituting the value of p in Equation (35), we get

y cx f c , which is the required solution.

Here, we neglect x f p because if we eliminate p between the given

Equation (35) and the factor x f p =0, we get another solution termed as

singular solution.

Note: When equation is in the Clairaut’s form, then the solution can be writtendown just by changing p to c in the given equation.

Example 4.16: Solve the differential equation sin cos cos sinpx y px y p andobtain the singular solution.

Solution: The given differential equation is sin cos cos sinpx y px y p

sin cos cos sinpx y px y p

sin px y p

1sinpx y p 1siny px p … (i)

which is a Clairaut’s form.

Thus the solution is 1siny cx c [Replacing p by c] … (ii)

Differentiating Equation (ii) partially with respect to c, we have

2

10

1x

c

2 1xc

x


NOTES


Substituting the value of c in Equation (ii), we get

22 1 1

1 sinx

y xx

, which is the required singular solution.

4.7 SINGULAR SOLUTIONS

A singular solution is a solution of differential equation which cannot be obtainedfrom its general solution by assigning any particular values to the arbitrary constants.

Discriminant

The discriminant of a polynomial is an expression which gives information aboutthe nature of the polynomial’s roots. We know that the discriminant of the quadratic

equation 2 0ax bx c is 2 4b ac . Here, if 2 4b ac > 0, the equation has two

real roots, if 2 4b ac = 0, the equation has equal roots which are real, and

if 2 4b ac < 0, the equation has two imaginary roots. But if equation has higher

degree than two, then the condition that the equation 0f x has two equal roots

is obtained by eliminating x between 0f x and 0f x .

p-Discriminant and c-Discriminant dy

pdx

Let , , 0f x y p … (36)

be the differential equation having

, , 0x y c … (37)

as the solution, where, c is an arbitrary constant.

Differentiating both sides of Equation (37) partially with respect to c, we get

, , 0x y cc

… (38)

Differentiating both sides of Equation (36) partially with respect to p, we get

, , 0f x y pp

… (39)

By eliminating c between the Equations (37) and (38), we get c-discriminant.Thus, c-discriminant represents the locus of each point which has equal values of

c in , , 0x y c .

By eliminating p between the Equations (36) and (39), we get p-discriminant.Thus, p-discriminant represents the locus of each point which has equal values ofp in

, , 0f x y p .

Note: (i) The envelope which is present in both c-discriminant and p-discriminantis known as the singular solution.

NOTES



(ii) If the singular solution of a differential equation , , 0f x y p whose

primitive is , , 0x y c is given by F , 0x y , then F ,x y will be a

factor of both the discriminates. (iii) In the Clairaut’s form of the differential equation both c-discriminant

and p-discriminant are same.

Method For Determining Singular Solution

Use the following steps to find the singular solutions of a differential equation

, , 0f x y p … (40)

Step 1: Find its complete solution , , 0x y c … (41)

where, c is an arbitrary constant.

Step 2: Find p-discriminant of Equation (40) and c-discriminant of Equation (41).Both contain the singular solution.

Now, when p-discriminant is equated to zero, it may include the followingas a factor:

Envelope, i.e., singular solution once (E) Cusp locus once (C) Tac locus, twice (T 2)

So p-discriminant 2ET C , which does not contain nodal locus and whenc-discriminant is equated to zero it may include the following as a factor:

Envelope, i.e, singular solution once (E) Cusp locus thrice (C3) Nodal locus twice (N 2)

So c-discriminant 2 3EN C

Example 4.17: Obtain the primitive and the singular solution of

2

2 4 0dy dy

x y xdx dx


2 4 0dy dy

x y xdx dx

… (i)

By putting dy

pdx

in Equation (i), we get

2 2 4 0xp yp x …(ii)

2

2

xp xy

p

Differentiating both sides, with respect to x, we get

2

2 2.

2 2

dy x dp p x dp

dx dx p p dx


NOTES


2

2 2. .

2 2

dp p dpx xp

dx p dxp

2

2 2

2 2

p x x dp

p p dx

2

22

44

2 2

p x dpp

p p dx

2 4 1 0x dp

pp dx

2 4 0p or 1 . 0x dp

p dx 2p or . 1

x dp

p dx

Now, for . 1x dp

p dx

dp dx

p x [Separating the variables]

Integrating both sides, we getlog log logp x c

where c is any arbitrary constant

log logp cx p cx … (iii)

Substituting the value of p from Equation (iii) in Equation (ii), we get

2 3 2 4 0c x ycx x 2 2 2 4 0c x yc … (iv)

which is the complete primitive of Equation (i),

From Equation (iv), we have p-discriminant (ET2C) is 2 24 16 0y x

2 24 0y x

Also from Equation (ii), c-discriminant (EN2C3) is 2 24 16 0y x

Since, 2 24y x is non-repeated common factor in p and c-discriminants and it

satisfies the differential Equation (ii).

Hence the singular solution of Equation (i) is

2 24 0y x 2y x

Check Your Progress

29. Reduce the differential equation 2px y x py p to Clairaut’s form

by substituting 2x u and 2y v . Also, find its complete primitive and its

singular solution.

30. Solve the differential equation 2y px p and obtain the singular solution.

31. Examine the equation 2 2 2 22 1y pxy p x m for singular solution.

NOTES



4.8 ORTHOGONAL TRAJECTORIES

A curve which cuts every member of a given family of curves according to adefinite law is called a trajectory of the given family.

Orthogonal Trajectory

A curve which cuts every member of a given family of curves at right angles iscalled an orthogonal trajectory of the given family.

Orthogonal trajectories of a given family of curves, themselves form a familyof curves.

We know that the angle between the two curves is equal to the angle betweenthe tangents at their common point of intersection. The slope of these tangents is

given by dy

dx.

When the curves intersects orthogonally,

Product of their slopes = – 1

Let the common point of intersection of two curves c1 and c

2 be P and the

angle between these curves at P is right angle, then

1 2 1 2

1c c c c

dy dy dy dx

dx dx dx dy

Oblique Trajectory

A curve which cuts every member of a given family of curves at certain angle otherthan right angle is called an oblique trajectory of the given family.

Differential Equation of Orthogonal Trajectories

Cartesian Coordinates

Step 1: Let the equation of the given family of curves be

f(x, y, c) = 0 … (42)where c is the parameter of the family.

Step 2: Differentiating Equation (42) with respect to x and eliminating c betweenEquation (42) and the resulting equation, we get the differential equation of thegiven family as

, , 0dy

F x ydx

… (43)

Step 3: Putting dy

dx=

dx

dy in Equation (43), we get

, , 0dx

F x ydy

… (44)

which is the differential equation of orthogonal trajectory.


NOTES


Step 4: Integrate Equation (44) to get the required equation of orthogonaltrajectories.

Example 4.18: Find the orthogonal trajectories of the family of co-axial circles x2

+ y2 + 2gx + c = 0, where g is a parameter and c is a constant.

Solution: The given equation is x2 + y2 + 2gx + c = 0 … (i)

Differentiating Equation (i) with respect to x, we get

2 2 2 0dy

x y gdx

0dy

x y gdx

… (ii)

Eliminating g between Equations (i) and (ii), we get the differential equation of the

given family of circles as 2 2 2 0dy

x y x x y cdx

2 2 2 0dy

y x xy cdx

… (iii)

By replacing dy

dx by

dx

dy

, we get

2 2 2 0dx

y x xy cdy

2 22

dxxy x c y

dy … (iv)

Let x2 = v. Then 2dx dv

xdy dy

The Equation (iv) becomes

2dvy v c y

dy

1dv cv y

dy y y

which is a linear differential equation in v.

1/

I.F.y dy

e = 1

1log

log logy y ye e e =

1

yThus, the required solution is

1 1cv y dy

y y y

1

vy

cy k

y , where k is an arbitrary constant

2v c y ky 2 2 0x y ky c

which is the orthogonal trajectories of the family of co-axial circles x2 + y2 + 2gx+ c = 0.

NOTES



Polar Co-Ordinates

Step 1: Let the equation of the given family of polar curves be

f(r, , c) = 0 … (45)

where c is the parameter of the family.

Step 2: Differentiating Equation (45) with respect to and eliminating c betweenEquation (45) and the resulting equation, we get the differential equation of thegiven family as

,θ, 0θ

drF r

d

… (46)

Step 3: From differential calculus, we know that d

rdr

is the tangent of the angle

between the tangent to a curve and the radius vector at any point (r, ). Since, thecurves intersects orthogonally,

1 2

1c c

d dr r

dr dr

, where c

1 and c

2 are two curves.

1

θ

d drr

dr r d

θ

dr

d= 2 d

rdr

Replace dr

d by 2 d

rdr

in (ii), we get

2, , 0d

F r rdr

… (47)

which is the differential equation of orthogonal trajectory.

Step 4: Integrate Equation (47) to get the required equation of the orthogonaltrajectories.

Example 4.19: Find the orthogonal trajectories of the curves sinn nr n a .

Solution: The given equation of the system of curve is sinn nr n a ... (i)

Taking log on both sides, we get

log logsin logn r n n a


1 1. . cos 0

sin

drn n n

r d n

1

cot 0dr

nr d

tand

r ndr

... (ii)

Replacingdr

d by 2 d

rdr

in (ii) we get

1tan

drn

r d

tan

drn d

r


NOTES


Integrating, we get 1tandr

n d cr , where 1c is any arbitrary constant.

1log logsec logr n c

n

[Putting 1 logc c ]

1/log log sec log

nr n c

1/log log sec

nr c n

1/sec

nr c n secn nr c n cosn nr n c

which is required equation of the orthogonal trajectories.

Check Your Progress

32. Find the orthogonal trajectories of 2 2

2 21

x y

a b

, where l is a

parameter.

33. Find the orthogonal trajectories of the system of curves rn = an cos nq.


1. y = c1ex + c

2e2x + c

3ex/2

2. 41 2 3cos sinxy e c c x x c c x x

3. 1 2cos sin sin log sec tan 1.y c x c x x x x

4. 3 3 21 2 3

1 1

8 8x x x xy c c x e c e xe x e .

5. 1 2

1cos sin sin cos3

4 16

xy c x c x x x

6. cos2

sin sin2 cos sin

n t ax e n t

n

2

'sin

2 cos

a nt

n

7.3

21 2 3

3 3cos sin

2 2

xxy c e e c x c x

15cos sin

26x x

8.2

1 2 3 4

3 3cos sin

2 2

x

y c c x e c x c x

4 5 3

23 312 20 3

x bx axa b bx

NOTES



9. 3 21 2 3 4

3

20x x xy c c x c e c e e

4

3 22 1 2 72sin cos

5 3 12 9 9

xx x x x

10. 3 21 2

12 4 3

8x xy c c x e e x x

11. 2 cos sinx xy ae b x c x e 3sin cos20

xxex x

12. 1 2

1 2cos2 sin 2 sin cos

3 9y c x c x x x x

13. 4 2 319 cos 4 sin

48x x x x x

14. An equation of the form,

11

0 1 11P ( ) P ( ) ... P ( ) P ( )

n nn n

n nn n


dx dx dx

where a, b, P0, P

1, P

2, …, P

n are constants and f(x) is either a constant or

a function of x.

This equation can be reduced to a homogeneous linear equation.

15.3

2 2 2

3

ya x x y xy c

16.2 2 2 12 tan

xx y a k

y

17. 2 2

13 0

2y c

x y

18. 2 2x y cx

19.3

3log

3

xy c

y

20.1

logx

xy cy xy

21. 2 2 xx y e c

22. 2 4 2 63 6 2x y xy y k

23.

3

26

yxy c

x

24. 2 2 22 2 2 0x y c x xy c


NOTES


25.2

2log 1

x xy

y y

c

26. x = tan p + c; y = p tan p + log cos p

27. 22y k x f k , where a k

28.1

2tan

1

px p

p

; 2

1

1y c

p

29. 2 0, 2 0, 2 0, 2 0x y x y x y x y

30. 24 0y x

31. 2 2 2 2y m x m

32. x2 + y2 – 2a2 log x = k

33. rn = kn sin n

4.10 SUMMARY

A linear differential equation of order n is of the form1 2

0 1 21 2P P P ....... P X

n n n

nn n n

d y d y d yy

dx dx dx

, where, P

0, P

1, P

2,….., P

n

and X are either the functions of x or constants.

D, which is used to replace the part d

dx of the symbol

dy

dx, is regarded as

differential operator.

If y = y1, y = y

2, …, y = y

n be the linear independent solutions of

1 20 1 2D D D ....... 0n n n

na a a a y , then

1 1 2 2 .... n ny c y c y c y where c1, c

2, c

3, ...,c

n are arbitrary constants,

is the general or complete solution of the differential equation1 2

0 1 21 2P P P ....... P X

n n n

nn n n

d y d y d yy

dx dx dx

.

The homogeneous equation 1

10 1 1

P P ... P Xn n

n nnn n

d y d yx x y

dx dx

can be

reduced into a linear equation with constant coefficients

0 1 1P D(D 1)...(D 1) P D(D 1)...(D 2) ... P D P Zn nn n y by changing the independent variable x to z and by putting x = ez, i.e., z =log x.

NOTES



An equation of the form 1

10 1 11

P ( ) P ( ) ... P ( ) P ( )n n

n nn nn n


dx dx dx

can be reduced to a linear equation with constant coefficients

10 1 1P D(D 1)...(D 1) P D(D 1)...(D 2) ... P D P

zn n

n n

e ab n b n b y f

b

by

putting (a + bx) = ez, i.e., z = log(a + bx).

The necessary and sufficient condition for the ordinary differential equation

M N 0dx dy to be exact is M N

y x

.

If M N 0dx dy is an exact differential equation, then its solution is given

by

constant

M (Terms of N not conatining )

y

dx x dy c .

The number of integrating factors of the differential equationM N 0dx dy is infinite.

The differential equation of first order and nth degree is generally representedby

1 2 31 2 3 1A A A A A 0n n n n

n np p p p p , where p denotes

dy

dxand

1 2 3 1A , A , A , , A , An n are functions of x and y.

The differential equation can be classified into three categorie, which are asfollows:

o Equations solvable for p, i.e., of the form ,p f x y

o Equations solvable for y, i.e., of the form ,y f x p

o Equations solvable for x, i.e., of the form ,x f y p

The general solution of 1 2 31 2 3 1A A A A A 0n n n n

n np p p p p

in case of solvable for p can be written as

1 2 3, , , , , ,f x y c f x y c f x y c , , 0nf x y c

A differential equation is said to be solvable for y, if it can be expressed as

,y f x p .

A differential equation is said to be solvable for x, if x can be expressed as

,x f y p .

The differential equation of the form y px f p is termed as Clairaut’ss

equation.


NOTES


A singular solution is a solution of differential equation which cannot beobtained from its general solution by assigning any particular values to thearbitrary constants.

The discriminant of a polynomial is an expression, which gives informationabout the nature of the polynomial’s roots.

c-discriminant represents the locus of each point, which has equal values of

c in , , 0x y c .

p-discriminant represents the locus of each point, which has equal values of

p in , , 0f x y p .

The differential equation of orthogonal trajectories in Cartesian coordinates

of the family f(x, y, z) = 0 is given by , , 0dx

F x ydy

.

The differential equation of orthogonal trajectories in polar coordinates of

the family f(r, , c) = 0 is given by 2, , 0

dF r r

dr

.

4.11 KEY TERMS

Linear differential equation: It is a differential equation in which thedependent variable and all its derivatives appear only in the first degree andare not multiplied together.

Inverse operator: 1

(D)f is known as the inverse operator of the

operator (D)f , if 1X

(D)y

f gives the solution of equation (D) Xf y .

Homogeneous linear equation: It is an equation of the form1

10 1 1

P P ... P Xn n

n nnn n

d y d yx x y

dx dx

, where P0, P

1, P

2, …, P

n are

constants and X is either constant or a function of x.

Exact differential equation: It is a differential equation which can be

obtained from its solution ,f x y c directly by differentiation without

any further multiplication and elimination.

Integrating factor: If the left side of the equation M N 0dx dy is not an

exact but it can be made exact by multiplying it with a function ,x y , then

,x y is called the integrating factor..

Orthogonal trajectory: It is a curve which cuts every member of a givenfamily of curves at right angles.

NOTES





1. Solve the following differential equations:

(i) 4

44 0

d yy

dx

(ii) 4 2D 13D 36 0y

(iii) 2D ( )D 0a b ab y

(iv) 6 6D 0a y

(v) 4 2 2 4D 2 D 0n n y

(vi)2

2sec

d yy x

dx

(vii)2

2 22

4 4 x xd y dyy e e

dxdx

2. Solve2

24 3 0

d x dxx

dtdt , given that for t = 0, x = 0 and 12

dx

dt .

3. Solve2

22

cosd x

b x k btdt

, given that x = 0 and 0dx

dt , when t = 0.


(i) 1 1 x ydyx e

dx

(ii) 2 0xy xyxe y dy ye dx

(iii) 2 20

xdy ydxxdx ydy

x y

(iv) 3 2 2 33 3 0x xy dx x y y dy

(v) ( cos sin ) (sin cos ) 0y x y y dx x x y x dy

(vi) 2[cos tan cos( )] [sin sec cos( )] 0x y x y dx x y x y dy

(vii)1

1 cos ( log sin ) 0y y dx x x x y dyx

(viii) cos (cos sin sin ) cos (cos sin sin ) 0x x a y dx y y a x dy


NOTES



(i) 2 3 3 0x ydx x y dy

(ii) 2 3 2 23 2 0xy y dx x y xy dy

(iii) 2 2 2 22 2 2 0y x y dx x x y dy

(iv) sin cos sin cos 0xy xy xy ydx xy xy xy xdy

(v) 2 2 0x y x dx xydy

(vi) 3 2

210

3 2 4

y xy dx x xy dy

(vii) 2 2 2 2 2 3 23 2 0xy x dx x y x y x y dy

(viii) 4 3 2 4 2 22 2 3 0y yxy e xy y dx x y e x y x dy

(ix) 34 2 3 5 0x ydx xdy y ydx xdy

(x) 23 4 2 6 0x ydx xdy y ydx xdy

6. Solve the differential equations for p:

(i) 2 0xp y x p y

(ii) 2 2 23 2 6 0xyp p x y xy

(iii) 2 2 cosh 1 0p p x

(iv)4 2

2 22 2

21 0

y y yy p p

x x x

(vi) 2 2 2( ) ( ) 0p p x xy y xy x y

7. Solve the differential equations for y:

(i) 2 2 0p py x

(ii) 2 2 0xp yp ax

(iii) 1tany x a p

(v) 3 2p mp a y mx

8. Solve the differential equations for x:

(i) 2 32y px y p

(ii) 3 24 8 0p xyp y

(iii) 2 2 2(1 )x a p

(vi) 3 ( 3) 0p p y x

NOTES



9. Solve the differential equation 2y px p p and obtain the singular solution.

10. Solve the differential equation logp px y and obtain the singular

solution.

11. Solve the differential equation 2 2 2y px a p b and obtain the singular

solution.



22

0d y dy

x x ydx dx

given that 1

y xx

is

one integral.


2 22

1 0d y dy

x x m ydx dx

given that

1sinm xy ce

is a solution.


(i) 2

2 2 22

2 2 2 0d y dy

x x x x x ydx dx

(ii)2

22 tan 5 secxd y dy

x y e xdx dx

(iii) 22

22

4 4 3 xd y dyx x y e

dx dx

(iv) 22

2 22

4 4 1 3 sin 2 5 6x xd y dyx x y e x e

dx dx

(v) 2

2 2

2 21 xd y dy

y xedx x dx x

(vi)2

2 2 2 3 22

(log ) 2 log 2 log 2(log ) (log )d y dy

x x x x x x y x xdx dx

(vii) 212

2 22

2 5xd y dy

x x y xedx dx

(viii) 22

22

4 4 1 5 3cos2xd y dyx x y e x

dx dx

(ix)2

22

2d y dyn y

dx x dx

(x)22 2 0

d dy dyx x y x y x y

dx dx dx

4. Solve the differential equation 3y px p and obtain the singular solution.

5. Reduce the equation 22 2 1px y px y p to Clairaut’s form by

putting x + y = u and xy = v and hence find its singular solution.


NOTES


6. Find the orthogonal trajectory of the following family of curves:(i) xy = c2 (ii) y2 = 4ax (iii) y = ax2 (iv) x2 + y2 = a2

7. Find the orthogonal trajectories of the family of curves px2 + qy2 = a2,where p and q are constants.

8. Find the orthogonal trajectories of the family of circles x2 + y2 + 2fy + 1 =0, where f is a parameter.

9. Find orthogonal trajectories of the series of logarithmic spirals r = a, wherea varies.

10. Find the orthogonal trajectories of x2 + y2 – ay = 0.

11. Find the orthogonal trajectories of the series of curves r = a + sin 5.

12. Find the orthogonal trajectories of 2

cosθ = αk

rr

, where is a

parameter.











NOTES

Linear Differential Equationsand Method of Variation of

Parameters


UNIT 5 LINEAR DIFFERENTIALEQUATIONS AND METHOD OFVARIATION OF PARAMETERS

Structure

5.0 Introduction5.1 Objectives5.2 Linear Differential Equations with Constant Coefficients5.3 Homogeneous Linear Ordinary Differential Equations5.4 Linear Differential Equation of Second Order5.5 Method of Variation of Parameters5.6 Answers to ‘Check Your Progress’5.7 Summary5.8 Key Terms5.9 Self-Assessment Questions and Exercises


5.0 INTRODUCTION

In mathematics, a differential equation has constant coefficients if only constantfunctions appear as coefficients in the associated homogeneous equation. Thelinear differential equation or a system of linear equations is generally the associatedhomogeneous equations having constant coefficients which may be solved byquadrature (mathematics), which means that the solutions may be expressed interms of integrals.

A homogeneous linear ordinary differential equation with constant coefficientsis an ordinary differential equation in which coefficients are constants (not functions),i.e., all terms are linear, and the entire differential equation is equal to zero, i.e., it isa homogeneous equation. Basically, a differential equation is homogeneous if it is ahomogeneous function of the unknown function and its derivatives. In the case oflinear differential equations, this means that there are no constant terms. The solutionsof any linear ordinary differential equation of any order may be deduced byintegration from the solution of the homogeneous equation obtained by removingthe constant term. The general solution of the homogeneous differential equationdepends on the roots of the characteristic quadratic equation.

The differential equations may contain one independent variable and two ormore than two dependent variables. The equation may be ordinary or partialdepending upon the ordinary or partial derivatives. To solve such equations, youwill require as many numbers of simultaneous differential equations as are thenumber of dependent variables.

Linear Differential Equationsand Method of Variation ofParameters

NOTES


In mathematics, variation of parameters, also known as variation of constants,is a general method to solve inhomogeneous linear ordinary differential equations.For first order inhomogeneous linear differential equations, it is usually possible tofind solutions through integrating factors or undetermined coefficients withconsiderably less effort, although those methods leverage heuristics that involveguessing and is not appropriate for all inhomogeneous linear differential equations.

In this unit, you will study about the linear differential equations with constantcoefficients, homogeneous linear ordinary differential equations, linear differentialequation of second order, transformation of equations by changing the dependentvariable and the independent variable, and method of variation of parameters.

5.1 OBJECTIVES


Discuss about the linear differential equations with constant coefficients

Understand the various theorems related to linear differential equations

Solve simultaneous linear differential equations with constant coefficients

Define what homogeneous linear ordinary differential equations are

Reduce a homogeneous linear equation into a linear equation with constantcoefficients

Reduce a linear differential equation to a homogeneous linear equation

Explain the characteristics of linear differential equation of second order

Solve a linear differential equation of second order

Analyse the transformation of equations by changing the dependent variableand the independent variable

Elaborate on the method of variation of parameters

5.2 LINEAR DIFFERENTIAL EQUATIONSWITH CONSTANT COEFFICIENTS

In mathematics, an Ordinary Differential Equation (ODE) is a differential equationcontaining one or more functions of one independent variable and the derivativesof those functions. Basically, an ordinary differential equation is a relation thatcontains function of only one independent variable, and one or more of than twodependent variables. The equation may be ordinary or partial depending upon theordinary or partial derivatives. To solve such equations, you will require as manynumbers of simultaneous differential equations as are the number of dependentvariables.

NOTES


Parameters


Methods of Solving Simultaneous Linear Differential Equations withConstant Coefficients

In this section, we shall discuss two methods for solving the simultaneous lineardifferential equation where x, and y are two dependent variables and t is theindependent variable.

Using Operator D

Let the symbolic form of the equations be F1(D)x + F

2(D)y = T

1 … (1)

and 1(D)x +

2(D)y = T

2 … (2)

where D denotes d

dt. Also, TT

1 and T

2 are functions of independent variable t and

F1(D), F

2(D),

1(D) and

2(D) are all rational integral functions of D with constant

coefficients.

Now, eliminate x from Equation (1) and Equation (2) by operating on bothsides of Equation (1) by

1(D) and Equation (2) by F

1(D), we get

F1(D)

1(D)x + F

2(D)

1(D)y =

1(D)T

1

1(D)F

1(D)x +

2(D)F

1(D)y = F

1(D)T

2

On subtracting these equations, we get

F2(D)

1(D)y

2(D)F

1(D)y =

1(D)T

1 F

1(D)T

2

g1(D) y = T (say)

Which is a linear equation in y and t. This equation can be solved to get the valueof y.

Now, by putting this value of y in Equation (1) or Equation (2), we get thevalue of x.

Note: Similarly, we can also eliminate y and get a linear differential equation in xand t which can be solved to get the value of x in terms of t. Further the value ofy can be obtained from Equation (1) or Equation (2) by putting the value of x.

Method of Differentiation

Sometimes, by differentiating one of the Equations (1) or (2) or both, we caneasily eliminate x or y. From resulting equation, after eliminating one dependentvariable, x or y can be solved to give the other dependent variable and then thevalue of the other variable can be obtained by putting these values in Equations (1)or (2).


NOTES


Example 5.1: Solve the simultaneous equations

4 3dx

x y tdt

and 2 5 tdyx y e

dt

Solution: The given equations are 4 3dx

x y tdt

… (i)

and 2 5 tdyx y e

dt … (ii)

By puttingd

dt= D in Equations (i) and (ii), we get

D 4 3x y t ... (iii)

and 2 D 5 tx y e ...(iv)

Eliminating y, we get D 4 D 5 6 D 5 3 tx t e

2D 9D 14 1 5 3 tx t e .

Its A.E. is 2D 9D 14 0 D 2, 7

C.F. = 2 71 2

t tc e c e

P.I. 2 2

1 1. 1 5 .3

14 9D D 14 9D Dtt e

1

21 9 1 31 D D . 1 5

14 14 14 14 9 1

tet

1 9 11 D ... 1 5

14 14 8tt e

1 9 1 1 31 1

1 5 .5 514 14 8 14 14 8

t tt e t e

2 71 2

5 1 31

14 8 196t t tx c e c e t e

2 71 2

5 12 7

14 8t t tdx

c e c e edt

By putting the values of x and dx

dtin Equation (i), we get

3y 2 71 2

10 5 31 1 12 3

7 14 49 8 2t t t tc e c e t t e e

2 71 2

1 3 27 52 3

3 7 98 8t t ty c e c e t e

and 2 21 2

5 31 1

14 196 8t t tx c e c e t e .

NOTES


Parameters


Example 5.2: Solve the simultaneous equations

2 2 2 3 tdx dyx y e

dt dt and 23 2 4 tdx dy

x y edt dt

Solution: The given Equations are 2 2 2 3 tdx dyx y e

dt dt … (i)

and 23 2 4 tdx dyx y e

dt dt … (ii)

By putting Dd

dt in the Equations (i) and (ii), we get

D 2D 2 2 3 tx y x y e

(D 2) 2(D 1) 3 tx y e … (iii)

and 2(3D 2) (D 1) 4 tx y e … (iv)

To eliminate y from Equations (iii) and (iv), multiply Equation (iv) by 2 and subtractfrom Equation (iii).

2(D 2) 2(D 1) 2(3D 2) 2(D 1) 3 8t tx y x y e e

2D 2 6D 4 3 8t tx e e

2( 5D 6) 3 8t tx e e

2(5D 6) 8 3t tx e e 26 8 3

5 5 5t tdx

x e edt

… (v)

Which is a linear differential equation of the form P Qdx

xdt

Where 26 8 3P and Q

5 5 5t te e

I.F. = 6 6

P5 5

dt tdte e e

Thus, the solution of Equation (v) is6 6

25 51

8 3

5 5

t tt tx e e e e dt c

6 16 11

5 5 51

8 3

5 5

t t txe e e dt c

16 11

5 51

8 5 3 5

5 16 5 11

t te e c

6 16 11 6

5 5 5 51

1 3

2 11

t t t tx e e e c e

6

2 51

1 3

2 11

tt tx e e c e

… (vi)

6

2 51

1 3 6(2)

2 11 5

tt tdxe e c e

dt

… (vii)


NOTES


From Equation (i), 2 2 2 3 tdx dyx y e

dt dt

Using Equations (vi) and (vii), we get

6 62 25 5

1 1

3 6 1 32 2 2 3

11 5 2 11

t tt t t t tdye e c e e e c e y e

dt

6

51

30 162 2

11 5

ttdy

y e c edt

6

51

15 8

11 5

ttdy

y e c edt

… (viii)

Which is a linear differential equation.

1I.F.

dt te e

Thus, the solution of Equation (viii) is

6

51 2

15 8

11 5

tt t tye e c e e dt c

1

2 51 2

15 8

11 5

tte c e dt c

12 5

1 2

158

22

tte c e c

6

51 2

158

22

tt ty e c e c e

Hence the required solutions of given equations are

62 5

1

1 3

2 11

tt tx e e c e

; 6

51 2

158

22

tt ty e c e c e

Example 5.3: Solve the simultaneous equations 0, 0dx dy

t y t xdt dt

given that x(1) = 1, y(–1) = 0.

Solution: The given equations are 0dx

t ydt

… (i)

and 0dy

t xdt

… (ii)

Differentiating Equation (i) with respect to t, we have2

20

d x dx dyt

dt dt dt

Multiplying by t, we get 2

22

0d x dx dy

t t tdt dt dt

… (iii)

Subtracting Equation (ii) from Equation (iii), we get2

22

0d x dx

t t xdt dt

… (iv)

Which is an homogeneous linear equation.

NOTES


Parameters


Put t = ez log t = z

Dd d

tdt dz

and 2

2

2D(D 1)

dt

dt

Equation (iv) becomes [D(D 1) (D 1)] 0x

2[D 1] 0x

Its A.E. is D 2 1 = 0 D2 = 1 D = 1

Thus the solution is 1 2z zx c e c e

1 21 2 1

cx c t c t c t

t … (v)

Differentiating Equation (v) with respect to t, we get 21 2

cdxc

dt t

By putting this value of dx

dt in Equation (i), we get 2

1 20

ct c y

t

21 0

cc t y

t 2

1

cy c t

t … (vi)

Given, x(1) = 1; y(1) = 0Putting t = 1, x = 1 in Equation (v), we have 1 = c

1 + c

2… (vii)

Putting t = 1, y = 0 in Equation (vi), we have 0 = c1 c

2… (viii)

Solving Equations (vii) and (viii), we get 1 2

1

2c c

Thus, the required solutions are1 1 1 1

;2 2

x t y tt t

.

Simultaneous Equations in a Different Form

If the equations are given in the form P1dx + Q

1dy + R

1dz = 0 … (3)

and P2dx + Q

2dy + R

2dz = 0 … (4)

where P1, P

2, Q

1, Q

2, R

1, R

2 are all function of x, y, z.

Dividing Equations (3) and (4) by dz, we get

1 1 1P Q R 0dydx

dz dz … (5)

and 2 2 2P Q R 0dydx

dz dz … (6)

Solving Equations (5) and (6), by cross-multiplication method, we get

1 2 2 1 1 2 2 1 1 2 2 1

1

Q R Q R R P R P P Q P Q

dydx

dz dz


NOTES


1 2 2 1 1 2 2 1 1 2 2 1Q R Q R R P R P P Q P Q

dydx dz

which is of the form P Q R

dydx dz … (7)

where P, Q, R are functions of x, y and z.

Thus, simultaneous equation of the type Equations (3) and (4) can alwaysbe put in the form Equation (5).

Methods for Solving the EquationP Q R

dx dy dz

First Method: Let the multipliers l, m, n be such that

P Q R P Q R

dy ldx mdy ndzdx dz

l m n

Choose l, m, n such that lP + mQ + nR = 0, and hence ldx + mdy + ndz = 0

If it is an exact differential equation say du, then on integrating, we get, onepart of the complete solution of Equation (5).

Again, if we choose another set of multipliers , ,l m n such that

' P 'Q 'R 0l m n we get ' ' ' 0l dx m dy n dz

Then, on integration, it will give another equation. The two equations thusobtained by using two sets of multipliers will form the complete solutions of givensimultaneous equations.

Note: Sometimes it may also happen that we choose multipliers l, m, n such that

ldx mdy ndz

lP mQ nR

is of the form that numerator is the exact differential co-efficient

of the denominator.

Second Method: The given equations are P Q R

dydx dz … (8)

First take any two members of P Q

dydx (say) and integrate it to get one of

the equation of the complete solution.

Again, take other two members Q R

dy dz (say) and integrate it also to get

another equation of the complete solution. These two equations so obtained formthe complete solution.

Example 5.4: Solve the simultaneous equations1cos( ) sin( )

dx dy dz

x y x y zz

.

Solution: The given equations are 1cos( ) sin( )

dx dy dz

x y x y zz

… (i)

NOTES


Parameters


Choosing 1, 1, 0 as multipliers, we get

1cos( ) sin( )

dx dy dz

x y x y zz

cos( ) sin( )

dx dy

x y x y

2 1 sin( ) cos( )

zdz dx dy

z x y x y

2

2 ( )

2( 1)2 sin

4

zdz d x y

zx y

2

1 2cosec ( ) ( )

1 42

zdzx y d x y

z


21

1 1log( 1) log tan ( ) log

2 42z x y c

2 1/ 21log( 1) log tan log

2 8

x yz c

2 1/ 2

1

( 1)log log

tan2 8

zc

x y

2 1 / 2

1

( 1)

tan2 8

zc

x y

… (ii)

Now, choosing 1, 1, 0 and 1, 1, 0 as multipliers in Equation (i), we get

cos( ) sin( ) cos( ) sin( )

dx dy dx dy

x y x y x y x y

cos( ) sin( )

( )cos( ) sin( )

x y x ydx dy dx dy

x y x y

cos( ) sin( )

( ) ( )sin( ) cos( )

x y x yd x y d x y

x y x y

Integrating both sides, we have 2log sin( ) cos( ) logx y x y x y c

2log sin( ) cos( ) ( ) logx y x y y x c

2log sin( ) cos( ) logy xx y x y e c

2sin( ) cos( ) y xx y x y e c … (iii)

Thus, Equations (ii) and (iii) together form the complete solution of the givenequations.

Example 5.5: Solve the simultaneous equations2 22

xdx dy dz

z yz y y z y z

.

Solution: The given equations are 2 22

xdx dy dz

z yz y y z y z

… (i)


NOTES


Choosing 1, y, z as multipliers, we get

2 22

xdx dy dz

z yz y y z y z

2 22 ( ) ( )

xdx ydy zdz

z yz y y y z z y z

0

xdx ydy zdz

xdx + ydy + zdz = 0

Integrating both sides, we get2 2 2

1

2 2 2 2

cx y z , where c

1 is any arbitrary

constant.

x2 + y2 + z2 = c1

… (ii)

From last two fractions of Equation (i), we have dy dz

y z y z

(y z) dy = (y + z) dz ydy (zdy + ydz) zdz = 0

Integrating both sides, we get 2 2

2

2 2 2

cy zyz , where c

2 is any arbitrary

constant.

y2 2yz z2 = c2

… (iii)


Example 5.6: Solvedx dy dz

y z z x x y

.

Solution: The given equations are dx dy dz

y z z x x y

… (i)

From Equation (i), we have

2

dx dy dy dz dx dy dz

y x z y x y z

Choosing the first two members, we have dx dy dy dz

y x z y

On integrating both sides, we get 1log log logy x z y c

1

y xc

z y

1( )x y c y z … (ii)

Again choosing the first and the last members, we have

2

1log log log

2x y x y z c

2

2x y x y z c …(iii)


NOTES


Parameters


Check Your ProgressSolving the following equations:

1. 7 0dx

x ydt

and 2 5 0dy

x ydt

.

2. 5 tdxx y e

dt and 23 tdy

x y edt

3. 2 , 2 , 2dx dy dz

y z xdt dt dt

.

4. ( ) ( ) ( )

adx bdy cdz

b c yz c a zx a b xy

5. 3 4 4 3 3 32 2 9 ( )

dx dy dz

y x x y x y z x y

6. ( )

dx dy dz

x y x y z

7. 2 2 2 2 2

dx dy dz

y x x y z

5.3 HOMOGENEOUS LINEAR ORDINARYDIFFERENTIAL EQUATIONS

Consider the following differential equations:

1.2

22

4 2 xd y dyx x y e

dx dx

2.3

33

(2 1) (2 1) 2 0d y dy

x x ydx dx

3.2

22 2 xd y dy

x x y edx dx

All these differential equations are linear since the dependent variable y andits derivatives occur only in the first degree and are not multiplied together. In thefirst equation the powers of x are same as the order of the derivatives associatedwith them. Such equations are called homogeneous linear equations. The secondequation in the given form is not homogeneous but can be changed to homogeneousby replacing 2x – 1 with X. The third equation is linear but it not homogeneous.

Homogeneous linear equation is an equation of the form1

10 1 1

P P ... P Xn n

n nnn n

d y d yx x y

dx dx

where P0, P

1, P

2, …, P

n are constants and X is either a constant or a function of x.


NOTES


Method of Solution


10 1 1

P P ... P Xn n

n nnn n

d y d yx x y

dx dx

…(9)

can be reduced into a linear equation with constant coefficients by changing theindependent variable x to z.

Taking x = ez Þ z = log x

Differentiating with respect to x, we get 1dz

dx x

Now, we have .dy dy dz

dx dz dx

1 dy

x dz

Thus,dy dy

xdx dz

…(10)

2

2

d y d dy

dx dx dx

1d dy

dx x dz

2

2 2

1 1. . .dy d y dz

x dz x dz dx

2 2

2 2 2 2 2

1 1 1. .d y dy d y dy

x dz x dz x dz dz

1dz

dx x

Thus, 2 2

22 2

d y d y dyx

dx dz dz …(11)

Proceeding likewise, we can have

3 3 23

3 3 23 2

d y d y d y dyx

dx dz dz dz …(12)

and so on.

Replacing d

dzby D in Equations (10), (11), (12), …

Ddy

x ydx

,

22 2

2D D = D(D 1)

d yx y y y

dx ,

33 3 2

3D 3D + 2D D(D 1)(D 2)

d yx y y y y

dx ,

……………………………….

D(D 1)(D 2)...(D 1)n

nn

d yx n y

dx

NOTES


Parameters


Substituting these values in Equation (9), we get

0 1 1P D(D 1)...(D 1) P D(D 1)...(D 2) ... P D P Zn nn n y

…(13)

where Z is a function of z to which X is transformed.

The Equation (13) is a linear equation with constant coefficients, which canbe solved for y in terms of z.

Suppose the solution of equation (v) is y = f(z), then the solution ofEquation (9) can be obtained as y = f(log x).

Example 5.7: Solve the linear differential equation2

2 22

3 logd y dy

x x y x xdx dx

.

Solution: Putting zx e logz x and Dd

dz in the given equation, we get

2D D 1 D 3 zy ze 2 2D 2D 3 zy ze

Then, the Auxiliary Equation (A.E.) is 2D 2D 3 0 Þ D 3, 1

C.F. 31 2

z zc e c e

And P.I. 22

1.

D 2D 3zze

2

2

1.

D 2 2 D 2 3ze z

1 1

X XD D

ax axe ef f a

2

2

1

D 2D 3ze z

2

2

1

2D D3 1

3 3

ze z

12 22D D

13 3 3

zez

21 2D1

3 3ze z

[By Binomial Theorem]

21 2

3 3ze z

Thus, the solution of the given equation is y = C.F. + P.I.

3 21 2

1 2

3 3z z zy c e c e e z

3 1 21 2

1 2log

3 3y c x c x x x

logze x z x


NOTES


Example 5.8: Solve the linear differential equation

2 2D 3 D 5 sin logx x y x .

Solution: Putting zx e logz x and . D D'd

x xdx

in the given equation,

we get

D' D' 1 3D' 5 siny z 2D' 4D' 5 siny z

Then, the A.E. is 2D' 4D' 5 0

4 16 20

D = 22

i

C.F. 21 2cos sinxe c z c z

And P.I. 2

1sin

D' 4D' 5z

1

sin1 4D' 5

z

1 1 1 D'. . sin

4 1 D' 1 + D'z

2

1 D'1. sin

4 1 D'z

1 D'1. sin

4 1 1z

1 1

sin D' sin sin cos8 8

z z z z

Thus, the solution of the given equation is y C.F. + P.I.

21 2

1cos sin sin cos

8zy e c z c z z z

21 2

1cos log sin log sin log cos log

8y x c x c x x x

logze x z x

Check Your Progress

8. Solve the differential equation 3 2

3 2

1d y d yx

dx dx x .


22

4 2 xd y dyx x y e

dx dx .

NOTES


Parameters


5.4 LINEAR DIFFERENTIAL EQUATION OFSECOND ORDER

Linear Differential Equation of Second Order is an equation of the form

2

2P Q R

d y dyy

dx dx

where P, Q and R are the functions of x.

Solution by Changing the Dependent Variable when One IntegralBelonging to the C.F. is Known

Let the linear differential equation of second order be

2

2P Q R

d y dyy

dx dx ...(14)

where P, Q and R are the functions of x only.

Let y = u be a known integral belonging to the C.F. of the Equation (14).Thus, its solution is

2

2P Q 0

d y dyy

dx dx ...(15)

Taking y = uv and differentiating with respect to x, we get

1 1

dyu v uv

dx and

2

2 1 1 222

d yu v u v uv

dx


2 1 1 2 1 12 P Q Ru v u v uv u v uv uv

2 1 1 2 12 P P Q Ruv u u v u u u v ...(16)

Since y = u is the solution of Equation (15), thus Equation (15) can bewritten as

2 1P Q 0u u u

Using this value in Equation (16), we get

2 1 12 P 0 Ruv u u v

2 1 1

2 RPv u v

u u

Now, putting 1v p and 2

dpv

dx , we get

1

2 RP

dpu p

dx u u

...(17)

This is a linear equation in p. Thus, its

I.F. = 12

Pu dxue

2P P P2log log 2. .dx dx dxu ue e e e u e


NOTES


Now, we have the solution of Equation (17) as

P P2 21

R. .

dx dxp u e u e dx c

u

P P P2 21R

dx dx dxp u e ue dx c u e

P P P2 21R

dx dx dxdvu e ue dx c u e

dx

dvp

dx


P P P2 21 2R

dx dx dxv u e ue dx dx c u e dx c

The complete solution of Equation (14) is given by y = uv

P P P2 21 2R

dx dx dxy u u e ue dx dx c u u e dx c u

Where c1 and c

2 are two arbitrary constants.

Determining the particular integral of 2

2P Q 0

d y dyy

dx dx

In case the integral of C.F. is not known while solving linear differential equationsof second degree, one of the following rules helps us in finding the particular integralof

2

2P Q 0

d y dyy

dx dx ...(18)

Rule 1: Let mxy e be the solution of Equation (18).

Differentiating with respect to x, we get

mxdyme

dx and

22

2mxd y

m edx

Putting these values in Equation (18), we get2 P Q 0mx mx mxm e me e

2 P Q 0mxm m e

2 P Q = 0m m

Thus, mxy e is the solution of Equation (18) if 2 P Q = 0m m .

Corollary: Taking m = 1, y = ex is a solution of Equation (18) if1 + P + Q = 0 .

Taking m = –1, y = e–x is a solution of Equation (18) if 1 – P + Q = 0.

Taking m = a, y = eax is a solution of Equation (18) if a2 + aP + Q = 0.

Rule 2: Let my x be the solution of Equation (18).

NOTES


Parameters



1mdymx

dx and

22

2( 1) md y

m m xdx


2 1( 1) P Q 0m m mm m x mx x

2( 1) P Q 0m m mx x

Thus, my x is the solution of Equation (18) if 2( 1) P Q 0m m mx x .

Corollary: Taking m = 1, y = x is a solution of Equation (18) if P + Qx = 0.

Taking m = 2, y = x2 is a solution of Equation (18) if 2 + 2Px +Qx2 = 0.

Example 5.9: Solve the differential equation

2

2cot 1 cot sin .xd y dy

x x y e xdx dx

Solution: The given differential equation is

2

2cot 1 cot sinxd y dy

x x y e xdx dx

... (i)

Comparing it with 2

2P Q R,

d y dyy

dx dx we get

P cot ,Q 1 cot ,R sinxx x e x

Now, P Q 1 cot 1 cot 1 0x x

Since P + Q + 1= 0, thus xy e is a part of C.F..

Taking xy ve ,

x xdy dve e v

dx dx and

2 2

2 22 .x x xd y d v dv

e e vedx dx dx

Substituting these values in Equation (i), we get

2

22 . cot 1 cot sinx x x x x x xd v dv dv

e e ve x e e v x ve e xdx dx dx

2

2+ 2 cot sin

d v dvx x

dx dx

2 cot sindp

x p xdx

where dv

pdx

and2

2

dp d v

dx dx ...(ii)

This is a linear differential equation in p, i.e., P ' Q 'dp

pdx

, where

P = 2 – cot x and Q = sin x. Thus, its I.F.

= P ' 2 cot 2 logsindx x dx x xe e e

2 2

logsin sin

x x

x

e e

e x


NOTES


Now, the solution of Equation (ii) is given by

p(I.F) 1Q'(I.F)dx c

2 2

1. sin .sin sin

x xe ep x dx c

x x

2 2

1.sin 2

x xe ep c

x

2

12 2

sin sin.

2

x

x x

e x xp c

e e

21

sinsin

2xdv x

c e xdx

dv

pdx

21

sinsin

2xx

dv c e x dx

Integrating both sides, we get 2

1

sinsin

2xx

dv c e x dx

2

12

cos2sin cos

2 5

xc exv x x c

Thus, the complete solution of Equation (i) is given by xy ve

12

1cos 2sin cos

2 5

xx xc e

y e x x x c e

Solution by Removing the First Derivative and Changing theDependent Variable

In case the integral of the C.F. is neither known nor can be found using the rules,there is a need of other method to find the solution of linear differential equation ofsecond order. Here, we will learn the method which is independent of integral ofC.F.

Consider the linear differential equation of second order

2

2P Q R

d y dyy

dx dx ...(19)

Change the dependent variable in the Equation (19) by putting y uv , whereu and v are the functions of x.

Now, dy dv du

u vdx dx dx

and 2 2 2

2 2 22 .

d y d v du dv d uu v

dx dx dx dx dx


2 2

2 22 . P Q( ) R

d v du dv d u dv duu v u v uv

dx dx dx dx dx dx

NOTES


Parameters


2 2

2 2P 2 P Q R

d v du dv d u duu u u v

dx dx dx dx dx

2 2

2 2

2 1 P RP . . . Q

d v du dv d u duv

dx u dx dx u dx u dx u

...(20)

Taking u such that the coefficient of first derivative dv

dx = 0 (i.e., removing

first derivative from Equation (20)), we get

2P + 0

du

u dx

P =

2

dudx

u


1 = P

2

dudx

u

1

log = P2

u dx

1

P2 =

dxu e

…(21)

Since 2

P + 0du

u dx , Equation (20) becomes

2 2

2 2

1 P R. . Q

d v d u duv

dx u dx u dx u

…(22)

From Equation (21),

1P

21 1

= P P2 2

dxdue u

dx

and 2

2 22

1 P P 1 1 P 1 1 P = P P

2 2 2 4 2

d u du d d du u u u u

dx dx dx dx dx


12P2 2

2

1 1 1 P P 1P P Q R

4 2 2

dxd v du u u v e

dx u dx u

12

P2 2 22

1 1 P 1P P Q R

4 2 2

dxd v dv e

dx dx

12P2 2

2

1 P 1Q P R

2 4

dxd v dv e

dx dx

2

2P Q

d vv

dx ...(23)

where 1

P2 21 P 1

P = Q . P ,Q = R2 4

dxde

dx

The Equation (23) is called the normal form of the Equation (19).Equation (23) can easily be integrated and then can be solved for v.

Thus, the general solution of Equation (19) is y = uv, which contains twoarbitrary constants.


NOTES


Example 5.10: Solve the differential equation2 2cos cos 0

d dyx y x

dx dx

.

Solution: The given differential equation is 2 2cos cos 0

d dyx y x

dx dx

2

2 22

cos 2sin cos cos 0d y dy

x x x y xdx dx

2

22 tan 0

d y dyx y

dx dx ...(i)

Comparing Equation (i) with 2

2P Q R

d y dyy

dx dx , we get

P = –2 tan x, Q = 1 and R = 0Putting y uv in Equation (i), the equation is transformed into

2

2P' Q

d vv

dx ...(ii)

where1 1

P 2 tan logsec2 2 secdx x dx xu e e e x

2 21 P 1 1 1P ' Q P 1 2 tan 4 tan

2 4 2 4

d dx x

dx dx

2 2 2 21 sec tan 1 1 tan tan 2x x x x

and1

P2Q' R 0

dxe

Putting these values in Equation (ii), it gets transformed into 2

22 0

d vv

dx

…(iii)

Symbolic form of this equation is 2D 2 0v

Its auxiliary equation is 2D 2 0 2D 2 D 2i

Now, the solution of Equation (iii) is given by1 2cos 2 sin 2v c x c x

Thus, the solution of Equation (i) is given by y uv

1 2sec cos 2 sin 2y x c x c x

Solution by Changing the Independent Variable

Consider the linear differential equation of second degree

2

2P Q R

d y dyy

dx dx ...(24)

Changing the independent variable x to z with the help of relation z = f(x) ,we get

.dy dy dz

dx dz dx

NOTES


Parameters


2

2

d y d dy

dx dx dx

22 2

2 2. .

d dy dz d y dz dy d z

dx dz dx dz dx dz dx

Substituting these values in the Equation (24), we get22 2

2 2. P . Q R

d y dz dy d z dy dzy

dz dx dz dx dz dx

2

2 2

2 2 22

P Q Rd z dz

d y dy ydx dxdz dzdz dz dz

dx dx dx

2

1 1 12P Q R

d y dyy

dz dz ...(25)

where

2

2

1 1 12 2 2

P Q RP ,Q ,R

d z dz

dx dxdz dz dz

dx dx dx

Here, P1, Q

1 and R

1 are the functions of x but can be expressed as the

functions of z with the help of the relation between z and x.

1. Choosing z such that the Coefficient of P1 is Zero.

2

2

1 2

PP 0

d z dz

dx dxdz

dx

2

2P 0

d z dz

dx dx

2

2P

d z dz

dx dx

2

2

P

d z

dxdz

dx


log Pdz

dxdx

Pdxdz

edx

Integrating again, we get Pdx

z e dx

For the relationPdx

z e dx

, P1 will be zero and Equation (25)

reduces to2

1 12Q R

d yy

dz ...(26)


NOTES


If Q1 is constant or a constant multiplied by 2

1

z, then Equation (26) can be

solved easily giving the value of y in terms of z. Then, by replacing z in terms of x,we get the general solution of Equation (24).

2. Choosing z such that the Coefficient of Q1 is Constant.

21Q a (say)

22

Qa

dz

dx

22 = Q

dza

dx

Qdz

adx

1

Qdz

dx a


1

Qz dxa

For the relation1

Qz dxa

, Q1 = a2 and Equation (25) reduces to

2

21 12

P Rd y dy

a ydz dz

...(27)

If P1 is a constant, then Equation (27) can be solved easily giving the value

of y in terms of z. Then, by replacing z in terms of x, we get the general solution ofEquation (24).

Example 5.11: Solve the differential equation

2

2 cos 22

3sin cot 2 sin sinxd y dyx x y x e x

dx dx .

Solution: The given differential equation is

2

2 cos 22

3sin cot 2 sin sinxd y dyx x y x e x

dx dx ...(i)

Comparing it with 2

2P Q R

d y dyy

dx dx , we get

2P 3sin cot , Q 2sinx x x and cos 2R sinxe x

Let P 3sin cotdx x x dxdz

e edx

3cos logsin 3cossin .x x xe x e

Integrating, we get

3cos 3cos1sin 3sin

3x xz e x dx e x dx

3cos1

3xe

On changing the independent variable x to z by the relation 3cos1

3xz e ,

Equation (i) reduces to the form2

1 1 12P Q R

d y dyy

dz dz …(ii)

NOTES


Parameters


where

2

1 1 2 2 2 22 3cos

Q 2sin 2 2P 0, Q

93sin x

x

zzdz x edx

and

cos 2

1 2 22 3cos

R sinR

sin

x

x

e x

dz x edx

7cos

7cos 3cos 36cos

xx x

x

ee e

e

7733

7 / 3

13

3z z

Substituting these values in Equation (ii), it becomes

723

7 / 32 2

2 1

9 3

d yy z

dz z

Multiplying by 2z , we get

122 3

7 / 32

2 1

9 3

d yz y z

dz

...(iii)

Putting logtz e z t

Dd d

zdz dt

and 2

22

D D 1d

zdz

Now, Equation (iii) reduces to

37 / 3

2 1D D 1

9 3

t

y e

Its auxiliary equation is 2 2D D 0

9

2 2 19D 9D 2 0 D ,

3 3

C.F. 2 1

2 / 3 1/ 33 31 2 1 2

t tc e c e c z c z

2 / 3 1/ 3

3cos 3cos1 2

1 1

3 3x xc e c e

2cos cos1 2A Ax xe e

and P.I. 37 / 3

2

1 12 3D D9

t

e

3 3

7 /3 7 / 3

1 1 9 11 1 2 3 6 39 3 9

t t

e e

11 3. 3cos3

1/ 3 1/ 3

1 1 1

36 3 6 3xz e

coscos

1/ 3 1/ 3

1 1.

66 3 3

xx e

e

Thus, the solution of Equation (i) iscos

2cos 2 cos1 2A A

6

xx x e

y e e

.


NOTES


Check Your Progress


2 2 32

2 2 xd y dyx x x x y x e

dx dx .

11. Solve the differential equation

2

2

2sin cos cos cos sin sin cos .

d y dyx x x x x y x x x x x

dx dx

12. Solve2

2 32

9 0,given that = isa part of solution.d y dy

x x y y xdx dx

13. Solve the differential equation2

2cot 2 tan sec

d y dyy x y x x

dx dx

.

14. Solve the differential equation2

2 1/ 3 / 3 4 / 3 2

1 1 1 60

4 6

d y dyy

dx x dx x x x

.


3 3 22

4 8 sind y dy

x x y x xdx dx

.


6 5 2 22

3d y dy

x x a y xdx dx

.

5.5 METHOD OF VARIATION OFPARAMETERS

In mathematics, variation of parameters, also known as variation of constants, is ageneral method to solve inhomogeneous linear ordinary differential equations. Forfirst order inhomogeneous linear differential equations, it is usually possible to findsolutions through integrating factors or undetermined coefficients with considerablyless effort, although those methods leverage heuristics that involve guessing and isnot appropriate for all inhomogeneous linear differential equations.

Here, we shall learn the method to find the complete primitive of a linearequation whose C.F. is known. In this method, the constants of the C.F. are takenas the functions of independent variables.

Consider the linear differential equation of second degree2

2P Q R

d y dyy

dx dx ...(28)

Let the C.F. of Equation (28) be

1 2y c u c v …(29)

where c1 and c

2 are two arbitrary constants.

Clearly, u and v are the integrals of2

2P Q 0

d y dyy

dx dx

NOTES


Parameters


P Q 0u u u ...(30)

and P Q 0v v v ...(31)

Let the constants c1 and c

2 in Equation (28) be the functions of x and the

complete primitive of Equation (28) be

1 2( ) ( )y c x u c x v ...(32)

Differentiating with respect to x, we get

1 1 2 2( ) ( ) ( ) ( )dy

c x u c x u c x v c x vdx

1 2 1 2( ) ( ) ( ) ( )dy

c x u c x v c x u c x vdx

Let c1(x) and c

2(x) satisfy the condition

1 2( ) ( )c x u c x v = 0 ...(33)

Thus, we get 1 2( ) ( )dy

c x u c x vdx

Again differentiating with respect to x, we get2

1 1 2 22( ) ( ) ( ) ( )

d yc x u c x u c x v c x v

dx


1 1 2 2 1 2 1 2( ) ( ) ( ) ( ) P ( ) ( ) Q ( ) ( ) Rc x u c x u c x v c x v c x u c x v c x u c x v

1 2 1 2( ) P Q ( ) P Q ( ) ( ) Rc x u u u c x v v v c x u c x v

Substituting the values from Equation (28) and Equation (29), we get

1 2 1 2( ) 0 ( ) 0 ( ) ( ) Rc x c x c x u c x v

1 2( ) ( ) Rc x u c x v 1 2( ) ( ) R = 0c x u c x v ...(34)

Solving the Equations (33) and (34) for 1 2( ), ( )c x c x , we get

1 2( ) ( ) 1

R R

c x c x

v u uv vu

1 2

R R( ) , ( )

v uc x c x

uv vu uv vu

Integrating, we get 1 2

R R( ) , ( )

W W

v uc x dx a c x dx b

where W = u v

uv vuu v

Putting these values in Equation (32), we getR R

W W

v uy u dx v dx

which is the particular solution of Equation (28).


NOTES


Example 5.12: Apply the method of variation of parameters to solve2

22

cosecd y

a y axdx

.


22

cosecd y

a y axdx

...(i)

Symbolic form of the equation is 2 2D coseca y ax

Its auxiliary equation is 2 2D 0a 2 2D a D ia

C.F. = A cos Bsinax ax

Let the complete solution of Equation (i) be cos siny u ax v ax ...(ii)

where u, v are unknown functions of x.

1 2y uy vy where 1 cosy ax and 2 siny ax

Let1 2

1 2

Wy y

y y

2 2cos sinW cos sin

sin cos

ax axa ax ax a

a ax a ax

Now 21

R

W

yu dx c where R cosecax

1

sin cosecax axdx c

a

1dx

a 1c

x

a

1c

and 1R

W

yv dx 2c

cos cosec ax ax

dxa

2ccot ax

dxa

2c 2

logsin ax

a

2c

Substituting the value of u and v in equation (ii), we get

1 22

logsincos sin

x axy c ax c ax

a a

1 2 2

logsincos sin cos sin

x axy c ax c ax ax ax

a a

which is the complete solution of Equation (i).

Check Your Progress

17. Apply the method of variation of parameters to solve2

25 6 2 xd y dy

y edx dx

.

18. Apply the method of variation of parameters to solve 2

22

secd y

n y nxdx

.

NOTES


Parameters



1. 61 2( cos sin )tx e c t c t ; y 6

1 2 1 2[( )cos ( )sin ]te c c t c c t

2. x = (c1 + c

2t)e4 +

24

25 36

tt e

e ; 4 21 2

1 71

25 36t t ty c c t e e e

3. 21 2 3cos( 3 )t tx c e c e t c ;

21 2 3

2cos 3

3t ty c e c e t c

and

21 2 3cos 3

3t tz c e c e t c

4. 2 2 21ax by cz c ; 2 2 2 2 2 2

2a x b y c z c

5. 1/ 3xyz a ; 2 2

y xb

x y

6. x + y = c1; 1 2logx c z c

7. 3 31x y c ; 3 3

2

6x y c

z

8. 21 2 3( log ) (log )

2

xy c c c x x x

9. 1 2 21 2

xy c x c x e x

10. 21 2

x x xy x e xe c xe c x

11. 2 1

1 1cos cos 2 sin 2

4 2y c x c x x x x

12. 3 32y kx c x

13. 1 2 1 2

tan sincos cos

2 2

x xy x c c x c c x x

14. 2/ 33/ 4 3 21 2

xy e c x c x

15. 2 2 21 2 sinx xy c e c e x

16. 1 22 2 2 2

1cos sin

2 2

a ay c c

x x a x

17. 2 31 2

x x xy c e c e e

18. 1 22

1logcos cos sin cos sin

xy nx nx nx c nx c nx

n n


NOTES


5.7 SUMMARY

In mathematics, an Ordinary Differential Equation (ODE) is a differentialequation containing one or more functions of one independent variable andthe derivatives of those functions.

Basically, an ordinary differential equation is a relation that contains functionof only one independent variable, and one or more of than two dependentvariables.

The equation may be ordinary or partial depending upon the ordinary orpartial derivatives. To solve such equations, you will require as many numbersof simultaneous differential equations as are the number of dependentvariables.

By eliminating x, we can get a linear differential equation in y and t whichwhen solved gives the value of y in terms of t. Further, the value of y can beobtained by putting the value of x in the given equation.

x or y can also be eliminated by differentiating one of the given equations orboth. Then, we can solve the resulting equation to get other variables.

If the symbolic form of the equations be F1(D)x + F

2(D)y = T

1

and 1(D)x +

2(D)y = T

2

where D denotes d

dt. Also, TT

1 and T

2 are functions of independent variable t

and F1(D), F

2(D),

1(D) and

2(D) are all rational integral functions of D

with constant coefficients.

If the multipliers l, m, n be such that

P Q R P Q R

dy ldx mdy ndzdx dz

l m n

Choose l, m, n such that lP + mQ + nR = 0, and hence ldx + mdy +ndz = 0

If the given equations are P Q R

dydx dz

First take any two members of P Q

dydx (say) and integrate it to get one of

the equation of the complete solution.

Again, take other two members Q R

dy dz (say) and integrate it also to get

another equation of the complete solution. These two equations so obtainedform the complete solution.

Homogeneous linear equation is an equation of the form1

10 1 1

P P ... P Xn n

n nnn n

d y d yx x y

dx dx

where P0, P

1, P

2, …, P

n are constants and X is either a constant or a

function of x.

NOTES


Parameters



10 1 1

P P ... P Xn n

n nnn n

d y d yx x y

dx dx

can be reduced into a linear equation with constant coefficients by changingthe independent variable x to z.

Linear differential equation of second order is an equation of the form

2

2P Q R

d y dyy

dx dx

where P, Q and R are the functions of x.

The techniques which will yield the solution for linear differential equation of

second order 2

2P Q R

d y dyy

dx dx , where P, Q and R are the functions of

x, are as follows: Changing the dependent variable when one integral belonging to the

C.F. is known Removing the first derivative and changing the dependent variable Changing the independent variable By using the method of variation of parameters

In mathematics, variation of parameters, also known as variation of constants,is a general method to solve inhomogeneous linear ordinary differentialequations.

For first order inhomogeneous linear differential equations, it is usuallypossible to find solutions through integrating factors or undeterminedcoefficients with considerably less effort, although those methods leverageheuristics that involve guessing and is not appropriate for all inhomogeneouslinear differential equations.

5.8 KEY TERMS

Linear differential equation: It is a differential equation in which thedependent variable and all its derivatives appear only in the first degree andare not multiplied together.

Ordinary Differential Equation (ODE): An Ordinary DifferentialEquation (ODE) is a differential equation containing one or more functionsof one independent variable and the derivatives of those functions.

Linear differential equations of second order: It is an equation of the

form 2

2P Q R

d y dyy

dx dx , where P, Q and R are the functions of x.


NOTES




1. What do you mean by linear differential equations with constant coefficients?

2. What are homogeneous linear ordinary differential equations/

3. Explain about the linear differential equation of second order giving examples.

4. Why the transformation of equations are done by changing the dependentvariable and the independent variable?

5. When is method of variation of parameters used?

6. Solve the following differential equations by the homogeneous equationmethod.

(i) 2 22 2 0xz yz dx yz xz dy x xy y dz

(ii) 2 3 2 2 2 3 2 2 0x y y y z dx xy x z x dy xy x y dz

(iii) 2 2 2 2 2 2 0yz x yz dx zx y zx dy xy z xy dz

(iv) 2 2 22 2 0z dx z yz dy y yz xz dz

(v) 2 2 2 2 22 2 0z xy y zdx z x xy zdy x y xy z dz


(i) 4

44 0

d yy

dx

(ii) 4 2D 13D 36 0y

(iii) 2D ( )D 0a b ab y

(iv) 6 6D 0a y

(v) 4 2 2 4D 2 D 0n n y

(vi)2

2sec

d yy x

dx

(vii)2

2 22

4 4 x xd y dyy e e

dxdx

(viii) 4D 1 sin 2y x

(ix) 2D 4D + 3 sin 3 cos 2y x x

(x) 3 2D 3D 4D 2 cosxy e x

NOTES


Parameters


(xi)2

2 22

4 4 8( sin 2 )xd y dyy x e x

dxdx

(xii) 3 2

23 2

6 1d y d y dy

xdxdx dx

8. Solve2

24 3 0

d x dxx

dtdt , given that for t = 0, x = 0 and 12

dx

dt .

9. Solve2

22

cosd x

b x k btdt

, given that x = 0 and 0dx

dt , when t = 0.


1. Discuss the characteristic features of linear differential equations with constantcoefficients and homogeneous linear ordinary differential equations givingappropriate examples.


(i) 2 2 2 2 2 22 2 2 0xyz y z yz dx x z xyz xz dy x y xy xyz dz

(ii) 2 2 22 2 0z dx z yz dy y yz xz dz

(iii) 2 2 2cos 1 sin 0dx dy dz

z z x z z z y xdt dt dt

(iv) 2 2 3 2 2 2 22 2 2 1 2 2 2 1x y xy xyz dx x x y x z xyz y z yz dy

2 3 2 1 0xy y y z dz

3. Solve the following simultaneous equations:

(i) 3 2 ; 5 3 0dx dy

x y x ydt dt

(ii)22

2 22 cos ; 2 sint tdy d yd x dx

x e t y e tdt dtdt dt

(iii) ; ' 'dx dy

ax by a x b ydt dt

(iv) 2 2cos 7sin ; 2 4cos 3sindx dy dx dy

y t t x t tdt dt dt dt

(v) 5 2 , 2 0dx dy

x y t x ydt dt

; given that x = y = 0 when t = 0

(vi)2

22 4 2 ;2 4 3 0

d y dz dy dzy x z

dx dx dx dx

(vii)2 2

22 2

4 5 ; 5 4 1d x d y

x y t x y tdt dt

(viii) ( ) ( ) ( )

mdyldx ndz

m n yz n l zx l m xy


NOTES


(ix) 2 2 2 2( ) ( ) ( )

dx dy dz

x y z y x z z x y

(x) 2 2 2 2 2

dx dy dz

x y z xy xz

(xi) 1 1

dx dy dz

y x z

(xii) 2 2 2( )

dx dy dz

y x xyz x y

(xiii) ( ) ( ) ( )(2 2 )

dydx dz

x x y y x y x y x y z

(xiv) 21 2 3 sin( 2 )

dx dy dz

x y x

4. Solve the differential equations by changing the dependent variable.

(i) 2

3/ 22 22

1 1d y dy

x x y x xdx dx

(ii) 2

22 1 1 0

d y dyx x x y

dx dx

(iii) 2

22 1 2 2 xd y dy

x x x y x edx dx

(iv) 2

2sin cos cos cos 0

d y dyx x x x x y x

dx dx

(v) 2

32

2 2d y dy

x x y xdx dx

(vi) 2

22

1 xd y dyx x y x e

dx dx

(vii)2

22

(1 cot ) cot sind y dy

x y x xdx dx


22

0d y dy

x x ydx dx

given that 1

y xx

is

one integral.


2 22

1 0d y dy

x x m ydx dx

given that

1sinm xy ce

is a solution.

7. Solve the following differential equations by removing first derivative:

(i) 2

2 2 22

2 2 2 0d y dy

x x x x x ydx dx

(ii)2

22 tan 5 secxd y dy

x y e xdx dx

NOTES


Parameters


(iii) 22

22

4 4 3 xd y dyx x y e

dx dx

(iv) 22

2 22

4 4 1 3 sin 2 5 6x xd y dyx x y e x e

dx dx

(v) 2

2 2

2 21 xd y dy

y xedx x dx x

(vi)2

2 2 2 3 22

(log ) 2 log 2 log 2(log ) (log )d y dy

x x x x x x y x xdx dx

(vii) 212

2 22

2 5xd y dy

x x y xedx dx

(viii) 22

22

4 4 1 5 3cos2xd y dyx x y e x

dx dx

(ix)2

22

2d y dyn y

dx x dx

(x)22 2 0

d dy dyx x y x y x y

dx dx dx

8. Solve the following differential equations by changing the independentvariable:

(i)2 2

2 4

2. 0

d y dy ay

dx x dx x

(ii) 2

2 3 32

4 1 4 2d y dy

x x x y xdx dx

(iii)2

22

cot 4 cosec 0d y dy

x y xdx dx

(iv) 2

2 2 2 3 32

11 4 4x xd y dy

x e y x x edx x dx

(v) 2 2

22

8181 . 0

9

d y dy xx y

dx x dx

(vi) 2

22

tan cos 0d y dy

x x ydx dx

(vii) 2

2 4 62

8 2 4x x xd y dye e y e

dx dx

(ix) 2

2

21 1 4coslog(1 )

d y dyx x y x

dx dx

9. Apply the method of variable of parameters to solve the following differentialequations:

(i)2

2cosec

d yy x

dx

(ii)2

2tan

d yy x

dx


NOTES


(iii)2

22 5 sec 2xd y dy

y e xdx dx

(iv)2

22

sinx xd yy e e

dx

(v) 2

2 22

1 4 1d y dy

x x x y xdx dx

(vi)2 3

2 26 9

xd y dy ey

dx dx x

(vii)2

2( 2) (2 5) 2 ( 1) xd y dyx x y x e

dx dx

10. Verify that y = x and y = x2 – 1 are linearly independent solutions of

2

22

1 2 2 0d y dy

x x ydx dx

. Find the general solution of

2

22 22

1 2 2 6 1d y dy

x x y xdx dx

.











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