DIFFERENTIAL take this opportunity to thank our Publisher M/s Himalaya Publishing House Pvt. Ltd., Sri Niraj Pandey, MD, Vijay Pandey, GM (Marketing) and Mr. G. Anil Kumar, Assistant

DIFFERENTIALEQUATIONS

(As per New CBCS Syllabus for B.Sc. Ist Year, IInd Semester ofAll the Universities in Telangana State w.e.f. 2016-17)

S. Saritha M.Sc. (Maths)Head, Dept. of Mathematics,

Indian Institute of Management and Commerce,Khairatabad, Hyderabad.

K. Ravi Kiran M.Sc. (Maths), B.Ed.Faculty, Dept. of Mathematics,

Indian Institute of Management and Commerce,Khairatabad, Hyderabad.

A. Vidya Jyothi M.Sc. (Maths)Head, Dept. of Mathematics,Sardar Patel Degree College,

Padma Rao Nagar, Secunderabad.

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PREFACE

This book is intended to provide a comprehensive presentation of problems for students pursuingB.Sc. (Semester-II) even without an exhaustive knowledge of Mathematics. The subject matter hasbeen presented in detail and simple form with clarity so as to enable the students to understand withtheir little efforts.

The book is designed for B.Sc. Ist Year, IInd Semester students that emphasize the solutiontechniques. Differential equations are interesting and important because they express relationshipswhich form the basis for developing ideas and studying phenomena of Science, Engineering,Economics, Medicine and every aspect of human knowledge without any exaggeration. The materialpresented in this book provides foundation for scientific, technological and industrial developments.Many researchers are engaged to develop refined methods for solutions of differential equations.

This book could be an appropriate textbook for an undergraduate course of Science andEngineering students also.

We express our sincere thanks to Sri K. Raghuveer, Principal, Indian Institute of Managementand Commerce, and Sri A. Ravi Kumar, Principal, Sardar Patel Degree College, Secunderabad whohad motivated us, and our colleagues Ms. Y. Sandhya, Ms. G. Sandhya, Smt P. Prashanthi,Vijayanand Goud, and B. Shyam Sunder who had been a great help at various stages while preparingthis book.

We take this opportunity to thank our Publisher M/s Himalaya Publishing House Pvt. Ltd.,Sri Niraj Pandey, MD, Vijay Pandey, GM (Marketing) and Mr. G. Anil Kumar, Assistant SalesManager, Hyderabad, and their staff for unflinching assistance and co-operation at all times whilepreparing this prestigious project.

We respectfully acknowledge that the critical comments, criticisms and constructive suggestionsmade for improvement for this book will be welcomed and shall be greatly appreciated and we shallsee that the same are incorporated in our revised editions.

Authors

SYLLABUSDIFFERENTIAL EQUATIONS

DSC- 1B BS:204Theory: 4 Credits and Practicals: 1 Credits

Theory: 4 hours/week and Practicals: 2 hours/week

Objective: The main aim of this course is to introduce the students to the techniques of solvingdifferential equations and to train to apply their skills in solving some of the problems of engineeringand science.

Outcomes: After learning the course, the students will be equipped with the various tools tosolve few types differential equations that arise in several branches of science.

Unit – IDifferential equations of first order and first degree:Exact differential equations – Integrating factors – Change in variables – Total differential equations –Simultaneous total differential equations – Equations of the form dx/P = dy/Q = dz/R.Differential equations first order but not of first degree:Equations solvable for y – Equations solvable for x – Equations that do not contain x (or y) –Clairaut’s equation.

Unit – IIHigher order linear differential equations: Solution of homogeneous linear differential equations withconstant coefficients – Solution of non-homogeneous differential equations P(D)y = Q(x) withconstant coefficients by means of polynomial operators when Q(x) = bxk, beax, eaxV, bcos(ax), bsin(ax).

Unit – IIIMethod of undetermined coefficients – Method of variation of parameters – Linear differentialequations with non constant coefficients – The Cauchy – Euler Equation.

Unit – IVPartial differential equations – Formation and solution – Equations easily integrable – Linearequations of first order – Non-linear equations of first order – Charpit’s method – Non-homogeneouslinear partial differential equations – Separation of variables.

CONTENTS

S.No. Page No.

Unit 1: Introduction 1 – 99Differential Equation – Order of a Differential Equation – Degree of aDifferential Equation – Solution of a Ordinary DifferentialEquations – Differential Equations of First Order and First Degree –Exact Equations – Integrating Factor – Linear Differential Equation –Bernoulli’s Equations – Change of Variables – Total DifferentialEquations – Simultaneous Total Differential Equations – Equations ofthe First Order but Not of First Degree – Equations Solvable for p –Differential Equations Solvable for y – Equations Solvable for x –Differential Equations that do not Contain x and y – Clairaut’sEquation – Practical Question Bank with Answers

Unit 2: Higher Order Linear Differential Equations 100 – 149Higher Order Homogeneous Linear Differential Equations – HigherOrder Non-homogeneous Linear Differential Equations – Rules forFinding Partially Integral in Some Special Cases – PracticalProblems – Practical Question Bank with Answers

Unit 3: Linear Differential Equations of Second Order 150 – 193Method of Undetermined Coefficients – Method of Variation ofParameters – Linear Differential Equations with Non-constantCoefficients – The Cauchy-Euler Equation – Legendre’s Equations –Practical Question Bank with Answers

Unit 4: Partial Differential Equations 194 – 242Partial Differential Equations – Order of a Partial DifferentialEquation – Formation of Partial Differential Equation – EquationsEasily Integrable – Solutions of a Partial Differential Equation –Linear Equations of the First Order – Non-linear Equations of FirstOrder – Charpit’s Method – Homogeneous Linear Equations withConstant Coefficients – Non-homogeneous Linear Partial DifferentialEquations – Separation of Variables – Practical Question Bank withAnswers

Unit

Introduction

Differential equations play a very significant role in almost all branches of science and engineering. These equations arise from various practical problems such as oscillations of mechanical and electrical system, bending of beams, variation of distance of moving body with respect to time and variation of current in an electric circuit, etc. Differential equations are formed with the help of a phenomena obeying a certain law or satisfying a certain geometrical property by establishing a relation between the variables, their derivatives and constants. The solutions, thus, obtained are interpreted in the light of the conditions of the problems.

Differential Equation Definition: An equation involving derivatives or differentials of one or more dependent

variables with respect to one or more independent variables is called a differential equation.

Types of Differential Equations (i) Ordinary differential equation (ii) Partial differential equation

Ordinary Differential Equation Definition: A differential equation involving derivatives with respect to a single independent

variable is called ordinary differential equation.

Example: (i) = 푥 + sin 푥

(ii) + + = 푒

Partial Differential Equations Definitions: A differential equation involving partial derivatives with respect to more than one

independent variable is called a partial differential equation.

Example: (i) = 푘

(ii) + + = 0

1

2 Differential Equations

Order of a Differential Equation Definition: The order of the highest order derivative involved in a differential equation is called

the order of the differential equation.

Degree of a Differential Equation Definition: Let 퐹 푥,푦,푦 ′,푦 ′′, … … …푦( ) = 0 be a differential equation of order 푛. If the given

differential equation is a polynomial in 푦( ), then the highest degree of 푦( ) is defined as the degree of the differential equation.

Example: (i) = sin푥

Here, the order is ′1′ and the degree is also one.

(ii) − + 푦 = 푐푠푐 푥

Here, the order is 2 and the degree is 3.

(iii) 푟 =

푟 = 1 +

푟 = 1 +

Here, the order is ′2′ and the degree also ′2 .

Solution of a Ordinary Differential Equation Definition: Any relation between the dependent and independent variables, when substituted in

the differential equation, reduces to an identity is called a solution or integral of the differential equation, complete primitive (or general solution).

Particular Solution and Singular Solution Definitions (i) General Solution: The solution of a differential equation of order ′푛′ containing ′푛′

independent arbitrary constants is called a general solution. (ii) Particular Solution: The solution of the differential equation obtained from its general

solution by giving particular values to the arbitrary constants is called a particular solution. (iii) Singular Solution: An equation 휓(푥,푦) = 0 is called a singular solution of the differential

equation on 퐹 푥,푦, 푦 ,푦 , … … …푦( ) = 0 if (a) 휓(푥,푦) = 0 is a solution of the given differential equation (b) 휓(푥,푦) = 0 does not contain arbitrary constant and (c) 휓(푥,푦) = 0 is not obtained by giving particular values to arbitrary constants in the

general solution.

Introduction 3

Formation of a Differential Equation Suppose we are given a family of curves containing ′푛′ arbitrary constants, then we can obtain an

푛 order differential equation whose solution is the given family as follows Let the given family of curves be

푄(푥,푦, 푐 , 푐 , … … … 푐 ) = 0 …(1) where 푐 , 푐 , … … … 푐 are 푛 arbitrary constants. Differentiating (1) successively 푛 times with respect to ′푥′, we get:

푓 (푥,푦, 푦 , 푐 , 푐 , … … … 푐 ) = 0 …(2)

푓 푥, 푦,푦( ),푦( ), 푐 , 푐 , … … … 푐 = 0 …(3)

… … … … … … … … … … … … … … … … … … … …. 푓 푥, 푦,푦( ),푦( ), … … … … ,푦( ),푐 , 푐 , … … … 푐 = 0 … … … … (푛 + 1)

Eliminating the 푛 arbitrary constants 푐 , 푐 , … … … 푐 from the above (푛 + 1) equations, we get the eliminant, the differential equation

푓 푥, 푦,푦( ),푦( ), … … … … ,푦( ) = 0

Note: A differential equation of 푛 order cannot have more than 푛 arbitrary constants in its solution.

Solved Problems Example 1: Find the differential equation of all circles which pass through the origin and whose

centre is on the 푥-axis. Solution: The equation of a circle passing through the origin and whose centre is on the 푥-axis is

given by 푥 + 푦 + 2푔푥 = 0 …(1) (푔 being an arbitrary constant) Differentiating equation (1) with respect to ′푥′, we get

2푥 + 2푦 + 2푔 = 0 …(2)

From (1) and (2), eliminating ′푔′, we get

2푥 + 2푦 − = 0

2푥푦 + 푥 − 푦 = 0

Example 2: Form a differential equation of which 푦 = 푒 (퐴 cos 푥 + 퐵 sin 푥) is a solution, 퐴 and 퐵 being arbitrary constants.

Solution: Given that 푦 = 푒 (퐴 cos 푥 + 퐵 sin 푥) …(1) Differentiating equation (1), we get 푦 = 푒 (−퐴 sin 푥 + 퐵 cos 푥) + 푒 (퐴 cos 푥 + 퐵 sin 푥) 푦 = 푒 (−퐴 sin 푥 + 퐵 cos 푥) + 푦 …(2) (from (1))


Again differentiating equation (2) with respect to ′푥′, we get 푦 = −푒 (퐴 cos 푥 + 퐵 sin푥) + 푒 (−퐴 sin푥 + 퐵 cos 푥) + 푦 푦 = −푦 + (푦 − 푦) + 푦 (From (1) and (2)) 푦 − 2푦 + 2푦 = 0

Example 3: Form the differential equation which has 푦 = 푎푥 + 푏푥 + 푐 as a solution where 푎, 푏 and 푐 are arbitrary constants.

Solution: Given 푦 = 푎푥 + 푏푥 + 푐 ...(1) Differentiating with respect to ′푥′, we get

= 2푎푥 + 푏 …(2)

Again, differentiating with respect to ′푥′, we get

= 2푎 …(3)

Differentiating with respect to ′푥′, we get

= 0 …(4)

∴ The required differential equation is = 0

Example 4: Eliminate 휆 from + = 1.

Solution: Given + = 1 …(1)

Differentiating (1) with respect to ′푥′, we get

+ = 0

= − => 푎 + 휆 = …(2)

Substituting (2) in (1), we get

+ = 1

− = 1

푥 푦 − 푥푦 = 푎 푦 Example 5: Form the differential equation of all circles with centers along the line 푦 = 푥 and of

given radius. Solution: Equation of circle is

(푥 − 훼) + (푦 − 훼) = 푟 where ′훼′ is arbitrary constant (푥 − 훼) + (푦 − 훼) = 푟 …(1) Differentiating (1) with respect to ′푥′, we get 2(푥 − α) + 2(푦 − α)푦 = 0 (푥 − 훼) = −(푦 − 훼)푦

Introduction 5

푥 − 훼 = −푦푦 + 훼푦

훼 =

푦 − 훼 = 푦 − =

∴ 푦 − 훼 = ( )

Similarly, 푥 − 훼 = ( )

Substituting the values of (푥 − 훼) and (푦 − 훼) in (1), we get

( )( )

+ ( )( )

= 푟

=> (푥 − 푦) (1 + 푦 ) = 푟 (1 + 푦 )

Example 6: Solve log = 푐푥.

Solution: Given log = 푐푥

= 푒 …(1) Differentiating (1) with respect to ′푥′, we get

= 푒 . 푐 …(2) Dividing (1) by (2), we get

= => 푐 =

log = 푥

log =

푦 log = 푥푦 − 푦

푥푦 − 푦 − 푦 log = 0

푥푦 − 푦 1 + log = 0

Example 7: Show that 푦 = 푎푒 + 푏푥푒 where 푎 and 푏 are arbitrary constants, is the solution of the differential equation 푦 ′′ − 4푦 ′ + 4푦 = 0.

Solution: Given 푦 = 푎푒 + 푏푥푒 …(1) Differentiating (1) twice with respect to ′푥′, we get 푦 = 2푎푒 + 푏(2푥푒 + 푒 ) 푦 = (푎 + 푏푥)2푒 + 푏푒 …(2) 푦 = (푎 + 푏푥)4푒 + 2푏푒 + 2푏푒 푦 = 4(푎 + 푏푥)푒 + 4푏푒 Now, 푦 − 4푦 + 4푦 = 4(푎 + 푏푥)푒 + 4푏푒 − 8(푎 + 푏푥)푒 − 4푏푒 + 4(푎 + 푏푥)푒

= 8(푎 + 푏푥)푒 − 8(푎 + 푏푥)푒 + 4푏푒 − 4푏푒 = 0 ∴ 푦 = 푎푒 + 푏푥푒 is the solution of 푦 − 4푦 + 4푦 = 0.


Example 8: Find the differential equation of the family of parabolas having vertex at the origin and foci on 푦-axis.

Solution: The equation of the family of parabolas with vertex at the origin and foci on 푦-axis is 푥 − 4푎푦 …(1) (where, ′푎′ is a parameter) Differentiating equation (1) with respect to ′푥′, we get

2푥 = 4푎

4푎 = => 4푎 = …(2)

From (1) and (2), we get

푥 = ∙ 푦

∴ The required differential equation is 푥 = 2푦

Example 9: Find the differential equation corresponding to 푦 = 푎푒 + 푏푒 + 푐푒 where 푎, 푏 and 푐 are parameters.

Solution: Given equation is 푦 = 푎푒 + 푏푒 + 푐푒 …(1) Differentiating equation (1) with respect to ′푥′, we get 푦 = 푎푒 + 2푏푒 + 3푐푒 푦 = (푎푒 + 푏푒 + 푐푒 ) + 푏푒 + 2푐푒 푦 = 푦 + 푏푒 + 2푐푒 …(2) (from (1)) Again, differentiating with respect to ′푥′, we get 푦 = 푦 + 2푏푒 + 6푐푒 푦 − 푦 = 2(푏푒 + 2푐푒 ) + 2푐푒 푦 − 푦 = 2(푦 − 푦) + 2푐푒 (from (2)) 푦 − 3푦 + 2푦 = 2푐푒 Again, differentiating with respect to ′푥′, we get 푦 − 3푦 + 2푦 = 6푐푒 푦 − 3푦 + 2푦 = 3(2푐푒 ) 푦 − 3푦 + 2푦 = 3(푦 − 3푦 + 2푦) (from (3)) ∴ The required differential equation is 푦 − 6푦 + 11푦 − 6푦 = 0

Exercise: (1) Form the differential for the following: (a) 푦 = 퐴푒 + 퐵푒 , 퐴 and 퐵 being arbitrary constants (b) 푦 = 푘 sin 푥 ,푘 parameter. (2) Find the equation of the family of parabolas 푦 = 4푎푥. (3) Find the differential equation of the family curves 푦 = 퐴푒 + 퐵푒 , for different values of

퐴 and 퐵.

Introduction 7

(4) Form the differential equation of the family of circles given by 푥 + 푦 + 2푎푥 + 2푏푦 + 푐 = 0 where 푎,푏 and 푐 are arbitrary constants.

Answers:

(1) (a) 푦 − 4푦 = 0, (b) 푦 = 푦 (1 − 푥 ) sin 푥 (2) 푦 = 2푥푦 (3) 푦 − 8푦 + 15푦 = 0 (4) 푦 (1 + (푦 ) ) = 3푦 (푦 )

Differential Equations of First Order and First Degree Definition: An equation of the form = 푓(푥,푦) is called a differential equation of the first

order and of the first degree. We now discuss various methods to solve such equations.

Variables Separable If the differential equation = 푓(푥,푦) can be expressed in the form = ( )

( ) or

= 푓(푥) ∙ 푔(푦) where 푓 and 푔 are continuous functions of a single variable, it is said to be of the form variables separable.

Method of Solving

→ Let the given equation = ( )( )

…(1)

It can be written by separating the variables as 푓(푥)푑푥 = 푔(푦)푑푦 Integrating on both sides, we get => ∫푓(푥)푑푥 = ∫푔(푦)푑푦 + 푐

Note: The constant of integration ′푐′ can be selected in any suitable form as , log 푐 , sin 푐 , 푒 or tan 푐.

Solved Problems Example 1: Solve (1 − 푥)푑푦 + (1 − 푦)푑푥 = 0.

Solution: Given differential equation can be written as + = 0

Integrating on both sides, we get

∫ + ∫ = ∫0

− log(1 − 푥) − log(1 − 푦) = − log 푐 −[log(1 − 푥) + log(1 − 푦)] = − log 푐 log(1 − 푥) + log(1 − 푦) = log 푐 (1 − 푥)(1 − 푦) = 푐


Example 2: Solve (푥푦 + 푥)푑푥 + (푦푥 + 푦)d푦 = 0. Solution: Given (푥푦 + 푥)푑푥 + (푦푥 + 푦)d푦 = 0

푥(1 + 푦 )푑푥 + 푦(1 + 푥 )푑푦 = 0

푑푥 + 푑푦 = 0


∫ 푑푥 + ∫ 푑푦 = ∫0

log(1 + 푥 ) + log(1 + 푦 ) = log 푐

log√1 + 푥 1 + 푦 = log 푐

√1 + 푥 1 + 푦 = 푐 (or) (1 + 푥 )(1 + 푦 ) = 푐

Example 3: Solve (푒 + 1) cos 푥 푑푥 + 푒 sin 푥 푑푦 = 0. Solution: Given (푒 + 1) cos 푥 푑푥 + 푒 sin푥 푑푦 = 0

Variables separable

푑푥 +( )

푑푦 = 0


∫ ( )( )

푑푥 = log 푓(푥) + 푐

∫ 푑푥 + ∫ ( )푑푦 = ∫0

log(sin 푥) + log(푒 + 1) = log 푐 log(sin 푥)(푒 + 1) = log 푐 sin 푥 (푒 + 1) = 푐

Example 4: Solve 3푒 tan 푦푑푥 + (1 − 푒 ) sec 푦푑푦 = 0. Solution: Given 3푒 tan 푦푑푥 + (1 − 푒 ) sec 푦푑푦 = 0

Variables separable

( )

푑푥 + 푑푦 = 0


3∫ ( )푑푥 + ∫ 푑푦 = ∫0

−3 log(1 − 푒 ) + log(tan 푦) = log 푐 log(1 − 푒 ) + log(tan 푦) = log 푐

log( )

= log 푐

tan 푦 = 푐(1 − 푒 )

Introduction 9

Example 5: Solve 1 + 푥 + 푦 + 푥 푦 + 푥푦 = 0.

Solution: Given 1 + 푥 + 푦 + 푥 푦 + 푥푦 = 0

Rewriting the given differential equation, we get

(1 + 푦 ) + 푥 (1 + 푦 ) + 푥푦 = 0

(1 + 푥 )(1 + 푦 ) + 푥푦 = 0

Variables separable

√ 푑푥 + = 0


∫ √ 푑푥 + ∫ = ∫0

( )

+ ∫( )

푑푥 + ∫( )

= 푐 …(1)

Now, ∫( )

= ∫ put 푥 = ,푑푥 = − 푑푡

= −∫√

= − log 푡 + √푡 + 1

= − log + + 1

= − log ( )

= log 푥 − log 1 + (1 + 푥 ) …(2)

Again ∫( )

= ∫√

(put 1 + 푥 = 푡, 2푥푑푥 = 푑푡, 2푑푥 = )

= ∫ 푡 푑푡

= 푡 = (1 + 푥 ) …(3)

Similarly, ∫( )

= (1 + 푦 ) …(4)

Using (2), (3) and (4) gives required solution

log 푥 − log 1 + (1 + 푥 ) + (1 + 푥 ) + (1 + 푦 ) = 푐

Example 6: Solve = (4푥 + 푦 + 1) .

Solution: Given = (4푥 + 푦 + 1) …(1)

Putting 4푥 + 푦 + 1 = 푡, we get

4 + =


= − 4

Substituting the above values in (1), we get

− 4 = 푡

= 푡 + 4

Variables separable

= 푑푥


∫ = ∫푑푥 + 푐

tan = 푥 + 푐

tan = 2(푥 + 푐)

= tan 2(푥 + 푐)

푡 = 2 tan 2(푥 + 푐) 4푥 + 푦 + 1 = 2 tan 2(푥 + 푐) is the general solution.

Example 7: Solve = 푥푦 + 푥 + 푦 + 1.

Solution: Given = 푥푦 + 푥 + 푦 + 1

= 푥(푦 + 1) + (푦 + 1)

= (푥 + 1)(푦 + 1)

Variables separable

= (푥 + 1)푑푥


∫ = ∫(푥 + 1)푑푥

log 푐 + log(푦 + 1) = + 푥

log(푦 + 1)푐 = + 푥

(푦 + 1)푐 = 푒

Exercise: Solve the following differential equations.

(1) =

(2) 푒 cot 푦푑푥 + (1 − 푒 ) cosec 푦푑푦 = 0

(3) = 푒 + 푥 푒

Introduction 11

(4) √1− 푥 sin 푥 푑푦 + 푦푑푥 = 0

(5) 푦 = 푥푒

(6) log = 푎푥 + 푏푦

(7) (푥푦 + 푥)푑푥 + (푦푥 + 푦)푑푦 = 0

Answers: (1) 푦 − 푥 = 푐(1 + 푦푥) (2) 푒 − 1 = 푐 cot 푦

(3) 푒 = 푒 + + 푐

(4) 푦 sin 푥 = 푐

(5) 푒 + 푒 = 푐 (6) 푏푒 + 푎푒 = 푐 (7) (푥 + 1)(푦 + 1) = 푐

Exact Differential Equations Definition: If 푀 and 푁 are functions of 푥 and 푦, the equation 푀푑푥 + 푁푑푦 = 0 is called exact

when there exists a function 푓(푥,푦) of 푥 and 푦, such that

푑 푓(푥,푦) = 푀푑푥 + 푁푑푦

푑푥 + 푑푦 = 푀푑푥 + 푁푑푦

Theorem: If 푀(푥,푦) and 푁(푥,푦) are two real valued functions which have continuous first partial derivatives on some rectangle 푅: |푥 − 푥 | ≤ 푎, |푦 − 푦 | ≤ 푏, then a necessary and sufficient condition for the differential equation 푀푑푥 + 푁푑푦 = 0 to be exact, is = in 푅.

Proof: (i) Necessary Condition Let the equation 푀푑푥 +푁푑푦 = 0 be an exact equation

We have to prove that =

By definition ∃ a function 푓(푥,푦) having continuous first and second partial derivatives such that 푑 푓(푥,푦) = 푀푑푥 + 푁푑푦 …(1)

Total definition 푑 푓(푥,푦) = 푑푥 + 푑푦 …(2)

From (1) and (2), we get 푀푑푥 +푁푑푦 = 푑푥 + 푑푦

so that 푀 = …(3)

푁 = …(4)


Differentiating equation (3) with respect to 푦 partially and (4) with respect to 푥 partially, we get

= …(5)

= ...(6)

For 푓(푥,푦), we have =

From (5) and (6), we get =

(ii) Sufficient Condition

Conversely suppose that =

We have to prove that 푀푑푥 +푁푑푦 = 0 is an exact equation For this, it is enough to prove ∃ a function 푓(푥,푦) ∋ 푑 푓(푥,푦) = 푀푑푥 +푁푑푦

Let us define 푢(푥,푦) = 푥∫ 푀푑푥 …(7)

[ x

denotes that while integrating 푦 is to be treated as constant]

Differentiating equation (7) with respect to ′푥′ partially, we get

∫ =

푀 = …(8) Differentiating on both sides with respect to ′푦′ partially, we get

=

Since = and = , we have

= =

Integrating on both sides with respect to ′푥′ treating 푦 as constant, we get

∫ 푑푥 = ∫ 푑푥 + 푐

푁 = + ∅(푦) …(9) From (8) and (9), we have

푀푑푥 +푁푑푦 = 푑푥 + + 휙(푦) 푑푦

= 푑푥 + 푑푦 + 휙(푦)푑푦

= 푑푢 + 휙(푦)푑푦 = 푑[푢 + ∫휙(푦)푑푦] which is an exact differential equation. ∴ 푀 푑푥 + 푁푑푦 = 0 is an exact differential equation.

Introduction 13

Method of Solving → Compare the given equation with 푀푑푥 +푁푑푦 = 0 and find out 푀 and 푁. Then find out

and .

→ If = , we conclude that the given equation is exact. If the given equation is exact, then the general solution of exact equation 푀푑푥 + 푁푑푦 = 0 is

푥∫ 푀푑푥 + ∫(term of 푁 not involving 푥)푑푦 = 푐.

Solved Problems Example 1: Solve 푦 sin 2푥 푑푥 − (1 + 푦 + cos 푥)푑푦 = 0. Solution: Given 푦 sin 2푥 푑푥 − (1 + 푦 + cos 푥)푑푦 = 0 …(1)

푀 = 푦 sin 2푥 , 푁 = −(1 + 푦 + cos 푥)

= sin 2푥, = −2 cos 푥 (− sin 푥) = sin 2푥

∴ =

Given equation is exact differential equation. The general solution of (1) is

푥∫ 푀푑푥 + ∫(term of 푁 not involving 푥)푑푦 = 푐

∫푦 sin 2푥 푑푥 + ∫(−1 − 푦 )푑푦 = 푐

푦 − − 푦 − = 푐

3푦 cos 2푥 − 6푦 − 2푦 = 푐

Example 2: Solve 푦 1 + + cos 푦 푑푥 + (푥 + log 푥 − 푥 sin 푦)푑푦 = 푐.

Solution: Given 푦 1 + + cos 푦 푑푥 + (푥 + log 푥 − 푥 sin 푦)푑푦 = 푐 …(1)

푀 = 푦 1 + + cos푦 , 푁 = 푥 + log 푥 − 푥 sin 푦

= 1 + − sin 푦 , = 1 + − sin 푦

=

∴ Equation (1) is exact differential equation. The general solution of (1) is


∫ 푦 1 + + cos 푦 푑푥 + ∫0 푑푦 = 푐

푦∫ 1 + 푑푥 + ∫ cos 푦푑푥 = 푐

푦(푥 + log 푥) + 푥 cos푦 = 푐


Example 3: Solve (푒 + 1) cos 푥 푑푥 + 푒 sin 푥 푑푦 = 0. Solution: Given (푒 + 1) cos 푥 푑푥 + 푒 sin푥 푑푦 = 0 …(1)

푀 = (푒 + 1) cos 푥 , 푁 = 푒 sin 푥

= 푒 cos 푥 , = 푒 cos 푥

=

Equation (1) is exact differential equation. The general solution of (1) is


∫(푒 + 1) cos 푥 푑푥 + ∫0푑푦 = 푐 (푒 + 1) sin 푥 = 푐

Example 4: Solve (푥 − 2푥푦 − 푦 )푑푥 − (푥 + 푦 + 2푥푦)푑푦 = 0. Solution: Given (푥 − 2푥푦 − 푦 )푑푥 − (푥 + 푦 + 2푥푦)푑푦 = 0 …(1)

푀 = 푥 − 2푥푦 − 푦 , 푁 = −푥 − 푦 − 2푥푦

= −2푥 − 2푦, = −2푥 − 2푦

=

∴ Equation (1) is exact differential equation. The general solution of (1) is


∫푥 − 2푥푦 − 푦 푑푥 + ∫−푦 푑푦 = 푐

− 2푦 ∙ − 푥푦 − = 푐

푥 − 3푥 푦 − 3푥푦 − 푦 = 푐

Example 5: Solve + = 0.

Solution: Here, 푀 = 푎푥 + ℎ푦 + 푔,푁 = ℎ푥 + 푏푦 + 푓

= ℎ, = ℎ

=

Given equation is an exact differential equation. The general solution of given equation is


∫(푎푥 + ℎ푦 + 푔)푑푥 + ∫(푏푦 + 푓)푑푦 = 푐

+ ℎ푥푦 + 푔푥 + + 푓푦 = 푐

푎푥 + 2ℎ푥푦 + 푏푦 + 2푔푥 + 2푓푦 + 푐 = 0

Introduction 15

Example 6: Solve 푥푑푥 + 푦푑푦 + = 0.

Solution: Given 푥푑푥 + 푦푑푦 + = 0

푥 − 푑푥 + 푦 + 푑푦 = 0 …(1)

Here, 푀 = 푥 − , 푁 = 푦 +

= 0 − ( ) ∙( )

, = 0 + ( ) ∙( )

=( )

, =( )

=( )

, =( )

=

Equation (1) is an exact differential equation. The general solution of equation (1) is


∫ 푥 − 푑푥 + ∫푦푑푦 = 푐

− tan + = 푐

푥 − 2 tan + 푦 = 2푐

Example 7: Solve 1 + 푒 푑푥 + 푒 1− 푑푦 = 0.

Solution: Given 1 + 푒 푑푥 + 푒 1− 푑푦 = 0 …(1)

Here, 푀 = 1 + 푒 , 푁 = 푒 1−

= 0 + 푒 ∙ , = 푒 ∙ − 푒 ∙ ∙ + 푒 ∙

= − 푒 , = − 푒 −

= − 푒

=

∴ Equation (1) is an exact differential equation. ∴ The general solution of equation (1) is


∫ 1 + 푒 푑푥 + ∫0 푑푦 = 푐

푥 + = 푐

푥 + 푦푒 = 푐


Example 8: Solve (sin 푥 cos푦 + 푒 )푑푥 + (cos 푥 sin 푦 + tan 푦)푑푦 = 0. Solution: Given (sin 푥 cos푦 + 푒 )푑푥 + (cos 푥 sin푦 + tan 푦)푑푦 = 0 …(1)

Here, 푀 = sin 푥 cos푦 + 푒 , 푁 = cos 푥 sin 푦 + tan 푦

= − sin푥 sin 푦 , = − sin 푥 sin푦

=

∴ Equation (1) is an exact differential equation. The general solution of equation (1) is


∫(sin 푥 cos 푦 + 푒 )푑푥 + ∫ tan 푦푑푦 = 푐

− cos 푥 cos푦 + + log(sec 푦) = 푐

Example 9: Solve 푦 1 + + cos 푦 푑푥 + (푥 + log 푥 − 푥 sin 푦)푑푦 = 0.

Solution: Given 푦 1 + + cos 푦 푑푥 + (푥 + log 푥 − 푥 sin 푦)푑푦 = 0 …(1)

Here, 푀 = 푦 1 + + cos 푦 , 푁 = 푥 + log 푥 − 푥 sin푦

= 1 + − sin 푦 , = 1 + − sin 푦

=

∴ Equation (1) is an exact differential equation. The general solution of equation (1) is


푦 ∫ 1 + 푑푥 + cos 푦 ∫1푑푥 + ∫0푑푦 = 푐

푦(푥 + log 푥) + 푥 cos 푦 = 푐 Example 10: Solve (푟 + sin 휃 − cos휃)푑푟 + 푟(sin휃 + cos 휃)푑휃 = 0. Solution: Given (푟 + sin 휃 − cos휃)푑푟 + 푟(sin휃 + cos 휃)푑휃 = 0 …(1)

Here, 푀 = 푟 + sin 휃 − cos휃 , 푁 = 푟(sin 휃 + cos 휃)

= sin휃 + cos 휃 , = sin 휃 + cos휃

=

Equation (1) is an exact differential equation. The general solution of (1) is


∫(푟 + sin휃 − cos 휃)푑푟 + ∫0푑휃 = 푐

+ 푟(sin 휃 − cos 휃) = 푐

Introduction 17

Example 11: Find the values of constant 휆 such that (2푥푒 + 3푦 ) + (3푥 + 휆푒 ) = 0 is exact. Further, for this values of 휆, solve the equation.

Solution: Given equation is (3푥 + 휆푒 )푑푥 + (2푥푒 + 3푦 )푑푦 = 0 …(1) 푀 = (3푥 + 휆푒 ), 푁 = 2푥푒 + 3푦

= 휆푒 , = 2푒

∴ Equation (1) is an exact differential equation. 휆푒 = 2푒 => 휆 = 2 The general solution of equation (1) is ∫(3푥 + 2푒 )푑푥 + ∫3푦 푑푦 = 푐 푥 + 2푒 푥 + 푦 = 푐


(1) (푥 + 푥푦 + 푎 푦)푑푥 + (푦 + 푦푥 − 푎 푥)푑푦 = 0 (2) (푥 + 푦 )(푥푑푥 + 푦푑푦) = 푥푑푦 − 푦푑푥 (3) (푥 + 푦 + 푥)푑푥 − (2푥 + 2푦 − 푦)푑푦 = 0 (4) (푥 − 4푥푦 − 2푦 )푑푥 + (푦 − 4푥푦 − 2푥 )푑푦 = 0 (5) (2푥푦 + 푦 − tan 푦)푑푥 + (푥 − 푥 tan 푦 + sec 푦)푑푦 = 0 (6) (2푥푦 cos 푥 − 2푥푦 + 1)푑푥 + (sin 푥 − 푥 )푑푦 = 0 (7) (sec 푥 tan 푥 tan 푦 − 푒 )푑푥 + sec 푥 sec 푦 푑푦 = 0 (8) 푦푒 푑푥 + (푥푒 + 2푦)푑푦 = 0

Answers: (1) 푥 + 2푥 푦 + 4푎 푦 + 푦 = 푐

(2) 푥 + 푦 + 2 tan

(3) 푥 + log(푥 + 푦 )− 2푦 = 푐

(4) 푥 + 푦 − 6푥푦(푥 + 푦) = 푐 (5) 푥 푦 + 푥푦 − 푥 tan 푦 + tan푦 = 푐 (6) 푦 sin 푥 − 푥 푦 + 푥 = 푐 (7) sec 푥 tan푦 − 푒 = 푐 (8) 푒 + 푦 = 푐

Equations Reducible to Exact Form Integrating factors: An equation of the form 푀푑푥 + 푁푑푦 = 0 is not exact, it can always be

made exact by multiplying by some function of 푥 and 푦. Such a multiplier is called an integrating factor.


Methods to Find Integrating Factors of 푴풅풙 + 푵풅풚 = ퟎ Method 1: By Inspection

An integrating factor of given equation 푀푑푥 +푁푑푦 = 0 can be found by inspection as explained below.

By rearranging the terms of given equation (or) by dividing with a suitable function of 푥 and 푦, the equation thus obtained will contain several parts integrable easily.

In this connection, the following exact differentials will be found useful.

(i) 푑(푥푦) = 푥푑푦 + 푦푑푥

(ii) 푑(log(푥푦)) =

(iii) 푑 =

(iv) 푑 =

(v) 푑 =

(vi) 푑 =

(vii) 푑 =

(viii) 푑 =

(ix) 푑 tan =

(x) 푑 tan =

(xi) 푑 =

(xii) 푑 =

(xiii) 푑 log =

(xiv) 푑 log =

(xv) 푑 = −

(xvi) 푑 − =

Solved Problems Example 1: Solve (1 + 푥푦)푦푑푥 + (1 − 푥푦)푥푑푦 = 0. Solution: Given(1 + 푥푦)푦푑푥 + (1 − 푥푦)푥푑푦 = 0

푥푑푦 + 푦푑푥 + 푥푦 푑푥 − 푥 푦푑푦 = 0 푑(푥푦) + 푥푦 푑푥 − 푥 푦푑푦 = 0

Introduction 19

Dividing by 푥 푦 , we get

−푑 + 푑푥 − 푑푦 = 0


−∫푑 + ∫ 푑푥 − ∫ 푑푦 = 푐

− + log 푥 − log푦 = 푐

− + log = 푐

log = 푐 +

Example 2: Solve (푥 푒 −푚푦 )푑푥 +푚푥푦푑푦 = 0. Solution: Given (푥 푒 −푚푦 )푑푥 +푚푥푦푑푦 = 0

Dividing by 푥 , we get

푒 푑푥 +푚 = 0

푒 푑푥 + = 0

푒 푑푥 + 푚푑 = 0


푒 + = 푐

2푥 푒 +푚푦 = 2푐푥 Example 3: Solve 푥푑푦 − 푦푑푥 = 푎(푥 + 푦 )푑푦.

Solution: Given equation is = 푎푑푦

푑 tan = 푎푑푦


∫푑 tan = ∫푎푑푦 + 푐

tan = 푎푦 + 푐

Theorem: The number of integrating factor for the equation 푀푑푥 +푁푑푦 = 0 is infinite. Proof: Let 휇 be an integrating factor. Then by definition 휇(푀푑푥 +푁푑푦) = 0 is an exact different equation. Therefore, there exists a function 푈(푥,푦) such that

푑푈 = 휇(푀푑푥 + 푁푑푦) ∴ 푢 = Constant is a solution Let 푓(푢) be any function of 푢. 푓(푢)푑푢 = 휇푓(푢)(푀푑푥 + 푁푑푦)


Since the term on left is an exact differential, the term on right must also be an exact differential. ∴ 휇푓(푢) is an integrating factor of the equation.

Since 푓(푢) is an arbitrary function of 푢, the number of integrating factor is infinite. Example 4: Solve 푎(푥푑푦 + 2푦푑푥) = 푥푦푑푦. Solution: Given 푎(푥푑푦 + 2푦푑푥) = 푥푦푑푦

The equation is not an exact differential equation. Dividing by 푥푦, we get

푎 + 2 = 푑푦


푎 ∫ + 2∫ = ∫푑푦 + 푐

푎 log푦 + 2 log 푥 = 푦 + 푐 푎 log푦푥 − 푦 = 푐

Example 5: Solve 푦푑푥 − 푥푑푦 + log 푥 푑푥 = 0. Solution: Given 푦푑푥 − 푥푑푦 + log 푥 푑푥 = 0

Given equation is not an exact differential equation. Dividing by 푥 , we get

− + 푑푥 = 0

∫−푑 + ∫ log 푥 푑푥 = ∫0

∫ log 푥 푑푥 − = 푐

=> − + ∫ 푑푥 − = 푐

− − = 푐

푦 + log 푥 + 1 + 푐푥 = 0 Example 6: Solve 푦(2푥푦 + 푒 )푑푥 − 푒 푑푦 = 0. Solution: Given 푦(2푥푦 + 푒 )푑푥 − 푒 푑푦 = 0

Dividing by 푦 , we get

2푥푑푥 + 푑 = 0

푑 푥 + = 0

푥 + = 푐

Example 7: Solve 푦푑푥 − 푥푑푦 + (1 + 푥 )푑푥 + 푥 sin 푦푑푦 = 0. Solution: Given 푦푑푥 − 푥푑푦 + (1 + 푥 )푑푥 + 푥 sin푦 푑푦 = 0

Dividing (1) by 푥 , we get

Introduction 21

=> + + 1 푑푥 + sin 푦 푑푦 = 0

=> −푑 + + 1 푑푥 + sin푦 푑푦 = 0

=> −∫푑 + ∫ + 1 푑푥 + ∫ sin 푦푑푦 = 푐

=> − − + 푥 − cos 푦 = 푐

General solution of (1) is 푥 − 푦 − 1 − 푥 cos 푦 = 푐푥 Example 8: Solve 푦(2푥 푦 + 푒 )푑푥 − (푒 + 푦 )푑푦 = 0. Solution: Given 푦(2푥 푦 + 푒 )푑푥 − (푒 + 푦 )푑푦 = 0

=> 2푥 푦 푑푥 + 푦푒 푑푥 − 푒 푑푦 − 푦 푑푦 = 0 …(1) Dividing by 푦 , we get

2푥 푑푥 + − 푦푑푦 = 0

=> 2푥 푑푥 + 푑 − 푦푑푦 = 0


2∫푥 푑푥 + ∫푑 − ∫푦푑푦 = 푐

+ − = 푐

Example 9: Solve 푦 sin 2푥 푑푥 = (1 + 푦 + cos 푥)푑푦. Solution: Given 푦 sin 2푥 푑푥 = (1 + 푦 + cos 푥)푑푦

=> 2푦 sin 푥 cos 푥 푑푦 − cos 푥 푑푦 = (1 + 푦 )푑푦 => −푑(푦 cos 푥) = (1 + 푦 )푑푦 => ∫−푑(푦 cos 푥) = ∫(1 + 푦 )푑푦

=> −푦 cos 푥 = 푦 + + 푐

=> 푦 + 3푦 + 3푦 cos 푥 = 푐

Example 10: Solve 푦 + cos푦 +√

푑푥 + (푥 − 푥 sin 푦 − 1)푑푦 = 0.

Solution: Given 푦 + cos푦 +√

푑푥 + (푥 − 푥 sin 푦 − 1)푑푦 = 0

(푦푑푥 + 푥푑푦) + (cos 푦푑푥 − 푥 sin 푦)푑푦 +√푑푥 − 푑푦 = 0

푑(푥푦) + 푑(푥 cos푦) +√푑푥 − 푑푦 = 0


∫푑(푥푦) + ∫푑(푥 cos 푦) + ∫√푑푥 − ∫푑푦 = 0

푥푦 + 푥 cos푦 + √푥 − 푦 = 푐



(1) 푥푑푦 − (푦 − 푥)푑푥 = 0 (2) 푦푑푥 − 푥푑푦 = 3푥 푒 푦 푑푥 (3) 푦(푥푦 + 푒 )푑푥 − 푒 푑푦 = 0 (4) (푥 + 푦 − 2푦)푑푦 = 2푥푑푥

(5) 푥푑푦 = 푦 + 푥 cos 푑푥

(6) (푥 + 푥푦 + 푎 푦)푑푥 + (푦 + 푦푥 − 푎 푥)푑푦 = 0 (7) (푥 + 푧) 푑푦 + 푦 (푑푥 + 푑푧) = 0 (8) (1 + 푥푦 )푑푥 + (1 + 푥 푦)푑푦 = 0 (9) sec 푥 tan 푦 푑푥 + sec 푦 tan 푥 푑푦 = 0

Answers: (1) + log 푥 = 푐

(2) 푥 = 푦푒 + 푐푦

(3) + = 푐

(4) 푦 = log(푥 + 푦 ) + 푐 (5) tan = log 푥 + 푐

(6) 푥 + 푦 + 2푎 tan = 푐

(7) 푦(푥 + 푧) = 푐(푥 + 푦 + 푧) (8) 2(푥 + 푦) + 푥 푦 = 푐 (9) tan 푥 ∙ tan 푦 = 푐 Note: When it is not possible to find out the integrating factor by inspection, it is found by certain rules given below.

The following table may help us in finding integrating factor for some particular cases of equation.

Case Condition on equations Integration factors (I.F.) 흁 I 푀 and 푁 are homogeneous functions of

푥 and 푦 휇 =1

푀푥 + 푁푦 ; 푀푥 + 푁푦 ≠ 0

II 푀 = 푦푓(푥,푦); 푁 = 푥푔(푥,푦) 휇 =1

푀푥 + 푁푦 ; 푀푥 + 푁푦 ≠ 0

III −

푁 = 푓(푥) 휇 = 푒∫ ( )

IV −

푀 = 푓(푦) 휇 = 푒∫ ( )

Introduction 23

Rule I: Theorem: If the given equation 푀푑푥 +푁푑푦 = 0 is a homogeneous and 푀푑푥 + 푁푑푦 ≠ 0, then is an integrating factor.

Proof: Given 푀푑푥 +푁푑푦 = 0 is a homogeneous differential equation

∴ 푀 = 푥 푓 ,푁 = 푥 푔 …(1)

By Euler Theorem on homogeneous functions, we have

푥 + 푦 = 푘푀 …(2)

and 푥 + 푦 = 푘푁 …(3)

Let us take 푀 = and 푁 =

Then =

=( )

( )

=( )

Similarly, =

=( )

( )

=( )

Now, − =( )

=( )

= ( ) ( )( )

= 0

∴ =

∴ 푀 푑푥 +푁 푑푦 = 0 is an exact differential equation.

=> (푀푑푥 +푁푑푦) = 0 is an exact differential equation

Hence, is an integrating factor of 푀푑푥 + 푁푑푦 = 0

Method of Solving → General equation is 푀푑푥 +푁푑푦 = 0 …(1)

Observe ≠

=> (1) is not exact → Observe 푀 and 푁 are homogeneous functions of same order


→ Find 푀푥 +푁푦 and observe it ≠ 0

Then is an integrating factor of equation (1)

→ Multiply equation (1) with integrating factor to transform into an exact equation of (1) 푀 푑푥 + 푁 푑푦 = 0 …(2) → Solve equation (2), we get the general solution of (1)

Example 1: Solve (푥 푦 − 2푥푦 )푑푥 − (푥 − 3푥 푦)푑푦 = 0 Solution: Here, 푀 = 푥 푦 − 2푥푦 , 푁 = −푥 + 3푥 푦

= 푥 − 4푥푦, = −3푥 + 6푥푦

≠

∴ Given equation is not exact. But homogeneous equation in 푥 and 푦 Now, 푀푥 +푁푦 = 푥(푥 푦 − 2푥푦 ) + 푦(−푥 + 3푥 푦) = 푥 푦 − 2푥 푦 − 푥 푦 + 3푥 푦 = 푥 푦 ≠ 0

∴ is an integrating factor of given equation.

Multiplying given equation with , it becomes exact

[(푥 푦 − 2푥푦 )푑푥 − (푥 − 3푥 푦)푑푦 = 0]

( )푑푥 − ( )푑푦 = 0 …(2)

General solution of (2) is

푥∫ 푀 푑푥 + ∫(term of 푁 not involving 푥)푑푦 = 푐

∫ ( )푑푥 − ∫− 푑푦 = 푐

∫ 푑푥 − ∫ 푑푥 + 3∫ 푑푦 = 푐

− 2 log 푥 + 3 log푦 = 푐

Example 2: Solve 푥 푦푑푥 − (푥 + 푦 )푑푦 = 0. Solution: Given equation is 푥 푦푑푥 − (푥 + 푦 )푑푦 = 0 …(1)

Here, 푀 = 푥 푦, 푁 = −푥 − 푦

= 푥 , = −3푥

≠

∴ Equation (1) is not exact. And homogeneous equation in 푥 and 푦 Now, 푀푥 +푁푦 = (푥 푦)푥 + (−푥 − 푦 )푦 = 푥 푦 − 푥 푦 − 푦 = −푦 ≠ 0

Introduction 25

Multiplying (1) with , it becomes exact equation

[푥 푦푑푥 − (푥 + 푦 )푑푦 = 0]

− 푑푥 + 푑푦 = 0 …(2)


푥∫ 푀 푑푥 + ∫(term of 푁 not involving 푥)푑푦 = 푐

−∫ 푑푥 + ∫ 푑푦 = 푐

− + log푦 = 푐

− + log푦 = 푐

Example 3: Solve (푥 + 푦 )푑푥 − 푥푦 푑푦 = 0. Solution: Given (푥 + 푦 )푑푥 − 푥푦 푑푦 = 0 …(1)

Here, 푀 = 푥 + 푦 , 푁 = −푥푦

= 4푦 , = −푦

≠

∴ Equation (1) is not exact. And homogeneous equation in 푥 and 푦 Now, 푀푥 +푁푦 = (푥 + 푦 )푥 + (−푥푦 )푦 = 푥 + 푥푦 − 푥푦 = 푥 ≠ 0

Multiplying (1) with , it becomes exact

[(푥 + 푦 )푑푥 − 푥푦 푑푦 = 0]

푑푥 − 푑푦 = 0 …(2)


∫ 푑푥 + 푦 ∫ − ∫0푑푦 = 푐

log 푥 + = 푐

4푥 log 푥 + 푦 = 푐푥 Example 4: Solve 푦 푑푥 + (푥 − 푥푦 − 푦 )푑푦 = 0. Solution: Given 푦 푑푥 + (푥 − 푥푦 − 푦 )푑푦 = 0 …(1)

Here, 푀 = 푦 , 푁 = 푥 − 푥푦 + 푦

= 2푦, = 2푥 − 푦

≠

∴ Equation (1) is not exact.


And homogeneous function of 푥 and 푦 Now, 푀푥 +푁푦 = (푦 )푥 + (푥 − 푥푦 + 푦 )푦 = 푥푦 + 푥 푦 − 푥푦 − 푦 = 푥 푦 − 푦 ≠ 0

Multiplying (1) with ( )

, it becomes exact

( )

[푦 푑푥 + (푥 − 푥푦 − 푦 )푑푦] = 0

( )

푑푥 +( )

푑푦 = 0 …(2)


푦 ∫ 푑푥 + ∫ 푑푦 = 푐

푦 ∙ log + log푦 = 푐

log 푦 = log 푐

(푥 − 푦)푦 = (푥 + 푦)푐

Example 5: Solve 푦 + 푥 = 푥푦 .

Solution: Given 푦 푑푥 + (푥 − 푥푦)푑푦 = 0 …(1) Here, 푀 = 푦 , 푁 = 푥 − 푥푦

= 2푦, = 2푥 − 푦

≠

∴ Equation (1) is not exact. And homogeneous function of 푥 and 푦 Now, 푀푥 +푁푦 = 푥푦 + (푥 푦 − 푥푦 ) = 푥 푦 ≠ 0

Multiplying (1) with , it becomes exact

[푦 푑푥 + (푥 − 푥푦)푑푦] = 0 …(2)


푦 ∫ 푑푥 + ∫ 푑푦 = 푐

+ log푦 = 푐

Exercise: Solve the following equations.

(1) 푥푦푑푥 − (푥 + 2푦 )푑푦 = 0

(2) 푦 − 푥 = 푥 + 푦

(3) 푟(휃 + 푟 )푑휃 − 휃(휃 + 2푟 )푑푟 = 0 (4) (3푥푦 − 푦 )푑푥 − (2푥 푦 − 푥푦 )푑푦 = 0 (5) (푥 + 푦 )푑푥 − 푥푦 푑푦 = 0

Introduction 27

Answers: (1) 4푦 log푦 − 푥 = 푐푦

(2) log 푥 + 푦 − tan = 푐

(3) – − log휃 + log 푟 = 푐

(4) 3 log 푥 − 2 log푦 + = 푐

(5) log 푥 − = 푐

Rule II: Theorem: If the equation 푀푑푥 +푁푑푦 = 0 is of the form 푦푓(푥푦)푑푥 + 푥푔(푥푦)푑푦 = 0 and

푀푥 −푁푦 ≠ 0, then is an integrating factor of 푀푑푥 + 푁푑푦 = 0.

Proof: Given equation is 푀푑푥 + 푁푑푦 = 0 …(1) Comparing (1) with 푦푓(푥푦)푑푥 + 푥푔(푥푦)푑푦 = 0, We have 푀 = 푦푓(푥푦), 푁 = 푥푔(푥푦)

Let 푥푦 = 푢 => = 푦, = 푥 …(2)

Let us take 푀 = , 푁 =

푀 = ( )( ) ( )

= ( )( )

푁 = ( )( ) ( )

= ( )( )

Let = 푓 and = 푔

Now =( ) ( ) ( ) ( ) ( )

( ) ( )

= ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

= ∙ 푥 ( ) ( ) ( ) ( )

( ) ( )=

( ) …(3)

=( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

= ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

=( )

…(4)

From (3) and (4), we have

= => 푀 푑푥 +푁 푑푦 = 0 is an exact equation

=> 푑푥 + 푑푦 = 0 is an exact equation

Hence, is an integrating factor of 푀푑푥 + 푁푑푦 = 0


Solved Problems Example 1: Solve 푦(푥 푦 + 푥 푦 + 푥푦 + 1)푑푥 + (푥 푦 − 푥 푦 − 푥푦 + 1)푥푑푦 = 0. Solution: Given equation

(푥 푦 + 푥 푦 + 푥푦 + 1)푦푑푥 + (푥 푦 − 푥 푦 − 푥푦 + 1)푥푑푦 = 0 …(1) Here, 푀 = 푥 푦 + 푥 푦 + 푥푦 + 푦, 푁 = 푥 푦 − 푥 푦 − 푥 푦 + 푥

= 4푥 푦 + 3푥 푦 + 2푥푦 + 1, = 4푥 푦 − 3푥 푦 − 2푥푦 + 1

≠

Equation (1) is not exact and it is in the form of 푦푓(푥푦)푑푥 + 푥푔(푥푦)푑푦 = 0

Then is an integrating factor of (1)

Now, 푀푥 −푁푦 = 푥 푦 + 푥 푦 + 푥 푦 + 푥푦 − 푥 푦 + 푥 푦 + 푥 푦 − 푥푦 = 2푥 푦 + 2푥 푦 = 2푥 푦 (푥푦 + 1) ≠ 0

Multiplying equation (1) with ( )

, it becomes exact

( )

[(푥 푦 + 푥 푦 + 푥푦 + 1)푦푑푥 + (푥 푦 − 푥 푦 − 푥푦 + 1)푥푑푦 = 0] …(2)

( ) ( )

( )+ ( )( )

( )= 0

( )( )( )

푑푥 + ( ) 푑푦 = 0 …(3)


∫ 푑푥 + ∫− 푑푦 = 푐

− − log푦 = 푐

푥푦 − − log푦 = 푐

Example 2: Solve 푦(1 + 푥푦)푑푥 + 푥(1 − 푥푦)푑푦 = 0. Solution: Given 푦(1 + 푥푦)푑푥 + 푥(1 − 푥푦)푑푦 = 0 …(1)

Here, 푀 = 푦 + 푥푦 , 푁 = 푥 − 푥 푦

= 1 + 2푥푦, = 1 − 2푥푦

≠



Now, 푀푥 −푁푦 = 푥푦 + 푥 푦 − 푥푦 + 푥 푦 = 2푥 푦 ≠ 0

Multiplying equation (1) with , it becomes exact

[(푦 + 푥푦 )푑푥 + (푥 − 푥 푦)푑푦 = 0] …(2)

Introduction 29


∫ 푑푥 + ∫ 푑푥 − ∫ 푑푦 = 푐


− + log = 2푐

Example 3: Solve (푥 푦 + 푥 푦 + 푥푦)푦푑푥 + (푥 푦 − 푥 푦 + 푥푦)푥푑푦 = 0. Solution: Given (푥 푦 + 푥 푦 + 푥푦)푦푑푥 + (푥 푦 − 푥 푦 + 푥푦)푥푑푦 = 0 …(1)

Here, 푀 = 푥 푦 + 푥 푦 + 푥푦 , 푁 = 푥 푦 − 푥 푦 + 푥 푦

= 5푥 푦 + 3푥 푦 + 2푥푦, = 5푥 푦 − 3푥 푦 + 2푥푦

≠



Now, 푀푥 −푁푦 = 푥 푦 + 푥 푦 + 푥 푦 − 푥 푦 + 푥 푦 − 푥 푦 = 2푥 푦 ≠ 0


[(푥 푦 + 푥 푦 + 푥푦 )푑푥 + (푥 푦 − 푥 푦 + 푥 푦 + 1)푑푦 = 0] …(2)

The general solution of (2) is

∫ 푥푦 + + 푑푥 + ∫− 푑푦 = 푐

+ log 푥 − − log푦 = 푐

+ log 푥 − log푦 − = 2푐

+ log − = 푐

Example 4: Solve (푥푦 sin 푥푦 + cos 푥푦)푦푑푥 + (푥푦 sin 푥푦 − cos 푥푦)푥푑푦 = 0 Solution: Given (푥푦 sin 푥푦 + cos 푥푦)푦푑푥 + (푥푦 sin 푥푦 − cos 푥푦)푥푑푦 = 0 …(1)

Here, 푀 = 푥푦 sin푥푦 + 푦 cos 푥푦 , 푁 = 푥 푦 sin 푥푦 − 푥 cos 푥푦

= 2푥푦 sin푥푦 + 푥푦 cos 푥푦 (푥) + cos 푥푦 + 푦(− sin 푥푦)(푥)

= 2푥푦 sin 푥푦 + 푥 푦 cos 푥푦 + cos 푥푦 − 푥푦 sin푥푦

= (푥 푦 + 1) cos 푥푦 + 푥푦 sin푥푦

= 푦2푥 sin 푥푦 + 푥 푦 cos 푥푦 (푦)− cos 푥푦 + 푥 sin 푥푦 . 푦

= 2푥푦 sin 푥푦 + 푥 푦 cos 푥푦 − cos 푥푦 + 푥푦 sin 푥푦

= 3푥푦 sin 푥푦 + (푥 푦 − 1) cos 푥푦


≠

Equation (1) is not exact. Now, 푀푥 −푁푦 = 푥 푦 sin푥푦 + 푥푦 cos 푥푦 − 푥 푦 sin 푥푦 + 푥푦 cos 푥푦 = 2푥푦 cos 푥푦 ≠ 0 Multiplying equation (1) with , it becomes exact equation

[(푥푦 sin 푥푦 + 푦 cos 푥푦)푑푥 + (푥 푦 sin 푥푦 − 푥 cos 푥푦)푑푦 = 0] …(2)


∫ ( ) 푑푥 + ∫− 푑푦 = 푐

∫ tan 푥푦 푑푥 + ∫ 푑푥 − ∫ 푑푦 = 푐

( ) + log 푥 − log푦 = 푐

푦 log ( ) + log = 2푐

log(sec 푥푦) = log 푐

푥 sec 푥푦 = 푐푦 Example 5: Solve (푥푦 + 2푥 푦 )푦푑푥 + (푥푦 − 푥 푦 )푥푑푦 = 0. Solution: Given (푥푦 + 2푥 푦 )푦푑푥 + (푥푦 − 푥 푦 )푥푑푦 = 0 …(1)

Here, 푀 = 푥푦 + 2푥 푦 , 푁 = 푥 푦 − 푥 푦

= 2푥푦 + 6푥 푦 , = 2푥푦 − 3푥 푦

≠

Equation (1) is not exact. Now, 푀푥 −푁푦 = 푥 푦 + 2푥 푦 − 푥 푦 + 푥 푦 = 3푥 푦 ≠ 0


[(푥푦 + 2푥 푦 )푑푥 + (푥 푦 − 푥 푦 )푑푦 = 0] …(2)


∫ 푑푥 + ∫ 푑푥 + ∫− 푑푦 = 푐


− + 2 log = 푐

Example 6: Solve (푥 푦 + 푥푦 + 1)푦푑푥 + (푥 푦 − 푥푦 + 1)푥푑푦 = 0. Solution: Given (푥 푦 + 푥푦 + 1)푦푑푥 + (푥 푦 − 푥푦 + 1)푥푑푦 = 0 …(1)

Here, 푀 = 푥 푦 + 푥푦 + 푦, 푁 = 푥 푦 − 푥 푦 + 푥

Introduction 31

= 3푥 푦 + 2푥푦 + 1, = 3푥 푦 − 2푥푦 + 1

≠

Equation (1) is not exact. Now, 푀푥 −푁푦 = 푥 푦 + 푥 푦 + 푥푦 − 푥 푦 + 푥 푦 − 푥푦 = 2푥 푦 ≠ 0


[(푥 푦 + 푥푦 + 푦)푑푥 + (푥 푦 − 푥 푦 + 푥)푑푦 = 0] …(2)


∫푑푥 + ∫ 푑푥 + ∫ 푑푥 − ∫ 푑푦 = 푐

+ log 푥 + − log푦 = 푐

− + 푥푦 + log = 푐


(1) (20푥 + 8푥푦 + 4푦 + 3푦 )푦푑푥 + 4(푥 + 푥푦 + 푦 + 푦 )푥푑푦 = 0 (2) (3푥 + 2푦 )푦푑푥 + 푥(4푥 + 6푦 )푑푦 = 0 (3) 푦(1 − 푥푦)푑푥 + 푥(1 + 푥푦)푑푦 = 0 (4) (2푥푦 + 1)푦푑푥 + (1 + 2푥푦 − 푥 푦 )푥푑푦 = 0

Answers:

(1) 4푥 + 2푥 푦 + 푥 푦 + 푥 푦 푦 = 푐

(2) 푥 푦 (푥 + 푦 ) = 푐

(3) 푥푦 log − 1 = 푐푥푦

(4) + + log푦 = 푐

Rule III: To find an integrating factor of 푴풅풙 +푵풅풚 = ퟎ

Theorem: If − is a function 푥 alone say 푓(푥), then 푒∫ ( ) is an integrating factor of 푀푑푥 +푁푑푦 = 0.

Proof: Given equation is 푀푑푥 + 푁푑푦 = 0 …(1)

and − = 푓(푥) so that

푁푓(푥) = − …(2)

Multiplying both sides of (1) by 푒∫ ( ) , we have 푀 푑푥 + 푁 푑푦 = 0 …(3)

where 푀 = 푀푒∫ ( ) and 푁 = 푁푒∫ ( )


Then = 푒∫ ( )

and = 푒∫ ( ) + 푁푒∫ ( ) ∙ 푓(푥)

= 푒∫ ( ) +푁푓(푥)

= 푒∫ ( ) + − (by (2))

= 푒∫ ( ) =

∴ The equation 푀 푑푥 +푁 푑푦 = 0 is exact.

Hence, 푒∫ ( ) is an integrating factor for the equation 푀푑푥 +푁푑푦 = 0. Method of Solving

→ Given equation is 푀푑푥 + 푁푑푦 = 0 …(1)

Observe ≠

=> Equation (1) not exact

→ Find − and observe it is a function of 푥 alone = 푓(푥) or real constant 푘

→ Then 푒∫ ( ) (or) 푒∫ is an integrating factor of (1) → Multiply equation (1) with integrating factor to transform it into exact equation of (1) i.e., 푀 푑푥 +푁 푑푦 = 0 …(2) → Solve equation (2), we get a general solution of (1)

Solved Problems Example 1: Solve (푥 + 푦 + 푥)푑푥 + 푥푦푑푦 = 0. Solution: Given (푥 + 푦 + 푥)푑푥 + 푥푦푑푦 = 0 …(1)

Here, 푀 = 푥 + 푦 + 푥, 푁 = 푥푦

= 2푦, = 푦

≠

Equation (1) is not exact.

Now, − = (2푦 − 푦) = = = 푓(푥) say

Integrating factor = 푒∫ ( ) = 푒∫ = 푒 = 푥 Multiplying equation (1) with ′푥 , it becomes exact 푥[(푥 + 푦 + 푥)푑푥 + 푥푦푑푦 = 0] …(2) General solution of (2) is ∫푥 푑푥 + 푦 ∫푥푑푥 + ∫푥 푑푥 + ∫0푑푦 = 푐

+ + = 푐

4푥 + 3푥 + 6푥 푦 = 푐

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