NCERT Solutions Class 12 Mathematics Chapter 9

The fundamentals of calculus are used to present a new topic called Differential Equations in NCERT Solutions for class 12 mathematics chapter 9. Students were previously taught how to differentiate a function f about an independent variable, that is, how to find f ′(x) for a given function f at each x in its domain of definition. Students were also taught how to find a function f whose derivative is the function g. An equation containing the derivative of the dependent variable concerning the independent variable can be expressed as a formal definition for differential equations using these concepts. These equations can also be classified as:

  • Ordinary Differential Equations - Differential Equations that involve derivatives of the dependent variable concerning only one independent variable.
  • Partial Differential Equations - Differential Equations that involve derivatives concerning many independent variables. 

These Equations have been explained in-depth with example problems in NCERT solutions for class 12 Mathematics chapter 9. Another aspect of calculus shown in NCERT answers for class 12 Mathematics chapter 9 is that students can effortlessly make a transition to solving the problems in this lesson once they are familiar with the various ways of differentiating and integrating a function. Differential equations in NCERT Solutions class 12 Mathematics chapter 9 have applications in determining population growth and decay, glucose absorption by the body, gauging epidemic spread, and solving various geometric questions involving various types of curves.

Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 9

This chapter introduces students to the fundamentals of differential equations. The various topics covered in this chapter assist students in comprehending the real-world applications of differential equations. Students can learn the definition of differential equations here, which can aid in their understanding of the concepts. The order and degree of differential equations are also taught to students. The creation of ordinary differential equations and the solution of a differential equation are two additional essential concepts addressed in this chapter. These are crucial elements for students to understand as they prepare for the main exam.

Every topic in NCERT Solutions Class 12 Mathematics Chapter 9 is equally essential because it focuses on distinct elements of differential equations. Furthermore, each section is linked to the next. Thus, students must devote equal and enough time to all sections of this chapter, so as to avoid any superficial knowledge or learning gaps. It also gives a good review of the previous topic matter covered in the session.

List of NCERT Solutions Class 12 Mathematics Chapter 9 Exercises

Because of Gottfried Wilhelm Freiherr Leibnitz (1646 - 1716), who gave the calculus identity to Mathematics, the concept of differential equations was born on November 11, 1675. Leibnitz was more engaged in finding a curve with stipulated tangents. As a result, he discovered a slew of related differential equation principles. The equations took on their current shape due to various researchers working on the topic's intricacy. Many of these facts and advice to help youngsters overcome boredom and study with excitement may be found in the NCERT solutions for class 12 Mathematics chapter 9.

Differential equations in NCERT Solutions Class 12 Mathematics Chapter 9 are challenging and have lengthy lessons. As a result, the only method to master the chapter is to review the contents frequently. It will be easier for students to progress through the chapter and build clear concepts if they have a strong foundation in the fundamentals. Extramarks has provided a full examination of all the exercise questions below to assist students in pursuing high-quality education.

Class 12 Mathematics Chapter No. 9 Ex 9.1 - 12 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.2 - 12 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.3 - 12 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.4 - 23 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.5 - 17 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.6 - 19 Questions

Class 12 Mathematics Chapter No. 9 Miscellaneous Ex - 18 Questions

NCERT Solutions Class 12 Mathematics Chapter 9 Formula List

NCERT solutions class 12 Mathematics chapter 9 solves differential equations using the formulas provided in earlier calculus lessons. This chapter focuses on formulating and solving differential equations under particular conditions. As a result, to successfully combine both lectures, it is critical to review the calculus concepts taught previously. There are some terminologies in the NCERT solutions for class 12 Mathematics chapter 9 that students should be familiar with. A  few of which are as under:

  • The General form of a differential equation: dy/dx = g(x), where y = f(x).
  • The general form of nth order derivative: dny/dxn.
  • The general form of a linear differential equation: dy/dx + Py = Q

Class 12 NCERT Mathematics Syllabus

Term - 1

Unit Name

 

Chapter Name
 

Relations and Function

Relations and Functions

Inverse Trigonometric Functions

Algebra

 

Matrices

Determinants

Calculus

 

Continuity and Differentiability

Application of Derivatives 

Linear Programming Linear Programming

Term - 2

Unit Name Chapter Name
 

Calculus

 

Integrals

Application of Integrals

Differential Equations

Vectors and Three-Dimensional Geometry  Vector Algebra

Three Dimensional Geometry 

Probability Probability 

 Subject experts at Extramarks create NCERT Solutions to assist students in understanding concepts more quickly and correctly. NCERT Solutions provide extensive, step-by-step explanations of textbook problems. All classes can benefit from such solutions –

  • NCERT Solutions class 1
  • NCERT Solutions class 2
  • NCERT Solutions class 3
  • NCERT Solutions class 4
  • NCERT Solutions class 5
  • NCERT Solutions class 6
  • NCERT Solutions class 7
  • NCERT Solutions class 8
  • NCERT Solutions class 9
  • NCERT Solutions class 10
  • NCERT Solutions class 11
  • NCERT Solutions class 12

NCERT CBSE Mathematics Exam Pattern

Duration of Marks 3 hours 15 minutes
Marks for Internal 20 marks
Marks for Theory 80 marks
Total Number of Questions 38 Questions
Very short answer question 20 Questions
Short answer questions 7 Questions
Long Answer Questions (4 marks each) 7 Questions
Long Answer Questions (6 marks each) 4 Questions

Key Features of NCERT Solutions for Class 12 Mathematics Chapter 9

Using the NCERT Solutions to learn the chapter, Differential Equations, students will be able to understand the following:

  • A Differential Equation's definition, order and degree.
  • General and Special Solutions. 
  • The formation of a differential equation with a given general solution. 
  • Solutions of homogeneous differential equations of the first order and first degree.
  • Solutions of differential equations using the separation of variables approach.

NCERT Exemplar Class 12 CBSE Mathematics 

NCERT Exemplars contain solutions and problems that help students prepare for their final exams. These example questions are a little more complex, and they cover every concept in each chapter of the Class 12 Mathematics subject.

Students will fully understand all the concepts covered in each chapter by practising these NCERT Exemplars for Mathematics Class 12. Each question in these materials is connected to topics covered in the CBSE Class 12 syllabus (2022-2023). They provide some of the best solutions to challenges that students confront. To match the ideas taught in each class and provide the greatest practising materials or worksheets for students, all of these questions reflect the question pattern found in NCERT books.

Q.1

Determine order and degree (if defined) of differential equation

d4ydx4 +sin(y”’) = 0

Ans.

d4ydx4+siny”’=0y””+siny”’=0The heighest order derivative present in the differential equationis d4ydx4. Therefore its order is 4.The given equation is not a polynomial equation in yand degreeof such a differential equation can not be defined.

Q.2

Determine order and degree if defined of differential equation: y’ + 5y = 0

Ans.

y’ + 5y=0 The heighest order derivative present in the differential equation is y’. Therefore its order is 1. The given equation is a polynomial equation in y’ so the heighest power raised by y’ is one. Thus, degree of the differential eqution is 1.

Q.3

Determine order and degree if defined of differential equation: ds dt 4 + 3s d 2 s dt 2 =0

Ans.

ds dt 4 + 3s d 2 s dt 2 =0 The heighest oder derivative present in differential equation is d 2 s dt 2 . Therefore, its order is 2. Since, given differential equation is a polynomial in d 2 s dt 2 and ds dt . The power raised by d 2 s dt 2  is 1, so the degree of equation is 1.

Q.4

Determine order and degree if defined of differential equation: d 2 y dx 2 2 + cos dy dx = 0

Ans.

d 2 y dx 2 2 +cos dy dx =0 The heighest oder derivative present in differential equation is d 2 y dx 2 . Therefore, its order is 2. Since, given differential equation is not a polynomial in d 2 y dx 2 and dy dx . So, its degree is not defined.

Q.5

Determine order and degree if defined of differential equation: d 2 y dx 2  = cos3x + sin3x

Ans.

d 2 y dx 2 =cos3x + sin3x d 2 y dx 2 cos3x sin3x=0 The heighest oder derivative present in differential equation is d 2 y dx 2 . Therefore, its order is 2. Since, given differential equation is a polynomial in d 2 y dx 2 . The power raised by d 2 y dx 2  is 1, so the degree of equation is 1.

Q.6

Determine order and degree if defined of differential equation: y”’ 2 + y” 3 + y’ 4 + y 5 = 0

Ans.

y”’ 2 + y” 3 + y’ 4 + y 5 =0 The heighest order derivative in given differential equation is y”’. So, its order is 3. This differential equation is a polynomial in y”’, y”, y’ and y. The heighest power raised by y”’ is 2, therefore degree of differential equation is 2.

Q.7

Determine order and degree if defined of differential equation: y”’ + 2y” + y’ = 0

Ans.

y”’ + 2y”+y‘=0The heighest order derivative in given differential equation is y”’.So, its order is 3.This differential equation is a polynomial in y”’, yand y.The heighest power raised by y”’ is 1, therefore degree of differential equation is 1.

Q.8

Determine order and degree if defined of differential equation: y’ + y =  e x

Ans.

y’+ y=e x Þy’+ y-e x =0 The heighest order derivative in given differential equation is y’. So, its order is 1. This differential equation is a polynomial in y’ and y. The heighest power raised by y’ is 1, therefore degree of differential equation is 1.

Q.9

Determine order and degree if defined of differential equation: y” + y’ 2 + 2y = 0

Ans.

y” + y’ 2 +2y=0 The heighest order derivative in given differential equation isy”. So, its order is 2. This differential equation is a polynomial in y”, y’ and y. The heighest power raised by y” is 1, therefore degree of differential equation is 1.

Q.10

Determine order and degree if defined of differential equation: y” + 2y’ + siny = 0

Ans.

y”+ 2y’+ siny=0 The heighest order derivative in given differential equation isy”. So, its order is 2. This differential equation is a polynomial in y” and y’. The heighest power raised by y” is 1, therefore degree of differential equation is 1.

Q.11

The degree of the differential equation ( d 2 y dx 2 ) 3 + ( dy dx ) 2 + sin( dy dx )+1=0is ( A ) 3 ( B) 2 ( C ) 1 ( D ) not​ defined MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGKbGaaeyzaiaabEga caqGYbGaaeyzaiaabwgacaqGGaGaae4BaiaabAgacaqGGaGaaeiDai aabIgacaqGLbGaaeiiaiaabsgacaqGPbGaaeOzaiaabAgacaqGLbGa aeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiBaiaabccaca qGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaaqa amaabmaabaWaaSaaaeaacaqGKbWaaWbaaSqabeaacaqGYaaaaOGaae yEaaqaaiaabsgacaqG4bWaaWbaaSqabeaacaqGYaaaaaaaaOGaayjk aiaawMcaamaaCaaaleqabaGaae4maaaakiaabUcadaqadaqaamaala aabaGaaeizaiaabMhaaeaacaqGKbGaaeiEaaaaaiaawIcacaGLPaaa daahaaWcbeqaaiaabkdaaaGccaqGRaGaaeiiaiaabohacaqGPbGaae OBamaabmaabaWaaSaaaeaacaqGKbGaaeyEaaqaaiaabsgacaqG4baa aaGaayjkaiaawMcaaiaabUcacaqGXaGaaGjbVlaab2dacaaMe8Uaae imaiaaykW7caaMc8UaaeyAaiaabohaaeaadaqadaqaaiaabgeaaiaa wIcacaGLPaaacaqGZaGaaCzcaiaaxMaadaqadaqaaiaabkeaaiaawI cacaGLPaaacaqGYaGaaGjbVlaaysW7caWLjaWaaeWaaeaacaqGdbaa caGLOaGaayzkaaGaaeymaiaaxMaacaWLjaWaaeWaaeaacaqGebaaca GLOaGaayzkaaGaaGjbVlaab6gacaqGVbGaaeiDaiaaygW7caqGGaGa aeizaiaabwgacaqGMbGaaeyAaiaab6gacaqGLbGaaeizaaaaaa@9D6C@

Ans.

( d 2 y d x 2 ) 3 + ( dy dx ) 2 + sin( dy dx )+1=0 Since given equation is not a polynomial equation in y’. Therefore, its degree is not defined. Hence, the correct answer is D. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaamaabmaabaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaa aOGaamyEaaqaaiaadsgacaWG4bWaaWbaaSqabeaacaaIYaaaaaaaaO GaayjkaiaawMcaamaaCaaaleqabaGaaG4maaaakiabgUcaRmaabmaa baWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG4baaaaGaayjkai aawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRGqabiaa=bcaciGG ZbGaaiyAaiaac6gadaqadaqaamaalaaabaGaamizaiaadMhaaeaaca WGKbGaamiEaaaaaiaawIcacaGLPaaacqGHRaWkcaaIXaGaeyypa0Ja aGimaaqaaiaadofacaWGPbGaamOBaiaadogacaWGLbGaaeiiaiaabE gacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabwgacaqGXbGaaeyD aiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMgacaqGZb Gaaeiiaiaab6gacaqGVbGaaeiDaiaabccacaqGHbGaaeiiaiaabcha caqGVbGaaeiBaiaabMhacaqGUbGaae4Baiaab2gacaqGPbGaaeyyai aabYgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGa ae4Baiaab6gacaqGGaGaaeyAaiaab6gacaqGGaGaaeyEaiaabEcaca qGUaGaaeivaiaabIgacaqGLbGaaeOCaiaabwgacaqGMbGaae4Baiaa bkhacaqGLbGaaeilaaqaaiaabMgacaqG0bGaae4CaiaabccacaqGKb GaaeyzaiaabEgacaqGYbGaaeyzaiaabwgacaqGGaGaaeyAaiaaboha caqGGaGaaeOBaiaab+gacaqG0bGaaeiiaiaabsgacaqGLbGaaeOzai aabMgacaqGUbGaaeyzaiaabsgacaqGUaaabaGaaeisaiaabwgacaqG UbGaae4yaiaabwgacaqGSaGaaeiiaiaabshacaqGObGaaeyzaiaabc cacaqGJbGaae4BaiaabkhacaqGYbGaaeyzaiaabogacaqG0bGaaeii aiaabggacaqGUbGaae4CaiaabEhacaqGLbGaaeOCaiaabccacaqGPb Gaae4CaiaabccacaqGebGaaeOlaaaaaa@C0CB@

Q.12

The order of the differential equation 2x 2 d 2 y dx 2 3 dy dx +y=0is ( A )3 ( B ) 2 ( C ) 0 ( D) not​ defined MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGVbGaaeOCaiaabsga caqGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAai aabwgacaqGGaGaaeizaiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGa aeyzaiaab6gacaqG0bGaaeyAaiaabggacaqGSbGaaeiiaiaabwgaca qGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbaabaGaaeOm aiaabIhadaahaaWcbeqaaiaabkdaaaGcdaWcaaqaaiaabsgadaahaa WcbeqaaiaabkdaaaGccaqG5baabaGaaeizaiaabIhadaahaaWcbeqa aiaabkdaaaaaaOGaaGjbVlaabobicaaMe8Uaae4mamaalaaabaGaae izaiaabMhaaeaacaqGKbGaaeiEaaaacaaMe8Uaae4kaiaaysW7caqG 5bGaaGjbVlaab2dacaaMe8UaaeimaiaaykW7caaMc8UaaeyAaiaabo haaeaadaqadaqaaiaabgeaaiaawIcacaGLPaaacaqGZaGaaCzcaiaa xMaadaqadaqaaiaabkeaaiaawIcacaGLPaaacaqGYaGaaCzcaiaaxM aadaqadaqaaiaaboeaaiaawIcacaGLPaaacaqGWaGaaCzcaiaaxMaa daqadaqaaiaabseaaiaawIcacaGLPaaacaqGUbGaae4Baiaabshaca aMb8UaaeiiaiaabsgacaqGLbGaaeOzaiaabMgacaqGUbGaaeyzaiaa bsgaaaaa@93FC@

Ans.

2 x 2 d 2 y d x 2 3 dy dx +y=0 The highest derivative of given differential equation is d 2 y d x 2 . Therefore, the order of given differential equation is 2. Thus, correct option is B. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOWaaSaaaeaa caWGKbWaaWbaaSqabeaacaaIYaaaaOGaamyEaaqaaiaadsgacaWG4b WaaWbaaSqabeaacaaIYaaaaaaakiabgkHiTiaaiodadaWcaaqaaiaa dsgacaWG5baabaGaamizaiaadIhaaaGaey4kaSIaamyEaiabg2da9i aaicdacaaMc8oabaGaamivaiaadIgacaWGLbGaaeiiaiaabIgacaqG LbGaaeyAaiaabEgacaqGObGaaeyzaiaabohacaqG0bGaaeiiaiaabs gacaqGLbGaaeOCaiaabMgacaqG2bGaaeyyaiaabshacaqGPbGaaeOD aiaabwgacaqGGaGaae4BaiaabAgacaqGGaGaae4zaiaabMgacaqG2b Gaaeyzaiaab6gacaqGGaGaaeizaiaabMgacaqGMbGaaeOzaiaabwga caqGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabggacaqGSbGaaeiiai aabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGa aeiiaiaabMgacaqGZbGaaeiiamaalaaabaGaamizamaaCaaaleqaba GaaGOmaaaakiaadMhaaeaacaWGKbGaamiEamaaCaaaleqabaGaaGOm aaaaaaGccaGGUaaabaGaamivaiaadIgacaWGLbGaamOCaiaadwgaca WGMbGaam4BaiaadkhacaWGLbGaaiilaiaabccacaqG0bGaaeiAaiaa bwgacaqGGaGaae4BaiaabkhacaqGKbGaaeyzaiaabkhacaqGGaGaae 4BaiaabAgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqG GaGaaeizaiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6 gacaqG0bGaaeyAaiaabggacaqGSbGaaeiiaiaabwgacaqGXbGaaeyD aiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMgacaqGZb GaaeiiaiaabkdacaqGUaaabaGaaeivaiaabIgacaqG1bGaae4Caiaa bYcacaqGGaGaae4yaiaab+gacaqGYbGaaeOCaiaabwgacaqGJbGaae iDaiaabccacaqGVbGaaeiCaiaabshacaqGPbGaae4Baiaab6gacaqG GaGaaeyAaiaabohacaqGGaGaaeOqaiaab6caaaaa@CE32@

Q.13

Verify that the given functions ( explicit or implicit )is a solution of the corresponding differential equation: y = e x + 1 : y” – y’ = 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqG9aGaaeiiaiaabwga daahaaWcbeqaaiaabIhaaaGccaqGRaGaaeiiaiaabgdacaaMc8UaaG PaVlaaxMaacaWLjaGaaGPaVlaabQdacaqGGaGaaCzcaiaaxMaacaqG GaGaaeyEaiaabEcacaqGNaGaaeiiaiaabobicaqGGaGaaeyEaiaabE cacaqGGaGaaeypaiaabccacaqGWaaaaaa@B232@

Ans.

y = e x + 1 Differentiating w.r.t. x, we get y’ = e x ( i ) Differentiating equation ( i ) w.r.t. x, we get y” = e x Substituting values of y’ and y” in L.H.S. of given differential equation y” – y’ = 0, we get y” – y’ =e x – e x = 0 = R.H.S. Thus,the given function is the solution of the corresponding differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq Gabeqaaa5gbaGaaeyEaiaabccacaqG9aGaaeiiaiaabwgadaahaaWc beqaaiaabIhaaaGccaqGGaGaae4kaiaabccacaqGXaaabaGaaeirai aabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0bGa aeyAaiaabggacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiaiaabEhaca qGUaGaaeOCaiaab6cacaqG0bGaaeOlaiaabccacaqG4bGaaeilaiaa bccacaqG3bGaaeyzaiaabccacaqGNbGaaeyzaiaabshaaeaacaqG5b Gaae4jaiaabccacaqG9aGaaeiiaiaabwgadaahaaWcbeqaaiaabIha aaGccaWLjaGaaeOlaiaab6cacaqGUaWaaeWaaeaacaqGPbaacaGLOa GaayzkaaaabaGaaeiraiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGa aeyzaiaab6gacaqG0bGaaeyAaiaabggacaqG0bGaaeyAaiaab6gaca qGNbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaa b+gacaqGUbGaaeiiamaabmaabaGaaeyAaaGaayjkaiaawMcaaiaabc cacaqG3bGaaeOlaiaabkhacaqGUaGaaeiDaiaab6cacaqGGaGaaeiE aiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0b aabaGaaeyEaiaabEcacaqGNaGaaeiiaiaab2dacaqGGaGaaeyzamaa CaaaleqabaGaaeiEaaaaaOqaaiaabofacaqG1bGaaeOyaiaabohaca qG0bGaaeyAaiaabshacaqG1bGaaeiDaiaabMgacaqGUbGaae4zaiaa bccacaqG2bGaaeyyaiaabYgacaqG1bGaaeyzaiaabohacaqGGaGaae 4BaiaabAgacaqGGaGaaeyEaiaabEcacaqGGaGaaeyyaiaab6gacaqG KbGaaeiiaiaabMhacaqGNaGaae4jaiaabccacaqGPbGaaeOBaiaabc cacaqGmbGaaeOlaiaabIeacaqGUaGaae4uaiaab6cacaqGGaGaae4B aiaabAgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGa GaaeizaiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6ga caqG0bGaaeyAaiaabggacaqGSbGaaeiiaaqaaiaabwgacaqGXbGaae yDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMhacaqG NaGaae4jaiaabccacaqGTaGaaeiiaiaabMhacaqGNaGaaeiiaiaab2 dacaqGGaGaaeimaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4z aiaabwgacaqG0baabaGaaeyEaiaabEcacaqGNaGaaeiiaiaab2caca qGGaGaaeyEaiaabEcacaqGGaGaaeypaiacycyGLbWaiGjGCaaaleqc ycyaiGjGcGaMagiEaaaakiaabccacaqGTaGaaeiiaiaabwgadaahaa WcbeqaaiaabIhaaaGccaqGGaGaaeypaiaabccacaqGWaGaaeiiaiaa b2dacaqGGaGaaeOuaiaab6cacaqGibGaaeOlaiaabofacaqGUaaaba GaaeivaiaabIgacaqG1bGaae4CaiaabYcacaqG0bGaaeiAaiaabwga caqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeOzai aabwhacaqGUbGaae4yaiaabshacaqGPbGaae4Baiaab6gacaqGGaGa aeyAaiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaaqaaiaabsgacaqGPbGaaeOzai aabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGa aeiBaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgaca qGVbGaaeOBaiaab6caaaaa@4B29@

Q.14

Verify that the given functions ( explicit or implicit )is a solution of the corresponding differential equation: y = x 2 +2x + C : y’ –2x –2 = 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqG9aGaaeiiaiaabIha daahaaWcbeqaaiaabkdaaaGccaqGRaGaaeOmaiaabIhacaqGGaGaae 4kaiaabccacaqGdbGaaGPaVlaaykW7caaMc8UaaCzcaiaabQdacaqG GaGaaCzcaiaaxMaacaqGGaGaaeyEaiaabEcacaqGGaGaae4eGiaabk dacaqG4bGaaeiiaiaabobicaqGYaGaaeiiaiaab2dacaqGGaGaaeim aaaaaa@B53B@

Ans.

y = x 2 +2x + C Differentiating w.r.t. x, we get y’ = 2x + 2 Substituting values of y’ in L.H.S. of given differential equation y’ –2x –2 = 0, we get y’ – 2x –2 = 2x + 2 –2x –2 = 0 = R.H.S. Thus,the given function is the solution of the corresponding differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMhacaqGGaGaaeypaiaabccacaqG4bWaaWbaaSqabeaa caqGYaaaaOGaae4kaiaabkdacaqG4bGaaeiiaiaabUcacaqGGaGaae 4qaaqaaiaabseacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwga caqGUbGaaeiDaiaabMgacaqGHbGaaeiDaiaabMgacaqGUbGaae4zai aabccacaqG3bGaaeOlaiaabkhacaqGUaGaaeiDaiaab6cacaqGGaGa aeiEaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgaca qG0baabaGaaeyEaiaabEcacaqGGaGaaeypaiaabccacaqGYaGaaeiE aiaabccacaqGRaGaaeiiaiaabkdaaeaacaqGtbGaaeyDaiaabkgaca qGZbGaaeiDaiaabMgacaqG0bGaaeyDaiaabshacaqGPbGaaeOBaiaa bEgacaqGGaGaaeODaiaabggacaqGSbGaaeyDaiaabwgacaqGZbGaae iiaiaab+gacaqGMbGaaeiiaiaabMhacaqGNaGaaeiiaiaabMgacaqG UbGaaeiiaiaabYeacaqGUaGaaeisaiaab6cacaqGtbGaaeOlaiaabc cacaqGVbGaaeOzaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOB aiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLb GaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaaabaGaaeyzaiaa bghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae yEaiaabEcacaqGGaGaae4eGiaabkdacaqG4bGaaeiiaiaabobicaqG YaGaaeiiaiaab2dacaqGGaGaaeimaiaabYcacaqGGaGaae4Daiaabw gacaqGGaGaae4zaiaabwgacaqG0baabaGaaeiiaiaabMhacaqGNaGa aeiiaiaab2cacaqGGaGaaeOmaiaabIhacaqGGaGaae4eGiaabkdaca qGGaGaaeypaiaabccacaqGYaGaaeiEaiaabccacaqGRaGaaeiiaiaa bkdacaqGGaGaae4eGiaabkdacaqG4bGaaeiiaiaabobicaqGYaaaba GaaCzcaiaaxMaacaqG9aGaaeiiaiaabcdacaqGGaGaaeypaiaabcca caqGsbGaaeOlaiaabIeacaqGUaGaae4uaiaab6caaeaacaqGubGaae iAaiaabwhacaqGZbGaaeilaiaabshacaqGObGaaeyzaiaabccacaqG NbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGaaeyDaiaab6 gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGPbGaae4C aiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4Caiaab+gacaqGSb GaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae4BaiaabAga caqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogacaqGVbGaaeOCai aabkhacaqGLbGaae4CaiaabchacaqGVbGaaeOBaiaabsgacaqGPbGa aeOBaiaabEgacaqGGaaabaGaaeizaiaabMgacaqGMbGaaeOzaiaabw gacaqGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabggacaqGSbGaaeii aiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUb GaaeOlaaaaaa@1803@

Q.15

Verify that the given functions ( explicit or implicit )is a solution of the corresponding differential equation: y = cos x + C : y’ + sinx = 0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqG9aGaaeiiaiaaboga caqGVbGaae4CaiaabccacaqG4bGaaeiiaiaabUcacaqGGaGaae4qai aaykW7caaMc8UaaGPaVlaaxMaacaaMc8UaaGPaVlaabQdacaqGGaGa aCzcaiaaxMaacaqGGaGaaeyEaiaabEcacaqGGaGaae4kaiaabccaca qGZbGaaeyAaiaab6gacaqG4bGaaeiiaiaab2dacaqGGaGaaeimaaaa aa@B923@

Ans.

y = cos x + C Differentiating w.r.t. x, we get y’ = –sinx Substituting values of y’ in L.H.S. of given differential equation y’ + sinx = 0, we get y’ + sinx = –sinx + sinx = 0 = R.H.S. Thus,the given function is the solution of the corresponding differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMhacaqGGaGaaeypaiaabccacaqGJbGaae4Baiaaboha caqGGaGaaeiEaiaabccacaqGRaGaaeiiaiaaboeaaeaacaqGebGaae yAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshacaqG PbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgacaqGGaGaae4Daiaab6 cacaqGYbGaaeOlaiaabshacaqGUaGaaeiiaiaabIhacaqGSaGaaeii aiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaabMhaca qGNaGaaeiiaiaab2dacaqGGaGaae4eGiaabohacaqGPbGaaeOBaiaa bIhaaeaacaqGtbGaaeyDaiaabkgacaqGZbGaaeiDaiaabMgacaqG0b GaaeyDaiaabshacaqGPbGaaeOBaiaabEgacaqGGaGaaeODaiaabgga caqGSbGaaeyDaiaabwgacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiai aabMhacaqGNaGaaeiiaiaabMgacaqGUbGaaeiiaiaabYeacaqGUaGa aeisaiaab6cacaqGtbGaaeOlaiaabccacaqGVbGaaeOzaiaabccaca qGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaa bAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshacaqGPbGaae yyaiaabYgacaqGGaaabaGaaeyzaiaabghacaqG1bGaaeyyaiaabsha caqGPbGaae4Baiaab6gacaqGGaGaaeyEaiaabEcacaqGGaGaae4kai aabccacaqGZbGaaeyAaiaab6gacaqG4bGaaeiiaiaab2dacaqGGaGa aeimaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgaca qG0baabaGaaeyEaiaabEcacaqGGaGaae4kaiaabccacaqGZbGaaeyA aiaab6gacaqG4bGaaeiiaiaab2dacaqGGaGaae4eGiaabohacaqGPb GaaeOBaiaabIhacaqGGaGaae4kaiaabccacaqGZbGaaeyAaiaab6ga caqG4bGaaeiiaiaab2dacaqGGaGaaeimaiaabccacaqG9aGaaeiiai aabkfacaqGUaGaaeisaiaab6cacaqGtbGaaeOlaaqaaiaabsfacaqG ObGaaeyDaiaabohacaqGSaGaaeiDaiaabIgacaqGLbGaaeiiaiaabE gacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabAgacaqG1bGaaeOB aiaabogacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMgacaqGZb GaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGZbGaae4BaiaabYga caqG1bGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOzai aabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4yaiaab+gacaqGYbGa aeOCaiaabwgacaqGZbGaaeiCaiaab+gacaqGUbGaaeizaiaabMgaca qGUbGaae4zaiaabccaaeaacaqGKbGaaeyAaiaabAgacaqGMbGaaeyz aiaabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGa GaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6ga caqGUaaaaaa@189F@

Q.16

Verify that the given functions ( explicit or implicit )is a solution of the corresponding differential equation: y = 1 + x 2 : y’ = xy 1 + x 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqG9aGaaeiiamaakaaa baGaaeymaiaabccacaqGRaGaaeiiaiaabIhadaahaaWcbeqaaiaabk daaaaabeaakiaaykW7caaMc8UaaGPaVlaaxMaacaWLjaGaaGPaVlaa ykW7caqG6aGaaeiiaiaaxMaacaWLjaGaaeiiaiaabMhacaqGNaGaae iiaiaab2dacaqGGaWaaSaaaeaacaqG4bGaaeyEaaqaaiaabgdacaqG GaGaae4kaiaabccacaqG4bWaaWbaaSqabeaacaqGYaaaaaaaaaaa@B755@

Ans.

y = 1 + x 2 Differentiating w.r.t. x, we get y’ = x 1 + x 2 Substituting values of y in R.H.S. of given differential equation y’ = xy 1 + x 2 , we get R.H.S. = xy 1 + x 2 = x( 1 + x 2 ) 1 + x 2 = x 1 + x 2 = y’ = L.H.S. Thus,the given function is the solution of the corresponding differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMhacaqGGaGaaeypaiaabccadaGcaaqaaiaabgdacaqG GaGaae4kaiaabccacaqG4bWaaWbaaSqabeaacaqGYaaaaaqabaaake aacaqGebGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOB aiaabshacaqGPbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgacaqGGa Gaae4Daiaab6cacaqGYbGaaeOlaiaabshacaqGUaGaaeiiaiaabIha caqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaa qaaiaabMhacaqGNaGaaeiiaiaab2dacaqGGaWaaSaaaeaacaqG4baa baWaaOaaaeaacaqGXaGaaeiiaiaabUcacaqGGaGaaeiEamaaCaaale qabaGaaeOmaaaaaeqaaaaaaOqaaiaabofacaqG1bGaaeOyaiaaboha caqG0bGaaeyAaiaabshacaqG1bGaaeiDaiaabMgacaqGUbGaae4zai aabccacaqG2bGaaeyyaiaabYgacaqG1bGaaeyzaiaabohacaqGGaGa ae4BaiaabAgacaqGGaGaaeyEaiaabccacaqGPbGaaeOBaiaabccaca qGsbGaaeOlaiaabIeacaqGUaGaae4uaiaab6cacaqGGaGaae4Baiaa bAgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaae izaiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG 0bGaaeyAaiaabggacaqGSbGaaeiiaaqaaiaabwgacaqGXbGaaeyDai aabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMhacaqGNaGa aeiiaiaab2dacaqGGaWaaSaaaeaacaqG4bGaaeyEaaqaaiaabgdaca qGGaGaae4kaiaabccacaqG4bWaaWbaaSqabeaacaqGYaaaaaaakiaa bYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baaba GaaeOuaiaab6cacaqGibGaaeOlaiaabofacaqGUaGaaeiiaiaab2da caqGGaWaaSaaaeaacaqG4bGaaeyEaaqaaiaabgdacaqGGaGaae4kai aabccacaqG4bWaaWbaaSqabeaacaqGYaaaaaaaaOqaaiaaxMaacaaM c8UaaeypaiaabccadaWcaaqaaiaabIhadaqadaqaamaakaaabaGaae ymaiaabccacaqGRaGaaeiiaiaabIhadaahaaWcbeqaaiaabkdaaaaa beaaaOGaayjkaiaawMcaaaqaaiaabgdacaqGGaGaae4kaiaabccaca qG4bWaaWbaaSqabeaacaqGYaaaaaaaaOqaaiaaxMaacaaMc8UaaGPa VlaaykW7caqG9aGaaeiiamaalaaabaGaaeiEaaqaamaakaaabaGaae ymaiaabccacaqGRaGaaeiiaiaabIhadaahaaWcbeqaaiaabkdaaaaa beaaaaaakeaacaWLjaGaaGPaVlaaykW7caaMc8Uaaeypaiaabccaca qG5bGaae4jaiaabccacaqG9aGaaeiiaiaabYeacaqGUaGaaeisaiaa b6cacaqGtbGaaeOlaaqaaiaabsfacaqGObGaaeyDaiaabohacaqGSa GaaeiDaiaabIgacaqGLbGaaeiiaiaabEgacaqGPbGaaeODaiaabwga caqGUbGaaeiiaiaabAgacaqG1bGaaeOBaiaabogacaqG0bGaaeyAai aab+gacaqGUbGaaeiiaiaabMgacaqGZbGaaeiiaiaabshacaqGObGa aeyzaiaabccacaqGZbGaae4BaiaabYgacaqG1bGaaeiDaiaabMgaca qGVbGaaeOBaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaa bwgacaqGGaGaae4yaiaab+gacaqGYbGaaeOCaiaabwgacaqGZbGaae iCaiaab+gacaqGUbGaaeizaiaabMgacaqGUbGaae4zaiaabccaaeaa caqGKbGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBai aabshacaqGPbGaaeyyaiaabYgacaqGGaGaaeyzaiaabghacaqG1bGa aeyyaiaabshacaqGPbGaae4Baiaab6gacaqGUaaaaaa@315E@

Q.17

MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqG9aGaaeiiaiaabgea caqG4bGaaGPaVlaaykW7caaMc8UaaCzcaiaaxMaacaaMc8UaaGPaVl aaxMaacaaMc8UaaGPaVlaaykW7caqG6aGaaeiiaiaaxMaacaWLjaGa aeiiaiaabIhacaqG5bGaae4jaiaabccacaqG9aGaaeiiaiaabMhaca WLjaGaaCzcamaabmaabaGaaeiEaiaabccacqGHGjsUcaqGGaGaaeim aaGaayjkaiaawMcaaaaaaa@BCAB@

Ans.

MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMhacaqGGaGaaeypaiaabccacaqGbbGaaeiEaaqaaiaa bseacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaae iDaiaabMgacaqGHbGaaeiDaiaabMgacaqGUbGaae4zaiaabccacaqG 3bGaaeOlaiaabkhacaqGUaGaaeiDaiaab6cacaqGGaGaaeiEaiaabY cacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaGa aeyEaiaabEcacaqGGaGaaeypaiaabccacaqGbbaabaGaae4uaiaabw hacaqGIbGaae4CaiaabshacaqGPbGaaeiDaiaabwhacaqG0bGaaeyA aiaab6gacaqGNbGaaeiiaiaabAhacaqGHbGaaeiBaiaabwhacaqGLb Gaae4CaiaabccacaqGVbGaaeOzaiaabccacaqG5bGaae4jaiaabcca caqGPbGaaeOBaiaabccacaqGmbGaaeOlaiaabIeacaqGUaGaae4uai aab6cacaqGGaGaae4BaiaabAgacaqGGaGaae4zaiaabMgacaqG2bGa aeyzaiaab6gacaqGGaGaaeizaiaabMgacaqGMbGaaeOzaiaabwgaca qGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabggacaqGSbGaaeiiaaqa aiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUb GaaeiiaiaabIhacaqG5bGaae4jaiaabccacaqG9aGaaeiiaiaabMha caqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaa qaaiaabYeacaqGUaGaaeisaiaab6cacaqGtbGaaeOlaiaabccacaqG 9aGaaeiiaiaabIhacaqG5bGaae4jaiaabccaaeaacaWLjaGaaeypai aabccacaqG4bWaaeWaaeaacaqGbbaacaGLOaGaayzkaaaabaGaaCzc aiaab2dacaqGGaGaaeyqaiaabIhacaqGGaGaaeypaiaabccacaqG5b Gaaeiiaiaab2dacaqGGaGaaeOuaiaab6cacaqGibGaaeOlaiaabofa caqGUaaabaGaaeivaiaabIgacaqG1bGaae4CaiaabYcacaqG0bGaae iAaiaabwgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqG GaGaaeOzaiaabwhacaqGUbGaae4yaiaabshacaqGPbGaae4Baiaab6 gacaqGGaGaaeyAaiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaaeii aiaabohacaqGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUb Gaaeiiaiaab+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabcca caqGJbGaae4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Bai aab6gacaqGKbGaaeyAaiaab6gacaqGNbGaaeiiaaqaaiaabsgacaqG PbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabM gacaqGHbGaaeiBaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiD aiaabMgacaqGVbGaaeOBaiaab6caaaaa@0A6F@

Q.18

MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqG9aGaaeiiaiaabIha caqGGaGaae4CaiaabMgacaqGUbGaaeiEaiaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaeOoaiaabccacaqGGaGaaeiEaiaabMhacaqGNaGa aeiiaiaab2dacaqGGaGaaeyEaiaabccacaqGRaGaaeiiaiaabIhaca qGGaWaaOaaaeaacaqG4bWaaWbaaSqabeaacaqGYaaaaOGaaeiiaiaa bobicaqGGaGaaeyEamaaCaaaleqabaGaiGgGbkdaaaaabeaakiaayk W7caaMc8+aaeWaaeaacaqG4bGaeyiyIKRaaeimaiaaykW7caqGHbGa aeOBaiaabsgacaqGGaGaaeiEaiaabccacaqG+aGaaeiiaiaabMhaca qGGaGaae4BaiaabkhacaqGGaGaaeiEaiaabccacaqG8aGaaeiiaiaa bobicaqG5baacaGLOaGaayzkaaaaaaa@D43D@

Ans.

MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMhacaqGGaGaaeypaiaabccacaqG4bGaae4CaiaabMga caqGUbGaaeiiaiaabIhaaeaacaqGebGaaeyAaiaabAgacaqGMbGaae yzaiaabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabshacaqG PbGaaeOBaiaabEgacaqGGaGaae4Daiaab6cacaqGYbGaaeOlaiaabs hacaqGUaGaaeiiaiaabIhacaqGSaGaaeiiaiaabEhacaqGLbGaaeii aiaabEgacaqGLbGaaeiDaaqaaiaabMhacaqGNaGaaeiiaiaab2daca qGGaGaaeiEaiaabogacaqGVbGaae4CaiaabccacaqG4bGaaeiiaiaa bUcacaqGGaGaae4CaiaabMgacaqGUbGaaeiEaaqaaiaabofacaqG1b GaaeOyaiaabohacaqG0bGaaeyAaiaabshacaqG1bGaaeiDaiaabMga caqGUbGaae4zaiaabccacaqG2bGaaeyyaiaabYgacaqG1bGaaeyzai aabohacaqGGaGaae4BaiaabAgacaqGGaGaaeyEaiaabEcacaqGGaGa aeyAaiaab6gacaqGGaGaaeitaiaab6cacaqGibGaaeOlaiaabofaca qGUaGaaeiiaiaab+gacaqGMbGaaeiiaiaabEgacaqGPbGaaeODaiaa bwgacaqGUbGaaeiiaiaabsgacaqGPbGaaeOzaiaabAgacaqGLbGaae OCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiBaiaabccaaeaa caqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBai aabccacaqG4bGaaeyEaiaabEcacaqGGaGaaeypaiaabccacaqG5bGa aeiiaiaabUcacaqGGaGaaeiEamaakaaabaGaaeiEamaaCaaaleqaba GaaeOmaaaakiaabccacaqGtaIaaeiiaiaabMhadaahaaWcbeqaaiaa bkdaaaaabeaakiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zai aabwgacaqG0baabaGaaeitaiaab6cacaqGibGaaeOlaiaabofacaqG UaGaaeiiaiaab2dacaqGGaGaaeiEaiaabMhacaqGNaGaaeiiaiaab2 dacaqGGaGaaeiEamaabmaabaGaaeiEaiaabogacaqGVbGaae4Caiaa bIhacaqGGaGaae4kaiaabccacaqGZbGaaeyAaiaab6gacaqG4baaca GLOaGaayzkaaaabaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG GaGaaeiiaiaabccacaqGGaGaaeiiaiaab2dacaqGGaGaaeiEamaaCa aaleqabaGaaeOmaaaakiaabogacaqGVbGaae4CaiaabIhacaqGGaGa ae4kaiaabccacaqG4bGaae4CaiaabMgacaqGUbGaaeiEaaqaaiaabk facaqGUaGaaeisaiaab6cacaqGtbGaaeOlaiaabccacaqG9aGaaeii aiaabIhacaqGZbGaaeyAaiaab6gacaqG4bGaaeiiaiaabUcacaqGGa GaaeiEamaakaaabaGaaeiEamaaCaaaleqabaGaaeOmaaaakiaabcca caqGtaIaaeiiamaabmaabaGaaeiEaiaabohacaqGPbGaaeOBaiaabI haaiaawIcacaGLPaaadaahaaWcbeqaaiaabkdaaaaabeaaaOqaaiaa bccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaae iiaiaabccacaqG9aGaaeiiaiaabIhacaqGZbGaaeyAaiaab6gacaqG 4bGaaeiiaiaabUcacaqGGaGaaeiEamaaCaaaleqabaGaaeOmaaaakm aakaaabaGaaeymaiaabccacaqGtaIaaeiiaiaabohacaqGPbGaaeOB amaaCaaaleqabaGaaeOmaaaakiaabIhaaSqabaaakeaacaqGGaGaae iiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG GaGaaeypaiaabccacaqG4bGaae4CaiaabMgacaqGUbGaaeiEaiaabc cacaqGRaGaaeiiaiaabIhadaahaaWcbeqaaiaabkdaaaGccaqGJbGa ae4BaiaabohacaqG4baabaGaae4uaiaab+gacaqGSaGaaeiiaiaabY eacaqGUaGaaeisaiaab6cacaqGtbGaaeOlaiaabccacaqG9aGaaeii aiaabkfacaqGUaGaaeisaiaab6cacaqGtbGaaeOlaaqaaiaabsfaca qGObGaaeyDaiaabohacaqGSaGaaeiDaiaabIgacaqGLbGaaeiiaiaa bEgacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabAgacaqG1bGaae OBaiaabogacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaabMgacaqG ZbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGZbGaae4BaiaabY gacaqG1bGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOz aiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4yaiaab+gacaqGYb GaaeOCaiaabwgacaqGZbGaaeiCaiaab+gacaqGUbGaaeizaiaabMga caqGUbGaae4zaiaabccaaeaacaqGKbGaaeyAaiaabAgacaqGMbGaae yzaiaabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqG GaGaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6 gacaqGUaaaaaa@850D@

Q.19

MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeiEaiaabMhacaqGGaGaaeypaiaabcca caqGSbGaae4BaiaabEgacaqG5bGaaeiiaiaabUcacaqGGaGaae4qai aabccacaqG6aGaaeiiaiaabMhacaqGNaGaaeiiaiaab2dacaqGGaWa aSaaaeaacaqG5bWaaWbaaSqabeaacaqGYaaaaaGcbaGaaeymaiaabc cacaqGtaIaaeiiaiaabIhacaqG5baaamaabmaabaGaaeiEaiaabMha cqGHGjsUcaqGWaaacaGLOaGaayzkaaaaaaa@B600@

Ans.

MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabIhacaqG5bGaaeiiaiaab2dacaqGGaGaaeiBaiaab+ga caqGNbGaaeyEaiaabccacaqGRaGaaeiiaiaaboeaaeaacaqGebGaae yAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshacaqG PbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgacaqGGaGaae4Daiaab6 cacaqGYbGaaeOlaiaabshacaqGUaGaaeiiaiaabIhacaqGSaGaaeii aiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaabIhaca qG5bGaae4jaiaabccacaqGRaGaaeiiaiaabMhacaqGGaGaaeypamaa laaabaGaaeymaaqaaiaabMhaaaGaaeyEaiaabEcacaqGGaGaae4kai aabccacaqGWaaabaGaaeiEaiaabMhacaqG5bGaae4jaiaabccacaqG RaGaaeiiaiaabMhadaahaaWcbeqaaiaabkdaaaGccaqGGaGaaeypai aabccacaqG5bGaae4jaaqaaiaabccacaqGGaGaaeiiaiaabccacaqG GaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG5bWaaWbaaSqabe aacaqGYaaaaOGaaeiiaiaab2dacaqGGaGaaeyEaiaabEcacaqGGaGa ae4eGiaabccacaqG4bGaaeyEaiaabMhacaqGNaaabaGaaeiiaiaabc cacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeii aiaabccacaqGGaGaaeiiaiaabccacaqG9aGaaeiiaiaabMhacaqGNa WaaeWaaeaacaqGXaGaaeiiaiaabobicaqGGaGaaeiEaiaabMhaaiaa wIcacaGLPaaaaeaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabc cacaqGGaGaaeiiaiaabccacaqGGaGaaeyEaiaabEcacaqGGaGaaeyp aiaabccadaWcaaqaaiaabMhadaahaaWcbeqaaiaabkdaaaaakeaada qadaqaaiaabgdacaqGGaGaae4eGiaabccacaqG4bGaaeyEaaGaayjk aiaawMcaaaaaaeaacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabs hacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOB aiaabccacaqGMbGaaeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVb GaaeOBaiaabccacaqGPbGaae4CaiaabccacaqG0bGaaeiAaiaabwga caqGGaGaae4Caiaab+gacaqGSbGaaeyDaiaabshacaqGPbGaae4Bai aab6gacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGa aeiiaiaabogacaqGVbGaaeOCaiaabkhacaqGLbGaae4Caiaabchaca qGVbGaaeOBaiaabsgacaqGPbGaaeOBaiaabEgacaqGGaaabaGaaeiz aiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0b GaaeyAaiaabggacaqGSbGaaeiiaiaabwgacaqGXbGaaeyDaiaabgga caqG0bGaaeyAaiaab+gacaqGUbGaaeOlaaaaaa@F6E3@

Q.20

MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaapaGaaeyEaiaabccacaqGtaIaaeiiaiaaboga caqGVbGaae4CaiaabMhacaqGGaGaaeypaiaabccacaqG4bGaaeiiai aabccacaqGGaGaaeiiaiaabQdacaqGGaGaaeiiaiaabccacaqGGaWa aeWaaeaacaqG5bGaae4CaiaabMgacaqGUbGaaeyEaiaabccacaqGRa GaaeiiaiaabogacaqGVbGaae4CaiaabMhacaqGGaGaae4kaiaabcca caqG4baacaGLOaGaayzkaaGaaeyEaiaabEcacaqGGaGaaeypaiaabc cacaqG5baaaaa@BC7B@

Ans.

MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadMha cqGHsislciGGJbGaai4BaiaacohacaWG5bGaeyypa0JaamiEaiaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaac6cacaGGUaGaaiOl amaabmaabaGaamyAaaGaayjkaiaawMcaaaqaaiaabseacaqGPbGaae OzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqG HbGaaeiDaiaabMgacaqGUbGaae4zaiaabccacaqG3bGaaeOlaiaabk hacaqGUaGaaeiDaiaab6cacaqGGaGaaeiEaiaabYcacaqGGaGaae4D aiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaGaaGPaVlaaykW7ca qG5bGaae4jaiabgUcaRiGacohacaGGPbGaaiOBaiaadMhacaGGUaGa amyEaiaacEcacqGH9aqpcaaIXaaabaGaamyEaiaacEcadaqadaqaai aaigdacqGHRaWkciGGZbGaaiyAaiaac6gacaWG5baacaGLOaGaayzk aaGaeyypa0JaaGymaaqaaiaaxMaacaWLjaGaamyEaiaacEcacqGH9a qpdaWcaaqaaiaaigdaaeaacaaIXaGaey4kaSIaci4CaiaacMgacaGG UbGaamyEaaaaaeaacaqGtbGaaeyDaiaabkgacaqGZbGaaeiDaiaabM gacaqG0bGaaeyDaiaabshacaqGPbGaaeOBaiaabEgacaqGGaGaaeOD aiaabggacaqGSbGaaeyDaiaabwgacaqGZbGaaeiiaiaab+gacaqGMb GaaeiiaiaabMhacaqGNaGaaeiiaiaabMgacaqGUbGaaeiiaiaabYea caqGUaGaaeisaiaab6cacaqGtbGaaeOlaiaabccacaqGVbGaaeOzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGKbGa aeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshaca qGPbGaaeyyaiaabYgacaqGGaaabaGaaeyzaiaabghacaqG1bGaaeyy aiaabshacaqGPbGaae4Baiaab6gacaqGGaWaaeWaaeaacaqG5bGaae 4CaiaabMgacaqGUbGaaeyEaiaabccacaqGRaGaaeiiaiaabogacaqG VbGaae4CaiaabMhacaqGGaGaae4kaiaabccacaqG4baacaGLOaGaay zkaaGaaeyEaiaabEcacaqGGaGaaeypaiaabccacaqG5bGaaeilaiaa bccacaqG3bGaaeyzaiaabccacaqGNbGaaeyzaiaabshaaeaadaqada qaaiaadMhaciGGZbGaaiyAaiaac6gacaWG5bGaey4kaSIaci4yaiaa c+gacaGGZbGaamyEaiabgUcaRiaadIhaaiaawIcacaGLPaaacaWG5b Gaai4jaiabg2da9maabmaabaGaamyEaiGacohacaGGPbGaaiOBaiaa dMhacqGHRaWkciGGJbGaai4BaiaacohacaWG5bGaey4kaSIaamiEaa GaayjkaiaawMcaamaalaaabaGaaGymaaqaamaabmaabaGaaGymaiab gUcaRiGacohacaGGPbGaaiOBaiaadMhaaiaawIcacaGLPaaaaaaaba GaaCzcaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7 cqGH9aqpdaqadaqaaiaadMhaciGGZbGaaiyAaiaac6gacaWG5bGaey 4kaSIaci4yaiaac+gacaGGZbGaamyEaiabgUcaRiaadMhacqGHsisl ciGGJbGaai4BaiaacohacaWG5baacaGLOaGaayzkaaWaaSaaaeaaca aIXaaabaWaaeWaaeaacaaIXaGaey4kaSIaci4CaiaacMgacaGGUbGa amyEaaGaayjkaiaawMcaaaaaaeaacaWLjaGaaCzcaiaaxMaacaWLja GaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaadaWadaqaaiaabAeacaqG YbGaae4Baiaab2gacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabs hacaqGPbGaae4Baiaab6gadaqadaqaaiaabMgaaiaawIcacaGLPaaa aiaawUfacaGLDbaaaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaGPaVl aaykW7caaMc8UaaGPaVlabg2da9maabmaabaGaamyEaiGacohacaGG PbGaaiOBaiaadMhacqGHRaWkcaWG5baacaGLOaGaayzkaaWaaSaaae aacaaIXaaabaWaaeWaaeaacaaIXaGaey4kaSIaci4CaiaacMgacaGG UbGaamyEaaGaayjkaiaawMcaaaaaaeaacaWLjaGaaCzcaiaaxMaaca WLjaGaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9iaadMhadaqadaqa aiGacohacaGGPbGaaiOBaiaadMhacqGHRaWkcaaIXaaacaGLOaGaay zkaaWaaSaaaeaacaaIXaaabaWaaeWaaeaacaaIXaGaey4kaSIaci4C aiaacMgacaGGUbGaamyEaaGaayjkaiaawMcaaaaaaeaacaWLjaGaaC zcaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlabg2da9iaa dMhacqGH9aqpcaWGsbGaaiOlaiaadIeacaGGUaGaam4uaiaac6caae aacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabshacaqGObGaaeyz aiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMb GaaeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabcca caqGPbGaae4CaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4Cai aab+gacaqGSbGaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqGGaGa ae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogaca qGVbGaaeOCaiaabkhacaqGLbGaae4CaiaabchacaqGVbGaaeOBaiaa bsgacaqGPbGaaeOBaiaabEgacaqGGaaabaGaaeizaiaabMgacaqGMb GaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabgga caqGSbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAai aab+gacaqGUbGaaeOlaaaaaa@D4B7@

Q.21

Verify that the given functions ( explicit or implicit )is a solution of the corresponding differential equation: x +y=ta n 1 y : y 2 y+ y 2 +1=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaaruGvLjhzH5wyaGqbb8aacaWF4bGaa8hiaiaa =TcacaWF5bGaeyypa0Jaa8hDaiaa=fgacaWFUbWaaWbaaSqabeaaca WFTaGaa8xmaaaakiaa=LhacaaMc8UaaGPaVlaaykW7caWLjaGaaCzc aiaaykW7caaMc8UaaGPaVlaaykW7caWF6aGaa8hiaiaaxMaacaWH5b WaaWbaaSqabeaacaWFYaaaaOGaa8xEaiaa=DcacaWFRaGaa8xEamaa CaaaleqabaGaa8Nmaaaakiaa=TcacaWFXaGaeyypa0Jaa8hmaaaaaa@BCFE@

Ans.

x +y= tan 1 y Differentiating w.r.t. x, we get 1+y’=( 1 1+ y 2 )y ( 1+y’ )( 1+ y 2 )=y 1+ y 2 +y+y y 2 =y 1+ y 2 +y y 2 =0 Thus,the given function is the solution of the corresponding differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caWLjaGaaCzcaiaaykW7caWG4bacbeGaa8hiaiabgUcaRiaadMhacq GH9aqpciGG0bGaaiyyaiaac6gadaahaaWcbeqaaiaac2cacaaIXaaa aOGaamyEaiaaykW7aeaacaqGebGaaeyAaiaabAgacaqGMbGaaeyzai aabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabshacaqGPbGa aeOBaiaabEgacaqGGaGaae4Daiaab6cacaqGYbGaaeOlaiaabshaca qGUaGaaeiiaiaabIhacaqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaa bEgacaqGLbGaaeiDaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8 UaaGPaVlaaykW7caaMc8UaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7 caqGXaGaae4kaiaabMhacaqGNaGaeyypa0ZaaeWaaeaadaWcaaqaai aaigdaaeaacaaIXaGaey4kaSIaamyEamaaCaaaleqabaGaiGjGikda aaaaaaGccaGLOaGaayzkaaGaamyEaiaacEcaaeaacaaMc8UaaGPaVl aaykW7caaMc8+aaeWaaeaacaqGXaGaey4kaSIaaGPaVlaabMhacaqG NaaacaGLOaGaayzkaaWaaeWaaeaacaaIXaGaey4kaSIaamyEamaaCa aaleqabaGaiGjGikdaaaaakiaawIcacaGLPaaacqGH9aqpcaWG5bGa ai4jaaqaaiaaigdacqGHRaWkcaWG5bWaaWbaaSqabeaacaaIYaaaaO Gaey4kaSIaamyEaiaacEcacqGHRaWkcaWG5bGaai4jaiaadMhadaah aaWcbeqaaiaaikdaaaGccqGH9aqpcaWG5bGaai4jaaqaaiaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGymaiab gUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG5bGaai 4jaiaadMhadaahaaWcbeqaaiaaikdaaaGccqGH9aqpcaaIWaaabaGa aeivaiaabIgacaqG1bGaae4CaiaabYcacaqG0bGaaeiAaiaabwgaca qGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeOzaiaa bwhacaqGUbGaae4yaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae yAaiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabohacaqG VbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaab+ gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae4B aiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqGKb GaaeyAaiaab6gacaqGNbGaaeiiaaqaaiaabsgacaqGPbGaaeOzaiaa bAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaae iBaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqG VbGaaeOBaiaab6caaaaa@0F43@

Q.22

Verify that the given functions ( explicit or implicit )is a solution of the corresponding differential equation: y= a 2 x 2 x( a,a ): x+y dy dx =0( y0 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVD0dg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaaqaaaaa aaaaWdbiaabAfacaqGLbGaaeOCaiaabMgacaqGMbGaaeyEaiaabcca caqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGMbGa aeyDaiaab6gacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabohaca qGGaWdamaabmaabaWdbiaabwgacaqG4bGaaeiCaiaabYgacaqGPbGa ae4yaiaabMgacaqG0bGaaeiiaiaab+gacaqGYbGaaeiiaiaabMgaca qGTbGaaeiCaiaabYgacaqGPbGaae4yaiaabMgacaqG0baapaGaayjk aiaawMcaa8qacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabohaca qGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaa b+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGJbGaae 4BaiaabkhacaqGYbGaaeyzaiaabohacaqGWbGaae4Baiaab6gacaqG KbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsgacaqGPbGaaeOzaiaabA gacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiB aiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVb GaaeOBaiaabQdaaeaaruGvLjhzH5wyaGqbb8aacaWF5bGaeyypa0Za aOaaaeaacaWHHbWaaWbaaSqabeaacaWHYaaaaOGaeyOeI0IaaCiEam aaCaaaleqabaGaaCOmaaaaaeqaaOGaaGPaVlaahIhacqGHiiIZdaqa daqaaiabgkHiTiaa=fgacaWFSaGaa8xyaaGaayjkaiaawMcaaiaayk W7caaMc8UaaGPaVlaaykW7caaMc8Uaa8Noaiaa=bcacaWLjaGaa8hE aiaa=TcacaWF5bWaaSaaaeaacaWFKbGaa8xEaaqaaiaa=rgacaWF4b aaaiaa=1dacaWFWaGaaGPaVlaaykW7daqadaqaaiaahMhacqGHGjsU caWHWaaacaGLOaGaayzkaaaaaaa@C593@

Ans.

y= a 2 x 2 Differentiating w.r.t. x, we get dy dx = x a 2 x 2 = x y Substituting the value of dy dx in L.H.S. of given differential equation x+y dy dx =0, we get L.H.S.=x+y dy dx =x+y( x y ) =xx =0=R.H.S. Thus,the given function is the solution of the corresponding differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaamyEaiabg2da9maakaaabaGaamyyamaaCaaaleqabaGaaG OmaaaakiabgkHiTiaadIhadaahaaWcbeqaaiaaikdaaaaabeaakiaa ykW7aeaacaqGebGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLb GaaeOBaiaabshacaqGPbGaaeyyaiaabshacaqGPbGaaeOBaiaabEga caqGGaGaae4Daiaab6cacaqGYbGaaeOlaiaabshacaqGUaGaaeiiai aabIhacaqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGa aeiDaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7daWcaaqaaiaa dsgacaWG5baabaGaamizaiaadIhaaaGaeyypa0ZaaSaaaeaacqGHsi slcaWG4baabaWaaOaaaeaacaWGHbWaaWbaaSqabeaacaaIYaaaaOGa eyOeI0IaamiEamaaCaaaleqabaGaaGOmaaaaaeqaaaaaaOqaaiaaxM aacaWLjaGaaCzcaiabg2da9maalaaabaGaeyOeI0IaamiEaaqaaiaa dMhaaaaabaGaae4uaiaabwhacaqGIbGaae4CaiaabshacaqGPbGaae iDaiaabwhacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiaiaabshacaqG ObGaaeyzaiaabccacaqG2bGaaeyyaiaabYgacaqG1bGaaeyzaiaabc cacaqGVbGaaeOzaiaabccadaWcaaqaaiaadsgacaWG5baabaGaamiz aiaadIhaaaGaaeiiaiaabMgacaqGUbGaaeiiaiaabYeacaqGUaGaae isaiaab6cacaqGtbGaaeOlaiaabccacaqGVbGaaeOzaiaabccacaqG NbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaabA gacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyy aiaabYgacaqGGaaabaGaaeyzaiaabghacaqG1bGaaeyyaiaabshaca qGPbGaae4Baiaab6gacaqGGaGaamiEaiabgUcaRiaadMhadaWcaaqa aiaadsgacaWG5baabaGaamizaiaadIhaaaGaeyypa0JaaGimaiaacY cacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaGa aeitaiaab6cacaqGibGaaeOlaiaabofacaqGUaGaeyypa0JaamiEai abgUcaRiaadMhadaWcaaqaaiaadsgacaWG5baabaGaamizaiaadIha aaaabaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpcaWG4b Gaey4kaSIaamyEamaabmaabaWaaSaaaeaacqGHsislcaWG4baabaGa amyEaaaaaiaawIcacaGLPaaaaeaacaWLjaGaaGPaVlaaykW7caaMc8 UaaGPaVlabg2da9iaadIhacqGHsislcaWG4baabaGaaCzcaiaaykW7 caaMc8UaaGPaVlaaykW7cqGH9aqpcaaIWaGaeyypa0JaamOuaiaac6 cacaWGibGaaiOlaiaadofacaGGUaaabaGaaeivaiaabIgacaqG1bGa ae4CaiaabYcacaqG0bGaaeiAaiaabwgacaqGGaGaae4zaiaabMgaca qG2bGaaeyzaiaab6gacaqGGaGaaeOzaiaabwhacaqGUbGaae4yaiaa bshacaqGPbGaae4Baiaab6gacaqGGaGaaeyAaiaabohacaqGGaGaae iDaiaabIgacaqGLbGaaeiiaiaabohacaqGVbGaaeiBaiaabwhacaqG 0bGaaeyAaiaab+gacaqGUbGaaeiiaiaab+gacaqGMbGaaeiiaiaabs hacaqGObGaaeyzaiaabccacaqGJbGaae4BaiaabkhacaqGYbGaaeyz aiaabohacaqGWbGaae4Baiaab6gacaqGKbGaaeyAaiaab6gacaqGNb GaaeiiaaqaaiaabsgacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaa bwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiBaiaabccacaqGLbGaae yCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaab6caaaaa @574C@

Q.23

The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0 (B) 2 (C) 3 (D) 4

Ans.

Since, number of constants in a differential equation of order n is equal to its order i.e., n.
Thus, number of arbitrary constants in a differential equation of fourth order are 4.
Thus, option D is correct.

Q.24

The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3 (B) 2 (C) 1 (D) 0

Ans.

In a particular solution, there are no arbitrary constants.

∴ Number of arbitrary constants = 0
Thus, option D is correct.

Q.25

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

x a + y b =1 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aadaWcaaqaaGqabiaa=HhaaeaacaWFHbaaaiaa=TcadaWcaaqaaiaa =LhaaeaacaWFIbaaaiaa=1dacaWFXaaaaa@3EBE@

Ans.

x a + y b =1 Differentiating both sides w.r.t. x, we get 1 a + 1 b dy dx =0 Againg,differentiating both sides w.r.t. x, we get 1 b d 2 y d x 2 =0 d 2 y d x 2 =0y=0 Hence, the required differential equation is y”=0. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWG 4baabaGaamyyaaaacqGHRaWkdaWcaaqaaiaadMhaaeaacaWGIbaaai abg2da9iaaigdaaeaacaqGebGaaeyAaiaabAgacaqGMbGaaeyzaiaa bkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabshacaqGPbGaae OBaiaabEgacaqGGaGaaeOyaiaab+gacaqG0bGaaeiAaiaabccacaqG ZbGaaeyAaiaabsgacaqGLbGaae4CaiaabccacaqG3bGaaeOlaiaabk hacaqGUaGaaeiDaiaab6cacaqGGaGaaeiEaiaabYcacaqGGaGaae4D aiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaWaaSaaaeaacaaIXa aabaGaamyyaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaWGIbaaamaa laaabaGaamizaiaadMhaaeaacaWGKbGaamiEaaaacqGH9aqpcaaIWa aabaGaaeyqaiaabEgacaqGHbGaaeyAaiaab6gacaqGNbGaaeilaiaa ykW7caqGKbGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaae OBaiaabshacaqGPbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgacaqG GaGaaeOyaiaab+gacaqG0bGaaeiAaiaabccacaqGZbGaaeyAaiaabs gacaqGLbGaae4CaiaabccacaqG3bGaaeOlaiaabkhacaqGUaGaaeiD aiaab6cacaqGGaGaaeiEaiaabYcacaqGGaGaae4DaiaabwgacaqGGa Gaae4zaiaabwgacaqG0baabaGaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8+aaSaaaeaacaaIXaaabaGaamOyaaaadaWcaaqaaiaads gadaahaaWcbeqaaiaaikdaaaGccaWG5baabaGaamizaiaadIhadaah aaWcbeqaaiacyciIYaaaaaaakiabg2da9iaaicdaaeaacqGHshI3ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVpaalaaabaGaamizamaaCaaa leqabaGaaGOmaaaakiaadMhaaeaacaWGKbGaamiEamaaCaaaleqaba GaiGjGikdaaaaaaOGaeyypa0JaaGimaiabgkDiElaadMhacaGGNaGa ai4jaiabg2da9iaaicdaaeaacaqGibGaaeyzaiaab6gacaqGJbGaae yzaiaabYcacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkhacaqG LbGaaeyCaiaabwhacaqGPbGaaeOCaiaabwgacaqGKbGaaeiiaiaabs gacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiD aiaabMgacaqGHbGaaeiBaiaabccacaqGLbGaaeyCaiaabwhacaqGHb GaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGPbGaae4Caiaabcca caqG5bGaae4jaiaabEcacaqG9aGaaeimaiaab6caaaaa@FBFD@

Q.26

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y 2 =a( b 2 x 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aaieqacaWF5bWaaWbaaSqabeaacaWHYaaaaOGaa8xpaiaa=fgadaqa daqaaiaahkgadaahaaWcbeqaaiaaikdaaaGccqGHsislcaWH4bWaaW baaSqabeaacaWHYaaaaaGccaGLOaGaayzkaaaaaa@429D@

Ans.

y2=a(b2x2)Differentiating both sides w.r.t. x, we get2ydydx=a(02x)2ydydx=2ax  ...(i)Againg,differentiating both sides w.r.t. x, we get2{(dydx)2+yd2ydx2}=2a    (dydx)2+yd2ydx2=a    ...(ii)Putting value ofa from equation(ii) to equation(i), we get        2ydydx=2{(dydx)2+yd2ydx2}x      ydydx=x(dydx)2+xyd2ydx2xyy+x(y)2yy=0Hence, the required differential equation is xyy”+x(y)2yy‘ = 0.

Q.27

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y=a e 3x +b e 2x MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aaieqacaWF5bGaa8xpaiaa=fgacaWFLbWaaWbaaSqabeaacaWFZaGa a8hEaaaakiaa=TcacaWFIbGaa8xzamaaCaaaleqabaGaeyOeI0Iaa8 Nmaiaa=Hhaaaaaaa@4369@

Ans.

y=a e 3x +b e 2x ( i ) Differentiating both sides w.r.t. x, we get y=3a e 3x 2b e 2x ( ii ) Againg,differentiating both sides w.r.t. x, we get y=9a e 3x +4b e 2x ( iii ) Adding equation( ii ) and 2×equation( i ),we get y+2y=3a e 3x 2b e 2x +2( a e 3x +b e 2x ) y+2y=3a e 3x 2b e 2x +2a e 3x +2b e 2x y+2y=5a e 3x a e 3x = y+2y 5 Substracting equation( ii ) from 3×equation( i ),we get 3yy=3a e 3x +3b e 2x ( 3a e 3x 2b e 2x ) 3yy=3a e 3x +3b e 2x 3a e 3x +2b e 2x 3yy=5b e 2x b e 2x = 3yy 5 Putting the values of ae 3x and be -2x in equation( iii ), we get y=9( y+2y 5 )+4( 3yy 5 ) 5y=9y+18y+12y4y 5y=5y+30y yy6y=0 Hence, the required differential equation is y”–y’–6y = 0. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyEaiabg2da 9iaadggacaWGLbWaaWbaaSqabeaacaaIZaGaamiEaaaakiabgUcaRi aadkgacaWGLbWaaWbaaSqabeaacqGHsislcaaIYaGaamiEaaaakiaa xMaacaWLjaGaaCzcaiaaxMaacaaMc8UaaiOlaiaac6cacaGGUaWaae WaaeaacaWGPbaacaGLOaGaayzkaaaabaGaaeiraiaabMgacaqGMbGa aeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabggaca qG0bGaaeyAaiaab6gacaqGNbGaaeiiaiaabkgacaqGVbGaaeiDaiaa bIgacaqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabohacaqGGaGaae 4Daiaab6cacaqGYbGaaeOlaiaabshacaqGUaGaaeiiaiaabIhacaqG SaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaai aaykW7caaMc8UaamyEaiaacEcacqGH9aqpcaaIZaGaamyyaiaadwga daahaaWcbeqaaiaaiodacaWG4baaaOGaeyOeI0IaaGOmaiaadkgaca 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Q.28

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y=e 2x a+bx

 

Ans.

y= e 2x ( a+bx ) Differentiating both sides w.r.t. x, we get y= e 2x ( 0+b )+2 e 2x ( a+bx ) y=b e 2x +2y y2y=b e 2x ( i ) Againg,differentiating both sides w.r.t. x, we get y2y=2b e 2x ( ii ) Dividing equation( ii ) by equation( i ), we get y2y y2y = 2b e 2x b e 2x y2y y2y =2 y2y=2( y2y ) y2y2y+4y=0 y4y+4y=0 Hence, the required differential equation is y” –4y’ + 4y = 0. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaxMaacaWLjaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7 caWG5bGaeyypa0JaamyzamaaCaaaleqabaGaaGOmaiaadIhaaaGcda qadaqaaiaadggacqGHRaWkcaWGIbGaaGPaVlaadIhaaiaawIcacaGL PaaaaeaacaqGebGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLb GaaeOBaiaabshacaqGPbGaaeyyaiaabshacaqGPbGaaeOBaiaabEga caqGGaGaaeOyaiaab+gacaqG0bGaaeiAaiaabccacaqGZbGaaeyAai aabsgacaqGLbGaae4CaiaabccacaqG3bGaaeOlaiaabkhacaqGUaGa aeiDaiaab6cacaqGGaGaaeiEaiaabYcacaqGGaGaae4Daiaabwgaca qGGaGaae4zaiaabwgacaqG0baabaGaaCzcaiaaxMaacaWLjaGaaGPa VlaaykW7caaMc8UaamyEaiaacEcacqGH9aqpcaWGLbWaaWbaaSqabe aacaaIYaGaamiEaaaakmaabmaabaGaaGimaiabgUcaRiaadkgaaiaa wIcacaGLPaaacqGHRaWkcaaIYaGaamyzamaaCaaaleqabaGaaGOmai aadIhaaaGcdaqadaqaaiaadggacqGHRaWkcaWGIbGaaGPaVlaadIha aiaawIcacaGLPaaaaeaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVl aaykW7caWG5bGaai4jaiabg2da9iaadkgacaWGLbWaaWbaaSqabeaa caaIYaGaamiEaaaakiabgUcaRiaaikdacaWG5baabaGaeyO0H4TaaC zcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaamyEaiaacEcacqGH sislcaaIYaGaamyEaiabg2da9iaadkgacaWGLbWaaWbaaSqabeaaca aIYaGaamiEaaaakiaaxMaacaWLjaGaaGPaVlaaxMaacaaMc8UaaGPa VlaaykW7caGGUaGaaiOlaiaac6cadaqadaqaaiaadMgaaiaawIcaca GLPaaaaeaacaqGbbGaae4zaiaabggacaqGPbGaaeOBaiaabEgacaqG SaGaaGPaVlaabsgacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaabw gacaqGUbGaaeiDaiaabMgacaqGHbGaaeiDaiaabMgacaqGUbGaae4z aiaabccacaqGIbGaae4BaiaabshacaqGObGaaeiiaiaabohacaqGPb GaaeizaiaabwgacaqGZbGaaeiiaiaabEhacaqGUaGaaeOCaiaab6ca caqG0bGaaeOlaiaabccacaqG4bGaaeilaiaabccacaqG3bGaaeyzai aabccacaqGNbGaaeyzaiaabshaaeaacaWLjaGaaCzcaiaaykW7caaM c8UaaGPaVlaadMhacaGGNaGaai4jaiabgkHiTiaaikdacaWG5bGaai 4jaiabg2da9iaaikdacaWGIbGaamyzamaaCaaaleqabaGaaGOmaiaa dIhaaaGccaaMc8UaaGPaVlaaxMaacaWLjaGaaCzcaiaaykW7caaMc8 UaaiOlaiaac6cacaGGUaWaaeWaaeaacaWGPbGaamyAaaGaayjkaiaa wMcaaaqaaiaabseacaqGPbGaaeODaiaabMgacaqGKbGaaeyAaiaab6 gacaqGNbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyA aiaab+gacaqGUbWaaeWaaeaacaqGPbGaaeyAaaGaayjkaiaawMcaai aabccacaqGIbGaaeyEaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGa aeiDaiaabMgacaqGVbGaaeOBamaabmaabaGaaeyAaaGaayjkaiaawM caaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG 0baabaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8+aaSaaae aacaWG5bGaai4jaiaacEcacqGHsislcaaIYaGaamyEaiaacEcaaeaa caWG5bGaai4jaiabgkHiTiaaikdacaWG5baaaiabg2da9maalaaaba GaaGOmaiaadkgacaWGLbWaaWbaaSqabeaacaaIYaGaamiEaaaaaOqa aiaadkgacaWGLbWaaWbaaSqabeaacaaIYaGaamiEaaaaaaaakeaaca WLjaGaaCzcaiaaaykW7caaMc8UaaGPaVlaaykW7daWcaaqaaiaadMha caGGNaGaai4jaiabgkHiTiaaikdacaWG5bGaai4jaaqaaiaadMhaca GGNaGaeyOeI0IaaGOmaiaadMhaaaGaeyypa0JaaGOmaaqaaiabgkDi ElaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWG5b Gaai4jaiaacEcacqGHsislcaaIYaGaamyEaiaacEcacqGH9aqpcaaI YaWaaeWaaeaacaWG5bGaai4jaiabgkHiTiaaikdacaWG5baacaGLOa GaayzkaaaabaGaeyO0H4TaamyEaiaacEcacaGGNaGaeyOeI0IaaGOm aiaadMhacaGGNaGaeyOeI0IaaGOmaiaadMhacaGGNaGaey4kaSIaaG inaiaadMhacqGH9aqpcaaIWaaabaGaeyO0H4TaaCzcaiaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaadMhacaGGNaGaai4jaiabgk HiTiaaisdacaWG5bGaai4jaiabgUcaRiaaisdacaWG5bGaeyypa0Ja aGimaaqaaiaabIeacaqGLbGaaeOBaiaabogacaqGLbGaaeilaiaabc cacaqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaabwgacaqGXbGaaeyD aiaabMgacaqGYbGaaeyzaiaabsgacaqGGaGaaeizaiaabMgacaqGMb GaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabgga caqGSbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAai aab+gacaqGUbGaaeiiaiaabMgacaqGZbGaaeiiaiaabMhacaqGNaGa ae4jaiaabccacaqGtaIaaeinaiaabMhacaqGNaGaaeiiaiaabUcaca qGGaGaaeinaiaabMhacaqGGaGaaeypaiaabccacaqGWaGaaeOlaaaa aa@CF90@

Q.29

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y= e x ( a cos x+b sinx ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aaieqacaWF5bGaa8xpaiaa=vgadaahaaWcbeqaaiaa=HhaaaGcdaqa daqaaiaa=fgacaWFGaGaa83yaiaa=9gacaWFZbGaa8hiaiaa=Hhaca WFRaGaa8Nyaiaa=bcacaWHZbGaaCyAaiaah6gacaaMc8UaaCiEaaGa ayjkaiaawMcaaaaa@4BA7@

Ans.

y= e x ( a cos x+b sinx )( i ) Differentiating both sides w.r.t. x, we get y= e x ( a sin x+b cosx )+ e x ( a cos x+b sinx ) y= e x ( a sin x+b cosx )+y[ Fromequation( i ) ] yy= e x ( a sin x+b cosx ) ( ii ) Againg,differentiating both sides w.r.t. x, we get yy= e x ( a cos xb sinx )+ e x ( a sin x+b cosx ) yy=y+yy [ From equation( ii ) ] y2y+2y=0 Hence, the required differential equation is y2y+2y=0. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8Mrpy0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaxMaacaWLjaGaaCzcaiaadMhacqGH9aqpcaWGLbWaaWba aSqabeaacaWG4baaaOWaaeWaaeaacaWGHbacbeGaa8hiaiGacogaca GGVbGaai4Caiaa=bcacaWG4bGaey4kaSIaamOyaiaa=bcaciGGZbGa aiyAaiaac6gacaaMc8UaamiEaaGaayjkaiaawMcaaiaac6cacaGGUa GaaiOlamaabmaabaGaamyAaaGaayjkaiaawMcaaaqaaiaabseacaqG PbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabM gacaqGHbGaaeiDaiaabMgacaqGUbGaae4zaiaabccacaqGIbGaae4B aiaabshacaqGObGaaeiiaiaabohacaqGPbGaaeizaiaabwgacaqGZb GaaeiiaiaabEhacaqGUaGaaeOCaiaab6cacaqG0bGaaeOlaiaabcca caqG4bGaaeilaiaabccacaqG3bGaaeyzaiaabccacaqGNbGaaeyzai aabshaaeaacaWLjaGaaCzcaiaaxMaacaWG5bGaai4jaiabg2da9iaa dwgadaahaaWcbeqaaiaadIhaaaGcdaqadaqaaiabgkHiTiaadggaca WFGaGaci4CaiaacMgacaGGUbGaa8hiaiaadIhacqGHRaWkcaWGIbGa a8hiaiGacogacaGGVbGaai4CaiaaykW7caWG4baacaGLOaGaayzkaa Gaey4kaSIaamyzamaaCaaaleqabaGaamiEaaaakmaabmaabaGaamyy aiaa=bcaciGGJbGaai4BaiaacohacaWFGaGaamiEaiabgUcaRiaadk gacaWFGaGaci4CaiaacMgacaGGUbGaaGPaVlaadIhaaiaawIcacaGL PaaaaeaacaWLjaGaaCzcaiaaxMaacaWG5bGaai4jaiabg2da9iaadw gadaahaaWcbeqaaiaadIhaaaGcdaqadaqaaiabgkHiTiaadggacaWF GaGaci4CaiaacMgacaGGUbGaa8hiaiaadIhacqGHRaWkcaWGIbGaa8 hiaiGacogacaGGVbGaai4CaiaaykW7caWG4baacaGLOaGaayzkaaGa ey4kaSIaamyEaiaaykW7caaMc8UaaGPaVpaadmaabaGaaeOraiaabk hacaqGVbGaaeyBaiaaykW7caaMc8UaaeyzaiaabghacaqG1bGaaeyy aiaabshacaqGPbGaae4Baiaab6gadaqadaqaaiaabMgaaiaawIcaca GLPaaaaiaawUfacaGLDbaaaeaacqGHshI3caWLjaGaaCzcaiaaykW7 caaMc8UaaGPaVlaaykW7caWG5bGaai4jaiabgkHiTiaadMhacqGH9a qpcaWGLbWaaWbaaSqabeaacaWG4baaaOWaaeWaaeaacqGHsislcaWG HbGaa8hiaiGacohacaGGPbGaaiOBaiaa=bcacaWG4bGaey4kaSIaam Oyaiaa=bcaciGGJbGaai4BaiaacohacaaMc8UaamiEaaGaayjkaiaa wMcaaiaaxMaacaWLjaGaaGPaVlaaxMaacaaMc8UaaGPaVlaaykW7ca GGUaGaaiOlaiaac6cadaqadaqaaiaadMgacaWGPbaacaGLOaGaayzk aaaabaGaaeyqaiaabEgacaqGHbGaaeyAaiaab6gacaqGNbGaaeilai aaykW7caqGKbGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGa aeOBaiaabshacaqGPbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgaca qGGaGaaeOyaiaab+gacaqG0bGaaeiAaiaabccacaqGZbGaaeyAaiaa bsgacaqGLbGaae4CaiaabccacaqG3bGaaeOlaiaabkhacaqGUaGaae iDaiaab6cacaqGGaGaaeiEaiaabYcacaqGGaGaae4DaiaabwgacaqG GaGaae4zaiaabwgacaqG0baabaGaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyEaiaacEcaca GGNaGaeyOeI0IaamyEaiaacEcacqGH9aqpcaWGLbWaaWbaaSqabeaa caWG4baaaOWaaeWaaeaacqGHsislcaWGHbGaa8hiaiGacogacaGGVb Gaai4Caiaa=bcacaWG4bGaeyOeI0IaamOyaiaa=bcaciGGZbGaaiyA aiaac6gacaaMc8UaamiEaaGaayjkaiaawMcaaiabgUcaRiaadwgada ahaaWcbeqaaiaadIhaaaGcdaqadaqaaiabgkHiTiaadggacaWFGaGa ci4CaiaacMgacaGGUbGaa8hiaiaadIhacqGHRaWkcaWGIbGaa8hiai GacogacaGGVbGaai4CaiaaykW7caWG4baacaGLOaGaayzkaaaabaGa aGPaVlaaykW7caaMc8UaaGPaVlaaxMaacaaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caWG5bGaai4jaiaacEcacqGHsislcaWG5bGaai4jaiabg2da9i abgkHiTiaadMhacqGHRaWkcaWG5bGaai4jaiabgkHiTiaadMhacaWL jaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaxMaadaWadaqaaiaabAeaca qGYbGaae4Baiaab2gacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaa bshacaqGPbGaae4Baiaab6gadaqadaqaaiaadMgacaWGPbaacaGLOa GaayzkaaaacaGLBbGaayzxaaaabaGaaGPaVlaaykW7caaMc8UaaGPa VlaaxMaacaaMc8UaaGPaVlaadMhacaGGNaGaai4jaiabgkHiTiaaik dacaWG5bGaai4jaiabgUcaRiaaikdacaWG5bGaeyypa0JaaGimaaqa aiaabIeacaqGLbGaaeOBaiaabogacaqGLbGaaeilaiaabccacaqG0b GaaeiAaiaabwgacaqGGaGaaeOCaiaabwgacaqGXbGaaeyDaiaabMga caqGYbGaaeyzaiaabsgacaqGGaGaaeizaiaabMgacaqGMbGaaeOzai aabwgacaqGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabggacaqGSbGa aeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gaca qGUbGaaeiiaiaabMgacaqGZbGaaeiiaiaadMhacaGGNaGaai4jaiab gkHiTiaaikdacaWG5bGaai4jaiabgUcaRiaaikdacaWG5bGaeyypa0 JaaGimaiaac6caaaaa@F573@

Q.30

Form the differential equation of the family of circles touching the y-axis at origin.

Ans.

The centre of the circle touching y-axis at origin lies on x-axis. Let the centre of circle be (a, 0).
Radius of circle will be ‘a’ because circle touches y-axis at origin. So, the equation of circle with centre (a, 0) and radius ‘a’ is as follows
(x – a)2 + y2 = a2
x2 – 2ax + a2 + y2 = a2
x2 + y2 = 2ax … (i)

Differentiating equation (i), w.r.t. x, we get

2x + 2yy’ = 2a

x + yy’ = a

Putting value of a in equation (i), we get

x2 + y2 = 2(x + yy’)x
= 2x2 + 2xyy’

y2 = x2 + 2xyy’

Thus, the required differential equation is 2xyy’ + x2 = y2.

Q.31

Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Ans.

The equation of parabolas having vertex at origin and axis along positive y-axis is as follows
x2 = 4ay …(i)
Differentiating w.r.t. x, we get
2x = 4ay’
or (2x/y’) = 4a
Putting value of 4a in equation (i), we get
x2 = (2x/y’)y

x2y’ = 2xy

or xy’ = 2y
or xy’ – 2y = 0

Thus, the required differential equation is
xy’ – 2y = 0.

Q.32

Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Ans.

Theequation of the family of ellipses having foci on y-axis and centre at origin is as follows x 2 b 2 + y 2 a 2 =1 ( i ) Differentiating w.r.t. x, we get 2x b 2 + 2yy’ a 2 =0 x b 2 + yy’ a 2 =0 ( ii ) Again,differentiating w.r.t. x, we get 1 b 2 + 1 a 2 ( yy”+y’y’ )=0 1 b 2 + 1 a 2 { yy”+ ( y’ ) 2 }=0 1 b 2 = 1 a 2 { yy”+ ( y’ ) 2 } Substituting the value of 1 b 2 in equation ( ii ), we get x[ 1 a 2 { yy”+ ( y’ ) 2 } ]+ yy’ a 2 =0 x{ yy”+ ( y’ ) 2 }+yy=0 xyyx ( y ) 2 +yy=0 xyy+x ( y ) 2 yy=0 Thus, the required differential equation is xyy” + x ( y’ ) 2 –yy’ = 0. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaaykW7caaMc8Uaaeyzaiaabgha caqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae4Bai aabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabAgacaqGHbGa aeyBaiaabMgacaqGSbGaaeyEaiaabccacaqGVbGaaeOzaiaabccaca qGLbGaaeiBaiaabYgacaqGPbGaaeiCaiaabohacaqGLbGaae4Caiaa bccacaqGObGaaeyyaiaabAhacaqGPbGaaeOBaiaabEgacaqGGaGaae Ozaiaab+gacaqGJbGaaeyAaiaabccacaqGVbGaaeOBaiaabckaaeaa caqG5bGaaeylaiaabggacaqG4bGaaeyAaiaabohacaqGGaGaaeyyai aab6gacaqGKbGaaeiiaiaabogacaqGLbGaaeOBaiaabshacaqGYbGa aeyzaiaabccacaqGHbGaaeiDaiaabccacaqGVbGaaeOCaiaabMgaca qGNbGaaeyAaiaab6gacaqGGaGaaeyAaiaabohacaqGGaGaaeyyaiaa bohacaqGGaGaaeOzaiaab+gacaqGSbGaaeiBaiaab+gacaqG3bGaae 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Q.33

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Ans.

The differential equation of the family of hyperbolas having foci on x-axis and centre at origin​ is as follows x 2 a 2 y 2 b 2 =1 ( i ) Differentiating​ w.r.t. x, we get 2x a 2 2yy’ b 2 =0 x a 2 yy’ b 2 =0 ( ii ) Again,differentiating w.r.t. x, we get 1 a 2 1 b 2 ( yy”+y’y’ )=0 1 a 2 1 b 2 { yy”+ ( y’ ) 2 }=0 1 a 2 = 1 b 2 { yy”+ ( y’ ) 2 } Substituting the value of 1 a 2 in equation ( ii ), we get x[ 1 b 2 { yy”+ ( y’ ) 2 } ] yy’ b 2 =0 x{ yy”+ ( y’ ) 2 }yy=0 xyy+x ( y ) 2 yy=0 Thus, the required differential equation is xyy” + x ( y’ ) 2 –yy’=0. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGKbGaaeyAaiaabAga caqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyai aabYgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGa ae4Baiaab6gacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgaca qGLbGaaeiiaiaabAgacaqGHbGaaeyBaiaabMgacaqGSbGaaeyEaiaa bccacaqGVbGaaeOzaiaabccacaqGObGaaeyEaiaabchacaqGLbGaae OCaiaabkgacaqGVbGaaeiBaiaabggacaqGZbGaaeiiaiaabIgacaqG HbGaaeODaiaabMgacaqGUbGaae4zaiaabccacaqGMbGaae4Baiaabo gacaqGPbaabaGaae4Baiaab6gacaqGGaGaaeiEaiaab2cacaqGHbGa aeiEaiaabMgacaqGZbGaaeiiaiaabggacaqGUbGaaeizaiaabccaca qGJbGaaeyzaiaab6gacaqG0bGaaeOCaiaabwgacaqGGaGaaeyyaiaa bshacaqGGcGaaGPaVlaab+gacaqGYbGaaeyAaiaabEgacaqGPbGaae 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Q.34

Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Ans.

The centre of the circle lies on y-axis. Let the centre of circle be (0, b).
So, the equation of circle with centre (0, b) and radius 3 is as follows
x2 + (y – b)2 = 32 … (i)

Differentiating equation ( i ), w.r.t. x, we get 2x+2( y b ) y=0 x+( y b ) y=0 ( yb )= x y

Putting value of a in equation ( i ), we get

x 2 + ( x y ) 2 =9 x 2 y 2 + x 2 =9y 2 ( y ) 2 ( x 2 9 )+ x 2 =0 This is the required differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aaceqaaiaacckacaGGGcGaaiiOaiaacckacaGGGcGaaiiOaiaaccka caGGGcGaaiiOaiaaykW7caqG4bWaaWbaaSqabeaacaqGYaaaaOGaey 4kaSYaaeWaaeaadaWcaaqaaiabgkHiTiaabIhaaeaacaaMc8UaaeyE aiaacMbicaGGGcaaaaGaayjkaiaawMcaamaaCaaaleqabaGaaeOmaa aakiabg2da9iaaiMdacaGGGcaabaGaaCzcaiaaykW7caaMc8UaaGPa VlaaykW7caWG4bWaaWbaaSqabeaacGaMaIOmaaaakiaadMhacaGGNa WaaWbaaSqabeaacaaIYaaaaOGaaGPaVlabgUcaRiaadIhadaahaaWc beqaaiacyciIYaaaaOGaeyypa0JaaGyoaiaadMhacaGGNaWaaWbaaS qabeaacaaIYaaaaaGcbaWaaeWaaeaacaWG5bGaai4jaaGaayjkaiaa wMcaamaaCaaaleqabaGaaGOmaaaakmaabmaabaGaamiEamaaCaaale qabaGaiGjGikdaaaGccqGHsislcaaI5aaacaGLOaGaayzkaaGaaGPa VlabgUcaRiaadIhadaahaaWcbeqaaiacyciIYaaaaOGaeyypa0JaaG imaaqaaiaadsfacaWGObGaamyAaiaadohacaqGGaGaaeyAaiaaboha caqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkhacaqGLbGaaeyCai aabwhacaqGPbGaaeOCaiaabwgacaqGKbGaaeiiaiaabsgacaqGPbGa aeOzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgaca qGHbGaaeiBaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaa bMgacaqGVbGaaeOBaiaab6caaaaa@A1C0@

Q.35

Which of the following differential equations has y = c1 ex + c2 ex as the general solution?(A)d2ydx2 + y = 0    (B)d2ydx2y = 0  (C)d2ydx2 + 1 = 0 (D)d2ydx21 = 0

Ans.

            y=c1ex+c2ex  ...(i)Differentiating w.r.t. x, we get      dydx=c1exc2exAgain,​ differentiating w.r.t. x, we get  d2ydx2=c1ex+c2ex          =yord2ydx2y=0Thus,​​ option B is correct.

Q.36

Which of the following differential equations has y = x as one of its particular solution? ( A ) d 2 y d x 2 x 2 dy dx +xy=x ( B ) d 2 y d x 2 x dy dx +xy=x ( C ) d 2 y d x 2 x 2 dy dx +xy=0 ( D ) d 2 y d x 2 +x dy dx +xy=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabEfacaqGObGaaeyAaiaabogacaqGObGaaeiiaiaab+ga caqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGMbGaae4Bai aabYgacaqGSbGaae4BaiaabEhacaqGPbGaaeOBaiaabEgacaqGGaGa aeizaiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6gaca qG0bGaaeyAaiaabggacaqGSbGaaeiiaiaabwgacaqGXbGaaeyDaiaa bggacaqG0bGaaeyAaiaab+gacaqGUbGaae4CaiaabccacaqGObGaae yyaiaabohacaqGGaGaaeyEaiaabccacaqG9aGaaeiiaiaabIhacaqG GaGaaeyyaiaabohacaqGGaaabaGaae4Baiaab6gacaqGLbGaaeiiai aab+gacaqGMbGaaeiiaiaabMgacaqG0bGaae4CaiaabccacaqGWbGa aeyyaiaabkhacaqG0bGaaeyAaiaabogacaqG1bGaaeiBaiaabggaca qGYbGaaeiiaiaabohacaqGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaa b+gacaqGUbGaae4paaqaamaabmaabaGaaeyqaaGaayjkaiaawMcaai aaykW7caaMc8+aaSaaaeaaieqacaWFKbWaaWbaaSqabeaacaWFYaaa aOGaa8xEaaqaaiaa=rgacaWF4bWaaWbaaSqabeaacaWFYaaaaaaaki abgkHiTiaa=HhadaahaaWcbeqaaiaa=jdaaaGcdaWcaaqaaiaahsga caWF5baabaGaaCizaiaahIhaaaGaa83kaiaa=HhacaWF5bGaa8xpai aa=HhacaaMc8UaaGPaVlaaykW7caaMc8UaaCzcaiaaxMaadaqadaqa aiaabkeaaiaawIcacaGLPaaacaaMc8UaaGPaVpaalaaabaGaa8hzam aaCaaaleqabaGaa8Nmaaaakiaa=LhaaeaacaWFKbGaa8hEamaaCaaa leqabaGaa8NmaaaaaaGccqGHsislcaWF4bWaaSaaaeaacaWHKbGaa8 xEaaqaaiaahsgacaWH4baaaiaa=TcacaWF4bGaa8xEaiaa=1dacaWF 4baabaWaaeWaaeaacaqGdbaacaGLOaGaayzkaaGaaGPaVlaaykW7da Wcaaqaaiaa=rgadaahaaWcbeqaaiaa=jdaaaGccaWF5baabaGaa8hz aiaa=HhadaahaaWcbeqaaiaa=jdaaaaaaOGaeyOeI0Iaa8hEamaaCa aaleqabaGaa8NmaaaakmaalaaabaGaaCizaiaa=LhaaeaacaWHKbGa aCiEaaaacaWFRaGaa8hEaiaa=LhacaWF9aGaa8hmaiaaxMaacaWLja GaaCzcamaabmaabaGaaeiraaGaayjkaiaawMcaaiaaykW7caaMc8+a aSaaaeaacaWFKbWaaWbaaSqabeaacaWFYaaaaOGaa8xEaaqaaiaa=r gacaWF4bWaaWbaaSqabeaacaWFYaaaaaaakiabgUcaRiaa=HhadaWc aaqaaiaahsgacaWF5baabaGaaCizaiaahIhaaaGaa83kaiaa=Hhaca WF5bGaa8xpaiaa=bdaaaaa@E777@

Ans.

We have y = x ( i ) Differentiating w.r.t. x, we get dy dx =1 ( ii ) Again,​ differentiating w.r.t. x, we get d 2 y d x 2 =0 ( iii ) Putting values of y, dy dx and d 2 y d x 2 in each given options, ( A )L.H.S.= d 2 y d x 2 x 2 dy dx +xy =0 x 2 ( 1 )+x( x ) =0R.H.S. Thus, it is not correct option. ( B )L.H.S.= d 2 y d x 2 x dy dx +xy =0x( x )+x( x ) =0R.H.S. Thus, it is not correct option. ( C )L.H.S.= d 2 y d x 2 x 2 dy dx +xy =0 x 2 ( 1 )+x( x ) =0=R.H.S. Thus, it is correct option. Hence, the correct solution is option C. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabEfacaqGLbGaaeiiaiaabIgacaqGHbGaaeODaiaabwga caqGGaGaaeyEaiaabccacaqG9aGaaeiiaiaabIhacaGGGcGaaiiOai aacckacaqGMaIaaeiiamaabmaabaGaaeyAaaGaayjkaiaawMcaaaqa aiaabseacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUb GaaeiDaiaabMgacaqGHbGaaeiDaiaabMgacaqGUbGaae4zaiaabcca caqG3bGaaeOlaiaabkhacaqGUaGaaeiDaiaab6cacaqGGaGaaeiEai aabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baa baGaaCzcaiaaxMaadaWcaaqaaiaadsgacaWG5baabaGaamizaiaadI haaaGaeyypa0JaaGymaiaaxMaacaWLjaGaaCzcaiaab6cacaqGUaGa aeOlamaabmaabaGaaeyAaiaabMgaaiaawIcacaGLPaaaaeaacaqGbb Gaae4zaiaabggacaqGPbGaaeOBaiaabYcacaaMb8Uaaeiiaiaabsga caqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDai aabMgacaqGHbGaaeiDaiaabMgacaqGUbGaae4zaiaabccacaqG3bGa aeOlaiaabkhacaqGUaGaaeiDaiaab6cacaqGGaGaaeiEaiaabYcaca qGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaGaaCzc aiaaxMaadaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG5b aabaGaamizaiaadIhadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0Ja aGimaiaaxMaacaWLjaGaaCzcaiaab6cacaqGUaGaaeOlamaabmaaba GaaeyAaiaabMgacaqGPbaacaGLOaGaayzkaaaabaGaaeiuaiaabwha caqG0bGaaeiDaiaabMgacaqGUbGaae4zaiaabccacaqG2bGaaeyyai aabYgacaqG1bGaaeyzaiaabohacaqGGaGaae4BaiaabAgacaqGGaGa aeyEaiaabYcacaqGGaWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgaca WG4baaaiaabccacaWGHbGaamOBaiaadsgacaqGGaWaaSaaaeaacaWG KbWaaWbaaSqabeaacaaIYaaaaOGaamyEaaqaaiaadsgacaWG4bWaaW baaSqabeaacaaIYaaaaaaakiaabccacaqGPbGaaeOBaiaabccacaqG LbGaaeyyaiaabogacaqGObGaaeiiaiaabEgacaqGPbGaaeODaiaabw gacaqGUbGaaeiiaiaab+gacaqGWbGaaeiDaiaabMgacaqGVbGaaeOB aiaabohacaqGSaaabaWaaeWaaeaacaqGbbaacaGLOaGaayzkaaGaaG PaVlaaykW7caqGmbGaaeOlaiaabIeacaqGUaGaae4uaiaac6cacqGH 9aqpdaWcaaqaaiaadsgadaahaaWcbeqaaiaaikdaaaGccaWG5baaba GaamizaiaadIhadaahaaWcbeqaaiaaikdaaaaaaOGaeyOeI0IaamiE amaaCaaaleqabaGaaGOmaaaakmaalaaabaGaamizaiaadMhaaeaaca WGKbGaamiEaaaacqGHRaWkcaWG4bGaamyEaaqaaiaaxMaacaWLjaGa eyypa0JaaGimaiabgkHiTiaadIhadaahaaWcbeqaaiaaikdaaaGcda qadaqaaiaaigdaaiaawIcacaGLPaaacqGHRaWkcaWG4bWaaeWaaeaa caWG4baacaGLOaGaayzkaaaabaGaaCzcaiaaxMaacqGH9aqpcaaIWa GaeyiyIKRaaeOuaiaab6cacaqGibGaaeOlaiaabofacaqGUaaabaGa aeivaiaabIgacaqG1bGaae4CaiaabYcacaqGGaGaaeyAaiaabshaca qGGaGaaeyAaiaabohacaqGGaGaaeOBaiaab+gacaqG0bGaaeiiaiaa bogacaqGVbGaaeOCaiaabkhacaqGLbGaae4yaiaabshacaqGGaGaae 4BaiaabchacaqG0bGaaeyAaiaab+gacaqGUbGaaeOlaaqaamaabmaa baGaaeOqaaGaayjkaiaawMcaaiaaykW7caqGmbGaaeOlaiaabIeaca qGUaGaae4uaiaab6cacqGH9aqpcaaMc8UaaGPaVpaalaaabaGaamiz amaaCaaaleqabaGaaGOmaaaakiaadMhaaeaacaWGKbGaamiEamaaCa aaleqabaGaaGOmaaaaaaGccqGHsislcaWG4bWaaSaaaeaacaWGKbGa amyEaaqaaiaadsgacaWG4baaaiabgUcaRiaadIhacaWG5baabaGaaC zcaiaaxMaacqGH9aqpcaaIWaGaeyOeI0IaamiEamaabmaabaGaamiE aaGaayjkaiaawMcaaiabgUcaRiaadIhadaqadaqaaiaadIhaaiaawI cacaGLPaaaaeaacaWLjaGaaCzcaiabg2da9iaaicdacqGHGjsUcaqG sbGaaeOlaiaabIeacaqGUaGaae4uaiaab6caaeaacaqGubGaaeiAai aabwhacaqGZbGaaeilaiaabccacaqGPbGaaeiDaiaabccacaqGPbGa ae4CaiaabccacaqGUbGaae4BaiaabshacaqGGaGaae4yaiaab+gaca qGYbGaaeOCaiaabwgacaqGJbGaaeiDaiaabccacaqGVbGaaeiCaiaa bshacaqGPbGaae4Baiaab6gacaqGUaaabaWaaeWaaeaacaqGdbaaca GLOaGaayzkaaGaaeitaiaab6cacaqGibGaaeOlaiaabofacaqGUaGa eyypa0ZaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaamyEaa qaaiaadsgacaWG4bWaaWbaaSqabeaacaaIYaaaaaaakiabgkHiTiaa dIhadaahaaWcbeqaaiaaikdaaaGcdaWcaaqaaiaadsgacaWG5baaba GaamizaiaadIhaaaGaey4kaSIaamiEaiaadMhaaeaacaWLjaGaaCzc aiabg2da9iaaicdacqGHsislcaWG4bWaaWbaaSqabeaacaaIYaaaaO WaaeWaaeaacaaIXaaacaGLOaGaayzkaaGaey4kaSIaamiEamaabmaa baGaamiEaaGaayjkaiaawMcaaaqaaiaaxMaacaWLjaGaeyypa0JaaG imaiabg2da9iaabkfacaqGUaGaaeisaiaab6cacaqGtbGaaeOlaaqa aiaabsfacaqGObGaaeyDaiaabohacaqGSaGaaeiiaiaabMgacaqG0b GaaeiiaiaabMgacaqGZbGaaeiiaiaabogacaqGVbGaaeOCaiaabkha caqGLbGaae4yaiaabshacaqGGaGaae4BaiaabchacaqG0bGaaeyAai aab+gacaqGUbGaaeOlaaqaaiaabIeacaqGLbGaaeOBaiaabogacaqG LbGaaeilaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4yaiaab+ gacaqGYbGaaeOCaiaabwgacaqGJbGaaeiDaiaabccacaqGZbGaae4B aiaabYgacaqG1bGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGPb Gaae4CaiaabccacaqGVbGaaeiCaiaabshacaqGPbGaae4Baiaab6ga caqGGaGaae4qaiaab6caaaaa@DA01@

Q.37

For the differential equations, find the general solution:

dy dx = 1cosx 1+cosx MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aadaWcaaqaaGqabiaa=rgacaWF5baabaGaa8hzaiaa=HhaaaGaeyyp a0ZaaSaaaeaacaWFXaGaeyOeI0Iaa83yaiaa=9gacaWFZbGaa8hEaa qaaiaa=fdacaWFRaGaa83yaiaa=9gacaWFZbGaa8hEaaaaaaa@4881@

Ans.

The given differential equation is: dy dx = 1cosx 1+cosx = 2 sin 2 x 2 2 cos 2 x 2 [ ∵Cos2x=12 sin 2 x=2 cos 2 x1 ] = tan 2 x 2 = sec 2 x 2 1 Separating the variables, we get dy=( sec 2 x 2 1 )dx Integrating both sides of the above equation, we get dy = ( sec 2 x 2 1 ) dx y=2tan x 2 x+C This is the required general solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAha caqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzai aabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaGa aeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gaca qGGaGaaeyAaiaabohacaqG6aaabaGaaGPaVlaaykW7daWcaaqaaiaa dsgacaWG5baabaGaamizaiaadIhaaaGaeyypa0ZaaSaaaeaacaaIXa GaeyOeI0Iaci4yaiaac+gacaGGZbGaamiEaaqaaiaaigdacqGHRaWk ciGGJbGaai4BaiaacohacaWG4baaaaqaaiaaxMaacqGH9aqpdaWcaa qaaiaaikdaciGGZbGaaiyAaiaac6gadaahaaWcbeqaaiaaikdaaaGc daWcaaqaaiaadIhaaeaacaaIYaaaaaqaaiaaikdaciGGJbGaai4Bai aacohadaahaaWcbeqaaiaaikdaaaGcdaWcaaqaaiaadIhaaeaacaaI YaaaaaaacaWLjaGaaCzcamaadmaabaGaeSynIeLaam4qaiaad+gaca WGZbGaaGOmaiaadIhacqGH9aqpcaaIXaGaeyOeI0IaaGOmaiGacoha caGGPbGaaiOBamaaCaaaleqabaGaaGOmaaaakiaadIhacqGH9aqpca aIYaGaci4yaiaac+gacaGGZbWaaWbaaSqabeaacaaIYaaaaOGaamiE aiabgkHiTiaaigdaaiaawUfacaGLDbaaaeaacaWLjaGaeyypa0Jaci iDaiaacggacaGGUbWaaWbaaSqabeaacaaIYaaaaOWaaSaaaeaacaWG 4baabaGaaGOmaaaaaeaacaWLjaGaeyypa0Jaci4CaiaacwgacaGGJb WaaWbaaSqabeaacaaIYaaaaOWaaSaaaeaacaWG4baabaGaaGOmaaaa cqGHsislcaaIXaaabaGaae4uaiaabwgacaqGWbGaaeyyaiaabkhaca qGHbGaaeiDaiaabMgacaqGUbGaae4zaiaabccacaqG0bGaaeiAaiaa bwgacaqGGaGaaeODaiaabggacaqGYbGaaeyAaiaabggacaqGIbGaae iBaiaabwgacaqGZbGaaeilaiaabccacaqG3bGaaeyzaiaabccacaqG NbGaaeyzaiaabshaaeaacaaMc8UaaGPaVlaabsgacaqG5bGaeyypa0 ZaaeWaaeaaciGGZbGaaiyzaiaacogadaahaaWcbeqaaiaaikdaaaGc daWcaaqaaiaadIhaaeaacaaIYaaaaiabgkHiTiaaigdaaiaawIcaca GLPaaacaWGKbGaamiEaaqaaiaabMeacaqGUbGaaeiDaiaabwgacaqG NbGaaeOCaiaabggacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiaiaabk gacaqGVbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgacaqGKbGaaeyz aiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLb GaaeiiaiaabggacaqGIbGaae4BaiaabAhacaqGLbGaaeiiaiaabwga caqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeilai aabccacaqG3bGaaeyzaiaabccacaqGNbGaaeyzaiaabshaaeaadaWd baqaaiaadsgacaWG5baaleqabeqdcqGHRiI8aOGaaGPaVlabg2da9m aapeaabaWaaeWaaeaaciGGZbGaaiyzaiaacogadaahaaWcbeqaaiaa ikdaaaGcdaWcaaqaaiaadIhaaeaacaaIYaaaaiabgkHiTiaaigdaai aawIcacaGLPaaaaSqabeqaniabgUIiYdGccaaMc8UaamizaiaadIha aeaacaaMc8UaaGPaVlaaykW7caaMc8UaamyEaiabg2da9iaaikdaci GG0bGaaiyyaiaac6gadaWcaaqaaiaadIhaaeaacaaIYaaaaiabgkHi TiaadIhacqGHRaWkcaWGdbaabaGaaeivaiaabIgacaqGPbGaae4Cai aabccacaqGPbGaae4CaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGa aeOCaiaabwgacaqGXbGaaeyDaiaabMgacaqGYbGaaeyzaiaabsgaca qGGaGaae4zaiaabwgacaqGUbGaaeyzaiaabkhacaqGHbGaaeiBaiaa bccacaqGZbGaae4BaiaabYgacaqG1bGaaeiDaiaabMgacaqGVbGaae OBaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaabwgacaqG GaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeizaiaabM gacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0bGaaeyA aiaabggacaqGSbGaaeiiaaqaaiaabwgacaqGXbGaaeyDaiaabggaca qG0bGaaeyAaiaab+gacaqGUbGaaeOlaaaaaa@63B0@

Q.38

For the differential equations, find the general solution:

dy dx = 4 y 2 ( 2<y<2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aadaWcaaqaaGqabiaa=rgacaWF5baabaGaa8hzaiaa=HhaaaGaeyyp a0ZaaOaaaeaacaWF0aGamGjGgkHiTiaa=LhadaahaaWcbeqaaiaa=j daaaaabeaakiaaxMaacaWLjaWaaeWaaeaacqGHsislcaWFYaGaa8hp aiaa=LhacaWF8aGaa8NmaaGaayjkaiaawMcaaaaa@4A3C@

Ans.

The given differential equation is: dy dx = 4 y 2 Separatingthe variables, we get dy 4 y 2 =dx Integrating both sides of above equation,we get dy 4 y 2 = dx sin 1 ( y 2 )=x+C y 2 =sin( x+C ) y=2sin( x+C ) This is the required general solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAha caqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzai aabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaGa aeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gaca qGGaGaaeyAaiaabohacaqG6aaabaGaaCzcaiaaykW7caaMc8+aaSaa aeaacaWGKbGaamyEaaqaaiaadsgacaWG4baaaiabg2da9maakaaaba GaaGinaiadycOHsislcaWG5bWaaWbaaSqabeaacaaIYaaaaaqabaaa keaacaqGtbGaaeyzaiaabchacaqGHbGaaeOCaiaabggacaqG0bGaae yAaiaab6gacaqGNbGaaGjbVlaabshacaqGObGaaeyzaiaabccacaqG 2bGaaeyyaiaabkhacaqGPbGaaeyyaiaabkgacaqGSbGaaeyzaiaabo hacaqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiD aaqaaiaaykW7caaMc8+aaSaaaeaacaWGKbGaamyEaaqaamaakaaaba GaaGinaiabgkHiTiaadMhadaahaaWcbeqaaiaaikdaaaaabeaaaaGc cqGH9aqpcaWGKbGaamiEaaqaaiaabMeacaqGUbGaaeiDaiaabwgaca qGNbGaaeOCaiaabggacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiaiaa bkgacaqGVbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgacaqGKbGaae yzaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeyyaiaabkgacaqG VbGaaeODaiaabwgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabs hacaqGPbGaae4Baiaab6gacaqGSaGaae4DaiaabwgacaqGGaGaae4z aiaabwgacaqG0baabaWaa8qaaeaadaWcaaqaaiaadsgacaWG5baaba WaaOaaaeaacaaI0aGaeyOeI0IaamyEamaaCaaaleqabaGaaGOmaaaa aeqaaaaaaeqabeqdcqGHRiI8aOGaaGPaVlabg2da9maapeaabaGaam izaiaadIhaaSqabeqaniabgUIiYdaakeaaciGGZbGaaiyAaiaac6ga daahaaWcbeqaaiabgkHiTiaaigdaaaGcdaqadaqaamaalaaabaGaam yEaaqaaiaaikdaaaaacaGLOaGaayzkaaGaeyypa0JaamiEaiabgUca RiaadoeaaeaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVpaalaaaba GaamyEaaqaaiaaikdaaaGaeyypa0Jaci4CaiaacMgacaGGUbWaaeWa aeaacaWG4bGaey4kaSIaam4qaaGaayjkaiaawMcaaaqaaiaaxMaaca aMc8UaaGPaVlaaykW7caaMc8UaamyEaiabg2da9iaaikdaciGGZbGa aiyAaiaac6gadaqadaqaaiaadIhacqGHRaWkcaWGdbaacaGLOaGaay zkaaaabaGaaeivaiaabIgacaqGPbGaae4CaiaabccacaqGPbGaae4C aiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaabwgacaqGXb GaaeyDaiaabMgacaqGYbGaaeyzaiaabsgacaqGGaGaae4zaiaabwga caqGUbGaaeyzaiaabkhacaqGHbGaaeiBaiaabccacaqGZbGaae4Bai aabYgacaqG1bGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGa aeOzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4zaiaabMgaca qG2bGaaeyzaiaab6gacaqGGaGaaeizaiaabMgacaqGMbGaaeOzaiaa bwgacaqGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabggacaqGSbaaba GaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6ga caqGUaaaaaa@3310@

Q.39

For the differential equations, find the general solution:

dy dx +y=1 ( y1 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aadaWcaaqaaGqabiaa=rgacaWF5baabaGaa8hzaiaa=HhaaaGaey4k aSIaaCyEaiabg2da9iaahgdacaWLjaGaaCzcamaabmaabaGaaCyEai abgcMi5kaahgdaaiaawIcacaGLPaaaaaa@46E0@

Ans.

The given differential equation is: dy dx +y=1 dy dx =1y Separating the variables, we get dy 1y =dx Integrating both sides of the given differential equation, we get dy 1y = dx log( 1y )=x+logC log( 1y )logC=x logC( 1y )=x C( 1y )= e x 1y= 1 C e x y=1 1 C e x =1A e x [ WhereA= 1 C ] This is the required general solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAha caqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzai aabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaGa aeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gaca qGGaGaaeyAaiaabohacaqG6aaabaGaaCzcamaalaaabaGaamizaiaa dMhaaeaacaWGKbGaamiEaaaacqGHRaWkcaWG5bGaeyypa0JaaGymaa qaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWL jaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7daWcaaqaaiaadsgacaWG5baabaGaamizaiaadIhaaaGaeyypa0Ja aGymaiabgkHiTiaadMhaaeaacaWGtbGaamyzaiaadchacaWGHbGaam OCaiaadggacaWG0bGaamyAaiaad6gacaWGNbGaaeiiaiaabshacaqG ObGaaeyzaiaabccacaqG2bGaaeyyaiaabkhacaqGPbGaaeyyaiaabk gacaqGSbGaaeyzaiaabohacaqGSaGaaeiiaiaabEhacaqGLbGaaeii aiaabEgacaqGLbGaaeiDaaqaaiaaxMaacaaMc8UaaGPaVlaaykW7ca aMc8+aaSaaaeaacaWGKbGaamyEaaqaaiaaigdacqGHsislcaWG5baa aiabg2da9iaadsgacaWG4baabaGaamysaiaad6gacaWG0bGaamyzai aadEgacaWGYbGaamyyaiaadshacaWGPbGaamOBaiaadEgacaqGGaGa aeOyaiaab+gacaqG0bGaaeiAaiaabccacaqGZbGaaeyAaiaabsgaca qGLbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqG0bGaaeiAaiaa bwgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaae izaiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6gacaqG 0bGaaeyAaiaabggacaqGSbGaaeiiaiaabwgacaqGXbGaaeyDaiaabg gacaqG0bGaaeyAaiaab+gacaqGUbGaaeilaiaabccacaqG3bGaaeyz aiaabccacaqGNbGaaeyzaiaabshaaeaacaWLjaGaaGPaVlaaykW7da WdbaqaamaalaaabaGaamizaiaadMhaaeaacaaIXaGaeyOeI0IaamyE aaaaaSqabeqaniabgUIiYdGccaaMc8Uaeyypa0Zaa8qaaeaacaWGKb GaamiEaaWcbeqab0Gaey4kIipaaOqaaiaaxMaacaWLjaGaaGPaVlab gkHiTiGacYgacaGGVbGaai4zamaabmaabaGaaGymaiabgkHiTiaadM haaiaawIcacaGLPaaacqGH9aqpcaWG4bGaey4kaSIaciiBaiaac+ga caGGNbGaam4qaaqaaiabgkDiElabgkHiTiGacYgacaGGVbGaai4zam aabmaabaGaaGymaiabgkHiTiaadMhaaiaawIcacaGLPaaacqGHsisl ciGGSbGaai4BaiaacEgacaWGdbGaeyypa0JaamiEaaqaaiabgkDiEl aaxMaacaWLjaGaciiBaiaac+gacaGGNbGaam4qamaabmaabaGaaGym aiabgkHiTiaadMhaaiaawIcacaGLPaaacqGH9aqpcqGHsislcaWG4b aabaGaeyO0H4TaaCzcaiaaykW7caaMc8UaaGPaVlaaxMaacaaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaam4qamaabmaaba GaaGymaiabgkHiTiaadMhaaiaawIcacaGLPaaacqGH9aqpcaWGLbWa aWbaaSqabeaacqGHsislcaWG4baaaaGcbaGaeyO0H4TaaCzcaiaayk W7caaMc8UaaGPaVlaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPa VlaaigdacqGHsislcaWG5bGaeyypa0ZaaSaaaeaacaaIXaaabaGaam 4qaaaacaWGLbWaaWbaaSqabeaacqGHsislcaWG4baaaaGcbaGaeyO0 H4TaaCzcaiaaykW7caaMc8UaaCzcaiaaxMaacaWLjaGaamyEaiabg2 da9iaaigdacqGHsisldaWcaaqaaiaaigdaaeaacaWGdbaaaiaadwga daahaaWcbeqaaiabgkHiTiaadIhaaaaakeaacaWLjaGaaCzcaiaaxM aacaaMc8UaaGPaVlaaykW7caaMc8UaaCzcaiaaykW7caaMc8UaaGPa VlaaykW7caaMc8Uaeyypa0JaaGymaiabgkHiTiaadgeacaWGLbWaaW baaSqabeaacqGHsislcaWG4baaaOGaaCzcaiaaxMaadaWadaqaaiaa dEfacaWGObGaamyzaiaadkhacaWGLbGaaGPaVlaadgeacqGH9aqpda WcaaqaaiaaigdaaeaacaWGdbaaaaGaay5waiaaw2faaaqaaiaadsfa caWGObGaamyAaiaadohacaqGGaGaaeyAaiaabohacaqGGaGaaeiDai aabIgacaqGLbGaaeiiaiaabkhacaqGLbGaaeyCaiaabwhacaqGPbGa aeOCaiaabwgacaqGKbGaaeiiaiaabEgacaqGLbGaaeOBaiaabwgaca qGYbGaaeyyaiaabYgacaqGGaGaae4Caiaab+gacaqGSbGaaeyDaiaa bshacaqGPbGaae4Baiaab6gacaqGGaGaae4BaiaabAgacaqGGaGaae iDaiaabIgacaqGLbGaaeiiaiaabEgacaqGPbGaaeODaiaabwgacaqG UbGaaeiiaiaabsgacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaabw gacaqGUbGaaeiDaiaabMgacaqGHbGaaeiBaiaabccaaeaacaqGLbGa aeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaab6caaa aa@CD14@

Q.40

For the differential equations, find the general solution:

se c 2 xtanydx+se c 2 ytanxdy=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aaieqacaWFZbGaa8xzaiaa=ngadaahaaWcbeqaaiaa=jdaaaGccaaM c8Uaa8hEaiaaykW7caWF0bGaa8xyaiaa=5gacaaMc8Uaa8xEaiaayk W7caWFKbGaa8hEaiaa=TcacaWFZbGaa8xzaiaa=ngadaahaaWcbeqa aiaa=jdaaaGccaWF5bGaaGPaVlaa=rhacaWFHbGaa8NBaiaa=Hhaca aMc8Uaa8hzaiaa=LhacaWF9aGaa8hmaaaa@58FB@

Ans.

The given differential equation is sec 2 xtany dx + sec 2 y tanxdy = 0 Dividing by tan x tan y, we get sec 2 xtanydx+ sec 2 ytanxdy tan x tan y =0 sec 2 x tanx dx+ sec 2 y tany dy=0 sec 2 x tanx dx= sec 2 y tany dy Integrating both sides of above equation, we get sec 2 x tanx dx= sec 2 y tany dy log tanx = -log tany + log C log tanx + log tany = log C log tanx tany = log C tanx tany = C This is the required solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAha caqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzai aabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaGa aeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gaca qGGaGaaeyAaiaabohaaeaacaqGZbGaaeyzaiaabogadaahaaWcbeqa aiaabkdaaaGccaaMc8UaaeiEaiaaykW7caaMe8UaaeiDaiaabggaca qGUbGaaGPaVlaabMhacaaMc8UaaeizaiaabIhacaqGGaGaae4kaiaa bccacaqGZbGaaeyzaiaabogadaahaaWcbeqaaiaabkdaaaGccaqG5b GaaeiiaiaaykW7caqG0bGaaeyyaiaab6gacaqG4bGaaGPaVlaabsga caqG5bGaaeiiaiaab2dacaqGGaGaaeimaaqaaiaabseacaqGPbGaae ODaiaabMgacaqGKbGaaeyAaiaab6gacaqGNbGaaeiiaiaabkgacaqG 5bGaaeiiaiaabshacaqGHbGaaeOBaiaabccacaqG4bGaaeiiaiaabs hacaqGHbGaaeOBaiaabccacaqG5bGaaeilaiaabccacaqG3bGaaeyz aiaabccacaqGNbGaaeyzaiaabshaaeaadaWcaaqaaiGacohacaGGLb Gaai4yamaaCaaaleqabaGaaGOmaaaakiaaykW7caWG4bGaaGPaVlGa cshacaGGHbGaaiOBaiaaykW7caWG5bGaaGPaVlaadsgacaWG4bGaey 4kaSIaci4CaiaacwgacaGGJbWaaWbaaSqabeaacaaIYaaaaOGaamyE aiaaykW7ciGG0bGaaiyyaiaac6gacaWG4bGaaGPaVlaadsgacaWG5b aabaGaaeiDaiaabggacaqGUbGaaeiiaiaabIhacaqGGaGaaeiDaiaa bggacaqGUbGaaeiiaiaabMhaaaGaeyypa0JaaGimaaqaaiabgkDiEl aaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua aGPaVlaaykW7daWcaaqaaiGacohacaGGLbGaai4yamaaCaaaleqaba GaaGOmaaaakiaaykW7caWG4baabaGaaGPaVlGacshacaGGHbGaaiOB aiaadIhaaaGaaGPaVlaaykW7caWGKbGaamiEaiabgUcaRmaalaaaba Gaci4CaiaacwgacaGGJbWaaWbaaSqabeaacaaIYaaaaOGaamyEaaqa aiGacshacaGGHbGaaiOBaiaaykW7caWG5baaaiaaykW7caWGKbGaam yEaiabg2da9iaaicdaaeaacqGHshI3caWLjaGaaCzcaiaaxMaacaaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7daWcaaqaaiGacohacaGGLbGaai4yamaa CaaaleqabaGaaGOmaaaakiaaykW7caWG4baabaGaaGPaVlGacshaca GGHbGaaiOBaiaadIhaaaGaaGPaVlaaykW7caWGKbGaamiEaiabg2da 9iabgkHiTmaalaaabaGaci4CaiaacwgacaGGJbWaaWbaaSqabeaaca aIYaaaaOGaamyEaaqaaiGacshacaGGHbGaaiOBaiaaykW7caWG5baa aiaaykW7caWGKbGaamyEaaqaaiaabMeacaqGUbGaaeiDaiaabwgaca qGNbGaaeOCaiaabggacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiaiaa bkgacaqGVbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgacaqGKbGaae yzaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeyyaiaabkgacaqG VbGaaeODaiaabwgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabs hacaqGPbGaae4Baiaab6gacaqGSaGaaeiiaiaabEhacaqGLbGaaeii aiaabEgacaqGLbGaaeiDaaqaaiaaykW7caaMc8UaaGPaVlaaxMaaca WLjaGaaCzcaiaaykW7daWdbaqaamaalaaabaGaci4CaiaacwgacaGG JbWaaWbaaSqabeaacaaIYaaaaOGaaGPaVlaadIhaaeaacaaMc8Uaci iDaiaacggacaGGUbGaamiEaaaaaSqabeqaniabgUIiYdGccaaMc8Ua amizaiaadIhacqGH9aqpcqGHsisldaWdbaqaamaalaaabaGaci4Cai aacwgacaGGJbWaaWbaaSqabeaacaaIYaaaaOGaamyEaaqaaiGacsha caGGHbGaaiOBaiaaykW7caWG5baaaaWcbeqab0Gaey4kIipakiaayk W7caWGKbGaamyEaaqaaiabgkDiElaaxMaacaWLjaGaaCzcaiaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaabYgacaqGVbGaae4zai aabccacaqG0bGaaeyyaiaab6gacaqG4bGaaeiiaiaab2dacaqGGaGa aeylaiaabYgacaqGVbGaae4zaiaabccacaqG0bGaaeyyaiaab6gaca qG5bGaaeiiaiaabUcacaqGGaGaaeiBaiaab+gacaqGNbGaaeiiaiaa boeaaeaacqGHshI3caWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaabY gacaqGVbGaae4zaiaabccacaqG0bGaaeyyaiaab6gacaqG4bGaaeii aiaabUcacaqGGaGaaeiBaiaab+gacaqGNbGaaeiiaiaabshacaqGHb GaaeOBaiaabMhacaqGGaGaaeypaiaabccacaqGSbGaae4BaiaabEga caqGGaGaae4qaaqaaiabgkDiElaaxMaacaWLjaGaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caqGSbGaae4BaiaabEgacaqGGaGaaeiDaiaa bggacaqGUbGaaeiEaiaabccacaqG0bGaaeyyaiaab6gacaqG5bGaae iiaiaab2dacaqGGaGaaeiBaiaab+gacaqGNbGaaeiiaiaaboeaaeaa cqGHshI3caWLjaGaaGPaVlaaykW7caaMc8UaaCzcaiaaxMaacaaMc8 UaaGPaVlaabshacaqGHbGaaeOBaiaabIhacaqGGaGaaeiDaiaabgga caqGUbGaaeyEaiaabccacaqG9aGaaeiiaiaaboeaaeaacaqGubGaae iAaiaabMgacaqGZbGaaeiiaiaabMgacaqGZbGaaeiiaiaabshacaqG ObGaaeyzaiaabccacaqGYbGaaeyzaiaabghacaqG1bGaaeyAaiaabk hacaqGLbGaaeizaiaabccacaqGZbGaae4BaiaabYgacaqG1bGaaeiD aiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOzaiaabccacaqG0b GaaeiAaiaabwgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6ga caqGGaGaaeizaiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzai aab6gacaqG0bGaaeyAaiaabggacaqGSbGaaeiiaiaabwgacaqGXbGa aeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeOlaaaaaa@3130@

Q.41

For the differential equations, find the general solution:

( e x + e x )dy( e x e x )dx=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aadGaMagWaaeacycicbeGaiGjG=vgadGaMaYbaaSqajGjGbGaMakac yc4F4baaaOGaiGjG=TcacGaMa+xzamacycihaaWcbKaMagacycOaiG jG=1cacGaMa+hEaaaaaOGaiGjGwIcacGaMaAzkaaGaiGjGykW7cGaM a+hzaiacyc4F5bGaeyOeI0YaaeWaaeaacaWFLbWaaWbaaSqabeaaca WF4baaaOGaeyOeI0Iaa8xzamaaCaaaleqabaGaa8xlaiaa=Hhaaaaa kiaawIcacaGLPaaacaaMc8Uaa8hzaiaa=HhacaWF9aGaa8hmaaaa@660D@

Ans.

The given differential equation is: ( e x + e x )dy( e x e x )dx=0 ( e x e x )dx=( e x + e x )dy dy= ( e x e x ) ( e x + e x ) dx Integrating both sides of above equation, we get dy = ( e x e x ) ( e x + e x ) dx y=log( e x + e x )+C This is the required general solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaadsfacaWGObGaamyzaiaabccacaqGNbGaaeyAaiaabAha caqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzai aabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaGa aeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gaca qGGaGaaeyAaiaabohacaqG6aaabaWaiGjGbmaabGaMakacyc4GLbWa iGjGCaaaleqcycyaiGjGcGaMaoiEaaaakiadycOHRaWkcGaMaoyzam acycihaaWcbKaMagacycOaiGjGc2cacGaMaoiEaaaaaOGaiGjGwIca cGaMaAzkaaGaiGjGykW7cGaMaoizaiacyc4G5bGaeyOeI0YaaeWaae aacaWGLbWaaWbaaSqabeaacaWG4baaaOGaeyOeI0IaamyzamaaCaaa leqabaGaaiylaiaadIhaaaaakiaawIcacaGLPaaacaaMc8Uaamizai aadIhacqGH9aqpcaaIWaaabaGaaCzcaiaaxMaacaaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7daqadaqaai aadwgadaahaaWcbeqaaiaadIhaaaGccqGHsislcaWGLbWaaWbaaSqa beaacaGGTaGaamiEaaaaaOGaayjkaiaawMcaaiaaykW7caWGKbGaam iEaiabg2da9macycyadaqaiGjGcGaMaoyzamacycihaaWcbKaMagac ycOaiGjGdIhaaaGccWaMaA4kaSIaiGjGdwgadGaMaYbaaSqajGjGbG aMakacycOGTaGaiGjGdIhaaaaakiacycOLOaGaiGjGwMcaaiacyciM c8UaiGjGdsgacGaMaoyEaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaCzcaiaa xMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlacyc4GKbGaiGjGdMhacqGH9aqpcGaMaIPaVpaalaaabaWaaeWa aeaacaWGLbWaaWbaaSqabeaacaWG4baaaOGaeyOeI0IaamyzamaaCa aaleqabaGaaiylaiaadIhaaaaakiaawIcacaGLPaaaaeaadGaMagWa aeacycOaiGjGdwgadGaMaYbaaSqajGjGbGaMakacyc4G4baaaOGamG jGgUcaRiacyc4GLbWaiGjGCaaaleqcycyaiGjGcGaMakylaiacyc4G 4baaaaGccGaMaAjkaiacycOLPaaaaaGaaGPaVlaadsgacaWG4baaba Gaamysaiaad6gacaWG0bGaamyzaiaadEgacaWGYbGaamyyaiaadsha caWGPbGaamOBaiaadEgacaqGGaGaaeOyaiaab+gacaqG0bGaaeiAai aabccacaqGZbGaaeyAaiaabsgacaqGLbGaae4CaiaabccacaqGVbGa aeOzaiaabccacaqGHbGaaeOyaiaab+gacaqG2bGaaeyzaiaabccaca qGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaa bYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baaba GaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 cGaMaIPaVlaaxMaacaWLjaGaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7daWdbaqaaiaadsgacaWG5baaleqabeqdcqGHRiI8aOGaeyypa0Ja aGPaVpaapeaabaWaaSaaaeaadaqadaqaaiaadwgadaahaaWcbeqaai aadIhaaaGccqGHsislcaWGLbWaaWbaaSqabeaacaGGTaGaamiEaaaa aOGaayjkaiaawMcaaaqaamacycyadaqaiGjGcGaMaoyzamacycihaa WcbKaMagacycOaiGjGdIhaaaGccWaMaA4kaSIaiGjGdwgadGaMaYba aSqajGjGbGaMakacycOGTaGaiGjGdIhaaaaakiacycOLOaGaiGjGwM caaaaaaSqabeqaniabgUIiYdGccaaMc8UaamizaiaadIhacaaMc8oa baGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caWLjaGaaCzcaiaaxMaacaWG5bGaeyyp a0JaciiBaiaac+gacaGGNbWaaeWaaeaacGaMaoyzamacycihaaWcbK aMagacycOaiGjGdIhaaaGccWaMaA4kaSIaiGjGdwgadGaMaYbaaSqa jGjGbGaMakacycOGTaGaiGjGdIhaaaaakiaawIcacaGLPaaacaaMc8 Uaey4kaSIaam4qaaqaaiaadsfacaWGObGaamyAaiaadohacaqGGaGa aeyAaiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkhaca qGLbGaaeyCaiaabwhacaqGPbGaaeOCaiaabwgacaqGKbGaaeiiaiaa bEgacaqGLbGaaeOBaiaabwgacaqGYbGaaeyyaiaabYgacaqGGaGaae 4Caiaab+gacaqGSbGaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqG GaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabE gacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabsgacaqGPbGaaeOz aiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHb GaaeiBaiaabccaaeaacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaa bMgacaqGVbGaaeOBaiaab6caaaaa@E760@

Q.42

For the differential equations, find the general solution:

dy dx =( 1+ x 2 )( 1+ y 2 ) MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aadaWcaaqaaGqabiaa=rgacaWF5baabaGaa8hzaiaa=HhaaaGaeyyp a0ZaaeWaaeaacaWFXaGaa83kaiaa=HhadaahaaWcbeqaaiaa=jdaaa aakiaawIcacaGLPaaadaqadaqaaiaa=fdacaWFRaGaa8xEamaaCaaa leqabaGaa8NmaaaaaOGaayjkaiaawMcaaaaa@4787@

Ans.

The given differential equation is: dy dx =( 1+ x 2 )( 1+ y 2 ) dy ( 1+ y 2 ) =( 1+ x 2 )dx Integrating both sides, we get dy ( 1+ y 2 ) = ( 1+ x 2 ) dx tan 1 y=x+ x 3 3 +C This is the general solution of given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAha caqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzai aabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaGa aeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gaca qGGaGaaeyAaiaabohacaqG6aaabaGaaCzcaiaaykW7caaMc8UaaGPa VlaaykW7caaMc8+aaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG4b aaaiabg2da9maabmaabaGaaGymaiabgUcaRiaadIhadaahaaWcbeqa aiaaikdaaaaakiaawIcacaGLPaaadaqadaqaaiaaigdacqGHRaWkca WG5bWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaaabaGaeyO0 H49aaSaaaeaacaWGKbGaamyEaaqaamaabmaabaGaaGymaiabgUcaRi aadMhadaahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaaaaGaeyyp a0ZaaeWaaeaacaaIXaGaey4kaSIaamiEamaaCaaaleqabaGaaGOmaa aaaOGaayjkaiaawMcaaiaadsgacaWG4baabaGaaeysaiaab6gacaqG 0bGaaeyzaiaabEgacaqGYbGaaeyyaiaabshacaqGPbGaaeOBaiaabE gacaqGGaGaaeOyaiaab+gacaqG0bGaaeiAaiaabccacaqGZbGaaeyA aiaabsgacaqGLbGaae4CaiaabYcacaqGGaGaae4DaiaabwgacaqGGa Gaae4zaiaabwgacaqG0baabaGaaGPaVlaaykW7caaMc8+aa8qaaeaa daWcaaqaaiaadsgacaWG5baabaWaaeWaaeaacaaIXaGaey4kaSIaam yEamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaaaaaSqabeqa niabgUIiYdGccaaMc8Uaeyypa0Zaa8qaaeaadaqadaqaaiaaigdacq GHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaaa leqabeqdcqGHRiI8aOGaaGPaVlaadsgacaWG4baabaGaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaabshacaqGHbGaaeOB amaaCaaaleqabaGaeyOeI0IaaGymaaaakiaadMhacqGH9aqpcaWG4b Gaey4kaSYaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIZaaaaaGcbaGa aG4maaaacqGHRaWkcaWGdbaabaGaaeivaiaabIgacaqGPbGaae4Cai aabccacaqGPbGaae4CaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGa ae4zaiaabwgacaqGUbGaaeyzaiaabkhacaqGHbGaaeiBaiaabccaca qGZbGaae4BaiaabYgacaqG1bGaaeiDaiaabMgacaqGVbGaaeOBaiaa bccacaqGVbGaaeOzaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaae OBaiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqG LbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaGaaeyzaiaabg hacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gacaqGUaaaaaa@032D@

Q.43

For the differential equations, find the general solution:

y logydxxdy=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aacaaMc8ocbeGaa8xEaiaa=bcacaWFSbGaa83Baiaa=DgacaWF5bGa aGPaVlaa=rgacaWF4bGaeyOeI0Iaa8hEaiaaykW7caWFKbGaa8xEai aa=1dacaWFWaaaaa@4A2C@

Ans.

The given differential equation is: y logydxxdy=0 y logydx=xdy dy y logy = dx x Integrating both sides, we get dy y logy = dx x loglogy=logx+logC loglogy=logCx logy=Cx y= e Cx This is the general solution of given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaadsfacaWGObGaamyzaiaabccacaqGNbGaaeyAaiaabAha caqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzai aabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaGa aeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gaca qGGaGaaeyAaiaabohacaqG6aaabaGaaGPaVlaaykW7caaMc8UaaGPa VlaadMhaieqacaWFGaGaciiBaiaac+gacaGGNbGaamyEaiaaykW7ca WGKbGaamiEaiabgkHiTiaadIhacaaMc8UaamizaiaadMhacqGH9aqp caaIWaaabaGaeyO0H4TaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caWG5bGaa8hiaiGacYgacaGGVbGaai4zaiaa dMhacaaMc8UaamizaiaadIhacqGH9aqpcaWG4bGaaGPaVlaadsgaca WG5baabaGaeyO0H4TaaCzcaiaaxMaacaaMc8UaaGPaVpaalaaabaGa amizaiaadMhaaeaacaWG5bGaa8hiaiGacYgacaGGVbGaai4zaiaadM haaaGaeyypa0ZaaSaaaeaacaWGKbGaamiEaaqaaiaadIhaaaaabaGa aeysaiaab6gacaqG0bGaaeyzaiaabEgacaqGYbGaaeyyaiaabshaca qGPbGaaeOBaiaabEgacaqGGaGaaeOyaiaab+gacaqG0bGaaeiAaiaa bccacaqGZbGaaeyAaiaabsgacaqGLbGaae4CaiaabYcacaqGGaGaae 4DaiaabwgacaqGGaGaae4zaiaabwgacaqG0baabaGaaCzcaiaaxMaa daWdbaqaamaalaaabaGaamizaiaadMhaaeaacaWG5bGaa8hiaiGacY gacaGGVbGaai4zaiaadMhaaaaaleqabeqdcqGHRiI8aOGaaGPaVlab g2da9maapeaabaWaaSaaaeaacaWGKbGaamiEaaqaaiaadIhaaaaale qabeqdcqGHRiI8aOGaaGPaVdqaaiabgkDiElaaxMaacaWLjaGaciiB aiaac+gacaGGNbGaciiBaiaac+gacaGGNbGaamyEaiabg2da9iGacY gacaGGVbGaai4zaiaadIhacqGHRaWkciGGSbGaai4BaiaacEgacaWG dbaabaGaeyO0H4TaaCzcaiaaxMaaciGGSbGaai4BaiaacEgaciGGSb Gaai4BaiaacEgacaWG5bGaeyypa0JaciiBaiaac+gacaGGNbGaam4q aiaadIhaaeaacqGHshI3caWLjaGaaCzcaiaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlGacYgacaGGVbGaai4zaiaadMhacqGH9aqp caWGdbGaamiEaaqaaiabgkDiElaaxMaacaWLjaGaaGPaVlaaykW7ca aMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaadMhacqGH9aqpcaWGLbWaaWbaaSqabeaaca WHdbGaaCiEaaaaaOqaaiaabsfacaqGObGaaeyAaiaabohacaqGGaGa aeyAaiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabEgaca qGLbGaaeOBaiaabwgacaqGYbGaaeyyaiaabYgacaqGGaGaae4Caiaa b+gacaqGSbGaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae 4BaiaabAgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqG GaGaaeizaiaabMgacaqGMbGaaeOzaiaabwgacaqGYbGaaeyzaiaab6 gacaqG0bGaaeyAaiaabggacaqGSbGaaeiiaiaabwgacaqGXbGaaeyD aiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeOlaaaaaa@5375@

Q.44

For the differential equations, find the general solution:

x 5 dy dx = y 5 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aacaWH4bWaaWbaaSqabeaacaWH1aaaaOWaaSaaaeaaieqacaWFKbGa a8xEaaqaaiaa=rgacaWF4baaaiabg2da9iabgkHiTiaahMhadaahaa Wcbeqaaiaaiwdaaaaaaa@42C3@

Ans.

The given differential equation is:x5dydx=y5      dyy5=dxx5Integrating both sides, we get    dyy5=dxx5    y5dy=x5dx  y44+C=x44    x4+y4=4C=CThis is the general solution of given differential equation.

Q.45

For the differential equation given below, find the general solution:

dy dx =si n 1 x MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aadaWcaaqaaGqabiaa=rgacaWF5baabaGaa8hzaiaa=HhaaaGaeyyp a0Jaa83Caiaa=LgacaWFUbWaaWbaaSqabeaacaWFTaGaa8xmaaaaki aa=Hhaaaa@434F@

Ans.

The given differential equation is:      dydx=sin1x      dy=sin1xdxIntegrating both sides, we get    dy=sin1xdx    y=sin1x.1dx    y=sin1x1dx(ddxsin1x1dx)dx    y=(sin1x).x(11x2x)dx+C    y=xsin1xxt×dt2x+C[Let 1x2=t2x=dtdx]    y=xsin1x+12t12dt+C    y=xsin1x+12t12(12)+C    y=xsin1x+1x2+CThis is the general solution of given differential equation.

Q.46

For the differential equation given below, find the general solution:

e x tan ydx+( 1 e x )se c 2 y dy=0 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aacaaMc8UaaCyzamaaCaaaleqabaacbeGaa8hEaaaakiaa=bcacaWF 0bGaa8xyaiaa=5gacaWFGaGaa8xEaiaaykW7caWFKbGaa8hEaiaa=T cadaqadaqaaiaahgdacqGHsislcaWFLbWaaWbaaSqabeaacaWF4baa aaGccaGLOaGaayzkaaGaaGPaVlaa=nhacaWFLbGaa83yamaaCaaale qabaGaa8Nmaaaakiaa=LhacaWFGaGaa8hzaiaa=LhacaWF9aGaa8hm aaaa@5541@

Ans.

The given differential equation is:ex tan ydx+(1ex)sec2ydy=0ex(1ex)dx+sec2ytanydy=0[Dividing both sides by(1ex)tany]Integrating both sides of above equation, we getex(1ex)dx+sec2ytanydy=Cext×dtex+sec2yz×dzsec2y=logC[Let t=1exandz=tanydtdx=exanddzdy=sec2y]logt+logz=logClog(zt)=logCtany1ex=Ctany=C(1ex)This is the required general solution of the given differential equation.

Q.47

For the differential equation given below, find a particular solution satisfying the given condition:

(x3+x2+x+1)dydx=2x2+x;​y=1 when x=0

Ans.

The given differential equation is(x3+x2+x+1)dydx=2x2+xdydx=2x2+x(x3+x2+x+1)      =2x2+x{x2(x+1)+1(x+1)}dy=2x2+x(x2+1)(x+1)dx...(i)Let2x2+x(x2+1)(x+1)=Ax+B(x2+1)+C(x+1)2x2+x=(Ax+B)(x+1)+C(x2+1)2x2+x=x2(A+C)+x(A+B)+(B+C)    A+C=2,  A+B=1 and B+C=0A=32,B=12andC=12    2x2+x(x2+1)(x+1)=3x12(x2+1)+12(x+1)Fromequation(i),wehavedy=3x12(x2+1)dx+12(x+1)dxIntegrating both sides, we get      dy=3x12(x2+1)dx+12(x+1)dxy=32xx2+1dx121x2+1dx+121x+1dx+Cy=32×12log(x2+1)12tan1x+12log(x+1)+Cy=14{3log(x2+1)+2log(x+1)}12tan1x+Cy=14{log(x2+1)3(x+1)2}12tan1x+C  ...(ii)Since,y=1 when x=01=14{log(0+1)3(0+1)2}12tan10+C1=CSubstituting C=1 in equation (iii), we gety=14{log(x2+1)3(x+1)2}12tan1x+1

Q.48

For the differential equation given below, find a particular solution satisfying the given condition:

x(x21)dydx=1;​ y=0 when x=2

Ans.

The given differential equation is x( x 2 1 ) dy dx =1 dy dx = 1 x( x 2 1 ) dy dx = 1 x( x1 )( x+1 ) dy= 1 x( x1 )( x+1 ) dx( i ) Let 1 x( x1 )( x+1 ) = A x + B x1 + C x+1 1=A( x 2 1 )+Bx( x+1 )+Cx( x1 ) Substituting x=0,1and1, we get A=1,  B=12  and  C=12    1x(x1)(x+1)=1x+12(x1)+12(x+1)From equation (i),wehavedy=1xdx+12(x1)dx+12(x+1)dxIntegrating both sides, we get      dy=1xdx+121x1dx+121x+1dx  y=logx+12log(x1)+12log(x+1)+logk  y=12log{k2(x1)(x+1)x2}  ...(ii)Now,  y=0  whenx=2  0=12log{k2(21)(2+1)22}  0=log{k2(1)(3)4}      log1=log{3k24}  1=3k24k2=43Substituting k2=43 in equation (ii), we get   y=12log{43(x1)(x+1)x2}y=12log{4(x21)3x2}=12log(x21x2)12log(34)

Q.49

For the differential equation given below, find a particular solution satisfying the given condition:

cos(dydx)=a(aR);​ y=1when x=0

Ans.

The given differential equation is: cos( dy dx )=a dy dx = cos 1 a dy= cos 1 adx Integrating both sides, we get dy = cos 1 a dx y=( cos 1 a )x+C( i ) Now,y=1 when x=0 1=( cos 1 a )0+C 1=C Substituting C = 1 in equation ( i ),we get y=( cos 1 a )x+1 y1 x = cos 1 acos( y1 x )=a MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAha caqGLbGaaeOBaiaabccacaqGKbGaaeyAaiaabAgacaqGMbGaaeyzai aabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabYgacaqGGaGa aeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gaca qGGaGaaeyAaiaabohacaqG6aaabaGaci4yaiaac+gacaGGZbWaaeWa aeaadaWcaaqaaiaadsgacaWG5baabaGaamizaiaadIhaaaaacaGLOa GaayzkaaGaeyypa0JaamyyaaqaaiaaxMaacaaMc8+aaSaaaeaacaWG KbGaamyEaaqaaiaadsgacaWG4baaaiabg2da9iGacogacaGGVbGaai 4CamaaCaaaleqabaGaeyOeI0IaaGymaaaakiaadggaaeaacaWLjaGa aGPaVlaadsgacaWG5bGaeyypa0Jaci4yaiaac+gacaGGZbWaaWbaaS qabeaacqGHsislcaaIXaaaaOGaamyyaiaaykW7caWGKbGaamiEaaqa aiaabMeacaqGUbGaaeiDaiaabwgacaqGNbGaaeOCaiaabggacaqG0b GaaeyAaiaab6gacaqGNbGaaeiiaiaabkgacaqGVbGaaeiDaiaabIga caqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabohacaqGSaGaaeiiai aabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aa8qaaeaaca WGKbGaamyEaaWcbeqab0Gaey4kIipakiaaykW7cqGH9aqpciGGJbGa ai4BaiaacohadaahaaWcbeqaaiabgkHiTiaaigdaaaGccaWGHbWaa8 qaaeaacaWGKbGaamiEaaWcbeqab0Gaey4kIipakiaaykW7aeaacaaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaayk W7caaMc8UaaGPaVlaaykW7caWG5bGaeyypa0ZaaeWaaeaaciGGJbGa ai4BaiaacohadaahaaWcbeqaaiabgkHiTiaaigdaaaGccaWGHbaaca GLOaGaayzkaaGaaGPaVlaadIhacqGHRaWkcaWGdbGaaGPaVlaaykW7 caGGUaGaaiOlaiaac6cadaqadaqaaiaadMgaaiaawIcacaGLPaaaae aacaqGobGaae4BaiaabEhacaqGSaGaaGPaVlaaysW7caqG5bGaaGjb Vlaab2dacaaMe8UaaeymaiaabccacaqG3bGaaeiAaiaabwgacaqGUb GaaeiiaiaabIhacaaMe8UaaeypaiaaysW7caqGWaaabaGaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaG PaVlaaykW7caaMc8UaaGymaiabg2da9maabmaabaGaci4yaiaac+ga caGGZbWaaWbaaSqabeaacqGHsislcaaIXaaaaOGaamyyaaGaayjkai aawMcaaiaaykW7caaIWaGaey4kaSIaam4qaaqaaiaaykW7caaMc8Ua aGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7ca aMc8UaaGPaVlaaigdacqGH9aqpcaWGdbaabaGaae4uaiaabwhacaqG IbGaae4CaiaabshacaqGPbGaaeiDaiaabwhacaqG0bGaaeyAaiaab6 gacaqGNbGaaeiiaiaaboeacaqGGaGaaeypaiaabccacaqGXaGaaeii aiaabMgacaqGUbGaaeiiaiaabwgacaqGXbGaaeyDaiaabggacaqG0b GaaeyAaiaab+gacaqGUbGaaeiiamaabmaabaGaaeyAaaGaayjkaiaa wMcaaiaabYcacaaMc8Uaae4DaiaabwgacaqGGaGaae4zaiaabwgaca qG0baabaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPa VlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaamyEaiabg2da9maabm aabaGaci4yaiaac+gacaGGZbWaaWbaaSqabeaacqGHsislcaaIXaaa aOGaamyyaaGaayjkaiaawMcaaiaaykW7caWG4bGaey4kaSIaaGymaa qaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaacaWG5bGa eyOeI0IaaGymaaqaaiaadIhaaaGaeyypa0Jaci4yaiaac+gacaGGZb WaaWbaaSqabeaacqGHsislcaaIXaaaaOGaamyyaiabgkDiElGacoga caGGVbGaai4CamaabmaabaWaaSaaaeaacaWG5bGaeyOeI0IaaGymaa qaaiaadIhaaaaacaGLOaGaayzkaaGaeyypa0Jaamyyaaaaaa@8AAE@

Q.50

For the differential equation given below, find a particular solution satisfying the given condition:

dydx=ytanx; ​y=1whenx=0

Ans.

The given differential equation is:    dydx=ytanx  dyy=tanxdxIntegrating both sides, we get        dyy=tanxdx      logy=logsecx+logC    y=secx.C  ...(i)Now,  y=1 when x=0            1=C.sec0            1=CSubstituting C = 1 in equation (i), we get          y=secx.1      y=secx

Q.51

Find the equation of a curve passing through the point (0, 0) and whose differential equation is y’ = ex sin x.

Ans.

The given differential equation is :      y= exsin x    dydx= exsin x Integrating both sides, we get    dy=exsin xdx+C        y=exsin xdx+C  ...(i)Let  I=exsin xdx  =sinxexdx(ddxsinxexdx)dx  =sinxex(cosxex)dx  =sinxex{cosxexdx(ddxcosxexdx)dx}  =sinxex{cosxex(sinx)exdx}  =sinxexexcosxexsin xdxI=sinxexexcosxI      2I=ex(sinxcosx)I=ex(sinxcosx)2From equation(i), we get        y=ex(sinxcosx)2+C...(i)Since, curve passes through the point (0,0).      0=e0(sin0cos0)2+C        0=1(01)2+CC=12 Putting value of C in equation (ii), weget          y=ex(sinxcosx)2+12      2y=ex(sinxcosx)+12y1=ex(sinxcosx)Hence, it is the required equation of the curve.

Q.52

For the differential equation xydydx=x+2y+2, find the solution curve passing through the point 1, -1.

Ans.

The given differential equation is :          xydydx=(x+2)(y+2)      ydy(y+2)= (x+2)xdx    (y+22)dy(y+2)= (x+2)xdx    1dy2(y+2)dy=(1+2x)dxIntegrating both sides, we get1dy2(y+2)dy=1dx+21xdx      y2log(y+2)=x+2logx+C        yxC=logx2+log(y+2)2        yxC=logx2(y+2)2  ...(ii)Since, the curve passes through (1,1).              11C=log(1)2(1+2)2                    2C=log(1)2    C=2Putting the value of C in equation(ii), we get        yx+2=logx2(y+2)2This is the required solution of the curve.

Q.53

Find the equation of the curve passing through the point (0, –2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Ans.

According to given condition,ydydx=x      y.dy=x.dxIntegrating both sides w.r.t. x, we get      ydy=xdx            y22=x22+C    y2=x2+2C    y2x2=2C...(i)The curve passes through the point (02).(2)202=2C      4=2CC=2Putting C = 2 in equation(i), we get    y2x2=4This is the required equation of the curve.

Q.54

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, –3). Find the equation of the curve given that it passes through (–2, 1).

Ans.

Slope(m1) of curve at point (x,y)=dydxslope(m2) of line segment joining (x,y)and (4,3)=y+3x+4According to given condition:m1=2m2dydx=2(y+3x+4)dyy+3=2(dxx+4)Integrating both sides, we get      dyy+3=21x+4dxlog(y+3)=2log(x+4)+logClog(y+3)=log(x+4)2Cy+3=C(x+4)2...(i)This is the general equation of the curve.The curve passes through the point(2,1).1+3=C(2+4)2      C=44=1Substituting C = 1 in equation(i), we gety+3=1(x+4)2y+3=(x+4)2This is the required equation of the curve.

Q.55

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Ans.

Let radius of spherical balloon be r units.Volume of balloon(V)=43πr3According to condition:dVdt=k      ddt(43πr3)=k    43π×3r2drdt=k  4πr2drdt=k  4πr2dr=kdtIntegrating both sides, we get4πr2dr=kdt4πr33=kt+C...(i)Now, r = 3 at t = 04π333=k(0)+C    36π=CAgain,r = 6 when t = 3 from equation(i), we get4π633=k(3)+36π        288π36π=3k        k=252π3  =84πHence, from equation (i), we have43πr3=84πt+36π      r3=63t+27      r=(63t+27)13Thus, the radius of balloon after t seconds is(63t+27)13.

Q.56

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if ` 100 double itself in 10 years (loge2= 0.6931).

Ans.

Let Principal be P, time be t years and rate of interest be r% p.a.Since, principal increases continuously at the rate of r% p.a.dPdt=r% ofP        =(r100)PdPP=(r100)dtIntegrating both sides, we get        dPP=r100dt      logeP=r100t+C  ...(i)Substituting P = 100 and t = 0 in equation(i), we get      loge100=r100×0+C    loge100=0+C        C=loge100 Again,putting P = 200 and t = 10 in equation(i), weget          loge200=r100×10+loge100loge200loge100=r10loge200100=r10      loge2=r10  0.6931=r10r=6.931Thus, the rate of interest is 6.93% p.a.

Q.57

In a bank, principal increases continuously at the rate of 5% per year. An amount of ₹ 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).

Ans.

Let principal be P, t be time and r be the rate of interest.Then,rate of principal increasing per year is:dPdt=(5100)P      =120PdPP=120dtIntergrating both sides, we get         dPP=120dt      logeP=120t+C    P=e(120t+C)        =eCe120t    P=Ke120t          ...(i)[Where,​ K=eC]Now,  t = 0 when P = 1000      1000=Ke120(0)      1000=KPutting the value of K in equation(i), we get    P=1000e120t  ...(ii)Now amount after 10 years,     P=1000e120(10)        =1000e0.5        =1000×1.648        =1648Hence,after 10 years the amount will be ₹ 1648.

Q.58

In a culture, the bacteria count is 1,00,000.
The number is increased by 10% in 2 hours.
In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Ans.

Let present number of bacteria be N at any instant t.Since, the rate of growth of the bacteria is proportional to N. dNdtαNdNdt=kNdNN=kdtIntegrating both sides, we get      dNN=kdt      logN=kt+C...(i)Let number of bacteria at t = 0 be N0,then    logN0=k(0)+C  C=logN0From equation(i), we have          logN=kt+logN0logNlogN0=kt    logNN0=kt...(ii)Since, bacteria increases 10% in 2 hours, so    N=(1+10100)N0  NN0=1110 Substituting t = 2 and NN0=1110 in equation(ii), we get    log1110=k(2)    k=12log(1110)From equation (ii), we have    logNN0=12log(1110)t    t=logNN012log(1110)=2logNN0log(1110)So, the time taken to increase the bacteria from 100000 to 200000=2log200000100000log(1110)     t=2log2log(1110)Hence,bacteria increases from 100000 to 200000 in 2log2log(1110)hrs.

Q.59

The general solution of the differential equation dydx = ex + y  is(A) ex + ey = C (B) ex + ey = C(C) ex + ey = C (D) ex + ey = C

Ans.

The differential equation isdydx=ex+y      =exeydyey=exdx    eydy=exdxIntegrating both sides, we geteydy=exdx    ey=ex+C1        ex+ey=C1        ex+ey=C[Where, C=C1]Thus, option A is correct.

Q.60

Show that the given differential equation is homogeneous and solve it.

(x2 + xy) dy = (x2 + y2) dx

Ans.

The given differential equation is:    (x2+xy)dy=(x2+y2)dx  dydx=(x2+y2)(x2+xy)    ...(i)Let         F(x,y)=(x2+y2)(x2+xy)Now,  f(λx,λy)=(λ2x2+λ2y2)(λ2x2+λ2xy)=λ2(x2+y2)λ2(x2+xy)=λ0F(x,y)Therefore, F(x,y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.Substituting y=vx then dydx=v+xdvdx in equation(i), we get    v+xdvdx=(x2+v2x2)(x2+xvx)=x2(1+v2)x2(1+v)xdvdx=(1+v2)(1+v)v=1+v2vv2(1+v)   xdvdx=1v1+v    1+v1vdv=dxx2(1v)1vdv=dxx    (21v1)dv=dxxIntegrating both sides, we get  (21v1)dv=dxx2log(1v)v=logx+logk                              v=2log(1v)logxlogk                              v=log{(1v)2xk}yx=log{(1yx)2xk}[Put v=yx]  (1yx)2xk=eyx  (xyx)2xk=eyx  (xy)2kx=eyx      (xy)2=xkeyx      (xy)2=Cxeyx[Let C=1k]This is the required solution of the given differential equation.

Q.61

Show that the given differential equation is homogeneous and solve each of them.

y= x+y x MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aaieqacaWF5bGaa83jaiaa=1dadaWcaaqaaiaa=HhacaWFRaGaa8xE aaqaaiaa=Hhaaaaaaa@3F27@

Ans.

The given differential equation is:        y=x+yx      dydx=x+yx  ...(i)Let  F(x, y)=x+yxNow,F(λx, λy)=λx+λyλx=λ(x+y)λx=x+yx=λ0F(x, y)Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.For solving equation (i), substitute y=vx and dydx=v+xdvdx, we get  v+xdvdx=x+vxx=x(1+v)x  v+xdvdx=(1+v)          xdvdx=(1+v)v           xdvdx=1    dv=dxxIntegrating both sides, we get      v=logx+C      yx=logx+C      y=xlogx+CxThis is the required solution of the given differential equation.

Q.62

Show that the given differential equation is homogeneous and solve it.

(x – y)dy – (x + y) dx = 0

Ans.

The given differential equation is:(xy)dy(x+y)dx=0        dydx=x+yxy...(i)Let F(x,y)=x+yxyNow,    F(λx,λy)=λx+λyλxλy    =λ(x+y)λ(xy)=λ0F(x,y)Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.For solving equation (i), substitute y=vx and dydx=v+xdvdx, we get    v+xdvdx=x+vxxvx    =x(1+v)x(1v)            xdvdx=1+vv+v21v    =1+v21v    1v1+v2dv=dxx    (11+v2v1+v2)dv=dxxIntegrating both sides, we get11+v2dvv1+v2dv=1xdx  tan1v12log(1+v2)=logx+C                          tan1(yx)=12log(1+yx22)+12.2logx+C                          tan1(yx)=12log(x2+y2x2.x2)+C                          tan1(yx)=12log(x2+y2)+CThis is the required solution of the given differential equation.

Q.63

Show that the given differential equation is homogeneous and solve it.

(x2 – y2)dx + 2xy dy = 0

Ans.

The given differential equation is:  (x2y2)dx+2xy dy=0        dydx=x2y22xy...(i)Let F(x,y)=x2y22xyNow,    F(λx,λy)=λ2x2λ2y22λ2xy    =λ2(x2y2)λ2(2xy)=λ0F(x,y)Therefore, F(x, y) is a homogenous function of degree zero.So, the given differential equation is a homogenous differential equation.For solving equation (i), substitute y =vx and dydx=v+xdvdx,we get    v+xdvdx=x2v2x22x(vx)  =(1v2)x22x2v  =(1v2)2v            xdvdx=(v21)2vv   =(v21)2v22v            xdvdx=(1+v2)2v2v(1+v2)dv=dxxIntegrating both sides, we get      2v(1+v2)dv=1xdx        log(1+v2)=logx+logC        log(1+v2)=log(Cx)    (1+y2x2)=Cx  (x2+y2)=CxThis is the required solution of the given differential equation.

Q.64

Show that the given differential equation is homogeneous and solve it.

x2dydx=x22y2+xy

Ans.

The given differential equation is:  x2dydx=x22y2+xy        dydx=x22y2+xyx2...(i) Let F(x,y)=x22y2+xyx2Now,F(λx,λy)=λ2x22λ2y2+λ2xyλ2x2=λ2(x22y2+xy)λ2(x2)=λ0F(x,y)Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.For solving equation (i), substitute y=vx and dydx=v+xdvdx,we getv+xdvdx=x22v2x2+xvxx2=(12v2+v)x2x2=12v2+vxdvdx=12v2+vvxdvdx=12v2dv12v2=dxxIntegrating both sides, we get 1 2 1 ( 1 2 ) 2 -v 2 dv= dx x 1 2 × 1 1 2 log( 1 2 +v 1 2 -v )=logx+C 1 2 2 log( 1 2 + y x 1 2 y x )=logx+C 1 2 2 log( x+ 2 y x- 2 y )=logx+C This is the required solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7daWcaaqaaiaabg daaeaacaqGYaaaamaapeaabaWaaSaaaeaacaqGXaaabaWaaeWaaeaa daWcaaqaaiaabgdaaeaadaGcaaqaaiaabkdaaSqabaaaaaGccaGLOa GaayzkaaWaaWbaaSqabeaacaqGYaaaaOGaaeylaiaabAhadaahaaWc beqaaiaabkdaaaaaaaqabeqaniabgUIiYdGccaaMc8UaaeizaiaabA hacaqG9aWaa8qaaeaadaWcaaqaaiaabsgacaqG4baabaGaaeiEaaaa aSqabeqaniabgUIiYdGccaaMc8oabaWaaSaaaeaacaqGXaaabaGaae OmaaaacaqGxdWaaSaaaeaacaqGXaaabaGaaeOmaiaabEnadaWcaaqa aiaabgdaaeaadaGcaaqaaiaabkdaaSqabaaaaaaakiaabYgacaqGVb Gaae4zamaabmaabaWaaSaaaeaadGaAaUaaaeacObOaiGgGbgdaaeac Ob4aiGgGkaaabGaAakacObyGYaaaleqcObiaaaGccGaAag4kaiacOb yG2baabaWaiGgGlaaabGaAakacObyGXaaabGaAaoacObOcaaqaiGgG cGaAagOmaaWcbKaAacaaaOGaaeylaiacObyG2baaaaGaayjkaiaawM caaiaab2dacaqGSbGaae4BaiaabEgacaqG4bGaae4kaiaaboeaaeaa caWLjaWaaSaaaeaacaqGXaaabaGaaeOmamaakaaabaGaaeOmaaWcbe aaaaGccaqGSbGaae4BaiaabEgadaqadaqaamaalaaabaWaiGgGlaaa bGaAakacObyGXaaabGaAaoacObOcaaqaiGgGcGaAagOmaaWcbKaAac aaaOGaiGgGbUcadaWcaaqaaiaabMhaaeaacaqG4baaaaqaamacOb4c aaqaiGgGcGaAagymaaqaiGgGdGaAaQaaaeacObOaiGgGbkdaaSqajG gGaaaakiaab2cadaWcaaqaaiaabMhaaeaacaqG4baaaaaaaiaawIca caGLPaaacaqG9aGaaeiBaiaab+gacaqGNbGaaeiEaiaabUcacaqGdb aabaGaaCzcamaalaaabaGaaeymaaqaaiaabkdadaGcaaqaaiaabkda aSqabaaaaOGaaeiBaiaab+gacaqGNbWaaeWaaeaadaWcaaqaaiaabI hacGaAag4kamacObOcaaqaiGgGcGaAagOmaaWcbKaAacGccaqG5baa baGaaeiEaiaab2cadGaAaQaaaeacObOaiGgGbkdaaSqajGgGaOGaae yEaaaaaiaawIcacaGLPaaacaqG9aGaaeiBaiaab+gacaqGNbGaaeiE aiaabUcacaqGdbaabaGaaeivaiaabIgacaqGPbGaae4Caiaabccaca qGPbGaae4CaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaa bwgacaqGXbGaaeyDaiaabMgacaqGYbGaaeyzaiaabsgacaqGGaGaae 4Caiaab+gacaqGSbGaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqG GaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabE gacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabsgacaqGPbGaaeOz aiaabAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHb GaaeiBaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMga caqGVbGaaeOBaiaab6caaaaa@08FD@

Q.65

Show that the given differential equation is homogeneous and solve it.

x dy y dx= x 2 + y 2 dx MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aaieqacaWF4bGaa8hiaiaa=rgacaWF5bGaeyOeI0Iaa8hiaiaa=Lha caWFGaGaa8hzaiaa=HhacaWF9aWaaOaaaeaacaWF4bWaaWbaaSqabe aacaWFYaaaaOGaey4kaSIaaCyEamaaCaaaleqabaGaaCOmaaaaaeqa aOGaaGPaVlaahsgacaWH4baaaa@4A9F@

Ans.

The given differential equation is:    xdyydx=x2+y2dx        xdy=x2+y2dx+ydx        dydx=x2+y2+yx...(i)Let F(x,y)=x2+y2+yxNow,    F(λx,λy)=λ2x2+λ2y2+λyλx     =λ(x2+y2+y)λx=λ0F(x,y)Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.For solving equation (i), substitute y=vx and dydx=v+xdvdx,we get    v+xdvdx=x2+v2x2+vxx  =(1+v2+v)xx    v+xdvdx=1+v2+v            xdvdx=1+v2      dv1+v2=dxxIntegrating both sides, we get          log|v+1+v2|=log|x|+logC          log|yx+1+y2x2|=logC|x|                |yx+1+y2x2|=C|x|                 |yx+1+y2x2|=C|x|                  y+x2+y2=Cx2This is the required solution of given differential equation.

Q.66

Show that the given differential equation is homogeneous and solve each of them.

xcosyx+ysinyxydx=ysinyxxcosyxxdy

Ans.

The given differential equation is:{xcos(yx)+ysin(yx)}ydx={ysin(yx)xcos(yx)}xdy      dydx={xcos(yx)+ysin(yx)}y{ysin(yx)xcos(yx)}x...(i)Let F(x,y)={xcos(yx)+ysin(yx)}y{ysin(yx)xcos(yx)}xNow,        F(λx,λy)={λxcos(λyλx)+λysin(λyλx)}λy{λysin(λyλx)λxcos(λyλx)}λx = λ 2 y{ x cos( y x )+y sin( y x ) } λ 2 x{ y sin( y x )x cos( y x ) } = λ 0 F( x,y ) Therefore, F( x,y ) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation. For solving equation ( i ), substitute y=vx and dy dx =v+x dv dx , we get v+x dv dx = vx{ x cos( vx x )+vx sin( vx x ) } x{ vx sin( vx x )x cos( vx x ) } = v{ cos( v )+v sin( v ) } { v sin( v ) cos( v ) } x dv dx = vcosv+ v 2 sinv v sinv cosv v x dv dx = vcosv+ v 2 sinv v 2 sinv+v cosv v sinv cosv x dv dx = 2vcosv v sinv cosv ( v sinv cosv ) dv vcosv = 2dx x Integrating both sides, we get v sinv vcosv dv cosv vcosv dv=2 1 x dx MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaxMaacaWLjaGaaCzcaiaaxMaacaaMc8UaaGPaVlaaykW7 caaMc8Uaeyypa0ZaaSaaaeaacqaH7oaBdaahaaWcbeqaaiaaikdaaa GccaWG5bWaaiWaaeaacaWG4bacbeGaa8hiaiGacogacaGGVbGaai4C amaabmaabaWaaSaaaeaacaWG5baabaGaamiEaaaaaiaawIcacaGLPa 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Q.67

Show that the given differential equation is homogeneous and solve each of them.

xdydxy+xsin(yx)=0

Ans.

The given differential equation is:                xdydxy+xsin(yx)=0      dydx=yxsin(yx)x...(i) Let F( x,y )= yxsin( y x ) x Now, F( λx,λy )= λyλxsin( λy λx ) λx = λ{ y-xsin( y x ) } λx 0 F( x,y ) Therefore, F( x,y ) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation. For solving equation ( i ), substitute y = vx and dy dx =v+x dv dx , we get v+x dv dx = vxxsin( vx x ) x = { vsinv }x x x dv dx =vsinvv dv sinv = dx x cosecvdv= dx x Integrating both sides, we get cosecv dv= 1 x dx log| cosecvcotv |=log| x |+logC log| cosecvcotv |=log C | x | cosec y x cot y x = C x 1 sin y x cos y x sin y x = C x x{ 1cos( y x ) }=Csin( y x ) This is the required solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVp aapeaabaGaci4yaiaac+gacaGGZbGaaiyzaiaacogacaWG2baaleqa beqdcqGHRiI8aOGaaGPaVlaadsgacaWG2bGaeyypa0JaeyOeI0Yaa8 qaaeaadaWcaaqaaiaaigdaaeaacaWG4baaaaWcbeqab0Gaey4kIipa kiaaykW7caWGKbGaamiEaaqaaiabgkDiElaaykW7ciGGSbGaai4Bai aacEgadaabdaqaaiGacogacaGGVbGaai4CaiaacwgacaGGJbGaamOD aiabgkHiTiGacogacaGGVbGaaiiDaiaadAhaaiaawEa7caGLiWoacq GH9aqpcqGHsislciGGSbGaai4BaiaacEgadaabdaqaaiaadIhaaiaa wEa7caGLiWoacqGHRaWkciGGSbGaai4BaiaacEgacaWGdbaabaGaey O0H4TaaGPaVlGacYgacaGGVbGaai4zamaaemaabaGaci4yaiaac+ga caGGZbGaaiyzaiaacogacaWG2bGaeyOeI0Iaci4yaiaac+gacaGG0b GaamODaaGaay5bSlaawIa7aiabg2da9iGacYgacaGGVbGaai4zamaa laaabaGaam4qaaqaamaaemaabaGaamiEaaGaay5bSlaawIa7aaaaae aacqGHshI3caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlGacogacaGGVbGaai4CaiaacwgacaGGJbWaaSaaaeaaca WG5baabaGaamiEaaaacqGHsislciGGJbGaai4BaiaacshadaWcaaqa aiaadMhaaeaacaWG4baaaiabg2da9maalaaabaGaam4qaaqaaiaadI haaaaabaGaeyO0H4TaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaM c8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaSaaaeaaca aIXaaabaGaci4CaiaacMgacaGGUbWaaSaaaeaacaWG5baabaGaamiE aaaaaaGaeyOeI0YaaSaaaeaaciGGJbGaai4BaiaacohadaWcaaqaai aadMhaaeaacaWG4baaaaqaaiGacohacaGGPbGaaiOBamaalaaabaGa amyEaaqaaiaadIhaaaaaaiabg2da9maalaaabaGaam4qaaqaaiaadI haaaaabaGaeyO0H4TaaCzcaiaaykW7caaMc8UaaGPaVlaadIhadaGa daqaaiaaigdacqGHsislciGGJbGaai4Baiaacohadaqadaqaamaala aabaGaamyEaaqaaiaadIhaaaaacaGLOaGaayzkaaaacaGL7bGaayzF aaGaeyypa0Jaam4qaiaaykW7caWGZbGaamyAaiaad6gadaqadaqaam aalaaabaGaamyEaaqaaiaadIhaaaaacaGLOaGaayzkaaaabaGaaeiv aiaabIgacaqGPbGaae4CaiaabccacaqGPbGaae4CaiaabccacaqG0b GaaeiAaiaabwgacaqGGaGaaeOCaiaabwgacaqGXbGaaeyDaiaabMga caqGYbGaaeyzaiaabsgacaqGGaGaae4Caiaab+gacaqGSbGaaeyDai aabshacaqGPbGaae4Baiaab6gacaqGGaGaae4BaiaabAgacaqGGaGa aeiDaiaabIgacaqGLbGaaeiiaiaabEgacaqGPbGaaeODaiaabwgaca qGUbGaaeiiaiaabsgacaqGPbGaaeOzaiaabAgacaqGLbGaaeOCaiaa bwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiBaiaabccacaqGLbGaae yCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaab6caaaaa @3F95@

Q.68

Show that the given differential equation is homogeneous and solve it.

ydx+xlog(yx)dy2xdy=0

Ans.

The given differential equation is: y d x + x l o g ( y x ) d y 2 x d y = 0 y d x + { x l o g ( y x ) 2 x } d y = 0 d y d x = y { 2 x x l o g ( y x ) } . . . ( i ) L e t F ( x , y ) = y { 2 x x l o g ( y x ) } N o w , F ( λ x , λ y ) = λ y { 2 λ x λ x l o g ( λ y λ x ) } = λ y λ { 2 x x l o g ( y x ) } = λ 0 F ( x , y ) Therefore, F ( x , y ) is a homogenous function of degree zero . S o , the given differential equation is a homogenous differential e q u a t i o n . For solving equation ( i ) , s u b s t i t u t e y = v x a n d d y d x = v + x d v d x , w e g e t v + x d v d x = v x { 2 x x l o g ( v x x ) } x d v d x = v 2 l o g v v x d v d x = v 2 v + v l o g v 2 l o g v x d v d x = v l o g v v 2 l o g v { 2 l o g v v ( l o g v 1 ) } d v = d x x { 1 + ( 1 l o g v ) v ( l o g v 1 ) } d v = d x x { 1 v ( l o g v 1 ) 1 v } d v = d x x I n t e g r a t i n g b o t h s i d e s , w e g e t 1 v ( l o g v 1 ) d v 1 v d v = 1 x d x l o g | l o g v 1 | l o g | v | = l o g | x | + l o g C l o g ( l o g v 1 ) v = l o g C x ( l o g v 1 ) v = C x x y ( l o g y x 1 ) = C x l o g ( y x ) 1 = C y T h i s i s t h e r e q u i r e d s o l u t i o n o f t h e g i v e n d i f f e r e n t i a l e q u a t i o n .

Q.69

Show that the given differential equation is homogeneous and solve each of them.

(1+exy)dx+exy(1xy)dy=0

Ans.

The given differential equation is:(1+exy)dx+exy(1xy)dy=0               dydx=(1+exy)exy(1xy)              dxdy=exy(1xy)(1+exy)...(i)Let F(x,y)=exy(1xy)(1+exy)Now,        F(λx,λy)=eλxλy(1λxλy)(1+eλxλy)          =exy(1xy)(1+exy)=λ0F(x, y)Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation. 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GHRaWkciGGSbGaai4BaiaacEgacaWGdbaaaaa@19D6@ log| x y + e x y |=log C y x y + e x y = C y x+y e x y =C This is the required solution of the given differential equation. 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Q.70

For the differential equation, find the particular situation satisfying the given condition:

(x + y) dy + (x – y) dx = 0; y = 1 when x = 1

Ans.

The given differential equation is:(x+y)dy+(xy)dx=0    (x+y)dy=(xy)dx      dydx=(yx)x+y  ...(i)Let F(x,y)=(yx)x+yNow,        F(λx, λy)=(λyλx)λx+λy        =λ(yx)λ(x+y)=λ0F(x, y)Therefore, F(x, y) is a homogenous function of degree zero. So,the given differential equation is a homogenous differential equation.For solving equation (i), substitute y =vx and dydx=v+xdvdx,we get                        v+xdvdx=vxxx+vx                      =(v1)x(1+v)x                                xdvdx=(v1)(1+v)v                      =v1vv2(1+v)                                xdvdx=(1+v2)(1+v)                    (1+v)(1+v2)dv=dxxIntegrating both sides, we get              11+v2dv+v1+v2dv=1xdx                tan1v+12log(1+v2)=logx+k                2tan1v+log(1+v2)=2logx+2k2tan1v+log(1+v2)+2logx=2k MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+ga caqGUbGaaeOlaaqaaiaabAeacaqGVbGaaeOCaiaabccacaqGZbGaae 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Q.71

For the differential equation, find the particular situation satisfying the given condition:

x2 dy + (xy + y2) dx = 0; y = 1 when x = 1

Ans.

The given differential equation is:    x2dy+(xy+y2)dx=0      x2dy=(xy+y2)dx              dydx=(xy+y2)x2  ...(i)Let         F(x, y)=(xy+y2)x2 Now,    F(λx,λy)=(λ2xy+λ2y2)λ2x2            =λ2(xy+y2)λ2x2=λ0F(x,y)Therefore, F(x, y) is a homogenous function of degree zero. So,the given differential equation is a homogenous differentialequation.For solving equation (i), substitute y =vx and dydx=v+xdvdx,we get      v+xdvdx=(xvx+v2x2)x2            =(v+v2)x2x2            xdvdx=vv2v            =v22v      dvv(v+2)=dxx12{(v+2)v}dvv(v+2)=dxx  12(1v1v+2)dv=dxxIntegrating both sides, we get121vdv121v+2dv=1xdx 12log|v|12log|v+2|=log|x|+logC        log(vv+2)=2log(Cx)          log(yxyx+2)=log(Cx)2        log(yy+2x)=log(Cx)2        yy+2x=(Cx)2        x2yy+2x=C2...(ii)Now,  y=1 when x=1        12(1)1+2(1)=C2            13=C2Substituting value of C2 in equation (ii), we get          x2yy+2x=13      3x2y=y+2x  y+2x=3x2yThis is the required solution of the given differential equation.

Q.72

For the differential equations, find the particular situation satisfying the given condition:

[xsin2(yx)y] dx+xdy=0; y=π when x=1

Ans.

The given differential equation is:    [xsin2(yx)y]dx+xdy=0      xdy=[xsin2(yx)y]dx              dydx=[xsin2(yx)y]x  ...(i)Let         F(x,y)=[xsin2(yx)y]xNow,    F(λx,λy)=[λxsin2(λyλx)λy]λx            =λ[xsin2(yx)y]λx=λ0F(x,y)Therefore, F(x, y) is a homogenous function of degree zero. So,the given differential equation is a homogenous differentialequation. For solving equation (i), substitute y=vx and dydx=v+xdvdx,we get      v+xdvdx=[xsin2(vxx)vx]x            xdvdx=sin2v+vv        cosec2vdv=dxxIntegrating both sides, we get      cosec2vdv=1xdx      cotv=log|x|+logC          cotyx=log|Cx|...(ii)Now, y=π4 at x=1      cot(π4)1=log|C.1|      1=log|C|C=eSubstituting value of C in equation (ii), we get          cotyx=log|ex|This is the required solution of the given differential equation.

Q.73

For the differential equations, find the particular situation satisfying the given condition:

dydxyx+cosec(yx)=0; y=0 when x=1

Ans.

The given differential equation is:dydxyx+cosec(yx)=0              dydx=yxcosec(yx)  ...(i)Let         F(x,y)=yxcosec(yx)Now,    F(λx,λy)=λyλxcosec(λyλx)            =yxcosec(yx)=λ0F(x,y)Therefore, F(x, y) is a homogenous function of degree zero. So,the given differential equation is a homogenous differentialequation.For solving equation (i), substitute y =vx and dydx=v+xdvdx,we get      v+xdvdx=vxxcosec(vxx)            xdvdx=vcosecvv    dvcosecv=dxxIntegrating both sides, we get  sinvdv=1xdx              cosv=log|x|+logC                  cos(yx)=log|Cx|...(ii) Now, y=0 at x=1          cos(01)=log|C.1|                  cos0=logC                          1=logC                              C=ePutting C = e in equation (ii), we get                  cos(yx)=log|ex|This is the required solution of the given differential equation.

Q.74

For the differential equation given below, find a particular situation satisfying the given condition:

2xy+y22x2dydx=0;​y=2whenx=1

Ans.

The given differential equation is:    2xy+y22x2dydx=0  2x2dydx=(2xy+y2)              dydx=(2xy+y2)2x2  ...(i)Let         F(x,y)=(2xy+y2)2x2Now,    F(λx,λy)=(λ22xy+λ2y2)2λ2x2            =λ2(2xy+y2)2λ2x2=λ0F(x,y) Therefore, F(x, y) is a homogenous function of degree zero. So,the given differential equation is a homogenous differentialequation.For solving equation (i), substitute y =vx and dydx=v+xdvdx,we get      v+xdvdx=(2xvx+v2x2)2x2            =(2v+v2)x22x2            xdvdx=2v+v22v            =v22          dvv2=dx2xIntegrating both sides, we get            2v2dv=1xdx      2v11=log|x|+C        2v=log|x|+C        2(yx)=log|x|+C         2xy=log|x|+C...(ii)Now,y=2 at x=1.        2(1)2=log|1|+CC=1Substituting C = –1 in equation(ii), we get        2xy=log|x|1    y=2x1log|x|(x0,xe)This is the required solution of the given differential equation.

Q.75

A homogeneous differential equation of the form dydx=h(xy)can be solved by making the substitution.(A)y=vx (B)v=yx (C)x=vy (D)x=v

Ans.

A homogeneous differential equation of the form dydx=h(xy)can be solved by making the substitutionas x=vy.Hence, the correct option is C.

Q.76

Which of the following is a homogeneous differential equation?
(A) (4x + 6y + 5)dy – (3y + 2x + 4) dx = 0

(B) (xy) dx – (x3 + y3) dy = 0

(C) (x3 + 2y2) dx + 2xy dy = 0

(D) y2 dx + (x2– xy – y2) dy = 0

Ans.

A function F(x, y) is said to be homogeneous function of degree n if F(λx, λy)= λn F(x, y) for any nonzero constant λ.

Let us check option D, y2dx+(x2xyy2)dy=0    (x2xyy2)dy=y2dxdydx=y2(x2xyy2)Let    F(x,y)=y2(x2xyy2)then        F(λx,λy)=λ2y2(λ2x2λ2xyλ2y2)    =λ2λ2y2(x2xyy2)=λ0F(x,y)Thus, the given differential equation in option D ishomogenous equation.

Q.77

Find the general solution of the differential equation given below.

dydx+2y=sinx

Ans.

The given differential equation is dydx+2y=sinxComparing it withdydx+Py=Q,we getP=2andQ=sinxSo,I.F.=ePdx    =e2dx    =e2xThe solution of given differential equation is given by the relation,y(I.F.)=(Q×I.F.)dx+C  ye2x=(sinx×e2x)dx+C  ...(i)Let  I=sinxe2xdx  =sinxe2xdx(ddxsinxe2xdx)dx  =sinx[e2x2]cosx.e2x2dx  =12sinxe2x12cosx.e2xdx  =12sinxe2x12{cosxe2xdx(ddxcosxe2xdx)dx}  =12sinxe2x12{cosx[e2x2](sinxe2x2)dx} I= 1 2 sinx e 2x 1 4 ( cosx e 2x +I ) I= 1 2 sinx e 2x 1 4 cosx e 2x 1 4 I 5 4 I= 1 4 e 2x ( 2sinxcosx ) I= 1 5 e 2x ( 2sinxcosx ) So, from equation( i ), we have y e 2x = 1 5 e 2x ( 2sinxcosx )+C y= 1 5 ( 2sinxcosx )+C e 2x This is the required general solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVJqaaiaa=LeacaaMc8Uaa8xpamaalaaabaGaa8xmaa qaaiaa=jdaaaGaa83Caiaa=LgacaWFUbGaa8hEaiaaykW7caWFLbWa aWbaaSqabeaacaWFYaGaa8hEaaaakiabgkHiTmaalaaabaGaa8xmaa qaaiaa=rdaaaWaaeWaaeaacaWFJbGaa83Baiaa=nhacaWF4bGaaGPa Vlaa=vgadaahaaWcbeqaaiaa=jdacaWF4baaaOGaa83kaiaa=Leaai aawIcacaGLPaaaaeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaa ykW7caaMc8UaaGPaVlaaykW7caWFjbGaaGPaVlaa=1dadaWcaaqaai aa=fdaaeaacaWFYaaaaiaa=nhacaWFPbGaa8NBaiaa=HhacaaMc8Ua a8xzamaaCaaaleqabaGaa8Nmaiaa=HhaaaGccqGHsisldaWcaaqaai aa=fdaaeaacaWF0aaaaiaa=ngacaWFVbGaa83Caiaa=HhacaaMc8Ua a8xzamaaCaaaleqabaGaa8Nmaiaa=HhaaaGccqGHsisldaWcaaqaai aa=fdaaeaacaWF0aaaaiaa=LeaaeaacaaMc8UaaGPaVlaaykW7caaM c8UaaGPaVpaalaaabaGaa8xnaaqaaiaa=rdaaaGaa8xsaiaaykW7ca WF9aWaaSaaaeaacaWFXaaabaGaa8hnaaaacaWFLbWaaWbaaSqabeaa caWFYaGaa8hEaaaakmaabmaabaGaa8Nmaiaa=nhacaWFPbGaa8NBai aa=HhacqGHsislcaWFJbGaa83Baiaa=nhacaWF4bGaaGPaVdGaayjk aiaawMcaaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl aaykW7caaMc8UaaGPaVlaa=LeacaaMc8Uaa8xpamaalaaabaGaa8xm aaqaaiaa=vdaaaGaa8xzamaaCaaaleqabaGaa8Nmaiaa=HhaaaGcda qadaqaaiaa=jdacaWFZbGaa8xAaiaa=5gacaWF4bGaeyOeI0Iaa83y aiaa=9gacaWFZbGaa8hEaiaaykW7aiaawIcacaGLPaaaaeaacaqGtb Gaae4BaiaabYcacaqGGaGaaeOzaiaabkhacaqGVbGaaeyBaiaabcca caqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBam aabmaabaGaaeyAaaGaayjkaiaawMcaaiaabYcacaqGGaGaae4Daiaa bwgacaqGGaGaaeiAaiaabggacaqG2bGaaeyzaaqaaiaaykW7caaMc8 UaaGPaVlaa=LhacaWFLbWaaWbaaSqabeaacaWFYaGaa8hEaaaakiaa =1dadaWcaaqaaiaa=fdaaeaacaWF1aaaaiaa=vgadaahaaWcbeqaai aa=jdacaWF4baaaOWaaeWaaeaacaWFYaGaa83Caiaa=LgacaWFUbGa a8hEaiabgkHiTiaa=ngacaWFVbGaa83Caiaa=HhacaaMc8oacaGLOa GaayzkaaGaa83kaiaa=neaaeaacqGHshI3caaMc8UaaGPaVlaaykW7 caaMc8UaaGPaVlaa=LhacaWF9aWaaSaaaeaacaWFXaaabaGaa8xnaa aadaqadaqaaiaa=jdacaWFZbGaa8xAaiaa=5gacaWF4bGaeyOeI0Ia a83yaiaa=9gacaWFZbGaa8hEaiaaykW7aiaawIcacaGLPaaacaWFRa Gaa83qaiaa=vgadaahaaWcbeqaaiaa=1cacaWFYaGaa8hEaaaaaOqa aiaabsfacaqGObGaaeyAaiaabohacaqGGaGaaeyAaiaabohacaqGGa GaaeiDaiaabIgacaqGLbGaaeiiaiaabkhacaqGLbGaaeyCaiaabwha caqGPbGaaeOCaiaabwgacaqGKbGaaeiiaiaabEgacaqGLbGaaeOBai aabwgacaqGYbGaaeyyaiaabYgacaqGGaGaae4Caiaab+gacaqGSbGa aeyDaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae4BaiaabAgaca qGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabEgacaqGPbGaaeODaiaa bwgacaqGUbGaaeiiaiaabsgacaqGPbGaaeOzaiaabAgacaqGLbGaae OCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiBaiaabccaaeaa caqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBai aab6caaaaa@5AAE@

Q.78

Find the general solution of the differential equation given below.

dydx+3y=e2x

Ans.

The given differential equation is dydx+3y=e2xComparing it with dydx+Py =Q, we getP=3 and Q=e2xSo,I.F.=ePdx    =e3dx    =e3x The solution of given differential equation is given by the relation, y( I.F. )= ( Q×I.F. ) dx+C y e 3x = ( e 2x × e 3x ) dx+C = e x dx+C y e 3x = e x +C y= e 2x +C e 3x This is the required general solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyzaiaabccacaqGZbGaae4BaiaabYga caqG1bGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOzai aabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGKbGa aeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshaca qGPbGaaeyyaiaabYgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaa bshacaqGPbGaae4Baiaab6gacaqGGaGaaeyAaiaabohacaqGGaGaae 4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeOyaiaabMhacaqG GaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkhacaqGLbGaaeiBaiaabg gacaqG0bGaaeyAaiaab+gacaqGUbGaaeilaaqaaiaabMhadaqadaqa aiaadMeacaGGUaGaamOraiaac6caaiaawIcacaGLPaaacqGH9aqpda WdbaqaamaabmaabaGaamyuaiabgEna0kaadMeacaGGUaGaamOraiaa c6caaiaawIcacaGLPaaaaSqabeqaniabgUIiYdGccaaMc8Uaamizai aadIhacqGHRaWkcaWGdbaabaGaaGPaVlaaykW7caaMc8UaamyEaiaa dwgadaahaaWcbeqaaiaaiodacaWG4baaaOGaeyypa0Zaa8qaaeaada qadaqaaiaadwgadaahaaWcbeqaaiabgkHiTiaaikdacaWG4baaaOGa ey41aqRaamyzamaaCaaaleqabaGaaG4maiaadIhaaaaakiaawIcaca GLPaaaaSqabeqaniabgUIiYdGccaaMc8UaamizaiaadIhacqGHRaWk caWGdbGaaGPaVdqaaiaaxMaacaaMc8UaaGPaVlaaykW7cqGH9aqpda WdbaqaaiaadwgadaahaaWcbeqaaiaadIhaaaaabeqab0Gaey4kIipa kiaaykW7caWGKbGaamiEaiabgUcaRiaadoeaaeaacaaMc8UaaGPaVl aaykW7caWG5bGaamyzamaaCaaaleqabaGaaG4maiaadIhaaaGccqGH 9aqpcaWGLbWaaWbaaSqabeaacaWG4baaaOGaey4kaSIaam4qaaqaai aaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Ua amyEaiabg2da9iaadwgadaahaaWcbeqaaiabgkHiTiaaikdacaWG4b aaaOGaey4kaSIaam4qaiaadwgadaahaaWcbeqaaiabgkHiTiaaioda caWG4baaaaGcbaGaaeivaiaabIgacaqGPbGaae4CaiaabccacaqGPb Gaae4CaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaabwga caqGXbGaaeyDaiaabMgacaqGYbGaaeyzaiaabsgacaqGGaGaae4zai aabwgacaqGUbGaaeyzaiaabkhacaqGHbGaaeiBaiaabccacaqGZbGa ae4BaiaabYgacaqG1bGaaeiDaiaabMgacaqGVbGaaeOBaiaabccaca qGVbGaaeOzaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4zaiaa bMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeizaiaabMgacaqGMbGaae OzaiaabwgacaqGYbGaaeyzaiaab6gacaqG0bGaaeyAaiaabggacaqG SbGaaeiiaaqaaiaabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAai aab+gacaqGUbGaaeOlaaaaaa@1CBE@

Q.79

Find the general solution of the differential equation given below.

dy dx + y x = x 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaake aadaWcaaqaaGqabiaa=rgacaWF5baabaGaa8hzaiaa=HhaaaGaa83k amaalaaabaGaa8xEaaqaaiaa=HhaaaGaa8xpaiaa=HhadaahaaWcbe qaaiaa=jdaaaaaaa@4232@

Ans.

The given differential equation is dydx+yx=x2Comparing it with dydx+Py=Q, we getP=1x and Q=x2So,I.F.=ePdx    =e1xdx    =elogx    =xThe solution of given differential equation is given by the relation,y(I.F.)=(Q×I.F.)dx+C  yx=(x2×x)dx+C = x 3 dx+C xy= x 4 4 +C This is the required general solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaaxMaacqGH9aqpdaWdbaqaaiaadIhadaahaaWcbeqaaiaa iodaaaaabeqab0Gaey4kIipakiaaykW7caWGKbGaamiEaiabgUcaRi aadoeaaeaacaaMc8UaaGPaVlaaykW7caWG4bGaamyEaiabg2da9maa laaabaGaamiEamaaCaaaleqabaGaaGinaaaaaOqaaiaaisdaaaGaey 4kaSIaam4qaaqaaiaabsfacaqGObGaaeyAaiaabohacaqGGaGaaeyA aiaabohacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkhacaqGLb GaaeyCaiaabwhacaqGPbGaaeOCaiaabwgacaqGKbGaaeiiaiaabEga caqGLbGaaeOBaiaabwgacaqGYbGaaeyyaiaabYgacaqGGaGaae4Cai aab+gacaqGSbGaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqGGaGa ae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabEgaca qGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabsgacaqGPbGaaeOzaiaa bAgacaqGLbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaae iBaiaabccaaeaacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMga caqGVbGaaeOBaiaab6caaaaa@908A@

Q.80

Find the general solution of the differential equation given below.

dydx+(secx)y = tanx(0x<π2)

Ans.

The given differential equation is dydx+(secx)y=tanx(0x<π2)Comparing it with dydx+Py=Q, we getP=secx and Q=tanxSo,I.F.=ePdx    =esecxdx    =elog(secx+tanx)    =secx+tanxThe solution of given differential equation is given by the relation,  y(I.F.)=(Q×I.F.)dx+C  y(secx+tanx)=(secx+tanx)tanxdx+C  y(secx+tanx)=secxtanxdx+tan2xdx+C    =secx+(sec2x1)dx+C    =secx+tanxx+CThis is the required general solution of the given differential equation.

Q.81

Find the general solution of the differential equation given below.

cos2xdydx+y=tanx(0x<π2)

Ans.

The given differential equation is cos2xdydx+y=tanx(0x<π2)dydx+(sec2x)y=sec2xtanxComparing it with dydx+Py=Q, we getP=sec2x and Q=sec2xtanxSo,I.F.=ePdx    =esec2xdx    =etanxThe solution of given differential equation is given by the relation,  y(I.F.)=(Q×I.F.)dx+C  yetanx=(sec2x  tanx)etanxdx+C  yetanx=tetsec2x  dtsec2x+C[Lett=tanxdtdx=sec2x]  yetanx=tetdt+C    =tetdt(ddttetdt)dt+C     =tet1.etdt+C    =tetet+C    =et(t1)+Cyetanx =etanx(tanx1)+C           y = (tanx1)+Cetanx This is the required general solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabsfacaqGObGaaeyAaiaabohacaqGGaGaaeyAaiaaboha caqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabkhacaqGLbGaaeyCai aabwhacaqGPbGaaeOCaiaabwgacaqGKbGaaeiiaiaabEgacaqGLbGa aeOBaiaabwgacaqGYbGaaeyyaiaabYgacaqGGaGaae4Caiaab+gaca qGSbGaaeyDaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae4Baiaa bAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabEgacaqGPbGaae ODaiaabwgacaqGUbGaaeiiaiaabsgacaqGPbGaaeOzaiaabAgacaqG LbGaaeOCaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiBaiaabc caaeaacaqGLbGaaeyCaiaabwhacaqGHbGaaeiDaiaabMgacaqGVbGa aeOBaiaab6caaaaa@79D8@

Q.82

Find the general solution of the differential equation given below.

xdydx+2y=x2logx

Ans.

The given differential equation is xdydx+2y=x2logxdydx+2yx=xlogxComparing it with dydx+Py=Q, we getP=2x and Q=xlogxSo,  I.F.=ePdx    =e2xdx    =e2logx    =elogx2    =x2 The solution of given differential equation is given by the relation,  y(I.F.)=(Q×I.F.)dx+C          yx2=xlogx.x2dx+C          yx2=x3logxdx+C     =logxx3dx(ddxlogxx3dx)dx+C    =logx.[x44](1x[x44])dx+C    =logx.[x44]14x3dx+C    =logx.[x44]14[x44]+C          yx2=x416(4logx1)+C            y=x216(4logx1)+Cx2This is the required general solution of the given differential equation.

Q.83

Find the general solution of the differential equation given below.

xlogxdydx + y=xlogx

Ans.

The given differential equation is xlogx dydx+y=2xlogx dydx+yxlogx=2x2Comparing it with dydx+Py=Q, we getP=1xlogx and Q=2x2So,I.F.=ePdx    =e1xlogxdx    =elog(logx)    =logxThe solution of given differential equation is given by therelation,y(I.F.)=(Q×I.F.)dx+Cylogx=(2x2×logx)dx+Cylogx=logx2x2dx(ddxlogx2x2dx)dx+Cylogx=2{logx(1x)1x×1xdx}+Cylogx=2{1xlogx+x2dx}+Cylogx=2(1xlogx+x11)+C ylogx=2xlogx2x+Cylogx=2x(logx+1)+CThis is the required general solution of the given differential equation.

Q.84

Find the general solution of the differential equation given below.

(1+x2)dy+2xydx=cotx dx (x0)

Ans.

The given differential equation is       (1+x2)dy+2xydx=cotxdx  (x0)      dydx+2x(1+x2)y=cotx(1+x2)Comparing it with dydx+Py=Q, we getP=2x(1+x2) and Q=cotx(1+x2)So,I.F.=ePdx    =e2x(1+x2)dx    =elog(1+x2)    =(1+x2)The solution of given differential equation is given by the

relation,

y( I.F. )= ( Q×I.F. ) dx+C y( 1+ x 2 )= { cot x ( 1+ x 2 ) ×( 1+ x 2 ) } dx+C y( 1+ x 2 )= cot x dx+C y( 1+ x 2 )=log| sinx |+C y= ( 1+ x 2 ) 1 log| sinx |+C ( 1+ x 2 ) 1 This is the required general solution of the given differential equation. MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9v8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaabMhadaqadaqaaiaadMeacaGGUaGaamOraiaac6caaiaa wIcacaGLPaaacqGH9aqpdaWdbaqaamaabmaabaGaamyuaiabgEna0k aadMeacaGGUaGaamOraiaac6caaiaawIcacaGLPaaaaSqabeqaniab gUIiYdGccaaMc8UaamizaiaadIhacqGHRaWkcaWGdbaabaGaamyEam aabmaabaGaaGymaiabgUcaRiaadIhadaahaaWcbeqaaiaaikdaaaaa kiaawIcacaGLPaaacqGH9aqpdaWdbaqaamaacmaabaWaaSaaaeaaci GGJbGaai4BaiaacshaieqacaWFGaGaamiEaaqaamaabmaabaGaaGym aiabgUcaRiaadIhadaahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPa aaaaGaey41aq7aaeWaaeaacaaIXaGaey4kaSIaamiEamaaCaaaleqa baGaaGOmaaaaaOGaayjkaiaawMcaaaGaay5Eaiaaw2haaaWcbeqab0 Gaey4kIipakiaaykW7caWGKbGaamiEaiabgUcaRiaadoeaaeaacaWG 5bWaaeWaaeaacaaIXaGaey4kaSIaamiEamaaCaaaleqabaGaaGOmaa aaaOGaayjkaiaawMcaaiabg2da9maapeaabaGaci4yaiaac+gacaGG 0bGaa8hiaiaadIhaaSqabeqaniabgUIiYdGccaaMc8UaamizaiaadI hacqGHRaWkcaWGdbaabaGaamyEamaabmaabaGaaGymaiabgUcaRiaa dIhadaahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaacqGH9aqpci GGSbGaai4BaiaacEgadaabdaqaaiGacohacaGGPbGaaiOBaiaadIha aiaawEa7caGLiWoacqGHRaWkcaWGdbaabaGaaCzcaiaaykW7caaMc8 UaaGPaVlaaykW7caWG5bGaeyypa0ZaaeWaaeaacaaIXaGaey4kaSIa amiEamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaamaaCaaale qabaGaeyOeI0IaaGymaaaakiGacYgacaGGVbGaai4zamaaemaabaGa ci4CaiaacMgacaGGUbGaamiEaaGaay5bSlaawIa7aiabgUcaRiaado eadaqadaqaaiaaigdacqGHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaa aaGccaGLOaGaayzkaaWaaWbaaSqabeaacqGHsislcaaIXaaaaaGcba GaaeivaiaabIgacaqGPbGaae4CaiaabccacaqGPbGaae4Caiaabcca caqG0bGaaeiAaiaabwgacaqGGaGaaeOCaiaabwgacaqGXbGaaeyDai aabMgacaqGYbGaaeyzaiaabsgacaqGGaGaae4zaiaabwgacaqGUbGa aeyzaiaabkhacaqGHbGaaeiBaiaabccacaqGZbGaae4BaiaabYgaca qG1bGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOzaiaa bccacaqG0bGaaeiAaiaabwgacaqGGaGaae4zaiaabMgacaqG2bGaae yzaiaab6gacaqGGaGaaeizaiaabMgacaqGMbGaaeOzaiaabwgacaqG YbGaaeyzaiaab6gacaqG0bGaaeyAaiaabggacaqGSbGaaeiiaaqaai aabwgacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGa aeOlaaaaaa@F796@

Q.85

Find the general solution of the differential equation given below.

xdydx+yx+xy cotx=0

Ans.

The given differential equation is xdydx+yx+xycotx=0      dydx+(1+xcotx<