NCERT Solutions Class 12 Mathematics Chapter 9

Q: How many problems are there in total in NCERT Solutions Class 12 Mathematics Chapter 9 Differential Equations?

NCERT Solutions Class 12 Mathematics Chapter 9 – Differential Equations has 113 questions separated into seven exercises. In addition to repetitive revision, a miscellaneous exercise consists of higher-order questions that push candidates to think outside of the box and help them apply their auxiliary mathematical skills. These are wonderful sums that aid students in gaining a comprehensive understanding of the subject.

Q: Why should I practise NCERT Solutions Class 12 Mathematics Chapter 9?

Students will gain confidence in solving questions on this topic by regularly practising the NCERT Solutions Class 12 Mathematics Chapter 9. They will be able to attempt the paper stress-free throughout any examination, whether it is a board or competitive exam, resulting in imminent success. Aside from that, by revisiting these solutions, students can build a solid Mathematical foundation that will serve them well throughout their lives.

Q: Is it necessary for one to practise all the NCERT Solutions for Class 12 mathematics Differential Equations?

Perfection requires a lot of practise. As a result, you should practise all the sums in the NCERT Solutions Class 12 Mathematics chapter 9 -Differential Equations at least twice. This chapter presents several new concepts built on previously taught material and, as a result, can be perplexing at times. Students not only pave the route to a higher exam result by solving each problem, but they also ensure that these topics come in use in real-life situations as well.

Q: What is the significance of NCERT Solutions for Class 12 Mathematics Chapter 9?

The board exams are based on NCERT Solutions Class 12 Mathematics Chapter 9. CBSE papers are based on the NCERT textbook format. Furthermore, these issues are organised so that they provide readers with a 360-degree view of the chapter. Extramarks recommend students use these to gain a thorough understanding of the subject and double-check their responses after solving a question.

Q: Are there answers to all the textbook problems in the NCERT Solutions for Class 12 Mathematics Chapter 9?

The NCERT Solutions for Class 12 Mathematics Chapter 9 are created by subject specialists at Extramarks and are accessible in the aforementioned links. These solutions are based entirely on the most recent CBSE Syllabus 2022-23, and they cover all of the key concepts for the first as well as the second term exams. The textbook questions are addressed in a step-by-step fashion based on marks weightage in the second-term exams.

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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:

Quick Links

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 Solutions – 12 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.2 Solutions – 12 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.3 Solutions – 12 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.4 Solutions – 23 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.5 Solutions – 17 Questions

Class 12 Mathematics Chapter No. 9 Ex 9.6 Solutions – 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

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
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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

$\frac{d^{4} y}{{dx}^{4}} + \sin (y ”’) = 0$

Ans.

$\begin{array}{l} \frac{d^{4} y}{{dx}^{4}} + \sin (y ”’) = 0 \Rightarrow y ”” + \sin (y ”’) = 0 \\ The heighest order derivative present in the differential equation \\ is \frac{d^{4} y}{{dx}^{4}} . Therefore its order is 4 . \\ The given equation is not a polynomial equation in y ‘ and degree \\ of such a differential equation can not be defined . \end{array}$

Q.2

$\begin{array}{l} Determine order and degree (if defined) of differential equation: \\ y’ + 5y = 0 \end{array}$

Ans.

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

Q.3

$\begin{array}{l} Determine order and degree (if defined) of differential equation: \\ {(\frac{ds}{dt})}^{4} + 3s \frac{d^{2} s}{{dt}^{2}} =0 \end{array}$

Ans.

$\begin{array}{l} {(\frac{ds}{dt})}^{4} + 3s \frac{d^{2} s}{{dt}^{2}} =0 \\ The heighest oder derivative present in differential equation is \\ \frac{d^{2} s}{{dt}^{2}} . Therefore, its order is 2 . \\ Since, given differential equation is a polynomial in \frac{d^{2} s}{{dt}^{2}} and \frac{ds}{dt} . \\ The power raised by \frac{d^{2} s}{{dt}^{2}} is 1, so the degree of equation is 1 . \end{array}$

Q.4

$\begin{array}{l} Determine order and degree (if defined) of differential equation: \\ {(\frac{d^{2} y}{{dx}^{2}})}^{2} + cos (\frac{dy}{dx}) = 0 \end{array}$

Ans.

$\begin{array}{l} {(\frac{d^{2} y}{{dx}^{2}})}^{2} +cos (\frac{dy}{dx}) =0 \\ The heighest oder derivative present in differential equation is \\ \frac{d^{2} y}{{dx}^{2}} . Therefore, its order is 2 . \\ Since, given differential equation is not a polynomial in \frac{d^{2} y}{{dx}^{2}} \\ and \frac{dy}{dx} . \\ So, its degree is not defined . \end{array}$

Q.5

$\begin{array}{l} Determine order and degree (if defined) of differential equation: \\ \frac{d^{2} y}{{dx}^{2}} = cos3x + sin3x \end{array}$

Ans.

$\begin{array}{l} \frac{d^{2} y}{{dx}^{2}} =cos3x + sin3x \Rightarrow \frac{d^{2} y}{{dx}^{2}} - cos3x - sin3x=0 \\ The heighest oder derivative present in differential equation is \\ \frac{d^{2} y}{{dx}^{2}} . Therefore, its order is 2 . \\ Since, given differential equation is a polynomial in \frac{d^{2} y}{{dx}^{2}} . \\ The power raised by \frac{d^{2} y}{{dx}^{2}} is 1, so the degree of equation is 1 . \end{array}$

Q.6

$\begin{array}{l} Determine order and degree (if defined) of differential equation: \\ {(y”’)}^{2} + {(y”)}^{3} + {(y’)}^{4} + {(y)}^{5} = 0 \end{array}$

Ans.

$\begin{array}{l} {(y”’)}^{2} + {(y”)}^{3} + {(y’)}^{4} + {(y)}^{5} =0 \\ The heighest order derivative in given differential equation is y”’ . \\ So, its order is 3 . \\ This differential equation is a polynomial in y”’, y”, y’ and y . \\ The heighest power raised by y”’ is 2, therefore degree of \\ differential equation is 2 . \end{array}$

Q.7

$\begin{array}{l} Determine order and degree (if defined) of differential equation: \\ y”’ + 2y” + y’ = 0 \end{array}$

Ans.

$\begin{array}{l} y ”’ + 2 y ”+ y ‘=0 \\ The heighest order derivative in given differential equation is y ”’ . \\ So, its order is 3 . \\ This differential equation is a polynomial in y ”’, y ” and y ‘ . \\ The heighest power raised by y ”’ is 1, therefore degree of \\ differential equation is 1 . \end{array}$

Q.8

$\begin{array}{l} Determine order and degree (if defined) of differential equation: \\ y’ + y = e^{x} \end{array}$

Ans.

$\begin{array}{l} {y’+ y=e}^{x} {Þy’+ y-e}^{x} =0 \\ The heighest order derivative in given differential equation is y’ . \\ So, its order is 1 . \\ This differential equation is a polynomial in y’ and y . \\ The heighest power raised by y’ is 1, therefore degree of \\ differential equation is 1 . \end{array}$

Q.9

$\begin{array}{l} Determine order and degree (if defined) of differential equation: \\ y” + {(y’)}^{2} + 2y = 0 \end{array}$

Ans.

$\begin{array}{l} y” + {(y’)}^{2} +2y=0 \\ The heighest order derivative in given differential equation isy” . \\ So, its order is 2 . \\ This differential equation is a polynomial in y”, y’ and y . \\ The heighest power raised by y” is 1, therefore degree of \\ differential equation is 1 . \end{array}$

Q.10

$\begin{array}{l} Determine order and degree (if defined) of differential equation: \\ y” + 2y’ + siny = 0 \end{array}$

Ans.

$\begin{array}{l} y”+ 2y’+ siny=0 \\ The heighest order derivative in given differential equation is y” . \\ So, its order is 2 . \\ This differential equation is a polynomial in y” and y’ . \\ The heighest power raised by y” is 1, therefore degree of \\ differential equation is 1 . \end{array}$

Q.11

$\begin{array}{l} The degree of the differential equation \\ {(\frac{d^{2} y}{{dx}^{2}})}^{3} + {(\frac{dy}{dx})}^{2} + sin (\frac{dy}{dx}) +1 = 0 is \\ (A) 3 (B) 2 (C) 1 (D) not ‹ defined \end{array}$

Ans.

$\begin{array}{l} {(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{2} + \sin (\frac{d y}{d x}) + 1 = 0 \\ S i n c e given equation is not a polynomial equation in y’ . Therefore, \\ its degree is not defined . \\ Hence, the correct answer is D . \end{array}$

Q.12

$\begin{array}{l} The order of the differential equation \\ {2x}^{2} \frac{d^{2} y}{{dx}^{2}} – 3 \frac{dy}{dx} + y = 0 is \\ (A) 3 (B) 2 (C) 0 (D) not ‹ defined \end{array}$

Ans.

$\begin{array}{l} 2 x^{2} \frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + y = 0 \\ T h e highest derivative of given differential equation is \frac{d^{2} y}{d x^{2}} . \\ T h e r e f o r e, the order of given differential equation is 2 . \\ Thus, correct option is B . \end{array}$

Q.13

$\begin{array}{l} Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: \\ {y = e}^{x} + 1 : y” – y’ = 0 \end{array}$

Ans.

$\begin{array}{l} {y = e}^{x} + 1 \\ Differentiating w .r .t . x, we get \\ {y’ = e}^{x} … (i) \\ Differentiating equation (i) w .r .t . x, we get \\ {y” = e}^{x} \\ Substituting values of y’ and y” in L .H .S . of given differential \\ equation y” – y’ = 0, we get \\ {y” – y’ =e}^{x} {– e}^{x} = 0 = R .H .S . \\ Thus,the given function is the solution of the corresponding \\ differential equation . \end{array}$

Q.14

$\begin{array}{l} Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: \\ {y = x}^{2} +2x + C : y’ –2x –2 = 0 \end{array}$

Ans.

$\begin{array}{l} {y = x}^{2} +2x + C \\ Differentiating w .r .t . x, we get \\ y’ = 2x + 2 \\ Substituting values of y’ in L .H .S . of given differential \\ equation y’ –2x –2 = 0, we get \\ y’ – 2x –2 = 2x + 2 –2x –2 \\ = 0 = R .H .S . \\ Thus,the given function is the solution of the corresponding \\ differential equation . \end{array}$

Q.15

$\begin{array}{l} Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: \\ y = cos x + C : y’ + sinx = 0 \end{array}$

Ans.

$\begin{array}{l} y = cos x + C \\ Differentiating w .r .t . x, we get \\ y’ = –sinx \\ Substituting values of y’ in L .H .S . of given differential \\ equation y’ + sinx = 0, we get \\ y’ + sinx = –sinx + sinx = 0 = R .H .S . \\ Thus,the given function is the solution of the corresponding \\ differential equation . \end{array}$

Q.16

$\begin{array}{l} Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: \\ y = \sqrt{{1 + x}^{2}} : y’ = \frac{xy}{{1 + x}^{2}} \end{array}$

Ans.

$\begin{array}{l} y = \sqrt{{1 + x}^{2}} \\ Differentiating w .r .t . x, we get \\ y’ = \frac{x}{\sqrt{{1 + x}^{2}}} \\ Substituting values of y in R .H .S . of given differential \\ equation y’ = \frac{xy}{{1 + x}^{2}}, we get \\ R .H .S . = \frac{xy}{{1 + x}^{2}} \\ = \frac{x (\sqrt{{1 + x}^{2}})}{{1 + x}^{2}} \\ = \frac{x}{\sqrt{{1 + x}^{2}}} \\ = y’ = L .H .S . \\ Thus,the given function is the solution of the corresponding \\ differential equation . \end{array}$

Q.17

Ans.

Q.18

Ans.

Q.19

Ans.

Q.20

Ans.

Q.21

$\begin{array}{l} Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: \\ x + y = t a n^{- 1} y : y^{2} y ‘ + y^{2} + 1 = 0 \end{array}$

Ans.

$\begin{array}{l} x + y = \tan^{- 1} y \\ Differentiating w .r .t . x, we get \\ 1+y’ = (\frac{1}{1 + y^{2}}) y ‘ \\ (1 + y’) (1 + y^{2}) = y ‘ \\ 1 + y^{2} + y ‘ + y ‘ y^{2} = y ‘ \\ 1 + y^{2} + y ‘ y^{2} = 0 \\ Thus,the given function is the solution of the corresponding \\ differential equation . \end{array}$

Q.22

$\begin{array}{l} Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: \\ y = \sqrt{a^{2} - x^{2}} x \in (- a, a) : x + y \frac{d y}{d x} = 0 (y \neq 0) \end{array}$

Ans.

$\begin{array}{l} y = \sqrt{a^{2} - x^{2}} \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = \frac{- x}{\sqrt{a^{2} - x^{2}}} \\ = \frac{- x}{y} \\ Substituting the value of \frac{d y}{d x} in L .H .S . of given differential \\ equation x + y \frac{d y}{d x} = 0, we get \\ L .H .S . = x + y \frac{d y}{d x} \\ = x + y (\frac{- x}{y}) \\ = x - x \\ = 0 = R . H . S . \\ Thus,the given function is the solution of the corresponding \\ differential equation . \end{array}$

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.

$\frac{x}{a} + \frac{y}{b} = 1$

Ans.

$\begin{array}{l} \frac{x}{a} + \frac{y}{b} = 1 \\ Differentiating both sides w .r .t . x, we get \\ \frac{1}{a} + \frac{1}{b} \frac{d y}{d x} = 0 \\ Againg, differentiating both sides w .r .t . x, we get \\ \frac{1}{b} \frac{d^{2} y}{d x^{2}} = 0 \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = 0 \Rightarrow y ‘ ‘ = 0 \\ Hence, the required differential equation is y”=0 . \end{array}$

Q.26 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

$y^{2} = a (b^{2} - x^{2})$

Ans.

$\begin{array}{l} y^{2} = a (b^{2} - x^{2}) \\ Differentiating both sides w . r . t . x, we get \\ 2 y \frac{dy}{dx} = a (0 - 2 x) \\ 2 y \frac{dy}{dx} = - 2 ax . .. (i) \\ Againg, differentiating both sides w . r . t . x, we get \\ 2 {{(\frac{dy}{dx})}^{2} + y \frac{d^{2} y}{{dx}^{2}}} = - 2 a \\ \Rightarrow {(\frac{dy}{dx})}^{2} + y \frac{d^{2} y}{{dx}^{2}} = - a . .. (ii) \\ Putting value of - a from equation (ii) to equation (i), we get \\ 2 y \frac{dy}{dx} = 2 {{(\frac{dy}{dx})}^{2} + y \frac{d^{2} y}{{dx}^{2}}} x \\ \Rightarrow y \frac{dy}{dx} = x {(\frac{dy}{dx})}^{2} + xy \frac{d^{2} y}{{dx}^{2}} \\ xyy ‘ ‘ + x {(y ‘)}^{2} - yy ‘ = 0 \\ Hence, the required differential equation is xyy ”+ x {(y ‘)}^{2} - yy ‘ = 0 . \end{array}$

Q.27 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

$y = a e^{3 x} + b e^{- 2 x}$

Ans.

$\begin{array}{l} y = a e^{3 x} + b e^{- 2 x} … (i) \\ Differentiating both sides w .r .t . x, we get \\ y ‘ = 3 a e^{3 x} - 2 b e^{- 2 x} … (i i) \\ Againg, differentiating both sides w .r .t . x, we get \\ y ‘ ‘ = 9 a e^{3 x} + 4 b e^{- 2 x} … (i i i) \\ Adding equation (ii) and 2×equation (i), we get \\ y ‘ + 2 y = 3 a e^{3 x} - 2 b e^{- 2 x} + 2 (a e^{3 x} + b e^{- 2 x}) \\ y ‘ + 2 y = 3 a e^{3 x} - 2 b e^{- 2 x} + 2 a e^{3 x} + 2 b e^{- 2 x} \\ y ‘ + 2 y = 5 a e^{3 x} \Rightarrow a e^{3 x} = \frac{y ‘ + 2 y}{5} \\ Substracting equation (ii) from 3×equation (i), we get \\ 3 y - y ‘ = 3 a e^{3 x} + 3 b e^{- 2 x} - (3 a e^{3 x} - 2 b e^{- 2 x}) \\ 3 y - y ‘ = 3 a e^{3 x} + 3 b e^{- 2 x} - 3 a e^{3 x} + 2 b e^{- 2 x} \\ 3 y - y ‘ = 5 b e^{- 2 x} \\ b e^{- 2 x} = \frac{3 y - y ‘}{5} \\ {Putting the values of ae}^{3x} and {be}^{-2x} in equation (iii), we get \\ y ‘ ‘ = 9 (\frac{y ‘ + 2 y}{5}) + 4 (\frac{3 y - y ‘}{5}) \\ \Rightarrow 5 y ‘ ‘ = 9 y ‘ + 18 y + 12 y - 4 y ‘ \\ \Rightarrow 5 y ‘ ‘ = 5 y ‘ + 30 y \\ \Rightarrow y ‘ ‘ - y ‘ - 6 y = 0 \\ Hence, the required differential equation is y”–y’–6y = 0 . \end{array}$

Q.28 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

${y=e}^{2x} (a+b x)$

Ans.

$\begin{array}{l} y = e^{2 x} (a + b x) \\ Differentiating both sides w .r .t . x, we get \\ y ‘ = e^{2 x} (0 + b) + 2 e^{2 x} (a + b x) \\ y ‘ = b e^{2 x} + 2 y \\ \Rightarrow y ‘ - 2 y = b e^{2 x} … (i) \\ Againg, differentiating both sides w .r .t . x, we get \\ y ‘ ‘ - 2 y ‘ = 2 b e^{2 x} … (i i) \\ Dividing equation (ii) by equation (i), we get \\ \frac{y ‘ ‘ - 2 y ‘}{y ‘ - 2 y} = \frac{2 b e^{2 x}}{b e^{2 x}} \\ \frac{y ‘ ‘ - 2 y ‘}{y ‘ - 2 y} = 2 \\ \Rightarrow y ‘ ‘ - 2 y ‘ = 2 (y ‘ - 2 y) \\ \Rightarrow y ‘ ‘ - 2 y ‘ - 2 y ‘ + 4 y = 0 \\ \Rightarrow y ‘ ‘ - 4 y ‘ + 4 y = 0 \\ Hence, the required differential equation is y” –4y’ + 4y = 0 . \end{array}$

Q.29 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

$y = e^{x} (a c o s x + b s i n x)$

Ans.

$\begin{array}{l} y = e^{x} (a \cos x + b \sin x) … (i) \\ Differentiating both sides w .r .t . x, we get \\ y ‘ = e^{x} (- a \sin x + b \cos x) + e^{x} (a \cos x + b \sin x) \\ y ‘ = e^{x} (- a \sin x + b \cos x) + y [From equation (i)] \\ \Rightarrow y ‘ - y = e^{x} (- a \sin x + b \cos x) … (i i) \\ Againg, differentiating both sides w .r .t . x, we get \\ y ‘ ‘ - y ‘ = e^{x} (- a \cos x - b \sin x) + e^{x} (- a \sin x + b \cos x) \\ y ‘ ‘ - y ‘ = - y + y ‘ - y [From equation (i i)] \\ y ‘ ‘ - 2 y ‘ + 2 y = 0 \\ Hence, the required differential equation is y ‘ ‘ - 2 y ‘ + 2 y = 0. \end{array}$

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.

$\begin{array}{l} The equation of the family of ellipses having foci on \\ y-axis and centre at origin is as follows \\ \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \dots (i) \\ Differentiating w .r .t . x, we get \\ \frac{2x}{b^{2}} + \frac{2yy’}{a^{2}} = 0 \\ \frac{x}{b^{2}} + \frac{yy’}{a^{2}} = 0 \dots (ii) \\ Again, differentiating w .r .t . x, we get \\ \frac{1}{b^{2}} + \frac{1}{a^{2}} (yy”+y’y’) = 0 \\ \frac{1}{b^{2}} + \frac{1}{a^{2}} {yy”+ {(y’)}^{2}} = 0 \\ \frac{1}{b^{2}} = - \frac{1}{a^{2}} {yy”+ {(y’)}^{2}} \\ Substituting the value of \frac{1}{b^{2}} in equation (ii), we get \\ x [- \frac{1}{a^{2}} {yy”+ {(y’)}^{2}}] + \frac{yy’}{a^{2}} = 0 \\ \Rightarrow - x {yy”+ {(y’)}^{2}} + y y ‘ = 0 \\ \Rightarrow - x y y ‘ ‘ - x {(y ‘)}^{2} + y y ‘ = 0 \\ \Rightarrow x y y ‘ ‘ + x {(y ‘)}^{2} - y y ‘ = 0 \\ Thus, the required differential equation is xyy” + x {(y’)}^{2} –yy’ = 0 . \end{array}$

Q.33 Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Ans.

$\begin{array}{l} The differential equation of the family of hyperbolas having foci \\ on x-axis and centre at origin ‹ is as follows \\ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \dots (i) \\ Differentiating ‹ w .r .t . x, we get \\ \frac{2x}{a^{2}} - \frac{2yy’}{b^{2}} = 0 \\ \frac{x}{a^{2}} - \frac{yy’}{b^{2}} = 0 \dots (ii) \\ Again, differentiating w .r .t . x, we get \\ \frac{1}{a^{2}} - \frac{1}{b^{2}} (yy”+y’y’) = 0 \\ \frac{1}{a^{2}} - \frac{1}{b^{2}} {yy”+ {(y’)}^{2}} = 0 \\ \frac{1}{a^{2}} = \frac{1}{b^{2}} {yy”+ {(y’)}^{2}} \\ Substituting the value of \frac{1}{a^{2}} in equation (ii), we get \\ x [\frac{1}{b^{2}} {yy”+ {(y’)}^{2}}] - \frac{yy’}{b^{2}} = 0 \\ \Rightarrow x {yy”+ {(y’)}^{2}} - y y ‘ = 0 \\ \Rightarrow x y y ‘ ‘ + x {(y ‘)}^{2} - y y ‘ = 0 \\ Thus, the required differential equation is xyy” + x {(y’)}^{2} –yy’=0 . \end{array}$

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)

$\begin{matrix} Differentiating equation (i), w . r . t . x, we get \\ 2x + 2 (y - b) y ’ = 0 \\ x + (y - b) y ’ = 0 \\ (y - b) = \frac{- x}{y ’} \end{matrix}$

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

$\begin{matrix} x^{2} + {(\frac{- x}{y ’})}^{2} = 9 \\ x^{2} y ‘^{2} + x^{2} = 9 y ‘^{2} \\ {(y ‘)}^{2} (x^{2} - 9) + x^{2} = 0 \\ T h i s is the required differential equation . \end{matrix}$

Q.35

$\begin{array}{l} Which of the following differential equations has \\ y = c_{1} e^{x} + c_{2} e^{- x} as the general solution ? \\ (A) \frac{d^{2} y}{{dx}^{2}} + y = 0 (B) \frac{d^{2} y}{{dx}^{2}} - y = 0 \\ (C) \frac{d^{2} y}{{dx}^{2}} + 1 = 0 (D) \frac{d^{2} y}{{dx}^{2}} - 1 = 0 \end{array}$

Ans.

$\begin{array}{l} y = c_{1} e^{x} + c_{2} e^{- x} . .. (i) \\ Differentiating w . r . t . x, we get \\ \frac{dy}{dx} = c_{1} e^{x} - c_{2} e^{- x} \\ Again, ‹ differentiating w . r . t . x, we get \\ \frac{d^{2} y}{{dx}^{2}} = c_{1} e^{x} + c_{2} e^{- x} \\ = y \\ or \frac{d^{2} y}{{dx}^{2}} - y = 0 \\ Thus, ‹‹ option B is correct . \end{array}$

Q.36

$\begin{array}{l} Which of the following differential equations has y = x as \\ one of its particular solution? \\ (A) \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = x (B) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + x y = x \\ (C) \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = 0 (D) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = 0 \end{array}$

Ans.

$\begin{array}{l} We have y = x … (i) \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = 1 … (ii) \\ Again, ‹ differentiating w .r .t . x, we get \\ \frac{d^{2} y}{d x^{2}} = 0 … (iii) \\ Putting values of y, \frac{d y}{d x} a n d \frac{d^{2} y}{d x^{2}} in each given options, \\ (A) L .H .S . = \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y \\ = 0 - x^{2} (1) + x (x) \\ = 0 \neq R .H .S . \\ Thus, it is not correct option . \\ (B) L .H .S . = \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + x y \\ = 0 - x (x) + x (x) \\ = 0 \neq R .H .S . \\ Thus, it is not correct option . \\ (C) L .H .S . = \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y \\ = 0 - x^{2} (1) + x (x) \\ = 0 = R .H .S . \\ Thus, it is correct option . \\ Hence, the correct solution is option C . \end{array}$

Q.37 For the differential equations, find the general solution:

$\frac{d y}{d x} = \frac{1 - c o s x}{1 + c o s x}$

Ans.

$\begin{array}{l} The given differential equation is: \\ \frac{d y}{d x} = \frac{1 - \cos x}{1 + \cos x} \\ = \frac{2 \sin^{2} \frac{x}{2}}{2 \cos^{2} \frac{x}{2}} [C o s 2 x = 1 - 2 \sin^{2} x = 2 \cos^{2} x - 1] \\ = \tan^{2} \frac{x}{2} \\ = \sec^{2} \frac{x}{2} - 1 \\ Separating the variables, we get \\ dy = (\sec^{2} \frac{x}{2} - 1) d x \\ Integrating both sides of the above equation, we get \\ \int d y = \int (\sec^{2} \frac{x}{2} - 1) d x \\ y = 2 \tan \frac{x}{2} - x + C \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.38 For the differential equations, find the general solution:

$\frac{d y}{d x} = \sqrt{4 - y^{2}} (- 2 < y < 2)$

Ans.

$\begin{array}{l} The given differential equation is: \\ \frac{d y}{d x} = \sqrt{4 - y^{2}} \\ Separating the variables, we get \\ \frac{d y}{\sqrt{4 - y^{2}}} = d x \\ Integrating both sides of above equation,we get \\ \int \frac{d y}{\sqrt{4 - y^{2}}} = \int d x \\ \sin^{- 1} (\frac{y}{2}) = x + C \\ \frac{y}{2} = \sin (x + C) \\ y = 2 \sin (x + C) \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.39 For the differential equations, find the general solution:

$\frac{d y}{d x} + y = 1 (y \neq 1)$

Ans.

$\begin{array}{l} The given differential equation is: \\ \frac{d y}{d x} + y = 1 \\ \frac{d y}{d x} = 1 - y \\ S e p a r a t i n g the variables, we get \\ \frac{d y}{1 - y} = d x \\ I n t e g r a t i n g both sides of the given differential equation, we get \\ \int \frac{d y}{1 - y} = \int d x \\ - \log (1 - y) = x + \log C \\ \Rightarrow - \log (1 - y) - \log C = x \\ \Rightarrow \log C (1 - y) = - x \\ \Rightarrow C (1 - y) = e^{- x} \\ \Rightarrow 1 - y = \frac{1}{C} e^{- x} \\ \Rightarrow y = 1 - \frac{1}{C} e^{- x} \\ = 1 - A e^{- x} [W h e r e A = \frac{1}{C}] \\ T h i s is the required general solution of the given differential \\ equation . \end{array}$

Q.40 For the differential equations, find the general solution:

$s e c^{2} x t a n y d x + s e c^{2} y t a n x d y = 0$

Ans.

$\begin{array}{l} The given differential equation is \\ {sec}^{2} x tan y {dx + sec}^{2} y tanx dy = 0 \\ Dividing by tan x tan y, we get \\ \frac{\sec^{2} x \tan y d x + \sec^{2} y \tan x d y}{tan x tan y} = 0 \\ \Rightarrow \frac{\sec^{2} x}{\tan x} d x + \frac{\sec^{2} y}{\tan y} d y = 0 \\ \Rightarrow \frac{\sec^{2} x}{\tan x} d x = - \frac{\sec^{2} y}{\tan y} d y \\ Integrating both sides of above equation, we get \\ \int \frac{\sec^{2} x}{\tan x} d x = - \int \frac{\sec^{2} y}{\tan y} d y \\ \Rightarrow log tanx = -log tany + log C \\ \Rightarrow log tanx + log tany = log C \\ \Rightarrow log tanx tany = log C \\ \Rightarrow tanx tany = C \\ This is the required solution of the given differential equation . \end{array}$

Q.41 For the differential equations, find the general solution:

$(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0$

Ans.

$\begin{array}{l} T h e given differential equation is: \\ (e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0 \\ (e^{x} - e^{- x}) d x = (e^{x} + e^{- x}) d y \\ d y = \frac{(e^{x} - e^{- x})}{(e^{x} + e^{- x})} d x \\ I n t e g r a t i n g both sides of above equation, we get \\ \int d y = \int \frac{(e^{x} - e^{- x})}{(e^{x} + e^{- x})} d x \\ y = \log (e^{x} + e^{- x}) + C \\ T h i s is the required general solution of the given differential \\ equation . \end{array}$

Q.42 For the differential equations, find the general solution:

$\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$

Ans.

$\begin{array}{l} The given differential equation is: \\ \frac{d y}{d x} = (1 + x^{2}) (1 + y^{2}) \\ \Rightarrow \frac{d y}{(1 + y^{2})} = (1 + x^{2}) d x \\ Integrating both sides, we get \\ \int \frac{d y}{(1 + y^{2})} = \int (1 + x^{2}) d x \\ {tan}^{- 1} y = x + \frac{x^{3}}{3} + C \\ This is the general solution of given differential equation . \end{array}$

Q.43 For the differential equations, find the general solution:

$y l o g y d x - x d y = 0$

Ans.

$\begin{array}{l} T h e given differential equation is: \\ y \log y d x - x d y = 0 \\ \Rightarrow y \log y d x = x d y \\ \Rightarrow \frac{d y}{y \log y} = \frac{d x}{x} \\ Integrating both sides, we get \\ \int \frac{d y}{y \log y} = \int \frac{d x}{x} \\ \Rightarrow \log \log y = \log x + \log C \\ \Rightarrow \log \log y = \log C x \\ \Rightarrow \log y = C x \\ \Rightarrow y = e^{C x} \\ This is the general solution of given differential equation . \end{array}$

Q.44 For the differential equations, find the general solution:

$x^{5} \frac{d y}{d x} = - y^{5}$

Ans.

$\begin{array}{l} The given differential equation is : \\ x^{5} \frac{dy}{dx} = - y^{5} \\ \Rightarrow \frac{dy}{y^{5}} = - \frac{dx}{x^{5}} \\ Integrating both sides, we get \\ \int \frac{dy}{y^{5}} = - \int \frac{dx}{x^{5}} \\ \Rightarrow \int y^{- 5} dy = - \int x^{- 5} dx \\ - \frac{y^{- 4}}{4} + C ‘ = \frac{x^{- 4}}{4} \\ x^{- 4} + y^{- 4} = 4 C ‘ = C \\ This is the general solution of given differential equation . \end{array}$

Q.45 For the differential equation given below, find the general solution:

$\frac{d y}{d x} = s i n^{- 1} x$

Ans.

$\begin{array}{l} The given differential equation is : \\ \frac{dy}{dx} = \sin^{- 1} x \\ \Rightarrow dy = \sin^{- 1} x dx \\ Integrating both sides, we get \\ \int dy = \int \sin^{- 1} x dx \\ \Rightarrow y = \int \sin^{- 1} x . 1 dx \\ \Rightarrow y = \sin^{- 1} x \int 1 dx - \int (\frac{d}{dx} \sin^{- 1} x \int 1 dx) dx \\ \Rightarrow y = (\sin^{- 1} x) . x - \int (\frac{1}{\sqrt{1 - x^{2}}} x) dx + C \\ \Rightarrow y = {xsin}^{- 1} x - \int \frac{x}{\sqrt{t}} \times \frac{dt}{- 2 x} + C [\begin{array}{l} Let 1 - x^{2} = t \\ \Rightarrow - 2 x = \frac{dt}{dx} \end{array}] \\ \Rightarrow y = {xsin}^{- 1} x + \frac{1}{2} \int t^{- \frac{1}{2}} dt + C \\ \Rightarrow y = {xsin}^{- 1} x + \frac{1}{2} \frac{t^{\frac{1}{2}}}{(\frac{1}{2})} + C \\ \Rightarrow y = {xsin}^{- 1} x + \sqrt{1 - x^{2}} + C \\ This is the general solution of given differential equation . \end{array}$

Q.46 For the differential equation given below, find the general solution:

$e^{x} t a n y d x + (1 - e^{x}) s e c^{2} y d y = 0$

Ans.

$\begin{array}{l} The given differential equation is: \\ e^{x} tan y dx + (1 - e^{x}) {sec}^{2} y d y = 0 \\ \frac{e^{x}}{(1 - e^{x})} d x + \frac{\sec^{2} y}{\tan y} d y = 0 [Dividing both sides by (1 - e^{x}) tany] \\ Integrating both sides of above equation, we get \\ \int \frac{e^{x}}{(1 - e^{x})} d x + \int \frac{\sec^{2} y}{\tan y} d y = C \\ \Rightarrow \int \frac{e^{x}}{t} \times \frac{d t}{- e^{x}} + \int \frac{\sec^{2} y}{z} \times \frac{d z}{\sec^{2} y} = \log C [\begin{array}{l} L e t t = 1 - e^{x} a n d z = \tan y \\ \Rightarrow \frac{d t}{d x} = - e^{x} a n d \frac{d z}{d y} = \sec^{2} y \end{array}] \\ \Rightarrow - \log t + \log z = \log C \\ \Rightarrow \log (\frac{z}{t}) = \log C \\ \Rightarrow \frac{\tan y}{1 - e^{x}} = C \\ \Rightarrow \tan y = C (1 - e^{x}) \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.47 For the differential equation given below, find a particular solution satisfying the given condition:

$(x^{3} + x^{2} + x + 1) \frac{d y}{d x} = 2 x^{2} + x; ‹ y = 1 w h e n x = 0$

Ans.

$\begin{array}{l} The given differential equation is \\ (x^{3} + x^{2} + x + 1) \frac{dy}{dx} = 2 x^{2} + x \\ \frac{dy}{dx} = \frac{2 x^{2} + x}{(x^{3} + x^{2} + x + 1)} \\ = \frac{2 x^{2} + x}{{x^{2} (x + 1) + 1 (x + 1)}} \\ dy = \frac{2 x^{2} + x}{(x^{2} + 1) (x + 1)} dx . .. (i) \\ Let \frac{2 x^{2} + x}{(x^{2} + 1) (x + 1)} = \frac{Ax + B}{(x^{2} + 1)} + \frac{C}{(x + 1)} \\ \Rightarrow 2 x^{2} + x = (Ax + B) (x + 1) + C (x^{2} + 1) \\ \Rightarrow 2 x^{2} + x = x^{2} (A + C) + x (A + B) + (B + C) \\ \Rightarrow A + C = 2, A + B = 1 and B + C = 0 \\ \Rightarrow A = \frac{3}{2}, B = - \frac{1}{2} and C = \frac{1}{2} \\ ∴ \frac{2 x^{2} + x}{(x^{2} + 1) (x + 1)} = \frac{3 x - 1}{2 (x^{2} + 1)} + \frac{1}{2 (x + 1)} \\ From equation (i), we have \\ dy = \frac{3 x - 1}{2 (x^{2} + 1)} dx + \frac{1}{2 (x + 1)} dx \\ Integrating both sides, we get \\ \int dy = \int \frac{3 x - 1}{2 (x^{2} + 1)} dx + \int \frac{1}{2 (x + 1)} dx \\ y = \frac{3}{2} \int \frac{x}{x^{2} + 1} dx - \frac{1}{2} \int \frac{1}{x^{2} + 1} dx + \frac{1}{2} \int \frac{1}{x + 1} dx + C \\ y = \frac{3}{2} \times \frac{1}{2} \log (x^{2} + 1) - \frac{1}{2} \tan^{- 1} x + \frac{1}{2} \log (x + 1) + C \\ y = \frac{1}{4} {3 \log (x^{2} + 1) + 2 \log (x + 1)} - \frac{1}{2} \tan^{- 1} x + C \\ y = \frac{1}{4} {\log {(x^{2} + 1)}^{3} {(x + 1)}^{2}} - \frac{1}{2} \tan^{- 1} x + C . .. (ii) \\ Since, y = 1 when x = 0 \\ 1 = \frac{1}{4} {\log {(0 + 1)}^{3} {(0 + 1)}^{2}} - \frac{1}{2} \tan^{- 1} 0 + C \\ 1 = C \\ Substituting C = 1 in equation (iii), we get \\ y = \frac{1}{4} {\log {(x^{2} + 1)}^{3} {(x + 1)}^{2}} - \frac{1}{2} \tan^{- 1} x + 1 \end{array}$

Q.48 For the differential equation given below, find a particular solution satisfying the given condition:

$x {(x}^{2} - 1) \frac{dy}{dx} = 1; ‹ y = 0 when x = 2$

Ans.

$\begin{array}{l} T h e given differential equation is \\ x (x^{2} - 1) \frac{d y}{d x} = 1 \\ \frac{d y}{d x} = \frac{1}{x (x^{2} - 1)} \\ \frac{d y}{d x} = \frac{1}{x (x - 1) (x + 1)} \\ d y = \frac{1}{x (x - 1) (x + 1)} d x … (i) \\ L e t \frac{1}{x (x - 1) (x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1} \\ \Rightarrow 1 = A (x^{2} - 1) + B x (x + 1) + C x (x - 1) \\ S u b s t i t u t i n g x = 0, 1 a n d - 1, we get \end{array}$ $\begin{array}{l} A = - 1, B = \frac{1}{2} and C = \frac{1}{2} \\ ∴ \frac{1}{x (x - 1) (x + 1)} = \frac{- 1}{x} + \frac{1}{2 (x - 1)} + \frac{1}{2 (x + 1)} \\ From equation (i), we have \\ dy = \frac{- 1}{x} dx + \frac{1}{2 (x - 1)} dx + \frac{1}{2 (x + 1)} dx \\ Integrating both sides, we get \\ \int dy = - \int \frac{1}{x} dx + \frac{1}{2} \int \frac{1}{x - 1} dx + \frac{1}{2} \int \frac{1}{x + 1} dx \\ \Rightarrow y = - logx + \frac{1}{2} \log (x - 1) + \frac{1}{2} \log (x + 1) + logk \\ \Rightarrow y = \frac{1}{2} \log {\frac{k^{2} (x - 1) (x + 1)}{x^{2}}} . .. (ii) \\ Now, y = 0 when x = 2 \\ 0 = \frac{1}{2} \log {\frac{k^{2} (2 - 1) (2 + 1)}{2^{2}}} \\ \Rightarrow 0 = \log {\frac{k^{2} (1) (3)}{4}} \\ \Rightarrow \log 1 = \log {\frac{3 k^{2}}{4}} \\ \Rightarrow 1 = \frac{3 k^{2}}{4} \Rightarrow k^{2} = \frac{4}{3} \\ Substituting k^{2} = \frac{4}{3} in equation (ii), we get \end{array}$ $\begin{array}{l} y = \frac{1}{2} \log {\frac{\frac{4}{3} (x - 1) (x + 1)}{x^{2}}} \\ \Rightarrow y = \frac{1}{2} \log {\frac{4 (x^{2} - 1)}{3 x^{2}}} = \frac{1}{2} \log (\frac{x^{2} - 1}{x^{2}}) - \frac{1}{2} \log (\frac{3}{4}) \end{array}$

Q.49 For the differential equation given below, find a particular solution satisfying the given condition:

$cos (\frac{dy}{dx}) = a (a \in R); ‹ y = 1 when x = 0$

Ans.

$\begin{array}{l} The given differential equation is: \\ \cos (\frac{d y}{d x}) = a \\ \frac{d y}{d x} = \cos^{- 1} a \\ d y = \cos^{- 1} a d x \\ Integrating both sides, we get \\ \int d y = \cos^{- 1} a \int d x \\ y = (\cos^{- 1} a) x + C … (i) \\ Now, y = 1 when x = 0 \\ 1 = (\cos^{- 1} a) 0 + C \\ 1 = C \\ Substituting C = 1 in equation (i), we get \\ y = (\cos^{- 1} a) x + 1 \\ \frac{y - 1}{x} = \cos^{- 1} a \Rightarrow \cos (\frac{y - 1}{x}) = a \end{array}$

Q.50 For the differential equation given below, find a particular solution satisfying the given condition:

$\frac{dy}{dx} = ytanx; ‹ y = 1 whenx = 0$

Ans.

$\begin{array}{l} The given differential equation is : \\ \frac{dy}{dx} = ytanx \\ \frac{dy}{y} = tanx dx \\ Integrating both sides, we get \\ \int \frac{dy}{y} = \int tanx dx \\ \log y = logsec x + logC \\ \Rightarrow y = secx . C . .. (i) \\ Now, y = 1 when x = 0 \\ 1 = C . \sec 0 \\ 1 = C \\ Substituting C = 1 in equation (i), we get \\ y = secx . 1 \\ \Rightarrow y = secx \end{array}$

Q.51 Find the equation of a curve passing through the point (0, 0) and whose differential equation is y’ = e^x sin x.

Ans.

$\begin{array}{l} The given differential equation is : \\ y ’ = e^{x} \sin x \\ \frac{dy}{dx} = e^{x} \sin x \end{array}$ $\begin{array}{l} Integrating both sides, we get \\ \int dy = \int e^{x} \sin x dx + C \\ y = \int e^{x} \sin x dx + C . .. (i) \\ Let I = \int e^{x} \sin x dx \\ = sinx \int e^{x} dx - \int (\frac{d}{dx} sinx \int e^{x} dx) dx \\ = sinx e^{x} - \int (cosx e^{x}) dx \\ = sinx e^{x} - {cosx \int e^{x} dx - \int (\frac{d}{dx} cosx \int e^{x} dx) dx} \\ = sinx e^{x} - {{cosxe}^{x} - \int (- sinx) e^{x} dx} \\ = sinx e^{x} - e^{x} cosx - \int e^{x} \sin x dx \\ I = sinx e^{x} - e^{x} cosx - I \\ 2 I = e^{x} (sinx - cosx) \\ I = \frac{e^{x} (sinx - cosx)}{2} \\ From equation (i), we get \\ y = \frac{e^{x} (sinx - cosx)}{2} + C . .. (i) \\ Since, curve passes through the point (0,0) . \\ ∴ 0 = \frac{e^{0} (\sin 0 - \cos 0)}{2} + C \\ 0 = \frac{1 (0 - 1)}{2} + C \Rightarrow C = \frac{1}{2} \end{array}$ $\begin{array}{l} Putting value of C in equation (ii), we get \\ y = \frac{e^{x} (sinx - cosx)}{2} + \frac{1}{2} \\ 2 y = e^{x} (sinx - cosx) + 1 \\ 2 y - 1 = e^{x} (sinx - cosx) \\ Hence, it is the required equation of the curve . \end{array}$

Q.52

$\begin{array}{l} For the differential equation xy \frac{dy}{dx} = (x +2) (y +2), find the \\ solution curve passing through the point (1, -1) . \end{array}$

Ans.

$\begin{array}{l} The given differential equation is : \\ xy \frac{dy}{dx} = (x + 2) (y + 2) \\ \frac{ydy}{(y + 2)} = \frac{(x + 2)}{x} dx \\ \frac{(y + 2 - 2) dy}{(y + 2)} = \frac{(x + 2)}{x} dx \\ 1 dy - \frac{2}{(y + 2)} dy = (1 + \frac{2}{x}) dx \\ Integrating both sides, we get \\ \int 1 dy - \int \frac{2}{(y + 2)} dy = \int 1 dx + 2 \int \frac{1}{x} dx \\ y - 2 \log (y + 2) = x + 2 logx + C \\ y - x - C = {logx}^{2} + \log {(y + 2)}^{2} \\ y - x - C = {logx}^{2} {(y + 2)}^{2} . .. (ii) \\ Since, the curve passes through (1, - 1) . \\ - 1 - 1 - C = \log {(1)}^{2} {(- 1 + 2)}^{2} \\ - 2 - C = \log {(1)}^{2} \\ \Rightarrow C = - 2 \\ Putting the value of C in equation (ii), we get \\ y - x + 2 = {logx}^{2} {(y + 2)}^{2} \\ This is the required solution of the curve . \end{array}$

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.

$\begin{array}{l} According to given condition, \\ y \frac{dy}{dx} = x \\ \Rightarrow y . dy = x . dx \\ Integrating both sides w . r . t . x, we get \\ \int y dy = \int x dx \\ \frac{y^{2}}{2} = \frac{x^{2}}{2} + C \\ y^{2} = x^{2} + 2 C \\ y^{2} - x^{2} = 2 C . .. (i) \\ The curve passes through the point (0 - 2) . \\ {(- 2)}^{2} - 0^{2} = 2 C \\ 4 = 2 C \Rightarrow C = 2 \\ Putting C = 2 in equation (i), we get \\ y^{2} - x^{2} = 4 \\ This is the required equation of the curve . \end{array}$

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.

$\begin{array}{l} Slope (m_{1}) of curve at point (x, y) = \frac{dy}{dx} \\ slope (m_{2}) of line segment joining (x, y) and (- 4, - 3) = \frac{y + 3}{x + 4} \\ According to given condition : \\ m_{1} = 2 m_{2} \\ \frac{dy}{dx} = 2 (\frac{y + 3}{x + 4}) \\ \frac{dy}{y + 3} = 2 (\frac{dx}{x + 4}) \\ Integrating both sides, we get \\ \int \frac{dy}{y + 3} = 2 \int \frac{1}{x + 4} dx \\ \log (y + 3) = 2 \log (x + 4) + logC \\ \log (y + 3) = \log {(x + 4)}^{2} C \\ y + 3 = C {(x + 4)}^{2} . .. (i) \\ This is the general equation of the curve . \\ The curve passes through the point (- 2, 1) . \\ 1 + 3 = C {(- 2 + 4)}^{2} \\ \Rightarrow C = \frac{4}{4} = 1 \\ Substituting C = 1 in equation (i), we get \\ y + 3 = 1 {(x + 4)}^{2} \\ \Rightarrow y + 3 = {(x + 4)}^{2} \\ This is the required equation of the curve . \end{array}$

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.

$\begin{array}{l} Let radius of spherical balloon be r units . \\ Volume of balloon (V) = \frac{4}{3} {πr}^{3} \\ According to condition : \\ \frac{dV}{dt} = k \\ \frac{d}{dt} (\frac{4}{3} {πr}^{3}) = k \\ \frac{4}{3} π \times 3 r^{2} \frac{dr}{dt} = k \\ 4 {πr}^{2} \frac{dr}{dt} = k \\ 4 {πr}^{2} dr = kdt \\ Integrating both sides, we get \\ 4 π \int r^{2} dr = \int k dt \\ 4 π \frac{r^{3}}{3} = kt + C . .. (i) \\ Now, r = 3 at t = 0 \\ 4 π \frac{3^{3}}{3} = k (0) + C \\ 36 π = C \\ Again, r = 6 when t = 3 from equation (i), we get \\ 4 π \frac{6^{3}}{3} = k (3) + 36 π \\ 288 π - 36 π = 3 k \\ k = \frac{252 π}{3} \\ = 84 π \\ Hence, from equation (i), we have \\ \frac{4}{3} {πr}^{3} = 84 πt + 36 π \\ r^{3} = 63 t + 27 \\ r = {(63 t + 27)}^{\frac{1}{3}} \\ Thus, the radius of balloon after t seconds is {(63 t + 27)}^{\frac{1}{3}} . \end{array}$

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 (log_e2= 0.6931).

Ans.

$\begin{array}{l} Let Principal be P, time be t years and rate of interest be r % p . a . \\ Since, principal increases continuously at the rate of r % p . a . \\ \frac{dP}{dt} = r % of P \\ = (\frac{r}{100}) P \\ \frac{dP}{P} = (\frac{r}{100}) dt \\ Integrating both sides, we get \\ \int \frac{dP}{P} = \int \frac{r}{100} dt \\ \log_{e} P = \frac{r}{100} t + C . .. (i) \\ Substituting P = 100 and t = 0 in equation (i), we get \\ \log_{e} 100 = \frac{r}{100} \times 0 + C \\ \log_{e} 100 = 0 + C \\ C = \log_{e} 100 \end{array}$ $\begin{array}{l} Again, putting P = 200 and t = 10 in equation (i), we get \\ \log_{e} 200 = \frac{r}{100} \times 10 + \log_{e} 100 \\ \log_{e} 200 - \log_{e} 100 = \frac{r}{10} \\ \log_{e} \frac{200}{100} = \frac{r}{10} \\ \log_{e} 2 = \frac{r}{10} \\ 0 .6931 = \frac{r}{10} \Rightarrow r = 6 .931 \\ Thus, the rate of interest is 6.93 % p . a . \end{array}$

Q.57 In a bank, principal increases continuously at the rate of 5% per year. An amount of â‚¹ 1000 is deposited with this bank, how much will it worth after 10 years (e^0.5 = 1.648).

Ans.

$\begin{array}{l} Let principal be P, t be time and r be the rate of interest . Then, \\ rate of principal increasing per year is : \\ \frac{dP}{dt} = (\frac{5}{100}) P \\ = \frac{1}{20} P \\ \Rightarrow \frac{dP}{P} = \frac{1}{20} dt \\ Intergrating both sides, we get \end{array}$ $\begin{array}{l} \int \frac{dP}{P} = \frac{1}{20} \int dt \\ \log_{e} P = \frac{1}{20} t + C \\ \Rightarrow P = e^{(\frac{1}{20} t + C)} \\ = e^{C} e^{\frac{1}{20} t} \\ \Rightarrow P = {Ke}^{\frac{1}{20} t} . .. (i) [Where, ‹ K = e^{C}] \\ Now, t = 0 when P = 1000 \\ 1000 = {Ke}^{\frac{1}{20} (0)} \\ 1000 = K \\ Putting the value of K in equation (i), we get \\ P = 1000 e^{\frac{1}{20} t} . .. (ii) \\ Now amount after 10 years, \\ P = 1000 e^{\frac{1}{20} (10)} \\ = 1000 e^{0 .5} \\ = 1000 \times 1 .648 \\ = 1648 \\ Hence, after 10 years the amount will be â‚¹ 1648 . \end{array}$

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.

$\begin{array}{l} Let present number of bacteria be N at any instant t . \\ Since, the rate of growth of the bacteria is proportional to N . \\ ∴ \frac{dN}{dt} αN \\ \frac{dN}{dt} = kN \\ \frac{dN}{N} = kdt \\ Integrating both sides, we get \\ \int \frac{dN}{N} = \int k dt \\ logN = kt + C . .. (i) \\ Let number of bacteria at t = 0 be N_{0}, then \\ {logN}_{0} = k (0) + C \\ \Rightarrow C = {logN}_{0} \\ From equation (i), we have \\ logN = kt + {logN}_{0} \\ logN - {logN}_{0} = kt \\ \log {\frac{N}{N}}_{0} = kt . .. (ii) \\ Since, bacteria increases 10% in 2 hours, so \\ N = (1 + \frac{10}{100}) N_{0} \\ \Rightarrow \frac{N}{N_{0}} = \frac{11}{10} \end{array}$ $\begin{array}{l} Substituting t = 2 and \frac{N}{N_{0}} = \frac{11}{10} in equation (ii), we get \\ \log \frac{11}{10} = k (2) \\ \Rightarrow k = \frac{1}{2} \log (\frac{11}{10}) \\ From equation (ii), we have \\ \log {\frac{N}{N}}_{0} = \frac{1}{2} \log (\frac{11}{10}) t \\ \Rightarrow t = \frac{\log {\frac{N}{N}}_{0}}{\frac{1}{2} \log (\frac{11}{10})} \\ = \frac{2 \log \frac{N}{N_{0}}}{\log (\frac{11}{10})} \\ So, the time taken to increase the bacteria from 100000 \\ to 200000 = \frac{2 \log \frac{200000}{100000}}{\log (\frac{11}{10})} \end{array}$ $\begin{array}{l} \Rightarrow t = \frac{2 \log 2}{\log (\frac{11}{10})} \\ Hence, bacteria increases from 100000 to 200000 in \frac{2 \log 2}{\log (\frac{11}{10})} hrs . \end{array}$

Q.59

$\begin{array}{l} The general solution of the differential equation \frac{dy}{dx} = e^{x + y} is \\ (A) e^{x} + e^{- y} = C (B) e^{x} + e^{y} = C \\ (C) e^{- x} + e^{y} = C (D) e^{- x} + e^{- y} = C \end{array}$

Ans.

$\begin{array}{l} The differential equation is \\ \frac{dy}{dx} = e^{x + y} \\ = e^{x} e^{y} \\ \frac{dy}{e^{y}} = e^{x} dx \\ \Rightarrow e^{- y} dy = e^{x} dx \\ Integrating both sides, we get \\ \int e^{- y} dy = \int e^{x} dx \\ - e^{- y} = e^{x} + C_{1} \\ e^{x} + e^{- y} = - C_{1} \\ e^{x} + e^{- y} = C [Where, C = - C_{1}] \\ Thus, option A is correct . \end{array}$

Q.60 Show that the given differential equation is homogeneous and solve it.

(x² + xy) dy = (x² + y²) dx

Ans.

$\begin{array}{l} The given differential equation is : \\ (x^{2} + xy) dy = (x^{2} + y^{2}) dx \\ \frac{dy}{dx} = \frac{(x^{2} + y^{2})}{(x^{2} + xy)} . .. (i) \\ Let F (x, y) = \frac{(x^{2} + y^{2})}{(x^{2} + xy)} \\ Now, f (λx, λy) = \frac{(λ^{2} x^{2} + λ^{2} y^{2})}{(λ^{2} x^{2} + λ^{2} xy)} \\ = \frac{λ^{2} (x^{2} + y^{2})}{λ^{2} (x^{2} + xy)} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ Substituting y = vx then \frac{dy}{dx} = v + x \frac{dv}{dx} in equation (i), we get \\ v + x \frac{dv}{dx} = \frac{(x^{2} + v^{2} x^{2})}{(x^{2} + xvx)} \\ = \frac{x^{2} (1 + v^{2})}{x^{2} (1 + v)} \\ x \frac{dv}{dx} = \frac{(1 + v^{2})}{(1 + v)} - v \\ = \frac{1 + v^{2} - v - v^{2}}{(1 + v)} \end{array}$ $\begin{array}{l} x \frac{dv}{dx} = \frac{1 - v}{1 + v} \Rightarrow \frac{1 + v}{1 - v} dv = \frac{dx}{x} \\ \Rightarrow \frac{2 - (1 - v)}{1 - v} dv = \frac{dx}{x} \Rightarrow (\frac{2}{1 - v} - 1) dv = \frac{dx}{x} \\ Integrating both sides, we get \\ \int (\frac{2}{1 - v} - 1) dv = \int \frac{dx}{x} \\ - 2 \log (1 - v) - v = logx + logk \\ v = - 2 \log (1 - v) - logx - logk \\ v = - \log {{(1 - v)}^{2} xk} \\ \frac{y}{x} = - \log {{(1 - \frac{y}{x})}^{2} xk} [Put v = \frac{y}{x}] \\ \Rightarrow {(1 - \frac{y}{x})}^{2} xk = e^{- \frac{y}{x}} \\ \Rightarrow {(\frac{x - y}{x})}^{2} xk = e^{- \frac{y}{x}} \\ \Rightarrow {(x - y)}^{2} \frac{k}{x} = e^{- \frac{y}{x}} \\ \Rightarrow {(x - y)}^{2} = \frac{x}{k} e^{- \frac{y}{x}} \\ \Rightarrow {(x - y)}^{2} = {Cxe}^{- \frac{y}{x}} [Let C = \frac{1}{k}] \\ This is the required solution of the given differential equation . \end{array}$

Q.61 Show that the given differential equation is homogeneous and solve each of them.

$y ‘ = \frac{x + y}{x}$

Ans.

$\begin{array}{l} The given differential equation is : \\ y ‘ = \frac{x + y}{x} \\ \Rightarrow \frac{dy}{dx} = \frac{x + y}{x} . .. (i) \\ Let F (x, y) = \frac{x + y}{x} \\ Now, F (λx, λy) = \frac{λx + λy}{λx} \\ = \frac{λ (x + y)}{λx} \\ = \frac{x + y}{x} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \\ we get \\ v + x \frac{dv}{dx} = \frac{x + vx}{x} \\ = \frac{x (1 + v)}{x} \\ v + x \frac{dv}{dx} = (1 + v) \\ \Rightarrow x \frac{dv}{dx} = (1 + v) - v \end{array}$ $\begin{array}{l} \Rightarrow x \frac{dv}{dx} = 1 \\ \Rightarrow dv = \frac{dx}{x} \\ Integrating both sides, we get \\ v = logx + C \\ \frac{y}{x} = logx + C \\ y = xlogx + Cx \\ This is the required solution of the given differential equation . \end{array}$

Q.62 Show that the given differential equation is homogeneous and solve it.

(x – y)dy – (x + y) dx = 0

Ans.

$\begin{array}{l} The given differential equation is : \\ (x - y) dy - (x + y) dx = 0 \\ \Rightarrow \frac{dy}{dx} = \frac{x + y}{x - y} . .. (i) \\ Let F (x, y) = \frac{x + y}{x - y} \\ Now, F (λx, λy) = \frac{λx + λy}{λx - λy} \\ = \frac{λ (x + y)}{λ (x - y)} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \end{array}$ $\begin{array}{l} we get \\ v + x \frac{dv}{dx} = \frac{x + vx}{x - vx} \\ = \frac{x (1 + v)}{x (1 - v)} \\ x \frac{dv}{dx} = \frac{1 + v - v + v^{2}}{1 - v} \\ = \frac{1 + v^{2}}{1 - v} \\ \frac{1 - v}{1 + v^{2}} dv = \frac{dx}{x} \\ (\frac{1}{1 + v^{2}} - \frac{v}{1 + v^{2}}) dv = \frac{dx}{x} \\ Integrating both sides, we get \\ \int \frac{1}{1 + v^{2}} dv - \int \frac{v}{1 + v^{2}} dv = \int \frac{1}{x} dx \\ \tan^{- 1} v - \frac{1}{2} \log (1 + v^{2}) = logx + C \\ \tan^{- 1} (\frac{y}{x}) = \frac{1}{2} \log (1 + {\frac{y}{x^{2}}}^{2}) + \frac{1}{2} . 2 logx + C \\ \tan^{- 1} (\frac{y}{x}) = \frac{1}{2} \log (\frac{x^{2} + y^{2}}{x^{2}} . x^{2}) + C \\ \tan^{- 1} (\frac{y}{x}) = \frac{1}{2} \log (x^{2} + y^{2}) + C \\ This is the required solution of the given differential equation . \end{array}$

Q.63 Show that the given differential equation is homogeneous and solve it.

(x² – y²)dx + 2xy dy = 0

Ans.

$\begin{array}{l} The given differential equation is : \\ (x^{2} - y^{2}) dx + 2 xy dy = 0 \\ \Rightarrow \frac{dy}{dx} = - \frac{x^{2} - y^{2}}{2 xy} . .. (i) \\ Let F (x, y) = - \frac{x^{2} - y^{2}}{2 xy} \\ Now, F (λx, λy) = - \frac{λ^{2} x^{2} - λ^{2} y^{2}}{2 λ^{2} xy} \\ = - \frac{λ^{2} (x^{2} - y^{2})}{λ^{2} (2 xy)} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . \\ So, the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \\ we get \\ v + x \frac{dv}{dx} = - \frac{x^{2} - v^{2} x^{2}}{2 x (vx)} \\ = - \frac{(1 - v^{2}) x^{2}}{2 x^{2} v} \\ = - \frac{(1 - v^{2})}{2 v} \\ x \frac{dv}{dx} = \frac{(v^{2} - 1)}{2 v} - v \end{array}$ $\begin{array}{l} = \frac{(v^{2} - 1) - 2 v^{2}}{2 v} \\ x \frac{dv}{dx} = \frac{- (1 + v^{2})}{2 v} \\ \frac{2 v}{(1 + v^{2})} dv = - \frac{dx}{x} \\ Integrating both sides, we get \\ \int \frac{2 v}{(1 + v^{2})} dv = - \int \frac{1}{x} dx \\ \log (1 + v^{2}) = - logx + logC \\ \Rightarrow \log (1 + v^{2}) = \log (\frac{C}{x}) \\ (1 + \frac{y^{2}}{x^{2}}) = \frac{C}{x} \\ (x^{2} + y^{2}) = Cx \\ This is the required solution of the given differential equation . \end{array}$

Q.64 Show that the given differential equation is homogeneous and solve it.

$x^{2} \frac{dy}{dx} = x^{2} - 2 y^{2} + xy$

Ans.

$\begin{array}{l} The given differential equation is : \\ x^{2} \frac{dy}{dx} = x^{2} - 2 y^{2} + xy \\ \frac{dy}{dx} = \frac{x^{2} - 2 y^{2} + xy}{x^{2}} . .. (i) \end{array}$ $\begin{array}{l} Let F (x, y) = \frac{x^{2} - 2 y^{2} + x y}{x^{2}} \\ Now, F (λ x, λ y) = \frac{λ^{2} x^{2} - 2 λ^{2} y^{2} + λ^{2} x y}{λ^{2} x^{2}} \\ = \frac{λ^{2} (x^{2} - 2 y^{2} + x y)}{λ^{2} (x^{2})} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = v x and \frac{d y}{d x} = v + x \frac{d v}{d x}, \\ we get \\ v + x \frac{d v}{d x} = \frac{x^{2} - 2 v^{2} x^{2} + x vx}{x^{2}} \\ = \frac{(1 - 2 v^{2} + v) x^{2}}{x^{2}} \\ = 1 - 2 v^{2} + v \\ x \frac{d v}{d x} = 1 - 2 v^{2} + v - v \\ x \frac{d v}{d x} = 1 - 2 v^{2} \\ \frac{d v}{1 - 2 v^{2}} = \frac{d x}{x} \\ Integrating both sides, we get \end{array}$ $\begin{array}{l} \frac{1}{2} \int \frac{1}{{(\frac{1}{\sqrt{2}})}^{2} {-v}^{2}} dv= \int \frac{dx}{x} \\ \frac{1}{2} × \frac{1}{2× \frac{1}{\sqrt{2}}} log (\frac{\frac{1}{\sqrt{2}} +v}{\frac{1}{\sqrt{2}} -v}) =logx+C \\ \frac{1}{2 \sqrt{2}} log (\frac{\frac{1}{\sqrt{2}} + \frac{y}{x}}{\frac{1}{\sqrt{2}} – \frac{y}{x}}) =logx+C \\ \frac{1}{2 \sqrt{2}} log (\frac{x+ \sqrt{2} y}{x- \sqrt{2} y}) =logx+C \\ This is the required solution of the given differential equation . \end{array}$

Q.65 Show that the given differential equation is homogeneous and solve it.

$x d y - y d x = \sqrt{x^{2} + y^{2}} d x$

Ans.

$\begin{array}{l} The given differential equation is : \\ xdy - ydx = \sqrt{x^{2} + y^{2}} dx \\ xdy = \sqrt{x^{2} + y^{2}} dx + ydx \\ \frac{dy}{dx} = \frac{\sqrt{x^{2} + y^{2}} + y}{x} . .. (i) \\ Let F (x, y) = \frac{\sqrt{x^{2} + y^{2}} + y}{x} \\ Now, F (λx, λy) = \frac{\sqrt{λ^{2} x^{2} + λ^{2} y^{2}} + λy}{λx} \end{array}$ $\begin{array}{l} = \frac{λ (\sqrt{x^{2} + y^{2}} + y)}{λx} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \\ we get \\ v + x \frac{dv}{dx} = \frac{\sqrt{x^{2} + v^{2} x^{2}} + vx}{x} \\ = \frac{(\sqrt{1 + v^{2}} + v) x}{x} \\ v + x \frac{dv}{dx} = \sqrt{1 + v^{2}} + v \\ x \frac{dv}{dx} = \sqrt{1 + v^{2}} \\ \frac{dv}{\sqrt{1 + v^{2}}} = \frac{dx}{x} \\ Integrating both sides, we get \\ \log | v + \sqrt{1 + v^{2}} | = \log | x | + logC \\ \log | \frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} | = logC | x | \\ | \frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} | = C | x | \end{array}$ $\begin{array}{l} | \frac{y}{x} + \sqrt{1 + \frac{y^{2}}{x^{2}}} | = C | x | \\ y + \sqrt{x^{2} + y^{2}} = {Cx}^{2} \\ This is the required solution of given differential equation . \end{array}$

Q.66 Show that the given differential equation is homogeneous and solve each of them.

$\{xcos (\frac{y}{x}) + ysin (\frac{y}{x})\} ydx = \{ysin (\frac{y}{x}) - xcos (\frac{y}{x})\} xdy$

Ans.

$\begin{array}{l} The given differential equation is : \\ {xcos (\frac{y}{x}) + ysin (\frac{y}{x})} ydx = {ysin (\frac{y}{x}) - xcos (\frac{y}{x})} xdy \\ \frac{dy}{dx} = \frac{{xcos (\frac{y}{x}) + ysin (\frac{y}{x})} y}{{ysin (\frac{y}{x}) - xcos (\frac{y}{x})} x} . .. (i) \\ Let F (x, y) = \frac{{xcos (\frac{y}{x}) + ysin (\frac{y}{x})} y}{{ysin (\frac{y}{x}) - xcos (\frac{y}{x})} x} \\ Now, F (λx, λy) = \frac{{λxcos (\frac{λy}{λx}) + λysin (\frac{λy}{λx})} λy}{{λysin (\frac{λy}{λx}) - λxcos (\frac{λy}{λx})} λx} \end{array}$ $\begin{array}{l} = \frac{λ^{2} y {x \cos (\frac{y}{x}) + y \sin (\frac{y}{x})}}{λ^{2} x {y \sin (\frac{y}{x}) - x \cos (\frac{y}{x})}} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = vx and \frac{d y}{d x} = v + x \frac{d v}{d x}, \\ we get \\ v + x \frac{d v}{d x} = \frac{vx {x \cos (\frac{vx}{x}) + vx \sin (\frac{vx}{x})}}{x {vx \sin (\frac{vx}{x}) - x \cos (\frac{vx}{x})}} \\ = \frac{v {\cos (v) + v \sin (v)}}{{v \sin (v) - \cos (v)}} \\ x \frac{d v}{d x} = \frac{v \cos v + v^{2} \sin v}{v \sin v - \cos v} - v \\ x \frac{d v}{d x} = \frac{v \cos v + v^{2} \sin v - v^{2} \sin v + v \cos v}{v \sin v - \cos v} \\ x \frac{d v}{d x} = \frac{2 v \cos v}{v \sin v - \cos v} \\ (v \sin v - \cos v) \frac{d v}{v \cos v} = \frac{2 d x}{x} \\ Integrating both sides, we get \\ \int \frac{v \sin v}{v \cos v} d v - \int \frac{\cos v}{v \cos v} d v = 2 \int \frac{1}{x} d x \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \log \sec v - \log | v | = 2 \log | x | + \log C \\ \Rightarrow \log (\frac{\sec v}{v}) = \log C x^{2} \\ \Rightarrow \frac{\sec \frac{y}{x}}{\frac{y}{x}} = C x^{2} \\ \Rightarrow x \sec \frac{y}{x} = C x^{2} y \\ \Rightarrow x y \cos (\frac{y}{x}) = \frac{1}{C} \\ = k [Let k = \frac{1}{C}] \\ \Rightarrow x y \cos (\frac{y}{x}) = k \\ This is the required solution of the given differential equation . \end{array}$

Q.67 Show that the given differential equation is homogeneous and solve each of them.

$x \frac{dy}{dx} - y + xsin (\frac{y}{x}) = 0$

Ans.

$\begin{array}{l} The given differential equation is : \\ x \frac{dy}{dx} - y + xsin (\frac{y}{x}) = 0 \\ \frac{dy}{dx} = \frac{y - xsin (\frac{y}{x})}{x} . .. (i) \end{array}$ $\begin{array}{l} Let F (x, y) = \frac{y - x \sin (\frac{y}{x})}{x} \\ Now, F (λ x, λ y) = \frac{λ y - λ x \sin (\frac{λ y}{λ x})}{λ x} \\ = \frac{λ {y-xsin (\frac{y}{x})}}{λx} {=λ}^{0} F (x,y) \\ Therefore, F (x,y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = v x and \frac{d y}{d x} = v + x \frac{d v}{d x}, \\ we get \\ v + x \frac{d v}{d x} = \frac{vx - x \sin (\frac{vx}{x})}{x} \\ = \frac{{v - \sin v} x}{x} \\ \Rightarrow x \frac{d v}{d x} = v - \sin v - v \\ \Rightarrow \frac{d v}{\sin v} = - \frac{d x}{x} \\ \Rightarrow cosec v d v = - \frac{d x}{x} \\ Integrating both sides, we get \end{array}$ $\begin{array}{l} \int cosec v d v = - \int \frac{1}{x} d x \\ \Rightarrow \log | cosec v - \cot v | = - \log | x | + \log C \\ \Rightarrow \log | cosec v - \cot v | = \log \frac{C}{| x |} \\ \Rightarrow cosec \frac{y}{x} - \cot \frac{y}{x} = \frac{C}{x} \\ \Rightarrow \frac{1}{\sin \frac{y}{x}} - \frac{\cos \frac{y}{x}}{\sin \frac{y}{x}} = \frac{C}{x} \\ \Rightarrow x {1 - \cos (\frac{y}{x})} = C s i n (\frac{y}{x}) \\ This is the required solution of the given differential equation . \end{array}$

Q.68 Show that the given differential equation is homogeneous and solve it.

$y dx + x \log (\frac{y}{x}) dy - 2 x dy = 0$

Ans.

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

Q.69 Show that the given differential equation is homogeneous and solve each of them.

${(1+e}^{\frac{x}{y}}) dx + e^{\frac{x}{y}} (1 - \frac{x}{y}) dy = 0$

Ans.

$\begin{array}{l} The given differential equation is : \\ (1 + e^{\frac{x}{y}}) dx + e^{\frac{x}{y}} (1 - \frac{x}{y}) dy = 0 \end{array}$ $\begin{array}{l} \frac{dy}{dx} = - \frac{(1 + e^{\frac{x}{y}})}{e^{\frac{x}{y}} (1 - \frac{x}{y})} \\ \frac{dx}{dy} = - \frac{e^{\frac{x}{y}} (1 - \frac{x}{y})}{(1 + e^{\frac{x}{y}})} . .. (i) \\ Let F (x, y) = - \frac{e^{\frac{x}{y}} (1 - \frac{x}{y})}{(1 + e^{\frac{x}{y}})} \\ Now, F (λx, λy) = - \frac{e^{\frac{λx}{λy}} (1 - \frac{λx}{λy})}{(1 + e^{\frac{λx}{λy}})} \\ = - \frac{e^{\frac{x}{y}} (1 - \frac{x}{y})}{(1 + e^{\frac{x}{y}})} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} For solving equation (i), substitute x = v y and \frac{d x}{d y} = v + y \frac{d v}{d y}, \\ we get \\ v + y \frac{d v}{d y} = - \frac{e^{\frac{vy}{y}} (1 - \frac{vy}{y})}{(1 + e^{\frac{vy}{y}})} \\ = - \frac{e^{v} (1 - v)}{(1 + e^{v})} \\ y \frac{d v}{d y} = - \frac{e^{v} (1 - v)}{(1 + e^{v})} - v \\ = \frac{v e^{v} - e^{v} - v - v e^{v}}{(1 + e^{v})} \\ = - \frac{(e^{v} + v)}{(1 + e^{v})} \\ \frac{(1 + e^{v})}{(e^{v} + v)} d v = - \frac{d y}{y} \\ Integrating both sides, we get \\ \int \frac{(1 + e^{v})}{(e^{v} + v)} d v = - \int \frac{1}{y} d y \\ \Rightarrow \log | v + e^{v} | = - \log y + \log C \end{array}$ $\begin{array}{l} \Rightarrow \log | \frac{x}{y} + e^{\frac{x}{y}} | = \log \frac{C}{y} \\ \Rightarrow \frac{x}{y} + e^{\frac{x}{y}} = \frac{C}{y} \\ \Rightarrow x + y e^{\frac{x}{y}} = C \\ This is the required solution of the given differential equation . \end{array}$

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.

$\begin{array}{l} The given differential equation is : \\ (x + y) dy + (x - y) dx = 0 \\ (x + y) dy = - (x - y) dx \\ \frac{dy}{dx} = \frac{(y - x)}{x + y} . .. (i) \\ Let F (x, y) = \frac{(y - x)}{x + y} \\ Now, F (λx, λy) = \frac{(λy - λx)}{λx + λy} \\ = \frac{λ (y - x)}{λ (x + y)} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \end{array}$ $\begin{array}{l} equation . \\ For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \\ we get \\ v + x \frac{dv}{dx} = \frac{vx - x}{x + vx} \\ = \frac{(v - 1) x}{(1 + v) x} \\ x \frac{dv}{dx} = \frac{(v - 1)}{(1 + v)} - v \\ = \frac{v - 1 - v - v^{2}}{(1 + v)} \\ x \frac{dv}{dx} = \frac{- (1 + v^{2})}{(1 + v)} \\ \frac{(1 + v)}{(1 + v^{2})} dv = - \frac{dx}{x} \\ Integrating both sides, we get \\ \int \frac{1}{1 + v^{2}} dv + \int \frac{v}{1 + v^{2}} dv = - \int \frac{1}{x} dx \\ \tan^{- 1} v + \frac{1}{2} \log (1 + v^{2}) = - logx + k \\ 2 \tan^{- 1} v + \log (1 + v^{2}) = - 2 logx + 2 k \\ 2 \tan^{- 1} v + \log (1 + v^{2}) + 2 logx = 2 k \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} 2 \tan^{- 1} \frac{y}{x} + \log (1 + {\frac{y}{x^{2}}}^{2}) x^{2} = 2 k \\ 2 \tan^{- 1} \frac{y}{x} + \log (x^{2} + y^{2}) = 2 k … (i i) \\ Now, putting y = 1 a t x = 1. \\ 2 \tan^{- 1} \frac{1}{1} + \log (1^{2} + 1^{2}) = 2 k \\ 2 \tan^{- 1} 1 + \log (2) = 2 k \\ 2 (\frac{π}{4}) + \log 2 = 2 k \\ \Rightarrow (\frac{π}{2}) + \log 2 = 2 k \\ Substituting the value of 2k in equation (ii), we get \\ 2 \tan^{- 1} \frac{y}{x} + \log (x^{2} + y^{2}) = (\frac{π}{2}) + \log 2 \\ This is the required solution of the given differential equation . \end{array}$

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.

$\begin{array}{l} The given differential equation is : \\ x^{2} dy + (xy + y^{2}) dx = 0 \\ x^{2} dy = - (xy + y^{2}) dx \\ \frac{dy}{dx} = - \frac{(xy + y^{2})}{x^{2}} . .. (i) \\ Let F (x, y) = - \frac{(xy + y^{2})}{x^{2}} \end{array}$ $\begin{array}{l} Now, F (λx, λy) = - \frac{(λ^{2} xy + λ^{2} y^{2})}{λ^{2} x^{2}} \\ = - \frac{λ^{2} (xy + y^{2})}{λ^{2} x^{2}} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \\ we get \\ v + x \frac{dv}{dx} = - \frac{(xvx + v^{2} x^{2})}{x^{2}} \\ = - \frac{(v + v^{2}) x^{2}}{x^{2}} \\ x \frac{dv}{dx} = - v - v^{2} - v \\ = - v^{2} - 2 v \\ \frac{dv}{v (v + 2)} = - \frac{dx}{x} \\ \frac{1}{2} \frac{{(v + 2) - v} dv}{v (v + 2)} = - \frac{dx}{x} \\ \frac{1}{2} (\frac{1}{v} - \frac{1}{v + 2}) dv = - \frac{dx}{x} \\ Integrating both sides, we get \\ \frac{1}{2} \int \frac{1}{v} dv - \frac{1}{2} \int \frac{1}{v + 2} dv = - \int \frac{1}{x} dx \end{array}$ $\begin{array}{l} \frac{1}{2} \log | v | - \frac{1}{2} \log | v + 2 | = - \log | x | + logC \\ \log (\frac{v}{v + 2}) = 2 \log (\frac{C}{x}) \\ \Rightarrow \log (\frac{\frac{y}{x}}{\frac{y}{x} + 2}) = \log {(\frac{C}{x})}^{2} \\ \Rightarrow \log (\frac{y}{y + 2 x}) = \log {(\frac{C}{x})}^{2} \\ \Rightarrow \frac{y}{y + 2 x} = {(\frac{C}{x})}^{2} \\ \Rightarrow \frac{x^{2} y}{y + 2 x} = C^{2} . .. (ii) \\ Now, y = 1 when x = 1 \\ \frac{1^{2} (1)}{1 + 2 (1)} = C^{2} \\ \Rightarrow \frac{1}{3} = C^{2} \\ Substituting value of C^{2} in equation (ii), we get \\ \frac{x^{2} y}{y + 2 x} = \frac{1}{3} \\ \Rightarrow 3 x^{2} y = y + 2 x \\ \Rightarrow y + 2 x = 3 x^{2} y \\ This is the required solution of the given differential equation . \end{array}$

Q.72 For the differential equations, find the particular situation satisfying the given condition:

${[xsin}^{2} (\frac{y}{x}) - y] dx + xdy = 0; y = \frac{π}{} when x = 1$

Ans.

$\begin{array}{l} The given differential equation is : \\ [{xsin}^{2} (\frac{y}{x}) - y] dx + xdy = 0 \\ xdy = - [{xsin}^{2} (\frac{y}{x}) - y] dx \\ \frac{dy}{dx} = - \frac{[{xsin}^{2} (\frac{y}{x}) - y]}{x} . .. (i) \\ Let F (x, y) = - \frac{[{xsin}^{2} (\frac{y}{x}) - y]}{x} \\ Now, F (λx, λy) = - \frac{[{λxsin}^{2} (\frac{λy}{λx}) - λy]}{λx} \\ = - \frac{λ [{xsin}^{2} (\frac{y}{x}) - y]}{λx} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \end{array}$ $\begin{array}{l} For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \\ we get \\ v + x \frac{dv}{dx} = - \frac{[{xsin}^{2} (\frac{vx}{x}) - vx]}{x} \\ x \frac{dv}{dx} = - \sin^{2} v + v - v \\ {cosec}^{2} v dv = - \frac{dx}{x} \\ Integrating both sides, we get \\ \int {cosec}^{2} v dv = - \int \frac{1}{x} dx \\ - cotv = - \log | x | + logC \\ \Rightarrow \cot \frac{y}{x} = \log | Cx | . .. (ii) \\ Now, y = \frac{π}{4} at x = 1 \\ \Rightarrow \cot \frac{(\frac{π}{4})}{1} = \log | C . 1 | \\ \Rightarrow 1 = \log | C | \Rightarrow C = e \\ Substituting value of C in equation (ii), we get \\ \cot \frac{y}{x} = \log | ex | \\ This is the required solution of the given differential equation . \end{array}$

Q.73 For the differential equations, find the particular situation satisfying the given condition:

$\frac{dy}{dx} - \frac{y}{x} + cosec (\frac{y}{x}) = 0; y = 0 when x = 1$

Ans.

$\begin{array}{l} The given differential equation is : \\ \frac{dy}{dx} - \frac{y}{x} + cosec (\frac{y}{x}) = 0 \\ \frac{dy}{dx} = \frac{y}{x} - cosec (\frac{y}{x}) . .. (i) \\ Let F (x, y) = \frac{y}{x} - cosec (\frac{y}{x}) \\ Now, F (λx, λy) = \frac{λy}{λx} - cosec (\frac{λy}{λx}) \\ = \frac{y}{x} - cosec (\frac{y}{x}) = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \\ we get \\ v + x \frac{dv}{dx} = \frac{vx}{x} - cosec (\frac{vx}{x}) \\ \Rightarrow x \frac{dv}{dx} = v - cosecv - v \\ \Rightarrow \frac{dv}{cosecv} = - \frac{dx}{x} \\ Integrating both sides, we get \\ - \int sinv dv = \int \frac{1}{x} dx \\ cosv = \log | x | + logC \\ \Rightarrow \cos (\frac{y}{x}) = \log | Cx | . .. (ii) \end{array}$ $\begin{array}{l} Now, y = 0 at x = 1 \\ \cos (\frac{0}{1}) = \log | C . 1 | \\ \Rightarrow \cos 0 = logC \\ \Rightarrow 1 = logC \\ \Rightarrow C = e \\ Putting C = e in equation (ii), we get \\ \cos (\frac{y}{x}) = \log | ex | \\ This is the required solution of the given differential equation . \end{array}$

Q.74 For the differential equation given below, find a particular situation satisfying the given condition:

${2xy+y}^{2} - 2 x^{2} \frac{dy}{dx} = 0; ‹ y = 2 whenx = 1$

Ans.

$\begin{array}{l} The given differential equation is : \\ 2 xy + y^{2} - 2 x^{2} \frac{dy}{dx} = 0 \\ - 2 x^{2} \frac{dy}{dx} = - (2 xy + y^{2}) \\ \frac{dy}{dx} = \frac{(2 xy + y^{2})}{2 x^{2}} . .. (i) \\ Let F (x, y) = \frac{(2 xy + y^{2})}{2 x^{2}} \\ Now, F (λx, λy) = \frac{(λ^{2} 2 xy + λ^{2} y^{2})}{2 λ^{2} x^{2}} \\ = \frac{λ^{2} (2 xy + y^{2})}{2 λ^{2} x^{2}} = λ^{0} F (x, y) \end{array}$ $\begin{array}{l} Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \\ we get \\ v + x \frac{dv}{dx} = \frac{(2 xvx + v^{2} x^{2})}{2 x^{2}} \\ = \frac{(2 v + v^{2}) x^{2}}{2 x^{2}} \\ x \frac{dv}{dx} = \frac{2 v + v^{2}}{2} - v \\ = \frac{v^{2}}{2} \\ \frac{dv}{v^{2}} = \frac{dx}{2 x} \\ Integrating both sides, we get \\ \int 2 v^{- 2} dv = \int \frac{1}{x} dx \\ \frac{2 v^{- 1}}{- 1} = \log | x | + C \\ - \frac{2}{v} = \log | x | + C \\ - \frac{2}{(\frac{y}{x})} = \log | x | + C \end{array}$ $\begin{array}{l} - \frac{2 x}{y} = \log | x | + C . .. (ii) \\ Now, y = 2 at x = 1. \\ - \frac{2 (1)}{2} = \log | 1 | + C \Rightarrow C = - 1 \\ Substituting C = -1 in equation (ii), we get \\ - \frac{2 x}{y} = \log | x | - 1 \\ \Rightarrow y = \frac{2 x}{1 - \log | x |} (x \neq 0, x \neq e) \\ This is the required solution of the given differential equation . \end{array}$

Q.75

$\begin{array}{l} A homogeneous differential equation of the form \\ \frac{dy}{dx} = h (\frac{x}{y}) can be solved by making the substitution . \\ (A) y = vx (B) v = yx (C) x = vy (D) x = v \end{array}$

Ans.

$\begin{array}{l} A homogeneous differential equation of the form \\ \frac{dy}{dx} = h (\frac{x}{y}) can be solved by making the substitution \\ as x = vy . \\ Hence, the correct option is C . \end{array}$

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

(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)= λⁿF(x, y) for any nonzero constant λ.

$\begin{array}{l} Let us check option D, \\ y^{2} dx + (x^{2} - xy - y^{2}) dy = 0 \\ (x^{2} - xy - y^{2}) dy = - y^{2} dx \\ \frac{dy}{dx} = - \frac{y^{2}}{(x^{2} - xy - y^{2})} \\ Let F (x, y) = - \frac{y^{2}}{(x^{2} - xy - y^{2})} \\ then F (λx, λy) = - \frac{λ^{2} y^{2}}{(λ^{2} x^{2} - λ^{2} xy - λ^{2} y^{2})} \\ = - λ^{2} \frac{λ^{2} y^{2}}{(x^{2} - xy - y^{2})} = λ^{0} F (x, y) \\ Thus, the given differential equation in option D is \\ homogenous equation . \end{array}$

Q.77 Find the general solution of the differential equation given below.

$\frac{dy}{dx} + 2 y = sinx$

Ans.

$\begin{array}{l} The given differential equation is \frac{dy}{dx} + 2 y = sinx \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = 2 andQ = sinx \\ So, I . F . = e^{\int Pdx} \\ = e^{\int 2 dx} \\ = e^{2 x} \\ The solution of given differential equation is given by the relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ {ye}^{2 x} = \int (sinx \times e^{2 x}) dx + C . .. (i) \\ Let I = \int {sinxe}^{2 x} dx \\ = sinx \int e^{2 x} dx - \int (\frac{d}{dx} sinx \int e^{2 x} dx) dx \\ = sinx [\frac{e^{2 x}}{2}] - \int cosx . \frac{e^{2 x}}{2} dx \\ = \frac{1}{2} sinx e^{2 x} - \frac{1}{2} \int cosx . e^{2 x} dx \\ = \frac{1}{2} sinx e^{2 x} - \frac{1}{2} {cosx \int e^{2 x} dx - \int (\frac{d}{dx} cosx \int e^{2 x} dx) dx} \\ = \frac{1}{2} sinx e^{2 x} - \frac{1}{2} {cosx [\frac{e^{2 x}}{2}] - \int (- sinx \frac{e^{2 x}}{2}) dx} \end{array}$ $\begin{array}{l} I = \frac{1}{2} s i n x e^{2 x} - \frac{1}{4} (c o s x e^{2 x} + I) \\ I = \frac{1}{2} s i n x e^{2 x} - \frac{1}{4} c o s x e^{2 x} - \frac{1}{4} I \\ \frac{5}{4} I = \frac{1}{4} e^{2 x} (2 s i n x - c o s x) \\ I = \frac{1}{5} e^{2 x} (2 s i n x - c o s x) \\ So, from equation (i), we have \\ y e^{2 x} = \frac{1}{5} e^{2 x} (2 s i n x - c o s x) + C \\ \Rightarrow y = \frac{1}{5} (2 s i n x - c o s x) + C e^{- 2 x} \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.78 Find the general solution of the differential equation given below.

$\frac{dy}{dx} + 3 y = e^{- 2 x}$

Ans.

$\begin{array}{l} The given differential equation is \frac{dy}{dx} + 3 y = e^{- 2 x} \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = 3 and Q = e^{- 2 x} \\ So, I . F . = e^{\int Pdx} \\ = e^{\int 3 dx} \\ = e^{3 x} \end{array}$ $\begin{array}{l} The solution of given differential equation is given by the relation, \\ y (I . F .) = \int (Q \times I . F .) d x + C \\ y e^{3 x} = \int (e^{- 2 x} \times e^{3 x}) d x + C \\ = \int e^{x} d x + C \\ y e^{3 x} = e^{x} + C \\ y = e^{- 2 x} + C e^{- 3 x} \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.79 Find the general solution of the differential equation given below.

$\frac{d y}{d x} + \frac{y}{x} = x^{2}$

Ans.

$\begin{array}{l} The given differential equation is \frac{dy}{dx} + \frac{y}{x} = x^{2} \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = \frac{1}{x} and Q = x^{2} \\ So, I . F . = e^{\int Pdx} \\ = e^{\int \frac{1}{x} dx} \\ = e^{logx} \\ = x \\ The solution of given differential equation is given by the relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ yx = \int (x^{2} \times x) dx + C \end{array}$ $\begin{array}{l} = \int x^{3} d x + C \\ x y = \frac{x^{4}}{4} + C \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.80 Find the general solution of the differential equation given below.

$\frac{dy}{dx} + (secx) y = tanx (0 \leq x < \frac{π}{2})$

Ans.

$\begin{array}{l} The given differential equation is \frac{dy}{dx} + (secx) y = tanx (0 \leq x < \frac{π}{2}) \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = secx and Q = tanx \\ So, I . F . = e^{\int Pdx} \\ = e^{\int secx dx} \\ = e^{\log (secx + tanx)} \\ = secx + tanx \\ The solution of given differential equation is given by the relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ y (secx + tanx) = \int (secx + tanx) tanx dx + C \\ y (secx + tanx) = \int secxtanx dx + \int \tan^{2} x dx + C \\ = secx + \int (\sec^{2} x - 1) dx + C \\ = secx + tanx - x + C \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.81 Find the general solution of the differential equation given below.

${cos}^{2} x \frac{dy}{dx} + y = \tan x (0 \leq x < \frac{π}{2})$

Ans.

$\begin{array}{l} The given differential equation is \cos^{2} x \frac{dy}{dx} + y = \tan x (0 \leq x < \frac{π}{2}) \\ \Rightarrow \frac{dy}{dx} + (\sec^{2} x) y = \sec^{2} xtan x \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = \sec^{2} x and Q = \sec^{2} xtan x \\ So, I . F . = e^{\int Pdx} \\ = e^{\int \sec^{2} x dx} \\ = e^{tanx} \\ The solution of given differential equation is given by the relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ {ye}^{tanx} = \int (\sec^{2} x tanx) e^{tanx} dx + C \\ {ye}^{tanx} = \int {te}^{t} \sec^{2} x \frac{dt}{\sec^{2} x} + C [\begin{array}{l} Let t = tanx \\ \Rightarrow \frac{dt}{dx} = \sec^{2} x \end{array}] \\ {ye}^{tanx} = \int {te}^{t} dt + C \\ = t \int e^{t} dt - \int (\frac{d}{dt} t \int e^{t} dt) dt + C \end{array}$ $\begin{array}{l} = {te}^{t} - \int 1 . e^{t} dt + C \\ = {te}^{t} - e^{t} + C \\ = e^{t} (t - 1) + C \\ {ye}^{tanx} = e^{tanx} (tanx - 1) + C \\ y = (tanx - 1) + {Ce}^{- tanx} \end{array}$ $\begin{array}{l} This is the required general solution of the given differential \\ equation . \end{array}$

Q.82 Find the general solution of the differential equation given below.

$x \frac{dy}{dx} + 2 y = x^{2} logx$

Ans.

$\begin{array}{l} The given differential equation is x \frac{dy}{dx} + 2 y = x^{2} logx \\ \Rightarrow \frac{dy}{dx} + 2 \frac{y}{x} = x logx \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = \frac{2}{x} and Q = x logx \\ So, I . F . = e^{\int Pdx} \\ = e^{\int \frac{2}{x} dx} \\ = {e^{2}}^{logx} \\ = e^{^{{logx}^{2}}} \\ = x^{2} \end{array}$ $\begin{array}{l} The solution of given differential equation is given by the relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ {yx}^{2} = \int x logx . x^{2} dx + C \\ {yx}^{2} = \int x^{3} logx dx + C \end{array}$ $\begin{array}{l} = logx \int x^{3} dx - \int (\frac{d}{dx} logx \int x^{3} dx) dx + C \\ = logx . [\frac{x^{4}}{4}] - \int (\frac{1}{x} [\frac{x^{4}}{4}]) dx + C \\ = logx . [\frac{x^{4}}{4}] - \frac{1}{4} \int x^{3} dx + C \\ = logx . [\frac{x^{4}}{4}] - \frac{1}{4} [\frac{x^{4}}{4}] + C \\ {yx}^{2} = \frac{x^{4}}{16} (4 logx - 1) + C \\ y = \frac{x^{2}}{16} (4 logx - 1) + {Cx}^{- 2} \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.83 Find the general solution of the differential equation given below.

$xlogx \frac{dy}{dx} + y = \frac{}{x} logx$

Ans.

$The given differential equation is xlogx \frac{dy}{dx} + y = \frac{2}{x} logx$ $\begin{array}{l} \Rightarrow \frac{dy}{dx} + \frac{y}{xlogx} = \frac{2}{x^{2}} \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = \frac{1}{xlogx} and Q = \frac{2}{x^{2}} \\ So, I . F . = e^{\int Pdx} \\ = e^{\int \frac{1}{xlogx} dx} \\ = e^{\log (logx)} \\ = logx \\ The solution of given differential equation is given by the \\ relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ ylogx = \int (\frac{2}{x^{2}} \times logx) dx + C \\ ylogx = logx \int \frac{2}{x^{2}} dx - \int (\frac{d}{dx} logx \int \frac{2}{x^{2}} dx) dx + C \\ ylogx = 2 {logx (- \frac{1}{x}) - \int \frac{1}{x} \times - \frac{1}{x} dx} + C \\ ylogx = 2 {- \frac{1}{x} logx + \int x^{- 2} dx} + C \\ ylogx = 2 (- \frac{1}{x} logx + \frac{x^{- 1}}{- 1}) + C \end{array}$ $\begin{array}{l} ylogx = - \frac{2}{x} logx - \frac{2}{x} + C \\ ylogx = - \frac{2}{x} (logx + 1) + C \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.84 Find the general solution of the differential equation given below.

${(1+x}^{2}) dy + 2 xydx = cotx dx (x \neq 0)$

Ans.

$\begin{array}{l} The given differential equation is \\ (1 + x^{2}) dy + 2 xydx = cotxdx (x \neq 0) \\ \Rightarrow \frac{dy}{dx} + \frac{2 x}{(1 + x^{2})} y = \frac{cotx}{(1 + x^{2})} \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = \frac{2 x}{(1 + x^{2})} and Q = \frac{cotx}{(1 + x^{2})} \\ So, I . F . = e^{\int Pdx} \\ = e^{\int \frac{2 x}{(1 + x^{2})} dx} \\ = e^{\log (1 + x^{2})} \\ = (1 + x^{2}) \\ The solution of given differential equation is given by the \end{array}$

relation,

$\begin{array}{l} y (I . F .) = \int (Q \times I . F .) d x + C \\ y (1 + x^{2}) = \int {\frac{\cot x}{(1 + x^{2})} \times (1 + x^{2})} d x + C \\ y (1 + x^{2}) = \int \cot x d x + C \\ y (1 + x^{2}) = \log | \sin x | + C \\ y = {(1 + x^{2})}^{- 1} \log | \sin x | + C {(1 + x^{2})}^{- 1} \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.85 Find the general solution of the differential equation given below.

$x \frac{d y}{d x} + y - x + x y c o t x = 0$

Ans.

$\begin{array}{l} The given differential equation is \\ x \frac{dy}{dx} + y - x + xycotx = 0 \\ \Rightarrow \frac{dy}{dx} + \frac{(1 + xcotx)}{x} y = 1 \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = \frac{(1 + xcotx)}{x} and Q = 1 \\ So, I . F . = e^{\int Pdx} \\ = e^{\int \frac{(1 + xcotx)}{x} dx} \end{array}$ $\begin{array}{l} = e^{\int (\frac{1}{x} + cotx) dx} \\ = e^{\log | x | + \log | sinx |} \\ = {e^{\log}}^{| xsinx |} \\ = xsinx \\ The solution of given differential equation is given by the \\ relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ y (xsinx) = \int {1 \times (xsinx)} dx + C \\ y (xsinx) = \int xsinx dx + C \\ y (xsinx) = x \int sinx dx - \int (\frac{d}{dx} x \int sinx dx) dx + C \\ y (xsinx) = - xcosx - \int (- cosx) dx + C \\ y (xsinx) = - xcosx + sinx + C \\ \Rightarrow y = \frac{1}{x} - cotx + \frac{C}{x} cosecx \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.86 Find the general solution of the differential equation given below.

$(x+y) \frac{dy}{dx} = 1$

Ans.

$\begin{array}{l} The given differential equation is (x + y) \frac{dy}{dx} = 1 \\ \Rightarrow \frac{dy}{dx} = \frac{1}{(x + y)} \end{array}$ $\begin{array}{l} \Rightarrow \frac{dx}{dy} = (x + y) \\ \Rightarrow \frac{dx}{dy} - x = y \\ Comparing it with \frac{dx}{dy} + Px = Q, we get \\ P = - 1 and Q = y \\ So, I . F . = e^{\int Pdy} \\ = e^{\int - 1 dy} = e^{- y} \\ The solution of given differential equation is given by the \\ relation, \\ x (I . F .) = \int (Q \times I . F .) dy + C \\ {xe}^{- y} = \int {y \times e^{- y}} dy + C \\ {xe}^{- y} = \int {ye}^{- y} dy + C \\ {xe}^{- y} = y \int e^{- y} dy - \int (\frac{d}{dy} y \int e^{- y} dy) dy + C \\ = - {ye}^{- y} + \int (1 . e^{- y}) dy + C \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = - {ye}^{- y} - e^{- y} + C \\ \Rightarrow x = - y - 1 + {Ce}^{y} (Dividing both sides by e^{- y}) \\ \Rightarrow x + y + 1 = {Ce}^{y} \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.87 Find the general solution of the differential equation given below.

$ydx+ {(x−y}^{2}) dy = 0$

Ans.

$\begin{array}{l} The given differential equation is ydx + (x - y^{2}) dy = 0 \\ \Rightarrow \frac{dy}{dx} = - \frac{y}{(x - y^{2})} \\ \Rightarrow \frac{dx}{dy} = - \frac{(x - y^{2})}{y} \\ \Rightarrow \frac{dx}{dy} = - (\frac{x}{y} - y) \\ \Rightarrow \frac{dx}{dy} + \frac{x}{y} = y \\ Comparing it with \frac{dx}{dy} + Px = Q, we get \\ P = \frac{1}{y} and Q = y \\ So, I . F . = e^{\int Pdy} \\ = e^{\int \frac{1}{y} dy} \\ = e^{logy} = y \\ The solution of given differential equation is given by the \\ relation, \\ x (I . F .) = \int (Q \times I . F .) dy + C \\ xy = \int {y \times y} dy + C \\ xy = \int y^{2} dy + C \end{array}$ $\begin{array}{l} x y = \frac{y^{3}}{3} + C \\ x = \frac{y^{2}}{3} + \frac{C}{y} \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.88 Find the general solution of the differential equation given below.

${(x+3y}^{2}) \frac{dy}{dx} = y (y > 0)$

Ans.

$\begin{array}{l} The given differential equation is (x + 3 y^{2}) \frac{dy}{dx} = y (y > 0) \\ \Rightarrow \frac{dy}{dx} = \frac{y}{(x + 3 y^{2})} \\ \Rightarrow \frac{dx}{dy} = \frac{(x + 3 y^{2})}{y} \\ \Rightarrow \frac{dx}{dy} = (\frac{x}{y} + 3 y) \\ \Rightarrow \frac{dx}{dy} - \frac{x}{y} = 3 y \\ Comparing it with \frac{dx}{dy} + Px = Q, we get \\ P = - \frac{1}{y} and Q = 3 y \\ So, I . F . = e^{\int Pdy} \end{array}$ $\begin{array}{l} = e^{\int - \frac{1}{y} dy} \\ = e^{- logy} \\ = \frac{1}{y} \\ The solution of given differential equation is given by the \\ relation, \\ x (I . F .) = \int (Q \times I . F .) dy + C \\ x \frac{1}{y} = \int {3 y \times \frac{1}{y}} dy + C \\ x \frac{1}{y} = \int 3 dy + C = 3 y + C \\ x = 3 y^{2} + Cy \\ This is the required general solution of the given differential \\ equation . \end{array}$

Q.89 For the differential equations, find a particular solution satisfying the given condition:

$\frac{dy}{dx} + 2 ytanx = sinx; y = 0 whenx = \frac{π}{}$

Ans.

$\begin{array}{l} The given differential equation is \frac{dy}{dx} + 2 ytanx = sinx; \\ y = 0 when x = \frac{π}{3} \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \end{array}$ $\begin{array}{l} P = 2 tanx and Q = sinx \\ So, I . F . = e^{\int Pdx} \\ = e^{\int 2 tanx dx} \\ = {e^{2}}^{logsecx} \\ = e^{^{{logsec}^{2} x}} \\ = \sec^{2} x \\ The solution of given differential equation is given by the relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ {ysec}^{2} x = \int (sinx \times \sec^{2} x) dx + C \\ {ysec}^{2} x = \int tanx . secx dx + C \\ = secx + C ‹ \\ y = cosx + {Ccos}^{2} x . .. (i) \\ This ‹ is the general equation of given differential equation . \\ Now, y = 0 when x = \frac{π}{3}, so \\ 0 = \cos \frac{π}{3} + {Ccos}^{2} \frac{π}{3} \\ 0 = \frac{1}{2} + C \frac{1}{4} \Rightarrow C = - 2 \\ Putting value of C in equation (i), we get \\ y = cosx - 2 \cos^{2} x \\ This ‹ is the particular solution of given differential equation . \end{array}$

Q.90 For the differential equation given below, find a particular solution satisfying the given condition:

${(1+x}^{2}) \frac{dy}{dx} + 2 xy = \frac{1}{1 + x^{2}}; y = 0 when x = 1$

Ans.

$\begin{array}{l} The given differential equation is (1 + x^{2}) \frac{dy}{dx} + 2 xy = \frac{1}{1 + x^{2}}; \\ y = 0 whenx = 1 \\ \Rightarrow \frac{dy}{dx} + \frac{2 x}{(1 + x^{2})} y = \frac{1}{{(1 + x^{2})}^{2}} \\ Comparing it with \frac{dx}{dy} + Px = Q, we get \\ P = \frac{2 x}{(1 + x^{2})} and Q = \frac{1}{{(1 + x^{2})}^{2}} \\ So, I . F . = e^{\int Pdy} \\ = e^{\int \frac{2 x}{(1 + x^{2})} dy} \\ = e^{\log (1 + x^{2})} \\ = (1 + x^{2}) \\ The solution of given differential equation is given by the \\ relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ y (1 + x^{2}) = \int {\frac{1}{{(1 + x^{2})}^{2}} \times (1 + x^{2})} dx + C \\ = \int \frac{1}{(1 + x^{2})} dx + C \end{array}$ $\begin{array}{l} y (1 + x^{2}) = \tan^{- 1} x + C . .. (i) \\ Now, y = 0 at x = 1 \\ 0 (1 + 1^{2}) = \tan^{- 1} 1 + C \\ 0 = \frac{π}{4} + C \Rightarrow C = - \frac{π}{4} \\ From equation (i), we have \end{array}$ $\begin{array}{l} y (1 + x^{2}) = \tan^{- 1} x - \frac{π}{4} \\ This ‹ is the particular solution of given differential equation . \end{array}$

Q.91 For the differential equation given below, find a particular solution satisfying the given condition:

$\frac{dy}{dx} - 3 ycotx = \sin 2 x; y = 2 whenx = \frac{π}{}$

Ans.

$\begin{array}{l} The given differential equation is \frac{dy}{dx} - 3 ycotx = \sin 2 x; \\ y = 2 whenx = \frac{π}{2} \\ Comparing it with \frac{dx}{dy} + Px = Q, we get \\ P = - 3 cotx and Q = \sin 2 x \\ So, I . F . = e^{\int Pdx} \\ = e^{\int - 3 cotxdx} \\ = e^{- 3 logsinx} \\ = \frac{1}{\sin^{3} x} = {cosec}^{3} x \end{array}$ $\begin{array}{l} The solution of given differential equation is given by the \\ relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ {ycosec}^{3} x = \int {\sin 2 x \times {cosec}^{3} x} dx + C \\ = \int 2 \frac{cosx}{\sin^{2} x} dx + C \\ = \int 2 cotx . cosecx dx + C \\ {ycosec}^{3} x = - 2 cosecx + C \\ y = - 2 \sin^{2} x + {Csin}^{3} x . .. (i) \\ Now, y = 2 at x = \frac{π}{2} \\ 2 = - 2 \sin^{2} \frac{π}{2} + {Csin}^{3} \frac{π}{2} \\ 2 = - 2 + C \Rightarrow C = 4 \\ From equation (i), we have \\ y = - 2 \sin^{2} x + 4 \sin^{3} x \\ This is the required particular solution of the given differential \\ equation . \end{array}$

Q.92 Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Ans.

$\begin{array}{l} Let slope of a curve f (x, y) at point (x, y) be \frac{dy}{dx} . \\ According to question, \end{array}$ $\begin{array}{l} \frac{dy}{dx} = x + y \\ \Rightarrow \frac{dy}{dx} - y = x \\ Comparing it with \frac{dx}{dy} + Px = Q, we get \\ P = - 1 and Q = x \\ So, I . F . = e^{\int Pdx} \\ = e^{\int - 1 dx} \\ = e^{- x} \\ The solution of given differential equation is given by the \\ relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ {ye}^{- x} = \int (x \times e^{- x}) dx + C \\ {ye}^{- x} = x \int e^{- x} dx - \int (\frac{d}{dx} x \int e^{- x} dx) dx + C \\ {ye}^{- x} = x (- e^{- x}) - \int 1 . (- e^{- x}) dx + C \\ {ye}^{- x} = - {xe}^{- x} + \int e^{- x} dx + C \\ {ye}^{- x} = - {xe}^{- x} - e^{- x} + C \\ \Rightarrow y = - x - 1 + {Ce}^{x} \end{array}$ $\begin{array}{l} x + y + 1 = {Ce}^{x} . .. (i) \\ Since, the curve passes through the origin, then \\ 0 + 0 + 1 = {Ce}^{0} \Rightarrow C = 1 \\ From equation (i), we get \\ x + y + 1 = e^{x} \\ Thus, the required equation passing through origin is \\ x + y + 1 = e^{x} . \end{array}$

Q.93 Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Ans.

$\begin{array}{l} Let slope of a curve f (x, y) at point (x, y) be \frac{dy}{dx} . \\ According to question, \\ \frac{dy}{dx} + 5 = x + y \\ \Rightarrow \frac{dy}{dx} - y = x - 5 \\ Comparing it with \frac{dx}{dy} + Px = Q, we get \\ P = - 1 and Q = x - 5 \\ So, I . F . = e^{\int Pdx} \end{array}$ $\begin{array}{l} = e^{\int - 1 dx} \\ = e^{- x} \\ The solution of given differential equation is given by the \\ relation, \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ {ye}^{- x} = \int (x - 5) \times e^{- x} dx + C \\ {ye}^{- x} = \int (x - 5) e^{- x} dx + C \\ {ye}^{- x} = {(x - 5) \int e^{- x} dx - \int (\frac{d}{dx} (x - 5) \int e^{- x} dx) dx} + C \\ {ye}^{- x} = {- (x - 5) e^{- x} + \int e^{- x} dx} + C \\ {ye}^{- x} = - (x - 5) e^{- x} - e^{- x} + C \\ {ye}^{- x} = (4 - x) e^{- x} + C \\ y = (4 - x) + {Ce}^{x} . .. (i) \\ The curve passes through (0,2), then \\ 2 = (4 - 0) + {Ce}^{0} \\ 2 = 4 + C \Rightarrow C = - 2 \\ Putting value of C in equation (i), we get \\ y = (4 - x) - 2 e^{x} \\ Thus, this is the required equation of the curve . \end{array}$

Q.94

$\begin{array}{l} The integrating factor of the differential equation \\ x \frac{dy}{dx} - y = 2 x^{2} is \end{array}$ ${(A)e}^{- x} (B) e^{- y} (C) \frac{1}{x} (D) x$

Ans.

$\begin{array}{l} The given differential equation is x \frac{dy}{dx} - y = 2 x^{2} \\ \Rightarrow \frac{dy}{dx} - \frac{y}{x} = 2 x \\ Comparing it with \frac{dy}{dx} + Py = Q, we get \\ P = - \frac{1}{x} and Q = 2 x \\ So, I . F . = e^{\int Pdx} \\ = e^{\int - \frac{1}{x} dx} = e^{- logx} \\ = \frac{1}{x} \\ Thus, the correct option is C . \end{array}$

Q.95

$\begin{array}{l} The Integrating Factor of the differential equation \\ (1 - y^{2}) \frac{dx}{dy} + yx = ay (- 1 < y < 1) is \\ (A) \frac{1}{y^{2} - 1} (B) \frac{1}{\sqrt{y^{2} - 1}} (C) \frac{1}{1 - y^{2}} (D) \frac{1}{\sqrt{1 - y^{2}}} \end{array}$

Ans.

$\begin{array}{l} The given differential equation is (1 - y^{2}) \frac{dx}{dy} + yx = ay (- 1 < y < 1) \\ \Rightarrow \frac{dx}{dy} + \frac{y}{(1 - y^{2})} x = a \frac{y}{(1 - y^{2})} \\ Comparing it with \frac{dx}{dy} + Px = Q, we get \\ P = \frac{y}{(1 - y^{2})} and Q = a \frac{y}{(1 - y^{2})} \\ So, I . F . = e^{\int Pdy} \\ = e^{\int \frac{y}{(1 - y^{2})} dy} \\ = e^{- \frac{1}{2} \log (1 - y^{2})} \\ = {e^{\log (1 - y^{2})}}^{^{- \frac{1}{2}}} \\ = {(1 - y^{2})}^{- \frac{1}{2}} \\ = \frac{1}{\sqrt{1 - y^{2}}} \\ Thus, the correct option is D . \end{array}$

Q.96 For the differential equations given below, indicate its order and degree (if defined).

$\begin{array}{l} (i) \frac{d^{2} y}{{dx}^{2}} + 5 x {(\frac{dy}{dx})}^{2} - 6 y = logx (ii) {(\frac{dy}{dx})}^{3} - 4 {(\frac{dy}{dx})}^{2} + 7 y = sinx \\ (iii) \frac{d^{4} y}{{dx}^{4}} - \sin (\frac{d^{3} y}{{dx}^{3}}) = 0 \end{array}$

Ans.

$\begin{array}{l} (i) The given differential equation is : \\ \frac{d^{2} y}{{dx}^{2}} + 5 x {(\frac{dy}{dx})}^{2} - 6 y = logx \\ \Rightarrow \frac{d^{2} y}{{dx}^{2}} + 5 x {(\frac{dy}{dx})}^{2} - 6 y - logx = 0 \\ The highest order derivative present in the given differential \\ equation is \frac{d^{2} y}{{dx}^{2}} . So, its order is 2 and highest power raised by \\ \frac{d^{2} y}{{dx}^{2}} is one . Hence its degree is one . \\ (ii) The given differential equation is : \\ {(\frac{dy}{dx})}^{3} - 4 {(\frac{dy}{dx})}^{2} + 7 y = sinx \\ \Rightarrow {(\frac{dy}{dx})}^{3} - 4 {(\frac{dy}{dx})}^{2} + 7 y - sinx = 0 \\ The highest order derivative present in the given differential \\ equation is \frac{dy}{dx} . So, its order is 1 and highest power raised by \\ \frac{dy}{dx} is three . Hence its degree is 3 . \end{array}$ $\begin{array}{l} (iii) The given differential equation is : \\ \frac{d^{4} y}{{dx}^{4}} - \sin (\frac{d^{3} y}{{dx}^{3}}) = 0 \\ The highest order derivative present in the given differential \\ equation is \frac{d^{4} y}{{dx}^{4}} . So, its order is 4 . \\ Since the given differential equation is not a polynomial equation . \\ Hence, its degree is not defined . \end{array}$

Q.97 For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$\begin{array}{l} (i) y = {ae}^{x} + {be}^{- x} + x^{2} : x \frac{d^{2} y}{{dx}^{2}} + 2 \frac{dy}{dx} - xy + x^{2} - 2 = 0 \\ (ii) y = e^{x} (acosx + bsinx) : \frac{d^{2} y}{{dx}^{2}} - 2 \frac{dy}{dx} + 2 y = 0 \\ (iii) y = xsin 3 x : \frac{d^{2} y}{{dx}^{2}} + 9 y - 6 \cos 3 x = 0 \\ (iv) x^{2} = 2 y^{2} logy : (x^{2} + y^{2}) \frac{dy}{dx} - xy = 0 \end{array}$

Ans.

$\begin{array}{l} (i) The given equation is: \\ y = a e^{x} + b e^{- x} + x^{2} … (i) \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x \end{array}$

Again, differentiating w.r.t. x, we get

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} = a e^{x} + b e^{- x} + 2 \\ Substituting the value of \frac{d^{2} y}{d x^{2}} and \frac{d y}{d x} in given differential \\ equation, we get: \\ L .H .S . = x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 \\ = x (a e^{x} + b e^{- x} + 2) + 2 (a e^{x} - b e^{- x} + 2 x) - x (a e^{x} + b e^{- x} + x^{2}) \\ + x^{2} - 2 \\ = a x e^{x} + b x e^{- x} + 2 x + 2 a e^{x} - 2 b e^{- x} + 4 x - a x e^{x} - b x e^{- x} - x^{3} \\ + x^{2} - 2 \\ = 2 a e^{x} - 2 b e^{- x} - x^{3} + x^{2} + 6 x - 2 \neq 0 \\ L .H .S . \neq R .H .S . \\ Thus, the given function is not a solution of corresponding \\ diffrential equation . \end{array}$ $\begin{array}{l} (i i) The given equation is: \\ y = e^{x} (a \cos x + b \sin x) … (i) \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = e^{x} \frac{d}{d x} (a \cos x + b \sin x) + (a \cos x + b \sin x) \frac{d}{d x} e^{x} \\ = e^{x} (- a \sin x + b \cos x) + (a \cos x + b \sin x) e^{x} \\ = e^{x} (- a \sin x + b \cos x + a \cos x + b \sin x) \\ Again, differentiating w .r .t . x, we get \\ \frac{d^{2} y}{d x^{2}} = e^{x} \frac{d}{d x} (- a \sin x + b \cos x + a \cos x + b \sin x) \end{array}$ $\begin{array}{l} + (- a \sin x + b \cos x + a \cos x + b \sin x) \frac{d}{d x} e^{x} \\ = e^{x} (- a \cos x - b s i n x - a \sin x + b \cos x) \\ + (- a \sin x + b \cos x + a \cos x + b \sin x) e^{x} \\ = e^{x} (\begin{array}{l} - a \cos x - b s i n x - a \sin x + b \cos x - a \sin x + b \cos x \\ + a \cos x + b \sin x \end{array}) \\ = e^{x} (- 2 a \sin x + 2 b \cos x) \\ Substituting the value of \frac{d^{2} y}{d x^{2}} and \frac{d y}{d x} in given differential \\ equation, we get: \\ L . H . S . = \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y \\ = e^{x} (- 2 a \sin x + 2 b \cos x) - 2 e^{x} (- a \sin x + b \cos x + a \cos x + b \sin x) \\ + 2 e^{x} (a \cos x + b \sin x) \\ = e^{x} (\begin{array}{l} - 2 a \sin x + 2 b \cos x + 2 a \sin x - 2 b \cos x - 2 a \cos x - 2 b \sin x \\ + 2 a \cos x + 2 b \sin x \end{array}) \\ = e^{x} (0) = 0 \\ L .H .S .=R .H .S . \\ Thus, the given function is a solution of corresponding \\ diffrential equation . \\ (i i i) The given equation is: \\ y = x \sin 3 x … (i) \\ Differentiating w .r .t . x, we get \\ \frac{d y}{d x} = x \frac{d}{d x} \sin 3 x + \sin 3 x \frac{d}{d x} x \end{array}$ $\begin{array}{l} = x .3 \cos 3 x + \sin 3 x .1 \\ = 3 x \cos 3 x + \sin 3 x \\ Again, differentiating w .r .t . x, we get \\ \frac{d^{2} y}{d x^{2}} = 3 x \frac{d}{d x} \cos 3 x + \cos 3 x \frac{d}{d x} 3 x + \frac{d}{d x} s i n 3 x \\ = - 9 x \sin 3 x + 3 \cos 3 x + 3 \cos 3 x \\ = - 9 x \sin 3 x + 6 \cos 3 x \\ Substituting the value of \frac{d^{2} y}{d x^{2}} and \frac{d y}{d x} in given differential \\ equation, we get: \\ L .H .S . = \frac{d^{2} y}{d x^{2}} + 9 y - 6 \cos 3 x \\ = - 9 x \sin 3 x + 6 \cos 3 x + 9 x \sin 3 x - 6 \cos 3 x \\ = 0 \\ L .H .S .=R .H .S . \\ Thus, the given function is a solution of corresponding \\ diffrential equation . \\ (i v) The given equation is: \\ x^{2} = 2 y^{2} \log y … (i) \\ Differentiating w .r .t . x, we get \\ 2 x = 2 y^{2} \frac{d}{d x} \log y + 2 \log y \frac{d}{d x} y^{2} \\ = 2 y^{2} \frac{1}{y} \frac{d y}{d x} + 2 \log y (2 y \frac{d y}{d x}) \\ = (2 y + 4 y \log y) \frac{d y}{d x} \end{array}$ $\begin{array}{l} x = (y + 2 y \log y) \frac{d y}{d x} \\ \frac{d y}{d x} = \frac{x}{y (1 + 2 \log y)} \\ Substituting the value of \frac{d y}{d x} in given differential \\ equation, we get: \\ L .H .S .= (x^{2} + y^{2}) \frac{d y}{d x} - x y \\ = (2 y^{2} \log y + y^{2}) \frac{x}{y (1 + 2 \log y)} - x y \\ = y^{2} (2 \log y + 1) \frac{x}{y (1 + 2 \log y)} - x y \\ = x y - x y \\ = 0 = R .H .S . \\ Thus, the given function is a solution of corresponding \\ diffrential equation . \end{array}$

Q.98 Form the differential equation representing the family of curves given by (x – a)² +2y² = a², where a is an arbitrary constant.

Ans.

$\begin{array}{l} The given equation is: \\ {(x - a)}^{2} + {2y}^{2} = a^{2} \\ x^{2} - 2 a x + a^{2} + 2 y^{2} = a^{2} \\ x^{2} - 2 a x + 2 y^{2} = 0 \end{array}$

Differentiating w.r.t. x, we get

$\begin{array}{l} 2 x - 2 a + 4 y \frac{d y}{d x} = 0 \\ \frac{d y}{d x} = \frac{2 a - 2 x}{4 y} \\ = \frac{a - x}{2 y} … (i i) \\ From equation (i), a = \frac{x^{2} + 2 y^{2}}{2 x} \\ Putting value of a in equation (ii), we get \\ \frac{d y}{d x} = \frac{(\frac{x^{2} + 2 y^{2}}{2 x}) - x}{2 y} \\ = \frac{x^{2} + 2 y^{2} - 2 x^{2}}{4 x y} \\ = \frac{2 y^{2} - x^{2}}{4 x y} \\ Which is the differential equation of the family of curves . \end{array}$

Q.99 Prove that x² – y² = c (x² + y²)² is the general solution of differential equation (x³ – 3xy²) dx = (y³ – 3x²y) dy, where c is a parameter.

Ans.

$\begin{array}{l} The given differential equation is : \\ (x^{3} - 3 {xy}^{2}) dx = (y^{3} - 3 x^{2} y) dy \end{array}$ $\begin{array}{l} \frac{dy}{dx} = \frac{(x^{3} - 3 {xy}^{2})}{(y^{3} - 3 x^{2} y)} \\ Let F (x, y) = \frac{(x^{3} - 3 {xy}^{2})}{(y^{3} - 3 x^{2} y)}, then \\ F (λx, λy) = \frac{(λ^{3} x^{3} - 3 λ^{3} {xy}^{2})}{(λ^{3} y^{3} - 3 λ^{3} x^{2} y)} \\ = \frac{λ^{3} (x^{3} - 3 {xy}^{2})}{λ^{3} (y^{3} - 3 x^{2} y)} = λ^{0} F (x, y) \\ Therefore, F (x, y) is a homogenous function of degree zero . So, \\ the given differential equation is a homogenous differential \\ equation . \\ For solving equation (i), substitute y = vx and \frac{dy}{dx} = v + x \frac{dv}{dx}, \\ we get \\ v + x \frac{dv}{dx} = \frac{(x^{3} - 3 v^{2} x^{3})}{(v^{3} x^{3} - 3 {vx}^{3})} \\ = \frac{(1 - 3 v^{2})}{(v^{3} - 3 v)} \\ x \frac{dv}{dx} = \frac{(1 - 3 v^{2})}{(v^{3} - 3 v)} - v \\ = \frac{1 - 3 v^{2} - v^{4} + 3 v^{2}}{(v^{3} - 3 v)} \end{array}$ $\begin{array}{l} = \frac{(1 - v^{4})}{(v^{3} - 3v)} \\ \frac{(v^{3} - 3v) d v}{(1 - v^{4})} = \frac{d x}{x} \\ Integrating both sides, we get \\ \int \frac{(v^{3} - 3v) d v}{(1 - v^{4})} = \int \frac{1}{x} d x … (i) \\ Let I = \int \frac{(v^{3} - 3v) d v}{(1 - v^{4})} \\ = \int \frac{v^{3} d v}{(1 - v^{4})} - 3 \int \frac{v d v}{(1 - v^{4})} \\ = - \frac{1}{4} \log (1 - v^{4}) - \frac{3}{4} \log | \frac{1 - v^{2}}{1 + v^{2}} | \\ From eqeuation (i), we get \\ - \frac{1}{4} \log (1 - v^{4}) - \frac{3}{4} \log | \frac{1 - v^{2}}{1 + v^{2}} | = \log | x | + \log C \\ - \frac{1}{4} \log {(1 - v^{4}) {| \frac{1 - v^{2}}{1 + v^{2}} |}^{3}} = \log C | x | \\ \log {(1 - v^{4}) {| \frac{1 - v^{2}}{1 + v^{2}} |}^{3}} = - 4 \log C | x | \end{array}$ $\begin{array}{l} \log {(1 - v^{4}) {(\frac{1 - v^{2}}{1 + v^{2}})}^{3}} = \log {(C | x |)}^{- 4} \\ \Rightarrow (1 - v^{4}) {(\frac{1 + v^{2}}{1 - v^{2}})}^{3} = {(C | x |)}^{- 4} \\ \Rightarrow (1 - \frac{y^{4}}{x^{4}}) {(\frac{1 + \frac{y^{2}}{x^{2}}}{1 - \frac{y^{2}}{x^{2}}})}^{3} = {(C | x |)}^{- 4} \\ \Rightarrow (\frac{x^{4} - y^{4}}{x^{4}}) {(\frac{\frac{x^{2} + y^{2}}{x^{2}}}{\frac{x^{2} - y^{2}}{x^{2}}})}^{3} = {(C ‘ | x |)}^{- 4} \\ \Rightarrow \frac{(x^{2} - y^{2}) (x^{2} + y^{2})}{x^{4}} {(\frac{x^{2} + y^{2}}{x^{2} - y^{2}})}^{3} = \frac{1}{{(C ‘ | x |)}^{4}} \\ \Rightarrow \frac{{(x^{2} + y^{2})}^{4}}{x^{4} {(x^{2} - y^{2})}^{2}} = \frac{1}{{(C ‘)}^{4} x^{4}} \\ \Rightarrow {(C ‘)}^{4} {(x^{2} + y^{2})}^{4} = {(x^{2} - y^{2})}^{2} \\ \Rightarrow (x^{2} - y^{2}) = {(C ‘)}^{2} {(x^{2} + y^{2})}^{2} \\ \Rightarrow (x^{2} - y^{2}) = C {(x^{2} + y^{2})}^{2} [Let C = {(C ‘)}^{2}] \\ Hence, the given result is proved . \end{array}$

Q.100 Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Ans.

$\begin{array}{l} The equation of circle with centre (a, a) and radius a which \\ touches the coordinate aes is : \\ {(x - a)}^{2} + {(y - a)}^{2} = a^{2} . .. (i) \\ Differentiating equation (i), w . r . t . x, we get \\ 2 (x - a) + 2 (y - a) y ‘ = 0 \\ \Rightarrow (x - a) + (y - a) y ‘ = 0 \\ \Rightarrow x - a + yy ‘ - ay ‘ = 0 \\ \Rightarrow x + yy ‘ - a (1 + y ‘) = 0 \\ \Rightarrow a = \frac{x + yy ‘}{(1 + y ‘)} \\ Substituting the value of a in equation (i), we get \\ {x - \frac{x + yy ‘}{(1 + y ‘)}}^{2} + {y - \frac{x + yy ‘}{(1 + y ‘)}}^{2} = {\frac{x + yy ‘}{(1 + y ‘)}}^{2} \\ \Rightarrow {\frac{x (1 + y ‘) - x - yy ‘}{(1 + y ‘)}}^{2} + {\frac{y (1 + y ‘) - x - yy ‘}{(1 + y ‘)}}^{2} = {\frac{x + yy ‘}{(1 + y ‘)}}^{2} \\ \Rightarrow {\frac{x + xy ‘ - x - yy ‘}{(1 + y ‘)}}^{2} + {\frac{y + yy ‘ - x - yy ‘}{(1 + y ‘)}}^{2} = {\frac{x + yy ‘}{(1 + y ‘)}}^{2} \end{array}$ $\begin{array}{l} \Rightarrow {\frac{x y ‘ - y y ‘}{(1 + y ‘)}}^{2} + {\frac{y - x}{(1 + y ‘)}}^{2} = {\frac{x + y y ‘}{(1 + y ‘)}}^{2} \\ \Rightarrow {(x - y)}^{2} y ‘^{2} + {(x - y)}^{2} = {(x + y y ‘)}^{2} \\ \Rightarrow {(x - y)}^{2} (y ‘^{2} + 1) = {(x + y y ‘)}^{2} \\ \Rightarrow {(x + y y ‘)}^{2} = {(x - y)}^{2} (y ‘^{2} + 1) \\ Which is the required differential equation . \end{array}$

Q.101

$\begin{array}{l} Find the general solution of the differential equation \\ \frac{dy}{dx} + \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} = 0 . \end{array}$

Ans.

$\begin{array}{l} The given differential equation is : \\ \frac{dy}{dx} + \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} = 0 \\ \Rightarrow \frac{dy}{dx} = - \sqrt{\frac{1 - y^{2}}{1 - x^{2}}} \\ \Rightarrow \frac{dy}{\sqrt{1 - y^{2}}} = - \frac{dx}{\sqrt{1 - x^{2}}} \\ Integrating both sides, we get \\ \sin^{- 1} y = - \sin^{- 1} x + C \\ \Rightarrow \sin^{- 1} x + \sin^{- 1} y = C \end{array}$ $\begin{array}{l} Thus, the general solution of the given differential equation is: \\ {sin}^{-1} x + {sin}^{-1} y = C \end{array}$

Q.102

$\begin{array}{l} Show that the general solution of the differential equation \\ \frac{dy}{dx} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0 is given by (x + y + 1) = A (1 - x - y - 2 xy), \\ where A is parameter . \end{array}$

Ans.

$\begin{array}{l} The differential equation is : \\ \frac{dy}{dx} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0 \\ \Rightarrow \frac{dy}{dx} = - \frac{y^{2} + y + 1}{x^{2} + x + 1} \\ \Rightarrow \frac{dy}{(y^{2} + y + 1)} = - \frac{dx}{x^{2} + x + 1} \\ \Rightarrow \frac{dy}{(y^{2} + y + 1)} + \frac{dx}{x^{2} + x + 1} = 0 \\ Integrating both sides, we get \\ \Rightarrow \int \frac{dy}{(y^{2} + y + 1)} + \int \frac{dx}{x^{2} + x + 1} = C \\ \Rightarrow \int \frac{dy}{{(y + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} + \int \frac{dx}{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = C \\ \Rightarrow \frac{1}{(\frac{\sqrt{3}}{2})} \tan^{- 1} (\frac{2 y + 1}{\sqrt{3}}) + \frac{1}{(\frac{\sqrt{3}}{2})} \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) = C \end{array}$ $\begin{array}{l} \Rightarrow \tan^{- 1} (\frac{2 y + 1}{\sqrt{3}}) + \tan^{- 1} (\frac{2 x + 1}{\sqrt{3}}) = \frac{\sqrt{3}}{2} C \\ \Rightarrow \tan^{- 1} {\frac{(\frac{2 y + 1}{\sqrt{3}}) + (\frac{2 x + 1}{\sqrt{3}})}{1 - (\frac{2 y + 1}{\sqrt{3}}) (\frac{2 x + 1}{\sqrt{3}})}} = \frac{\sqrt{3}}{2} C \\ \Rightarrow \tan^{- 1} {\frac{\sqrt{3} (2 y + 1 + 2 x + 1)}{3 - (2 y + 1) (2 x + 1)}} = \frac{\sqrt{3}}{2} C \\ \Rightarrow \tan^{- 1} {\frac{\sqrt{3} (2 y + 2 x + 2)}{3 - (4 xy + 2 x + 2 y + 1)}} = \frac{\sqrt{3}}{2} C \\ \Rightarrow \tan^{- 1} {\frac{2 \sqrt{3} (y + x + 1)}{2 (1 - 2 xy - x - y)}} = \frac{\sqrt{3}}{2} C \\ \Rightarrow {\frac{\sqrt{3} (y + x + 1)}{(1 - 2 xy - x - y)}} = \tan (\frac{\sqrt{3}}{2} C) \\ \Rightarrow (y + x + 1) = \frac{1}{\sqrt{3}} \tan (\frac{\sqrt{3}}{2} C) (1 - 2 xy - x - y) \\ \Rightarrow (y + x + 1) = A (1 - 2 xy - x - y) \\ [Where A = \frac{1}{\sqrt{3}} \tan (\frac{\sqrt{3}}{2} C)] \\ Thus, the given result is proved . \end{array}$ $\begin{array}{l} \end{array}$

Q.103

$\begin{array}{l} Find the equation of the curve passing through the point \\ (0, \frac{π}{4}) whose differential equation is \\ sinxcosy dx + cosxsiny dy = 0 . \end{array}$

Ans.

$\begin{array}{l} The given differential equation is \\ sinxcosydx + cosxsinydy = 0 \\ Dividing equation by \cos x \cos y, we get \\ \frac{sinxcosydx + cosxsinydy}{\cos x \cos y} = 0 \\ \Rightarrow tanxdx + tanydy = 0 \\ Integrating both sides, we get \\ \int tanx dx + \int tany dy = logC \\ logsecx + logsecy = logC \\ \Rightarrow logsecxsecy = logC \\ secxsecy = C . .. (i) \\ The curve passes through (0, \frac{π}{4}), we get \\ \sec 0 \sec \frac{π}{4} = C \\ \Rightarrow 1 . \sqrt{2} = C \Rightarrow C = \sqrt{2} \\ Putting C = \sqrt{2} in equation (i), we get \\ secxsecy = \sqrt{2} \Rightarrow cosy = \frac{secx}{\sqrt{2}} \\ Which is the required equation of the curve . \end{array}$

Q.104 Find the particular solution of the differential equation (1 + e^2x) dy + ( 1 + y²) e^x dx = 0, given that y = 1 when x = 0.

Ans.

$\begin{array}{l} The given differential equation is \\ (1 + e^{2 x}) dy + (1 + y^{2}) e^{x} dx = 0 \\ \Rightarrow \frac{dy}{(1 + y^{2})} + \frac{e^{x} dx}{(1 + e^{2 x})} = 0 \\ Integrating both sides, we get \\ \int \frac{dy}{(1 + y^{2})} + \int \frac{e^{x} dx}{{1 + {(e^{x})}^{2}}} = 0 \\ \tan^{- 1} y + \tan^{- 1} e^{x} = C . .. (i) \\ Now, y = 1 at x = 0 \\ \tan^{- 1} 1 + \tan^{- 1} e^{0} = C \\ \Rightarrow \frac{π}{4} + \tan^{- 1} 1 = C \\ \Rightarrow \frac{π}{4} + \frac{π}{4} = C \Rightarrow C = \frac{π}{2} \\ Putting C = \frac{π}{2} in equation (i), we get \\ \tan^{- 1} y + \tan^{- 1} e^{x} = \frac{π}{2} \\ This is the required particular question of given differential \\ equation . \end{array}$

Q.105

${Solve the differential equation y e}^{\frac{x}{y}} dx = ({xe}^{\frac{x}{y}} + y^{2}) dy (y \neq 0) .$

Ans.

$\begin{array}{l} The given differential equation is : \\ {ye}^{\frac{x}{y}} dx = ({xe}^{\frac{x}{y}} + y^{2}) dy (y \neq 0) \\ \Rightarrow {ye}^{\frac{x}{y}} \frac{dx}{dy} = ({xe}^{\frac{x}{y}} + y^{2}) \\ \Rightarrow {ye}^{\frac{x}{y}} \frac{dx}{dy} - {xe}^{\frac{x}{y}} = y^{2} \\ \Rightarrow (y \frac{dx}{dy} - x) e^{\frac{x}{y}} = y^{2} \\ \Rightarrow \frac{(y \frac{dx}{dy} - x)}{y^{2}} e^{\frac{x}{y}} = 1 . .. (i) \\ Let e^{\frac{x}{y}} = t \\ Differentiating w . r . t . y, we get \\ \frac{d}{dx} e^{\frac{x}{y}} = \frac{dt}{dx} \Rightarrow e^{\frac{x}{y}} (\frac{y \frac{d}{dy} x - x \frac{d}{dy} y}{y^{2}}) = \frac{dt}{dx} \end{array}$ $\begin{array}{l} \Rightarrow e^{\frac{x}{y}} (\frac{y \frac{dx}{dy} - x}{y^{2}}) = \frac{dt}{dx} . .. (ii) \\ From equation (i) and equation (ii), we get \\ \frac{dt}{dx} = 1 \Rightarrow dt = dx \\ Integrating both sides, we get \\ \int dt = \int dx \\ t = x + C \\ \Rightarrow e^{\frac{x}{y}} = x + C \\ This is the general solution of the given differential equation . \end{array}$

Q.106 Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = – 1, when x = 0.

Ans.

$\begin{array}{l} The differential equation is : \\ (x - y) (dx + dy) = dx - dy \\ (x - y) dx + (x - y) dy = dx - dy \\ \Rightarrow (x - y + 1) dy = (1 - x + y) dx \\ \Rightarrow \frac{dy}{dx} = \frac{(1 - x + y)}{(x - y + 1)} \\ \Rightarrow \frac{dy}{dx} = \frac{1 - (x - y)}{1 + (x - y)} . .. (i) \end{array}$ $\begin{array}{l} L e t v = x - y \Rightarrow \frac{d v}{d x} = 1 - \frac{d y}{d x} \Rightarrow \frac{d y}{d x} = 1 - \frac{d v}{d x} \\ Substituting value of (x - y) and \frac{d y}{d x} in equation (i), we get \\ 1 - \frac{d v}{d x} = \frac{1 - v}{1 + v} \\ \Rightarrow \frac{d v}{d x} = 1 - \frac{1 - v}{1 + v} \\ = \frac{1 + v - 1 + v}{1 + v} \\ \Rightarrow \frac{d v}{d x} = \frac{2 v}{1 + v} \\ \Rightarrow \frac{1 + v}{v} d v = 2 d x \\ Integrating both sides, we get \\ \int (\frac{1}{v} + 1) d v = 2 \int d x \\ \Rightarrow \log | v | + v = 2 x + C \\ \Rightarrow \log | x - y | + x - y = 2 x + C \\ \Rightarrow \log | x - y | = x + y + C … (i i) \\ N o w, y = - 1 at x = 0 \\ \Rightarrow \log | 1 | = 0 - 1 + C \Rightarrow C = 1 \\ Substituting C = 1 in equation (ii), we get \\ \log | x - y | = x + y + 1 \\ This is the required particular solution of the given \\ differential equation . \end{array}$

Q.107

$Solve the differential equation [\frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}] \frac{dx}{dy} = 1 (x \neq 0)$

Ans.

$\begin{array}{l} The differential equation is : \\ [\frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}] \frac{dx}{dy} = 1 (x \neq 0) \\ \Rightarrow \frac{dy}{dx} = [\frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}] \\ \Rightarrow \frac{dy}{dx} + \frac{y}{\sqrt{x}} = \frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} \\ Comparing the given equation with \frac{dy}{dx} + Py = Q, we get \\ P = \frac{1}{\sqrt{x}} and Q = \frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} \\ Now, I . F . = e^{\int P dx} \\ = e^{\int \frac{1}{\sqrt{x}} dx} = e^{2 \sqrt{x}} \\ The general solution of the given differential equation is : \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ \Rightarrow y (e^{2 \sqrt{x}}) = \int (\frac{e^{- 2 \sqrt{x}}}{\sqrt{x}} \times e^{2 \sqrt{x}}) dx + C \\ = \int (\frac{1}{\sqrt{x}}) dx + C \end{array}$ $\begin{array}{l} y (e^{2 \sqrt{x}}) = 2 \sqrt{x} + C \\ This is the general solution of the given equation . \end{array}$

Q.108

$\begin{array}{l} Find a particular solution of the differential equation \\ \frac{dy}{dx} + ycotx = 4 xcosecx (x \neq 0), given thaty = 0 \\ when x = \frac{π}{2} . \end{array}$

Ans.

$\begin{array}{l} The differential equation is : \\ \frac{dy}{dx} + ycotx = 4 xcosecx (x \neq 0) \\ This is a linear differential equation in the form of \\ \frac{dy}{dx} + Py = Q, where P = cotx and Q = 4 xcosecx \\ Now, I . F . = e^{\int P dx} \\ = e^{\int cotx dx} \\ = e^{logsinx} \\ = sinx \\ The general solution of given differential equation is : \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ \Rightarrow ysinx = \int (4 xcosecx \times sinx) dx + C \\ = \int 4 x dx + C \end{array}$ $\begin{array}{l} = 4 \frac{x^{2}}{2} + C \\ \Rightarrow y \sin x = 2 x^{2} + C … (i) \\ Which is general solution of the given differential equation . \\ Now, y = 0 when x = \frac{π}{2} \\ From equation (i), we have \\ (0) \sin \frac{π}{2} = 2 {(\frac{π}{2})}^{2} + C \Rightarrow C = - \frac{π^{2}}{2} \\ Putting value of C in equation (i), we get \\ y \sin x = 2 x^{2} - \frac{π^{2}}{2} \\ This is the particular solution of the given differential equation . \end{array}$

Q.109

$\begin{array}{l} Find a particular solution of the differential equation \\ (x + 1) \frac{dy}{dx} = 2 e^{- y} - 1, given that y = 0 when x = 0 . \end{array}$

Ans.

$\begin{array}{l} The differential equation is : \\ (x + 1) \frac{dy}{dx} = 2 e^{- y} - 1 \\ \Rightarrow \frac{dy}{(2 e^{- y} - 1)} = \frac{dx}{(x + 1)} \\ Integrating both sides, we get \\ \int \frac{dy}{(2 e^{- y} - 1)} = \int \frac{1}{(x + 1)} dx \end{array}$ $\begin{array}{l} \Rightarrow \int \frac{e^{y} d y}{(2 - e^{y})} = \int \frac{1}{(x + 1)} d x \\ \Rightarrow - \log (2 - e^{y}) = \log | x + 1 | + \log C \\ \Rightarrow \log {(2 - e^{y})}^{- 1} = \log | C (x + 1) | \\ \Rightarrow {(2 - e^{y})}^{- 1} = C (x + 1) \\ \Rightarrow (2 - e^{y}) = \frac{1}{C (x + 1)} … (i) \\ Now, y = 0 when x = 0 \\ (2 - e^{0}) = \frac{1}{C (0 + 1)} \\ \Rightarrow (2 - 1) = \frac{1}{C} \\ \Rightarrow C = 1 \\ Putting value C in equation (i), we get \\ (2 - e^{y}) = \frac{1}{1 (x + 1)} \\ \Rightarrow e^{y} = 2 - \frac{1}{x + 1} \\ = \frac{2 x + 2 - 1}{x + 1} \\ y = \log | \frac{2 x + 1}{x + 1} |, x \neq - 1 \\ This is the particular solution of the given differential equation . \end{array}$

Q.110 The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time.
If the population of the village was 20,000 in 1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Ans.

$\begin{array}{l} Let number of people in a village at any instant (t) be P . \\ Then, according to question : \\ \frac{dP}{dt} α P \\ \frac{dP}{dt} = kP [Where k is a proportionality constant .] \\ \frac{dP}{P} = kdt \\ Integrating both sides, we get \\ \int \frac{dP}{P} = k \int dt \\ logP = kt + C . .. (i) \\ In the year 1999, t = 0 and P = 20,000 \\ ∴ \log 20,000 = k (0) + C \\ C = \log 20,000 \\ Putting value of C in equation (i), we get \\ logP = kt + \log 20,000 . .. (ii) \\ In the year 2004, t = 5 years and P = 25,000, then from equation (ii), \\ we have \\ \log 25,000 = k (5) + \log 20,000 \end{array}$ $\begin{array}{l} \log \frac{25,000}{20,000} = 5 k \\ k = \frac{1}{5} \log (\frac{5}{4}) \\ Putting value of k in equation (ii), we have \\ \log P = \frac{1}{5} \log (\frac{5}{4}) t + \log 20,000 \\ In the year 2009, t = 10 years \\ \log P = 10 \times \frac{1}{5} \log (\frac{5}{4}) + \log 20,000 \\ \Rightarrow \log P = 2 \log (\frac{5}{4}) + \log 20,000 \\ \Rightarrow \log P = \log {(\frac{5}{4})}^{2} + \log 20,000 \\ = \log {{(\frac{5}{4})}^{2} \times 20, 000} \\ P = \frac{5}{4} \times \frac{5}{4} \times 20, 000 = 31250 \\ Therefore, the population of the village in year 2009 is 31250 . \end{array}$

Q.111

$\begin{array}{l} The general solution of the differential equation \\ \frac{ydx - xdy}{y} = 0 is \\ (A) xy = C (B) x = {Cy}^{2} (C) y = Cx (D) y = {Cx}^{2} \end{array}$

Ans.

$\begin{array}{l} The differential equation is : \\ \frac{ydx - xdy}{y} = 0 \\ \Rightarrow \frac{dx}{x} - \frac{dy}{y} = 0 \\ Integrating both sides, we get \\ \int \frac{1}{x} dx - \int \frac{1}{y} dy = logC ‘ \\ logx - logy = logC ‘ \\ \Rightarrow \log (\frac{x}{y}) = logC ‘ \\ \Rightarrow \frac{x}{y} = C ‘ \\ \Rightarrow y = \frac{1}{C ‘} x \\ \Rightarrow y = Cxwhere, C = \frac{1}{C ‘} \\ This is the general solution of the given differential equation . \\ Hence, the correct option is C . \end{array}$

Q.112

$\begin{array}{l} The general solution of a differential equation of the type \\ \frac{dy}{dx} + P_{1} x = Q_{1} is \\ (A) {ye}^{\int P_{1} dy} = \int (Q_{1} e^{\int P_{1} dy}) dy + C \\ (B) y e^{\int P_{1} dX} = \int (Q_{1} e^{\int P_{1} dX}) dx + C \\ (C) x e^{\int P_{1} dy} = \int (Q_{1} e^{\int P_{1} dy}) dy + C \\ (D) {xe}^{\int P_{1} dx} = \int (Q_{1} e^{\int P_{1} dx}) dx + C \end{array}$

Ans.

$\begin{array}{l} The differential equation of the type \\ \frac{dx}{dy} + P_{1} x = Q_{1} is a linear differential equation . \\ I . F . = e^{\int P_{1} dy} \\ The general solution of given differential equation is : \\ x (I . F .) = \int (Q \times I . F .) dy + C \\ x e^{\int P_{1} dy} = \int (Q_{1} \times e^{\int P_{1} dy}) dy + C \\ Therefore, the correct option is C . \end{array}$

Q.113

$\begin{array}{l} The general solution of the differential equation \\ e^{x} dy + ({ye}^{x} + 2 x) dx = 0 is \\ (A) {xe}^{y} + x^{2} = C (B) x e^{y} + y^{2} = C \\ (C) {ye}^{x} + x^{2} = C (D) {ye}^{y} + x^{2} = C \end{array}$

Ans.

$\begin{array}{l} The differential equation is \\ e^{x} dy + ({ye}^{x} + 2 x) dx = 0 \\ \Rightarrow e^{x} \frac{dy}{dx} + {ye}^{x} = - 2 x \\ \Rightarrow \frac{dy}{dx} + y = - 2 {xe}^{- x} \\ This is a linear differential equation in the form of \\ \frac{dy}{dx} + Py = Q, where P = 1 and Q = - 2 {xe}^{- x} \\ Now, I . F . = e^{\int P dx} \\ = e^{\int 1 dx} \\ = e^{x} \\ The general solution of given differential equation is : \\ y (I . F .) = \int (Q \times I . F .) dx + C \\ \Rightarrow {ye}^{x} = \int (- 2 {xe}^{- x} \times e^{x}) dx + C \\ = - 2 \int x dx + C \\ = - 2 \frac{x^{2}}{2} + C \\ \Rightarrow {ye}^{x} = - x^{2} + C \\ \Rightarrow {ye}^{x} + x^{2} = C \\ Hence, the correct option is C . \end{array}$

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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1. How many problems are there in total in NCERT Solutions Class 12 Mathematics Chapter 9 Differential Equations?

NCERT Solutions Class 12 Mathematics Chapter 9 – Differential Equations has 113 questions separated into seven exercises. In addition to repetitive revision, a miscellaneous exercise consists of higher-order questions that push candidates to think outside of the box and help them apply their auxiliary mathematical skills. These are wonderful sums that aid students in gaining a comprehensive understanding of the subject.

2. Why should I practise NCERT Solutions Class 12 Mathematics Chapter 9?

Students will gain confidence in solving questions on this topic by regularly practising the NCERT Solutions Class 12 Mathematics Chapter 9. They will be able to attempt the paper stress-free throughout any examination, whether it is a board or competitive exam, resulting in imminent success. Aside from that, by revisiting these solutions, students can build a solid Mathematical foundation that will serve them well throughout their lives.

3. Is it necessary for one to practise all the NCERT Solutions for Class 12 mathematics Differential Equations?

Perfection requires a lot of practise. As a result, you should practise all the sums in the NCERT Solutions Class 12 Mathematics chapter 9 -Differential Equations at least twice. This chapter presents several new concepts built on previously taught material and, as a result, can be perplexing at times. Students not only pave the route to a higher exam result by solving each problem, but they also ensure that these topics come in use in real-life situations as well.

4. What is the significance of NCERT Solutions for Class 12 Mathematics Chapter 9?

The board exams are based on NCERT Solutions Class 12 Mathematics Chapter 9. CBSE papers are based on the NCERT textbook format. Furthermore, these issues are organised so that they provide readers with a 360-degree view of the chapter. Extramarks recommend students use these to gain a thorough understanding of the subject and double-check their responses after solving a question.

5. Are there answers to all the textbook problems in the NCERT Solutions for Class 12 Mathematics Chapter 9?

The NCERT Solutions for Class 12 Mathematics Chapter 9 are created by subject specialists at Extramarks and are accessible in the aforementioned links. These solutions are based entirely on the most recent CBSE Syllabus 2022-23, and they cover all of the key concepts for the first as well as the second term exams. The textbook questions are addressed in a step-by-step fashion based on marks weightage in the second-term exams.

NCERT Solutions Class 12 Mathematics Chapter 9

Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 9

List of NCERT Solutions Class 12 Mathematics Chapter 9 Exercises

Class 12 NCERT Mathematics Syllabus

Term – 1

Term – 2

NCERT CBSE Mathematics Exam Pattern

Key Features of NCERT Solutions for Class 12 Mathematics Chapter 9

NCERT Exemplar Class 12 CBSE Mathematics

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

FAQs (Frequently Asked Questions)

1. How many problems are there in total in NCERT Solutions Class 12 Mathematics Chapter 9 Differential Equations?

2. Why should I practise NCERT Solutions Class 12 Mathematics Chapter 9?

3. Is it necessary for one to practise all the NCERT Solutions for Class 12 mathematics Differential Equations?

4. What is the significance of NCERT Solutions for Class 12 Mathematics Chapter 9?

5. Are there answers to all the textbook problems in the NCERT Solutions for Class 12 Mathematics Chapter 9?

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