NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations (Ex 9.5) Exercise 9.5

The Central Board of Secondary Education (CBSE) Board is responsible for conducting board examinations of Class 12 students. The Class 12 Board examinations are crucial for deciding the future of Class 12 students. All the subjects of Class 12 should be prepared well for scoring higher marks in the board examinations. Nowadays, it has become important to access online resources to prepare for exams. All the online resources should be reviewed from time to time during the preparation period. Students are required to make a study timetable for learning the chapters of a subject.

Mathematical concepts can sometimes be challenging to learn. To learn Mathematics chapters, it is important to understand the concepts and keep practising questions at regular intervals.

Mathematics is an essential component of human intellect and logic. It promotes logical thinking and mental rigidity and is useful for building mental discipline. Additionally, studying Mathematics is crucial for understanding other subjects like Physics, Social Science, and even Music and Arts. Mathematical concepts are significant in finding solutions for real-world problems. It is necessary to learn the fundamentals of Mathematics in school. The CBSE Board students of Class 12 need to study some broad concepts of Mathematics like Integrals, Differential Equations, Probability, etc. All the students must go through each of the Class 12 Mathematics topics to prepare effectively for the Mathematics board exam.

The students of Class 12 are advised to thoroughly revise the chapters of Mathematics to secure higher marks in the exam. All the topics and subtopics should be consistently practised. It is necessary to practise chapters to enhance the knowledge of students. Students are expected to solve the problems regularly to score higher marks in the board exam of Mathematics. Revision is crucial in a subject like Mathematics. Some concepts of Mathematics are used in other concepts as well. It becomes necessary to revise them from time to time. Regularly practising exercise problems can help students understand the concepts.It is also useful for enhancing the memory of formulas, definitions, theorems, etc.

Before preparing for the mathematics board exam, students should review the mathematics syllabus.

The Class 12 Mathematics Syllabus can be accessed from the Extramarks’ website and mobile application. The students of Class 12 Central Board of Secondary Education (CBSE) Board are required to get an overview of all the topics and subtopics covered in the Mathematics syllabus. The syllabus is crucial for the students to prepare a study plan for learning Mathematics topics.

It is advisable that students practise the sample papers and past years’ papers in Mathematics to improve their preparation for the exam. Practising sample papers and previous years’ papers helps students understand the mathematics marking scheme.

With the help of sample papers and previous years’ papers in mathematics, students can learn to answer questions from the perspective of the board exam.

All the revision notes of Class 12 chapters need to be consistently reviewed. Students can access the revision notes of Mathematics from the Extramarks’ website and mobile application. It is important to prepare a study plan for learning Mathematics chapters.. All the chapters are important from the perspective of the exam. Students need to solve the exercise problems after learning the concepts. In Mathematics, it is critical to learn definitions, formulas, theorems, graphs, etc. to understand the concepts effectively.

The topic covered in Chapter 9 of Class 12 Mathematics is Differential Equations. It is necessary to recall the concepts of differentiation from Class 11 while learning the Differential Equations topics. Students must practise the exercise questions after understanding the concepts of Differential Equations. Exercise 9.5 is based on the topic of Homogeneous Differential Equations. The questions of Class 12th Exercise 9.5 are important from the board exam perspective. Class 12 students of the CBSE Board are expected to solve the Exercise 9.5 questions regularly. The students can solve the Ex 9.5 Class 12 questions with the help of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5. Students can access the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 from the Extramarks’ website and mobile application in PDF format. Students must go through the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 one by one and practise Exercise 9.5 Class 12 problems.

Students are required to include the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 in their learning process. Some of the topics in Chapter 9 are difficult to understand. It is necessary to focus more on difficult topics. Students of the Class 12 Central Board of Secondary Education (CBSE) are supposed to make a list of all the difficult topics and spend more time preparing them.

There are a total of 17 questions in Exercise 9.5.All the questions are based on the Homogeneous Differential Equations topic. The students need to understand the requirements of each question before solving them. All the required formulas, definitions, and theorems need to be applied to solve the Exercise 9.5 questions effectively. Students must go through each of the topics in Class 12 Mathematics from the perspective of the exam. The topicsshould be studied one by one to build the concepts gradually.

The mathematical concepts learned at the Class 12 level are significant for understanding the topics at the next level. Class 12 Mathematics concepts help prepare for competitive exams like JEE Mains, JEE Advance, etc. It is important to practise the Class 12 Mathematics concepts thoroughly to learn the advanced concepts of Mathematics in various courses like Engineering, Bachelor of Science, etc. Students should focus on improving their problem-solving skills to solve Mathematics related questions.

Students must solve the Exercise 9.5 questions. The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 help solve the Exercise 9.5 questions. Students are advised to utilise the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 during the board exam preparation of Mathematics.

Questions asked in the Mathematics board exam of Class 12 can be daunting to solve. Students can practice well for the Mathematics exam with the help of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5. Students of the Class 12 Central Board of Secondary Education must access the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 from the Extramarks’ website and mobile application in PDF format. The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 are accessible in PDF format and can be viewed in offline mode as well.

NCERT Solutions are made available by Extramarks for all CBSE Board classes. From Extramarks’ website and mobile application, students can download the NCERT solutions in PDF format. Students can get help from Extramarks in getting ready for competitive exams like the JEE Mains, JEE Advanced, CUET, NEET, etc. On Extramarks’ website and mobile application, students may access all the study materials necessary for gearing up for the different competitive tests. The Extramarks’ website and mobile application provide access to study materials for students belonging to the ICSE, CBSE, and various state boards. To get better grades in examinations, students must continually refer to all the study resources.

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations (Ex 9.5) Exercise 9.5

The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 can be downloaded from the Extramarks’ website and mobile application in PDF format. Students having problems in solving the questions of Exercise 9.5 are advised to refer to the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5. Practising Exercise 9.5 questions with the help of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 helps prepare effectively for the board exam of Mathematics. All the solutions of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 are prepared by expert Mathematics teachers.

The solutions help boost the confidence of students for the board exam of Mathematics. All the topics need to be covered in detail. Students are advised to keep revising the topics to get in-depth knowledge of the concepts. All the topics of Chapter 9 are crucial from the board exam point of view.

Students are encouraged to make use of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 to prepare well for the Mathematics examination. All the solutions given in the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 can be found just below their respective questions.

Mathematics is a subject that can fetch higher marks in exams. Students should practise well for the upcoming board exam of Mathematics to score higher marks. All the students are required to go through the topics consistently and practise the exercise questions. The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 are beneficial for enhancing the answer writing capacity of the students. Students of the Class 12 CBSE Board should keep referring to the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 to boost their preparations for the board exam of Mathematics.

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Question 17 of Exercise 9.5 requires students to find the Homogeneous Differential Equation from the given equations. To solve question 17 efficiently it is important to make use of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5.

Practice is crucial in preparing for the board exam in Mathematics. All the exercise questions related to the Differential Equations chapter should be practised well before the Mathematics exam. Students are advised to learn the topics thoroughly before solving the exercise questions. All points of a topic should be revised several times.Students are advised to keep studying the theory of the chapters and increase their understanding of them. Students must improve their writing speed to score well in the Mathematics exam. Students must solve the exercise questions from time to time to increase their problem-solving speed in Mathematics.

It is crucial for students to learn the appropriate implementation of formulas and theorems. Students are advised to solve the problems regularly to learn the exact methods of solving a problem. Students should make use of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 to learn how to apply the correct formulas to the questions and get the desired results. The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 are beneficial for enhancing the problem-solving skills of students.

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Important Topics

Important topics in Chapter 9 need to be covered very carefully. Students should practise the questions on significant topics more often. The important topics are significant for scoring higher marks in the board exam of Mathematics. Each chapter’s major questions must be answered by students. Students must use the Extramarks website and mobile application to obtain the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5. Experienced Mathematics teachers at Extramarks have prepared the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5. The solutions are useful for breaking complex problems into easier-to-understand, simpler, and smaller steps. It is suggested that students create a thorough list of all the topics and study them regularly to perform well in the Mathematics exam.

To increase students’ self-confidence, NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 is beneficial. Students of the Central Board of Secondary Education (CBSE) Board in Class 12 are advised to review the CBSE Mathematics Syllabus. It is advisable to create a separate strategy for each topic in Mathematics. The Syllabus can be used to create a learning plan for the various chapters.

The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 are useful for effectively grasping the Homogeneous Differential Equations topic. Applying formulas to Exercise 9.5 questions can be learned by using the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5. The Differential Equations chapter’s exercises should all be consistently practised.

Tips to solve NCERT Class 12 Maths Chapter 9 Exercise 9.5

The NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 can be easily accessed from the Extramarks’ website and mobile application in PDF format.

Access NCERT Solutions for Class 12 Maths Chapter 9 –Differential Equations

NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Exercise 9.5

It is crucial to access the online study materials from credible sources to prepare effectively for the Class 12 Mathematics Exam. The resources available on Extramarks’ website and mobile application are very reliable and trustworthy. Students of Class 12 CBSE Board can access the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5.

To solve the questions of Exercise 9.5, it is necessary to keep revising the concepts of Homogeneous Differential Equations at regular intervals.Students are advised to practise the questions based on the Homogeneous Differential Equations topic. There are solved examples given in the NCERT textbook of Class 12 Mathematics related to the concepts of Chapter 9. The solved examples are meant for students to get an idea of how questions can be solved related to each topic. Students need to thoroughly review the solved examples from time to time. The exercise questions are asked based on concepts and should be readily practised by the students. The questions in Exercise 9.5 can be solved with the help of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5.

Students are advised to follow the study plan for Mathematics regularly to be on the right track for Mathematics exam preparation. Students need to access all the study materials that help in preparing effectively for the Mathematics exam. All the study materials are available on the Extramarks website and mobile application. Students need to keep referring to the study materials while preparing for the Mathematics exam.

NCERT Solutions for Class 12 Maths PDF Download

The NCERT solutions are extremely useful as study resources for the mathematics exam.The Class 12 students of the Central Board of Secondary Education can download the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 from the learning app in PDF format. Students are advised to make use of the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.5 to enhance their ongoing preparations for the board exam of Mathematics.

The NCERT textbooks are the main source of studying for Class 12 chapters. It is recommended that students rely on the NCERT textbooks to study Class 12 Mathematics chapters. The students are advised to keep evaluating their preparation for the Mathematics exams at regular intervals. Evaluation is crucial for identifying the challenges, and students can overcome those challenges with the help of better planning.

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Q.1 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.2 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.3 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.4 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.5 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.6 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.7 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.8 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.9 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.10 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.11 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.12 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.13 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.14 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.15 For the differential equation given below, find a particular situation satisfying the given condition:

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

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.16

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

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.17 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.

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

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NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations (Ex 9.5) Exercise 9.5

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