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NCERT Solutions for Exercise 9.2 Class 12 Maths Chapter 9- Differential Equations

NCERT Solutions for Exercise 9.2 Class 12 Maths Chapter 9- Differential Equations

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:09 AM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2

NCERT Solutions for Exercise 9.2 Class 12 Maths Chapter 9 Differential Equations are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for exercise 9.2 Class 12 Maths chapter 9 gives questions and solutions around the concepts general and particular solutions of differential equations. After exercise 9.1 the NCERT explains the concepts of arbitrary constants, general solutions and particular solutions. A few sample question based on topic 9.3 is given in the NCERT books to understand the ideas given in the book.

This Story also Contains
  1. NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2
  2. Access NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2
  3. Differential Equations Class 12 Chapter 9 Exercise: 9.2
  4. More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2
  5. Also Read| Differential Equations Class 12th Notes
  6. Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2
  7. Key Features Of NCERT Solutions for Exercise 9.2 Class 12 Maths Chapter 9
  8. NCERT Solutions Subject Wise
  9. Subject Wise NCERT Exemplar Solutions

The exercise 9.2 Class 12 Maths are questions for practice. Solving Class 12 Maths chapter 9 exercise 9.2 gives an idea of concepts studied after the first exercise. Class 12th Maths chapter 6 exercise 9.2 are solved here by expert mathematics faculties. Not only the exercise 9.2 Class 12 Maths are solved. 12th class Maths exercise 9.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Access NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2

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Differential Equations Class 12 Chapter 9 Exercise: 9.2

Question:1 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = e^x + 1 \qquad :\ y'' -y'=0

Answer:

Given,

y = e^x + 1

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x

Again, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y' }{\mathrm{d} x} = \frac{\mathrm{d}e^x }{\mathrm{d} x} = e^x

\implies y'' = e^x

Substituting the values of y’ and y'' in the given differential equations,

y'' - y' = e x - e x = 0 = RHS.

Therefore, the given function is the solution of the corresponding differential equation.

Question:2 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = x^2 + 2x + C\qquad:\ y' -2x - 2 =0

Answer:

Given,

y = x^2 + 2x + C

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x^2 + 2x + C) = 2x + 2

Substituting the values of y’ in the given differential equations,

y' -2x - 2 =2x + 2 - 2x - 2 = 0= RHS .

Therefore, the given function is the solution of the corresponding differential equation.

Question:3. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = \cos x + C\qquad :\ y' + \sin x = 0

Answer:

Given,

y = \cos x + C

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(cosx + C) = -sinx

Substituting the values of y’ in the given differential equations,

y' - \sin x = -sinx -sinx = -2sinx \neq RHS .

Therefore, the given function is not the solution of the corresponding differential equation.

Question:4. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = \sqrt{1 + x^2}\qquad :\ y' = \frac{xy}{1 + x^2}

Answer:

Given,

y = \sqrt{1 + x^2}

Now, differentiating both sides w.r.t. x,

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

Substituting the values of y in RHS,

\frac{x\sqrt{1+x^2}}{1 + x^2} = \frac{x}{\sqrt{1+x^2}} = LHS .

Therefore, the given function is a solution of the corresponding differential equation.

Question:5 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = Ax\qquad :\ xy' = y\;(x\neq 0)

Answer:

Given,

y = Ax

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(Ax) = A

Substituting the values of y' in LHS,

xy' = x(A) = Ax = y = RHS .

Therefore, the given function is a solution of the corresponding differential equation.

Question:6. Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y = x\sin x\qquad :\ xy' = y + x\sqrt{x^2 - y^2}\ (x\neq 0\ \textup{and} \ x > y\ or \ x < -y)

Answer:

Given,

y = x\sin x

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(xsinx) = sinx + xcosx

Substituting the values of y' in LHS,

xy' = x(sinx + xcosx)

Substituting the values of y in RHS.

\\xsinx + x\sqrt{x^2 - x^2sin^2x} = xsinx + x^2\sqrt{1-sinx^2} = x(sinx+xcosx) = LHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:7 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

xy = \log y + C\qquad :\ y' = \frac{y^2}{1 - xy}\ (xy\neq 1)

Answer:

Given,

xy = \log y + C

Now, differentiating both sides w.r.t. x,

\\ y + x\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(logy) = \frac{1}{y}\frac{\mathrm{d}y }{\mathrm{d} x}\\ \\ \implies y^2 + xyy' = y' \\ \\ \implies y^2 = y'(1-xy) \\ \\ \implies y' = \frac{y^2}{1-xy}

Substituting the values of y' in LHS,

y' = \frac{y^2}{1-xy} = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:8 In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

y - cos y = x \qquad :(\ y\sin y + \cos y + x) y' = y

Answer:

Given,

y - cos y = x

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d}y }{\mathrm{d} x} +siny\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x}(x) = 1

\implies y' + siny.y' = 1

\implies y'(1 + siny) = 1

\implies y' = \frac{1}{1+siny}

Substituting the values of y and y' in LHS,

(\ (x+cosy)\sin y + \cos y + x) (\frac{1}{1+siny})

= [x(1+siny) + cosy(1+siny)]\frac{1}{1+siny}

= (x + cosy) = y = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:9 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

x + y = \tan^{-1}y\qquad :\ y^2y' + y^2 + 1 = 0

Answer:

Given,

x + y = \tan^{-1}y

Now, differentiating both sides w.r.t. x,

\\ 1 + \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{1 + y^2}\frac{\mathrm{d} y}{\mathrm{d} x}\\ \\ \implies1+y^2 = y'(1-(1+y^2)) = -y^2y' \\ \implies y' = -\frac{1+y^2}{y^2}

Substituting the values of y' in LHS,

y^2(-\frac{1+y^2}{y^2}) + y^2 + 1 = -1- y^2+ y^2 +1 = 0 = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:10 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)\qquad : \ x + y \frac{dy}{dx} = 0\ (y\neq 0)

Answer:

Given,

y = \sqrt{a^2 - x^2}

Now, differentiating both sides w.r.t. x,

\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{\mathrm{d}}{\mathrm{d} x}(\sqrt{a^2 - x^2}) = \frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}}

Substituting the values of y and y' in LHS,

x + y \frac{dy}{dx} = x + (\sqrt{a^2 - x^2})(\frac{-x}{\sqrt{a^2 - x^2}}) = 0 = RHS

Therefore, the given function is a solution of the corresponding differential equation.

Question:11 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

Answer:

(D) 4

The number of constants in the general solution of a differential equation of order n is equal to its order.

Question:12 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

Answer:

(D) 0

In a particular solution of a differential equation, there is no arbitrary constant.

More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2

Two examples are given before exercise 9.2 Class 12 Maths. These examples and the Class 12 Maths chapter 9 exercise 9.2 questions are to verify the given function is a solution of the given differential equation. There are 12 questions out of which two are multiple-choice questions. Multiple choice questions of NCERT solutions for Class 12 Maths chapter 9 exercise 9.2 are to find the number of arbitrary constants in general and particular solutions of a differential equation.

Also Read| Differential Equations Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2

  • The solutions of exercise 9.2 Class 12 Maths are curated by mathematics subject matter experts and are according to the CBSE pattern.
  • The NCERT solutions for Class 12 Maths chapter 9 exercise 9.2 can be used to prepare for the CBSE board exam and also for different state boards that follow NCERT Syllabus.
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Key Features Of NCERT Solutions for Exercise 9.2 Class 12 Maths Chapter 9

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 9.2 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 9.2, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 9.2 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 9.2 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 9.2 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 9.2 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

1. What is the number of examples given in the topic 9.3 of Class 12 NCERT Mathematics Book?

Two examples are given to understand the concepts explained in exercise 9.3.

2. How many exercises are there in NCERT solutions for Class 12 Maths chapter 9?

A total of 7 exercises including the miscellaneous are there in the chapter differential equation.

3. How many questions are covered in exercise 9.2?

12 questions are discussed in the Class 12th Maths chapter 6 exercise 9.2 .

4. Whether the chapter differential equations is important for the Class 12 exam?

Yes. Solving Exercise 9.2 Class 12 Maths is necessary to prepare for the board exam of Class 12.

5. What are some basics knowledge required to solve the questions in differential equations?

The knowledge of differentiation and integration is required.

6. What is the topic after exercise 9.2?

Topic 9.4 is about how to form a differential equation if the general solutions are given.

7. Why solve exercise 9.1?

It is necessary to understand the basics before proceeding with the chapter. Exercise 9.1 gives a good knowledge of the degree and order of differential equations. So it is necessary to solve this exercise

8. Is it necessary to solve exercise 9.2?

Yes. To understand the concepts discussed in topic 9.3 it is necessary to solve exercise 9.2  

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Divyan 06 August,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

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Yuvan 01 September,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

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