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.

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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## Differential Equations Class 12 Chapter 9 Exercise: 9.2

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

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.

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

Given,

$\dpi{100} y = x^2 + 2x + C$

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

$y = \sqrt{a^2 - x^2}\ x\in (-a,a)\qquad : \ x + y \frac{dy}{dx} = 0\ (y\neq 0)$

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.

(D) 4

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

(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.

## 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.

## Subject Wise NCERT Exemplar Solutions

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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### Questions related to CBSE Class 12th

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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.

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Hello student,

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• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
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• 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.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

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 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

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 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

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 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9