NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.2

Q: 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.

Q: 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.

Q: How many questions are covered in exercise 9.2?

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

Q: 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.

Q: What are some basics knowledge required to solve the questions in differential equations?

The knowledge of differentiation and integration is required.

Q: 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.

Q: 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

Q: 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

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

Edited By Komal Miglani | Updated on May 08, 2025 02:27 PM IST | #CBSE Class 12th

Suppose you are watching the way a drug dissolves in your blood after some time. The rate of variation of the concentration can be described by a differential equation. These real-life situations, in which the rate of change is dependent on the quantity itself, are the best examples of how the general and particular solutions of differential equations are used.

NCERT Solutions for Class 12 Maths Chapter 9 – Exercise 9.2 are explained in a plain, step-by-step manner so that the students can easily understand these important concepts. The NCERT solutions are prepared by experienced mathematics faculty members of Careers360, strictly following the CBSE 2025–26 syllabus. In NCERT solution for Class 12 Maths Chapter 9 Exercise 9.2 students learn the difference between a general solution (with arbitrary constants) and a particular solution (obtained after assigning definite constants), differentiating and integrating to move between solutions and equations, and apply the basic concepts learned in Exercise 9.1 to more practical solution problem scenarios. Proper practice of these exercises not only increases confidence for board exams but is also beneficial for competitive exams such as the JEE.

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This Story also Contains

Class 12 Maths Chapter 9 Exercise 9.2 Solutions: Download PDF
NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.2
Question:1 Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
Topics covered in Chapter 9 Differential Equation: Exercise 9.2
NCERT Solutions Subject Wise
Subject Wise NCERT Exemplar Solutions

Class 12 Maths Chapter 9 Exercise 9.2 Solutions: Download PDF

This material provides easy solutions to all the questions of Exercise 9.2 of Differential Equations. The PDF can be downloaded by the students for practice and enhancement of the chapter for board and and competitive exams

Download PDF

NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.2

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

$y = e^{x} + 1 : y^{″} - y^{'} = 0$

Answer:

Given,

$y = e^{x} + 1$

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

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

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

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

$⟹ y^{″} = e^{x}$

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:

$y = x^{2} + 2 x + C : y^{'} - 2 x - 2 = 0$

Answer:

Given,

$y = x^{2} + 2 x + C$

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

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

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

$y^{'} - 2 x - 2 = 2 x + 2 - 2 x - 2 = 0 = R H S$ .

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:

$y = \cos x + C : y^{'} + \sin x = 0$

Answer:

Given,

$y = \cos x + C$

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

$\frac{d y}{d x} = \frac{d}{d x} (c o s x + C) = - s i n x$

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

$y^{'} - \sin x = - s i n x - s i n x = - 2 s i n x \neq R H S$ .

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:

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

Answer:

Given,

$y = \sqrt{1 + x^{2}}$

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

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

Substituting the values of y in RHS,

$\frac{x \sqrt{1 + x^{2}}}{1 + x^{2}} = \frac{x}{\sqrt{1 + x^{2}}} = L H S$ .

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:

$y = A x : x y^{'} = y (x \neq 0)$

Answer:

Given,

$y = A x$

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

$\frac{d y}{d x} = \frac{d}{d x} (A x) = A$

Substituting the values of y' in LHS,

$x y^{'} = x (A) = A x = y = R H S$ .

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:

$y = x \sin x : x y^{'} = y + x \sqrt{x^{2} - y^{2}} (x \neq 0 and x > y o r x < - y)$

Answer:

Given,

$y = x \sin x$

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

$\frac{d y}{d x} = \frac{d}{d x} (x s i n x) = s i n x + x c o s x$

Substituting the values of y' in LHS,

$x y^{'} = x (s i n x + x c o s x)$

Substituting the values of y in RHS.

$x s i n x + x \sqrt{x^{2} - x^{2} s i n^{2} x} = x s i n x + x^{2} \sqrt{1 - s i n x^{2}} = x (s i n x + x c o s x) = L H S$

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:

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

Answer:

Given,

$x y = \log y + C$

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

$y + x \frac{d y}{d x} = \frac{d}{d x} (l o g y) = \frac{1}{y} \frac{d y}{d x}$ \ \

$⟹ y^{2} + x y y^{'} = y^{'}$

$⟹ y^{2} = y^{'} (1 - x y)$

$⟹ y^{'} = \frac{y^{2}}{1 - x y}$

Substituting the values of y' in LHS,

$y^{'} = \frac{y^{2}}{1 - x y} = R H S$

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:

$y - c o s y = x : (y \sin y + \cos y + x) y^{'} = y$

Answer:

Given,

$y - c o s y = x$

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

$\frac{d y}{d x} + s i n y \frac{d y}{d x} = \frac{d}{d x} (x) = 1$

$⟹$ y' + siny.y' = 1

$⟹$ y'(1 + siny) = 1

$⟹ y^{'} = \frac{1}{1 + s i n y}$

Substituting the values of y and y' in LHS,

$((x + c o s y) \sin y + \cos y + x) (\frac{1}{1 + s i n y})$

$= [x (1 + s i n y) + c o s y (1 + s i n y)] \frac{1}{1 + s i n y}$

= (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:

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

Answer:

Given,

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

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

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

$⟹ 1 + y^{2} = y^{'} (1 - (1 + y^{2})) = - y^{2} y^{'}$

$⟹ y^{'} = - \frac{1 + y^{2}}{y^{2}}$

Substituting the values of y' in LHS,

$y^{2} (- \frac{1 + y^{2}}{y^{2}}) + y^{2} + 1 = - 1 - y^{2} + y^{2} + 1 = 0 = R H S$

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:

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

Answer:

Given,

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

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

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

Substituting the values of y and y' in LHS,

$x + y \frac{d y}{d x} = x + (\sqrt{a^{2} - x^{2}}) (\frac{- x}{\sqrt{a^{2} - x^{2}}}) = 0 = R H S$

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 constan

Also check -

Topics covered in Chapter 9 Differential Equation: Exercise 9.2

Topic	Description	Example
General Solution	A solution with general constants that determine a class of curves.	$y = C e^{x}$ is a general solution of $\frac{d y}{d x} = y$
Particular Solution	A specific solution obtained by giving a particular value to the arbitrary constant(s).	If $y = C e^{x}$ and $y = 2$ when $x = 0$ , then $C = 2$ , so $y =$ $2 e^{x}$
Arbitrary Constants	Constants that are used in integration. Used to represent general solutions.	$C$ in $y = C e^{- x}$
Formulating Differential Equations	Making of a differential equation from a known general solution by eliminating the arbitrary constants.	Given: $y = A e^{x} + B e^{- x}$ , eliminate $A$ and $B$ to form the DE.

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Frequently Asked Questions (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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Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

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Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

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Explore Alternative Options:
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Focus on NEET 2025 Preparation:
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- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
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- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,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,

Qualifications:
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Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

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

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

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

Class 12 Maths Chapter 9 Exercise 9.2 Solutions: Download PDF

NCERT Solutions Class 12 Maths Chapter 9: Exercise 9.2

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

Topics covered in Chapter 9 Differential Equation: Exercise 9.2

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