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NCERT Solutions for Exercise 9.3 Class 12 Maths Chapter 9- Differential Equations

NCERT Solutions for Exercise 9.3 Class 12 Maths Chapter 9- Differential Equations

Edited By Ramraj Saini | Updated on Dec 04, 2023 08:15 AM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.3

NCERT Solutions for Exercise 9.3 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 9.3 Class 12 Maths chapter 9 comes under topic 9.4 of NCERT Class 12 Mathematics Book. The questions in exercise 9.3 Class 12 Maths are related to the concepts of forming a Differential Equation that represents a given family of curves. Differential equations related to the family of lines, parabolas, ellipse, circles etc are discussed. The procedures to form the differential equation that represents the family of a curve are detailed before the Class 12th Maths chapter 6 exercise 9.3. The other concepts discussed in the unit are detailed in the following exercises.

This Story also Contains
  1. NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.3
  2. Access NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3
  3. Differential Equations Class 12 Chapter 9 Exercise: 9.3
  4. More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3
  5. Also Read| Differential Equations Class 12th Notes
  6. Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3
  7. The Class 12 Maths chapter 9 exercise 9.3 are curated by mathematics faculties and can be used for the preparation of the CBSE board exams.
  8. Rather than exam point of view solving exercise will give an insight to the concepts studied.
  9. Key Features Of NCERT Solutions for Exercise 9.3 Class 12 Maths Chapter 9
  10. Also see-
  11. NCERT Solutions Subject Wise
  12. Subject Wise NCERT Exemplar Solutions

12th class Maths exercise 9.3 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 9.3

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Differential Equations Class 12 Chapter 9 Exercise: 9.3

Question:1 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

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

Answer:

Given equation is

\frac{x}{a} + \frac{y}{b} = 1
Differentiate both the sides w.r.t x
\frac{d\left ( \frac{x}{a}+\frac{y}{b} \right )}{dx}=\frac{d(1)}{dx}
\frac{1}{a}+\frac{1}{b}.\frac{dy}{dx} = 0\\ \frac{dy}{dx} = -\frac{b}{a}
Now, again differentiate it w.r.t x
\frac{d^2y}{dx^2} =0
Therefore, the required differential equation is \frac{d^2y}{dx^2} =0 or y^{''} =0

Question:2 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y^2 = a(b^2 - x^2)

Answer:

Given equation is
y^2 = a(b^2 - x^2)
Differentiate both the sides w.r.t x
\frac{d\left ( y^2 \right )}{dx}=\frac{d(a(b^2-x^2))}{dx}
2y\frac{dy}{dx}= -2ax\\ \\ y.\frac{dy}{dx}= -ax\\ \\ y.y^{'}=-ax -(i)
Now, again differentiate it w.r.t x
y^{'}.y^{'}+y.y^{''}= -a\\ (y^{'})^2+y.y^{''}=-a -(ii)
Now, divide equation (i) and (ii)
\frac{(y^{'})^2+y.y^{''}}{y.y^{'}}= \frac{-a}{-ax}\\ \\ x(y^{'})^2+x.y.y^{''}=y.y^{'}\\ \\ x(y^{'})^2+x.y.y^{''}-y.y^{'}=0
Therefore, the required differential equation is x(y^{'})^2+x.y.y^{''}-y.y^{'}=0

Question:3 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. y = ae^{3x} + b e^{-2x}

Answer:

Given equation is
y = ae^{3x} + b e^{-2x} -(i)
Differentiate both the sides w.r.t x
\frac{d\left ( y \right )}{dx}=\frac{d(ae^{3x}+be^{-2x})}{dx}
y^{'}=\frac{dy}{dx}= 3ae^{3x}-2be^{-2x}\\ \\ -(ii)
Now, again differentiate w.r.t. x
y^{''}= \frac{d^2y}{dx^2} = 9ae^{3x}+4be^{-2x} -(iii)
Now, multiply equation (i) with 2 and add equation (ii)
2(ae^{3x}+be^{-2x})+(3a-2be^{-x}) = 2y+y^{'}\\ 5ae^{3x} = 2y+y^{'}\\ ae^{3x}= \frac{2y+y^{'}}{5} -(iv)
Now, multiply equation (i) with 3 and subtract from equation (ii)
3(ae^{3x}+be^{-2x})-(3a-2be^{-x}) = 3y-y^{'}\\ 5be^{-2x} = 3y-y^{'}\\ be^{-2x}= \frac{3y-y^{'}}{5} -(v)
Now, put values from (iv) and (v) in equation (iii)
y^{''}= 9.\frac{2y+y^{'}}{5}+4.\frac{3y-y^{'}}{5}\\ \\ y^{''}= \frac{18y+9y^{'}+12y-4y^{'}}{5}\\ \\ y^{''}= \frac{5(6y-y^{'})}{5}=6y-y^{'}\\ \\ y^{''}+y^{'}-6y=0

Therefore, the required differential equation is y^{''}+y^{'}-6y=0

Question:4 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b. y = e^{2x}(a+bx)

Answer:

Given equation is
y = e^{2x}(a+bx) -(i)
Now, differentiate w.r.t x
\frac{dy}{dx}= \frac{d(e^{2x}(a+bx))}{dx}= 2e^{2x}(a+bx)+e^{2x}.b -(ii)
Now, again differentiate w.r.t x
y^{''}= \frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx} = 4e^{2x}(a+bx)+2be^{2x}+2be^{2x}= 4e^{2x}(a+bx)+4be^{2x} -(iii)
Now, multiply equation (ii) with 2 and subtract from equation (iii)
4e^{2x}(a+bx)+4be^{2x}-2\left ( 2e^{2x}(a+bx)+be^{2x} \right )=y^{''}-2y^{'}\\ \\ 2be^{2x} = y^{''}-2y^{'}\\ \\ be^{2x}= \frac{y^{''}-2y^{'}}{2} -(iv)
Now,put the value in equation (iii)
y^{''}=4y+4.\frac{y^{''}-2y^{'}}{2}\\ \\ y^{''}= 4y+2y^{''}-4y^{'}\\ \\ y^{''}-4y^{'}+4y=0
Therefore, the required equation is y^{''}-4y^{'}+4y=0

Question:5 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y=e^x(a\cos x + b\sin x)

Answer:

Given equation is
y=e^x(a\cos x + b\sin x) -(i)
Now, differentiate w.r.t x
\frac{dy}{dx}= \frac{d(e^{x}(a\cos x+b\sin x))}{dx}= e^{x}(a\cos x+b\sin x)+e^x(-a\sin x+b\cos x ) -(ii)
Now, again differentiate w.r.t x
y^{''}= \frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx} =e^{x}(a\cos x+b\sin x)+e^x(-a\sin x+b\cos x ) +e^x(-a\sin x+b\cos x )+e^x(-a\cos x-b\sin x)
=2e^x(-a\sin x+b\cos x ) -(iii)
Now, multiply equation (i) with 2 and multiply equation (ii) with 2 and add and subtract from equation (iii) respectively
we will get

y^{''}-2y^{'}+2y = 0
Therefore, the required equation is y^{''}-2y^{'}+2y = 0

Question:6 Form the differential equation of the family of circles touching the y-axis at origin.

Answer:

1628484808301 If the circle touches y-axis at the origin then the centre of the circle lies at the x-axis
Let r be the radius of the circle
Then, the equation of a circle with centre at (r,0) is
(x-r)^2+(y-0)^2 = r^2
x^2+r^2-2xr+y^2=r^2\\ x^2+y^2-2xr=0 -(i)
Now, differentiate w.r.t x
2x+2y\frac{dy}{dx}-2r=0\\ y\frac{dy}{dx}\Rightarrow yy^{'}+x-r=0
yy^{'}+x=r -(ii)
Put equation (ii) in equation (i)
x^2+y^2=2x(yy^{'}+x)\\ y^2=2xyy^{'}+x^2
Therefore, the required equation is y^2=2xyy^{'}+x^2

Question:7 Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Answer:

Equation of perabola having vertex at origin and axis along positive y-axis is
x^2= 4ay (i)
Now, differentiate w.r.t. c
2x= 4a\frac{dy}{dx}\\ \\ \frac{dy}{dx} =y^{'}= \frac{x}{2a}
a=\frac{x}{2y^{'}} -(ii)
Put value from equation (ii) in (i)
x^2= 4y.\frac{x}{2y^{'}}\\ xy^{'}-2y = 0
Therefore, the required equation is xy^{'}-2y = 0

Question:8 Form the differential equation of the family of ellipses having foci on y-axis and centre at origin.

Answer:

Equation of ellipses having foci on y-axis and centre at origin is
\frac{x^2}{b^2}+\frac{y^2}{a^2} = 1 -
Now, differentiate w..r.t. x
\frac{2x}{b^2}+\frac{2y}{a^2}.\frac{dy}{dx}=0\\ -(i)
Now, again differentiate w.r.t. x
\frac{2}{b^2}+\frac{2}{a^2}.y^{'}.y^{'}+\frac{2y}{a^2}.y^{''}=0\\ \\ \frac{1}{b^2}=-\frac{1}{a^2}\left ( (y^{'})^2+yy^{''} \right ) -(ii)
Put value from equation (ii) in (i)
Our equation becomes
\frac{2y}{a^2}y^{'}-\frac{2x}{a^2}\left ( (y^{'})^2+yy^{''} \right )=0\\ \\ 2yy^{'}-2(y^{'})^2x+2yy^{''}x=0\\ \\ xyy^{''}-x(y^{'})^2+yy^{'}= 0
Therefore, the required equation is xyy^{''}-x(y^{'})^2+yy^{'}= 0

Question:9 Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.

Answer:

Equation of hyperbolas having foci on x-axis and centre at the origin
\frac{x^2}{b^2}+\frac{y^2}{a^2} = 1
Now, differentiate w..r.t. x
\frac{2x}{b^2}+\frac{2y}{a^2}.\frac{dy}{dx}=0\\ -(i)
Now, again differentiate w.r.t. x
\frac{2}{b^2}+\frac{2}{a^2}.y^{'}.y^{'}+\frac{2y}{a^2}.y^{''}=0\\ \\ \frac{1}{b^2}=-\frac{1}{a^2}\left ( (y^{'})^2+yy^{''} \right ) -(ii)
Put value from equation (ii) in (i)
Our equation becomes
\frac{2y}{a^2}y^{'}-\frac{2x}{a^2}\left ( (y^{'})^2+yy^{''} \right )=0\\ \\ 2yy^{'}-2(y^{'})^2x+2yy^{''}x=0\\ \\ xyy^{''}-x(y^{'})^2+yy^{'}= 0
Therefore, the required equation is xyy^{''}-x(y^{'})^2+yy^{'}= 0

Question:10 Form the differential equation of the family of circles having centre on y-axis and radius 3 units.

Answer:

1628484856150 Equation of the family of circles having centre on y-axis and radius 3 units
Let suppose centre is at (0,b)
Now, equation of circle with center (0,b) an radius = 3 units
(x-0)^2+(y-b)^2=3^2 \ \ \ \ \ \ \ \ \ \ \ -(i)\\ x^2+y^2+b^2-2yb = 9
Now, differentiate w.r.t x
we get,
2x+2yy^{'}-2by^{'}= 0\\ 2x+2y(y-b)= 0\\ (y-b)=\frac{-x}{y^{'}} \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
Put value fro equation (ii) in (i)
(x-0)^2+(\frac{-x}{y^{'}})^2=3^2 \\ x^2+\frac{x^2}{(y^{'})^2}=9\\ x^2(y^{'})^2+x^2=9(y^{'})^2\\ \\ (x^2-9)(y^{'})^2+x^2 = 0
Therefore, the required differential equation is (x^2-9)(y^{'})^2+x^2 = 0

Question:11 Which of the following differential equations has y = c_1e^x + c_2e^{-x} as the general solution?

(A) \frac{d^2y}{dx^2} + y = 0

(B) \frac{d^2y}{dx^2} - y = 0

(C) \frac{d^2y}{dx^2} +1 = 0

(D) \frac{d^2y}{dx^2} -1 = 0

Answer:

Given general solution is
y = c_1e^x + c_2e^{-x}
Differentiate it w.r.t x
we will get
\frac{dy}{dx} = c_1e^x-c_2e^{-x}
Again, Differentiate it w.r.t x
\frac{d^2y}{dx^2} = c_1e^x+c_2e^{x}=y\\ \frac{d^2y}{dx^2} - y = 0
Therefore, (B) is the correct answer

Question:12 Which of the following differential equations has y = x as one of its particular solution?

(A) \frac{d^2y}{dx^2} - x^2\frac{dy}{dx} + xy =x

(B) \frac{d^2y}{dx^2} + x\frac{dy}{dx} + xy =x

(C) \frac{d^2y}{dx^2} - x^2\frac{dy}{dx} + xy =0

(D) \frac{d^2y}{dx^2} + x\frac{dy}{dx} + xy =0

Answer:

Given equation is
y = x
Now, on differentiating it w.r.t x
we get,
\frac{dy}{dx} = 1
and again on differentiating it w.r.t x
we get,

\frac{d^2y}{dx^2} = 0
Now, on substituting the values of \frac{d^2y}{dx^2} , \frac{dy}{dx} \ and \ y in all the options we will find that only option c which is \frac{d^2y}{dx^2} - x^2\frac{dy}{dx} + xy =0 satisfies
Therefore, the correct answer is (C)

More About NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3

Examples 4 to 8 are given before the exercise 9.3 Class 12 Maths to get an idea of the concepts discussed in the NCERT Book of Class 12 Maths chapter topic 9.4. And 12 questions are given in the Class 12 Maths chapter 9 exercise 9.3 for practice and these are solved here by mathematics expert faculties. It is better to try to solve the questions without looking for solutions. If any doubts arise while solving use NCERT solutions for Class 12 Maths chapter 9 exercise 9.3.

Also Read| Differential Equations Class 12th Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.3

  • The Class 12 Maths chapter 9 exercise 9.3 are curated by mathematics faculties and can be used for the preparation of the CBSE board exams.

  • Rather than exam point of view solving exercise will give an insight to the concepts studied.

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Key Features Of NCERT Solutions for Exercise 9.3 Class 12 Maths Chapter 9

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 9.3 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 9.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 9.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 9.3 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 9.3 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 9.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. How many questions are solved in the NCERT solutions for Class 12 Maths chapter 9 exercise 9.3?

12 questions are explained through exercise 9.3 Class 12 Maths.

2. What is the number of multiple-choice questions in the Class 12 Maths chapter 9 exercise 9.3?

Two objective questions with 4 choices are given in the Class 12th Maths chapter 6 exercise 9.3.

3. What type of curve is discussed in example 6 of Class 12 chapter 9?

Ellipse.

4. What are the concepts to be covered before solving exercise 9.3?

Students must be able to differentiate the given function. For this students should be aware of basic differentiation. 

5. Is the chapter differential equations important for CBSE board students?

Yes, as NCERT is followed by CBSE students it is important to cover this chapter.

6. Whether the concepts of differential equations are used in physics?

Yes, to solve certain problems or to represent a certain type of motions differential equations are used.

7. What are the topics covered before exercise 9.3 of Class 12 Maths?

The concepts of order and degree of differential equations and general and particular solutions are covered.

8. Which topic is covered in exercise 9.3?

The NCERT Class 12 exercise 9.3 covers the topic of forming differential equations of a family of curves.

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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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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,

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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
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.

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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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kumardurgesh1802 10 July,2024
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Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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
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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
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