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

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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.

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NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.3
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Differential Equations Class 12 Chapter 9 Exercise: 9.3
More About 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.
Key Features Of NCERT Solutions for Exercise 9.3 Class 12 Maths Chapter 9
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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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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) $\frac{d^2y}{dx^2} - y = 0$

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

(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)

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.

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NCERT Solutions Subject Wise

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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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Yes, you can switch from CBSE to CHSE Odisha in Class 12th, but there are a few conditions:

Eligibility : Check if CHSE Odisha allows lateral entry into Class 12th. Generally, students can transfer from CBSE to state boards with prior approval.
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Prachi Kumari 30 December,2024

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!

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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.

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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

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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 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

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

NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.3

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Differential Equations Class 12 Chapter 9 Exercise: 9.3

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