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

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 in the examples given before Class 12 Maths chapter 9 exercise 9.3 and also in the NCERT Solutions for Class 12 Maths chapter 9 exercise 9.3. 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.

• Differential Equations exercise 9.1

• Differential Equations exercise 9.2

• Differential Equations exercise 9.4

• Differential Equations exercise 9.5

• Differential Equations exercise 9.6

• Differential Equations miscellaneous exercise

## 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.

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

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.

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

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. $y = ae^{3x} + b e^{-2x}$

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. $y = e^{2x}(a+bx)$

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.

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

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.

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.

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.

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.

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.

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

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

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.

## 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.

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 9

• NCERT Solutions for Class 12 Maths Chapter 9

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology