# NCERT Solutions Class 12 Maths Chapter 9 Differential Equation Exercise 9.4

Since the beginning of civilisation itself, the numerous disciplines of Science including Mathematics, Physics, Chemistry, Biology etc. have sparked interest and curiosity among people across the world. People have raised questions and have attempted to seek answers through discoveries, experimentation, and calculations of natural and other phenomena, seeking accuracy. Mathematics in particular, is often regarded as one of the oldest scientific disciplines to be taught and researched. One of the reasons behind this prominence of Mathematics could be that it is a subject whose knowledge is fundamental to the efficient organisation and smooth functioning of a majority of human activities and a variety of man-made operations. A multitude of associated disciplines and sub-disciplines make up the composition of Mathematics as a subject. This underlines the vast untapped potential for research contained within this subject, which is yet to be explored.

Thus, Mathematics has emerged as an indispensable academic subject. It has been a part of academic instruction since the earliest levels of school education and gets more versatile as one rises in grades. Students who wish to pursue a career in Mathematics which is indeed a field with many lucrative career options, should choose the Mathematics stream in classes 11 and 12. Mathematics occupies an important portion of the prescribed NCERT academic curriculum for Class 12.

TThe mathematics syllabus prescribed for Class 12 is geared towards solidifying the fundamentals of students and preparing them for competitive exams in the future. The importance of dependable learning resources like the NCERT Solutions which can act as reliable reference materials, is noteworthy in this context. The NCERT academic curriculum for Class 12 is composed of thirteen chapters that cover various crucial themes. These chapters are organised into a sequence to facilitate learning. Chapter 9 covers the theme of Differential Equations. The Ex 9.3 Class12 is one of the multiple comprehensive prescribed exercises that make up the entirety of Chapter 9. This exercise is extremely important for examinations and hence has to be understood and practised regularly. The Extramarks’ website has compiled the NCERT Solutions For Class 12 Maths Chapter 9 Exercise 9.2 to facilitate students’ learning processes. To ensure high quality content, the NCERT Solutions for Class 12 Maths, Chapter 9, Exercise 9.3 have been compiled in collaboration with experts.The NCERT Solutions Class 12 Maths Chapter 9 Exercise 9.3 could be effectively utilised as an efficient resource for self-study and revision by students. The nuances of the latest updated NCERT academic syllabus for Mathematics have been carefully considered in the preparation of the NCERT Solutions Class 12 Maths Chapter 9 Exercise 9.3.

## NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations (Ex 9.3) Exercise 9.3

Differential Equations have caused considerable apprehension and hesitation among students, much to the chagrin of their instructors and the students themselves.This attitude has not only proven to be a hindrance in terms of the standards of academic performance but also in terms of retaining the interest of students in the content being delivered by the teachers.

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### Access NCERT Solution for Class 12 Maths Chapter 9 – Differential Equations

Differential Equations are unquestionably an important part of the entire NCERT academic curriculum for mathematics in Class 12.However, students and teachers have frequently expressed concern that differential equations is a complex and distinct academic topic with its own set of peculiarities.It is a sophisticated topic with heavy and conceptually versatile content. This is precisely the reason why Differential Equations is also the focal point of a wide variety of academic research in Mathematics.

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### What Does Differential Equation Mean?

According to the NCERT Solutions Class 12 Maths Chapter 9 Exercise 9.3, a Differential Equation can be defined as an equation which involves the derivatives of a function or many functions. The ones which define the rate of change of a function at a point are referred to as Derivatives of that function.  The most important objective of studying differential equations is to study solutions that satisfy the equations and the properties of the solutions. The relevance, definitions, and comprehensive methods of solving and identifying differential equations have been covered in NCERT Solutions Class 12 Maths Chapter 9 Exercise 9.3. The NCERT Solutions Class 12 Maths Chapter 9 Exercise 9.3 provide detailed, step-by-step solutions to the problems which comprise Exercise 9.3. The formulae used to do the calculations have also been mentioned and covered in detail to ensure that students have total clarity about the logical sequence of the calculations. Extramarks is providing the NCERT Solutions Class 12 Maths Chapter 9 Exercise 9.3 to complement students’ hard work and perseverance.

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

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

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

Ans

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{x}{a}+\frac{y}{b}=1\\ \text{Differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0\\ \text{Againg,}\text{\hspace{0.17em}}\text{differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{b}\frac{{d}^{2}y}{d{x}^{2}}=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}=0⇒y‘‘=0\\ \text{Hence, the required differential equation is y”=0}\text{.}\end{array}$

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

${y}^{2}=a\left({b}^{2}-{x}^{2}\right)$

Ans

$\begin{array}{l}{\mathrm{y}}^{2}=\mathrm{a}\left({\mathrm{b}}^{2}-{\mathrm{x}}^{2}\right)\\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a}\left(0-2\mathrm{x}\right)\\ 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=-2\mathrm{ax} \mathrm{ }...\left(\mathrm{i}\right)\\ \mathrm{Againg},\mathrm{ }\mathrm{differentiating}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ 2\left\{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}+\mathrm{y}\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}\right\}=-2\mathrm{a}\\ ⇒ {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}+\mathrm{y}\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=-\mathrm{ }\mathrm{a} ...\left(\mathrm{ii}\right)\\ \mathrm{Putting}\mathrm{value}\mathrm{of}–\mathrm{a}\mathrm{from}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{to}\mathrm{equation}\left(\mathrm{i}\right), \mathrm{we}\mathrm{get}\\ 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=2\left\{{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}+\mathrm{y}\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}\right\}\mathrm{x}\\ ⇒ \mathrm{ }\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}+\mathrm{xy}\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}\\ \mathrm{xyy}‘‘+\mathrm{x}{\left(\mathrm{y}‘\right)}^{2}-\mathrm{yy}‘=0\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{xyy}”+\mathrm{x}{\left(\mathrm{y}‘\right)}^{\mathrm{2}}–\mathrm{yy}‘ = 0\mathrm{.}\end{array}$

Q.3 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

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

Ans

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=a{e}^{3x}+b{e}^{-2x}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\dots \left(i\right)\\ \text{Differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘=3a{e}^{3x}-2b{e}^{-2x}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{Againg,}\text{\hspace{0.17em}}\text{differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ y‘‘=9a{e}^{3x}+4b{e}^{-2x}\text{}\text{\hspace{0.17em}}\text{}\text{}\dots \left(iii\right)\\ \text{Adding equation}\left(\text{ii}\right)\text{and 2×equation}\left(\text{i}\right)\text{,}\text{\hspace{0.17em}}\text{we get}\\ y‘+2y=3a{e}^{3x}-2b{e}^{-2x}+2\left(a{e}^{3x}+b{e}^{-2x}\right)\\ y‘+2y=3a{e}^{3x}-2b{e}^{-2x}+2a{e}^{3x}+2b{e}^{-2x}\\ y‘+2y=5a{e}^{3x}⇒a{e}^{3x}=\frac{y‘+2y}{5}\\ \text{Substracting equation}\left(\text{ii}\right)\text{from 3×equation}\left(\text{i}\right)\text{,}\text{\hspace{0.17em}}\text{we get}\\ 3y-y‘=3a{e}^{3x}+3b{e}^{-2x}-\left(3a{e}^{3x}-2b{e}^{-2x}\right)\\ \text{\hspace{0.17em}}3y-y‘=3a{e}^{3x}+3b{e}^{-2x}-3a{e}^{3x}+2b{e}^{-2x}\\ \text{\hspace{0.17em}}3y-y‘=5b{e}^{-2x}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}b{e}^{-2x}=\frac{3y-y‘}{5}\\ {\text{Putting the values of ae}}^{\text{3x}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{be}}^{\text{-2x}}\text{in equation}\left(\text{iii}\right)\text{, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘‘=9\left(\frac{y‘+2y}{5}\right)+4\left(\frac{3y-y‘}{5}\right)\\ ⇒5y‘‘=9y‘+18y+12y-4y‘\\ ⇒5y‘‘=5y‘+30y\\ ⇒y‘‘-y‘-6y=0\\ \text{Hence, the required differential equation is y”–y’–6y = 0}\text{.}\end{array}$

Q.4 Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

${\text{y=e}}^{\text{2x}}\left(\text{a+b}\text{\hspace{0.17em}}\text{x}\right)$

Ans

$\begin{array}{l}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y={e}^{2x}\left(a+b\text{\hspace{0.17em}}x\right)\\ \text{Differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘={e}^{2x}\left(0+b\right)+2{e}^{2x}\left(a+b\text{\hspace{0.17em}}x\right)\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘=b{e}^{2x}+2y\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘-2y=b{e}^{2x}\text{}\text{}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\\ \text{Againg,}\text{\hspace{0.17em}}\text{differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘‘-2y‘=2b{e}^{2x}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{Dividing equation}\left(\text{ii}\right)\text{by equation}\left(\text{i}\right)\text{, we get}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{y‘‘-2y‘}{y‘-2y}=\frac{2b{e}^{2x}}{b{e}^{2x}}\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{y‘‘-2y‘}{y‘-2y}=2\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘‘-2y‘=2\left(y‘-2y\right)\\ ⇒y‘‘-2y‘-2y‘+4y=0\\ ⇒\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘‘-4y‘+4y=0\\ \text{Hence, the required differential equation is y” –4y’ + 4y = 0}\text{.}\end{array}$

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

$y={e}^{x}\left(acosx+bsin\text{\hspace{0.17em}}x\right)$

Ans

$\begin{array}{l}\text{}\text{}\text{}y={e}^{x}\left(a\mathrm{cos}x+b\mathrm{sin}\text{\hspace{0.17em}}x\right)\dots \left(i\right)\\ \text{Differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{}\text{}y‘={e}^{x}\left(-a\mathrm{sin}x+b\mathrm{cos}\text{\hspace{0.17em}}x\right)+{e}^{x}\left(a\mathrm{cos}x+b\mathrm{sin}\text{\hspace{0.17em}}x\right)\\ \text{}\text{}\text{}y‘={e}^{x}\left(-a\mathrm{sin}x+b\mathrm{cos}\text{\hspace{0.17em}}x\right)+y\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{From}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{equation}\left(\text{i}\right)\right]\\ ⇒\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘-y={e}^{x}\left(-a\mathrm{sin}x+b\mathrm{cos}\text{\hspace{0.17em}}x\right)\text{}\text{}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{Againg,}\text{\hspace{0.17em}}\text{differentiating both sides w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘‘-y‘={e}^{x}\left(-a\mathrm{cos}x-b\mathrm{sin}\text{\hspace{0.17em}}x\right)+{e}^{x}\left(-a\mathrm{sin}x+b\mathrm{cos}\text{\hspace{0.17em}}x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘‘-y‘=-y+y‘-y\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\left[\text{From equation}\left(ii\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y‘‘-2y‘+2y=0\\ \text{Hence, the required differential equation is}y‘‘-2y‘+2y=0.\end{array}$

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

Ans

The centre of the circle touching y-axis at origin lies on x-axis. Let the centre of circle be (a, 0).
Radius of circle will be ‘a’ because circle touches y-axis at origin. So, the equation of circle with centre (a, 0) and radius ‘a’ is as follows
(x – a)2 + y2 = a2
x2 – 2ax + a2 + y2 = a2
x2 + y2 = 2ax … (i)

Differentiating equation (i), w.r.t. x, we get

2x + 2yy’ = 2a

x + yy’ = a

Putting value of a in equation (i), we get

x2 + y2 = 2(x + yy’)x
= 2x2 + 2xyy’

y2 = x2 + 2xyy’

Thus, the required differential equation is 2xyy’ + x2 = y2.

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

Ans

The equation of parabolas having vertex at origin and axis along positive y-axis is as follows
x2 = 4ay …(i)
Differentiating w.r.t. x, we get
2x = 4ay’
or (2x/y’) = 4a
Putting value of 4a in equation (i), we get
x2 = (2x/y’)y

x2y’ = 2xy

or xy’ = 2y
or xy’ – 2y = 0

Thus, the required differential equation is
xy’ – 2y = 0.

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

Ans

$\begin{array}{l}\text{The}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{equation of the family of ellipses having foci on}\\ \text{y-axis and centre at origin is as follows}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\frac{{\text{x}}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{}+\text{}\frac{{\text{y}}^{\text{2}}}{{\text{a}}^{\text{2}}}=\text{1}\text{}\text{}\dots \text{}\left(\text{i}\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\frac{\text{2x}}{{\text{b}}^{\text{2}}}+\frac{\text{2yy’}}{{\text{a}}^{\text{2}}}=\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{x}}{{\text{b}}^{\text{2}}}+\frac{\text{yy’}}{{\text{a}}^{\text{2}}}=\text{0}\text{}\text{}\dots \text{}\left(\text{ii}\right)\\ \text{Again,}\text{\hspace{0.17em}}\text{differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\frac{\text{1}}{{\text{b}}^{\text{2}}}+\frac{\text{1}}{{\text{a}}^{\text{2}}}\left(\text{yy”+y’y’}\right)=\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\frac{\text{1}}{{\text{b}}^{\text{2}}}+\frac{\text{1}}{{\text{a}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}=\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{{\text{b}}^{\text{2}}}=-\frac{\text{1}}{{\text{a}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}\\ \text{Substituting the value of}\frac{\text{1}}{{\text{b}}^{\text{2}}}\text{in equation}\left(\text{ii}\right)\text{, we get}\\ \text{\hspace{0.17em}}\text{x}\left[-\frac{\text{1}}{{\text{a}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}\right]+\frac{\text{yy’}}{{\text{a}}^{\text{2}}}=\text{0}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-x\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}+yy‘=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-xyy‘‘-x{\left(y‘\right)}^{2}+yy‘=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}xyy‘‘+x{\left(y‘\right)}^{2}-yy‘=0\\ \text{Thus, the required differential equation is xyy” + x}{\left(\text{y’}\right)}^{\text{2}}\text{–yy’ = 0}\text{.}\end{array}$

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

Ans

$\begin{array}{l}\text{The differential equation of the family of hyperbolas having foci}\\ \text{on x-axis and centre at}\text{\hspace{0.17em}}\text{origin}\text{â€‹}\text{is as follows}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\text{}\text{}\dots \text{}\left(\text{i}\right)\\ \text{Differentiating}\text{â€‹}\text{w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}\frac{\text{2x}}{{\text{a}}^{\text{2}}}-\frac{\text{2yy’}}{{\text{b}}^{\text{2}}}=\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{x}}{{\text{a}}^{\text{2}}}-\frac{\text{yy’}}{{\text{b}}^{\text{2}}}=\text{0}\text{}\text{}\dots \text{}\left(\text{ii}\right)\\ \text{Again,}\text{\hspace{0.17em}}\text{differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{{\text{a}}^{\text{2}}}-\frac{\text{1}}{{\text{b}}^{\text{2}}}\left(\text{yy”+y’y’}\right)=\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\frac{\text{1}}{{\text{a}}^{\text{2}}}-\frac{\text{1}}{{\text{b}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}=\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{{\text{a}}^{\text{2}}}=\frac{\text{1}}{{\text{b}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}\\ \text{Substituting the value of}\frac{\text{1}}{{\text{a}}^{\text{2}}}\text{in equation}\left(\text{ii}\right)\text{, we get}\\ \text{\hspace{0.17em}}\text{x}\left[\frac{\text{1}}{{\text{b}}^{\text{2}}}\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}\right]-\frac{\text{yy’}}{{\text{b}}^{\text{2}}}=\text{0}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\left\{\text{yy”+}{\left(\text{y’}\right)}^{\text{2}}\right\}-yy‘=0\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}xyy‘‘+x{\left(y‘\right)}^{2}-yy‘=0\\ \text{Thus, the required differential equation is xyy” + x}{\left(\text{y’}\right)}^{\text{2}}\text{–yy’=0}\text{.}\end{array}$

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

Ans

The centre of the circle lies on y-axis. Let the centre of circle be (0, b).
So, the equation of circle with centre (0, b) and radius 3 is as follows
x2 + (y – b)2 = 32 … (i)

$\begin{array}{c}\text{Differentiating equation}\left(\text{i}\right),\text{w}.\text{r}.\text{t}.\text{x},\text{we get}\\ \text{2x}+\text{2}\left(\text{y}–\text{b}\right)\text{y}’=0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{x}+\left(\text{y}–\text{b}\right)\text{y}’=0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{y}-\text{b}\right)=\frac{-\text{x}}{\text{\hspace{0.17em}}\text{y}’}\\ \end{array}$

Putting value of a in equation ( i ), we get

$\begin{array}{c}\text{\hspace{0.17em}}{\text{x}}^{\text{2}}+{\left(\frac{-\text{x}}{\text{\hspace{0.17em}}\text{y}’}\right)}^{\text{2}}=9\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}^{2}y{‘}^{2}\text{\hspace{0.17em}}+{x}^{2}=9y{‘}^{2}\\ {\left(y‘\right)}^{2}\left({x}^{2}-9\right)\text{\hspace{0.17em}}+{x}^{2}=0\\ This\text{is the required differential equation}\text{.}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{differential}\mathrm{equations}\mathrm{has}\\ \mathrm{y}={\mathrm{c}}_{\mathrm{1}}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_{\mathrm{2}}{\mathrm{e}}^{–\mathrm{x}}\mathrm{as}\mathrm{the}\mathrm{general}\mathrm{solution}?\\ \left(\mathrm{A}\right)\mathrm{ }\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{2}}}+\mathrm{y}= 0 \left(\mathrm{B}\right)\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{2}}}-\mathrm{y}= 0 \mathrm{ }\\ \left(\mathrm{C}\right)\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{2}}}+ 1 = 0\left(\mathrm{D}\right)\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{\mathrm{2}}}-1= 0\end{array}$

Ans

$\begin{array}{l} \mathrm{ }\mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-\mathrm{x}} ...\left(\mathrm{i}\right)\\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}-{\mathrm{c}}_{2}\mathrm{ }{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{Again},\mathrm{â€‹}\mathrm{differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.\mathrm{x},\mathrm{we}\mathrm{get}\\ \mathrm{ }\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{2}}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_{2}\mathrm{ }{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{ }=\mathrm{y}\\ \mathrm{or}\mathrm{ }\frac{{\mathrm{d}}^{\mathrm{2}}\mathrm{y}}{{\mathrm{dx}}^{2}}-\mathrm{y}=\mathrm{0}\\ \mathrm{Thus},â€‹â€‹\mathrm{option}\mathrm{B}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.12

$\begin{array}{l}\text{Which of the following differential equations has y = x as}\\ \text{one of its particular solution?}\\ \left(\text{A}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}-{x}^{2}\frac{dy}{dx}+xy=x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\left(\text{B}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}-x\frac{dy}{dx}+xy=x\\ \left(\text{C}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}-{x}^{2}\frac{dy}{dx}+xy=0\text{}\text{}\text{}\left(\text{D}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}+x\frac{dy}{dx}+xy=0\end{array}$

Ans

$\begin{array}{l}\text{We have y = x}\text{…}\left(\text{i}\right)\\ \text{Differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{}\frac{dy}{dx}=1\text{}\text{}\text{}\dots \left(\text{ii}\right)\\ \text{Again,}\text{â€‹}\text{differentiating w}\text{.r}\text{.t}\text{. x, we get}\\ \text{}\text{}\frac{{d}^{2}y}{d{x}^{2}}=0\text{}\text{}\text{}\dots \left(\text{iii}\right)\\ \text{Putting values of y,}\frac{dy}{dx}\text{}and\text{}\frac{{d}^{2}y}{d{x}^{2}}\text{in each given options,}\\ \left(\text{A}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{L}\text{.H}\text{.S}.=\frac{{d}^{2}y}{d{x}^{2}}-{x}^{2}\frac{dy}{dx}+xy\\ \text{}\text{}=0-{x}^{2}\left(1\right)+x\left(x\right)\\ \text{}\text{}=0\ne \text{R}\text{.H}\text{.S}\text{.}\\ \text{Thus, it is not correct option}\text{.}\\ \left(\text{B}\right)\text{\hspace{0.17em}}\text{L}\text{.H}\text{.S}\text{.}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{d}^{2}y}{d{x}^{2}}-x\frac{dy}{dx}+xy\\ \text{}\text{}=0-x\left(x\right)+x\left(x\right)\\ \text{}\text{}=0\ne \text{R}\text{.H}\text{.S}\text{.}\\ \text{Thus, it is not correct option}\text{.}\\ \left(\text{C}\right)\text{L}\text{.H}\text{.S}\text{.}=\frac{{d}^{2}y}{d{x}^{2}}-{x}^{2}\frac{dy}{dx}+xy\\ \text{}\text{}=0-{x}^{2}\left(1\right)+x\left(x\right)\\ \text{}\text{}=0=\text{R}\text{.H}\text{.S}\text{.}\\ \text{Thus, it is correct option}\text{.}\\ \text{Hence, the correct solution is option C}\text{.}\end{array}$