Chapter 9 Differential Equations

NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

Class 12 Maths Chapter 9, Differential Equations, introduces students to the order and degree of differential equations, formation of differential equations from families of curves, and methods for solving them. The chapter covers important methods such as variable separable, homogeneous, and linear differential equations. A clear understanding of these concepts helps students solve questions in the CBSE Class 12 Maths syllabus and builds a foundation for further studies in subjects such as physics, economics, and engineering.

ALLEN's NCERT Solutions for Class 12 Maths Chapter 9 covers the NCERT textbook questions while adhering to the most recent CBSE syllabus. Detailed solutions describe how to build differential equations, determine their order and degree, and use the proper method to solve them. Students can enhance their accuracy, comprehend typical question kinds, and get ready for CBSE board exams and competitive examinations by practicing these solutions on a regular basis.

1.0Key Concepts of Class 12 Maths Chapter 9 Differential Equations

Class 12 Maths Chapter 9, Differential Equations, introduces equations involving derivatives and explains how they are formed and solved. The main concepts covered in NCERT Solutions for Class 12 Maths Chapter 9 include:

Differential Equation: Understand equations that contain derivatives of dependent variables with respect to independent variables.
Order and Degree: Learn how to identify the order and degree of a differential equation after writing it in the required form.
General and Particular Solutions: Understand how arbitrary constants are used to obtain the general solution and how given conditions help find a particular solution.
Formation of Differential Equations: Learn how to form a differential equation by eliminating arbitrary constants from a given relation.
Methods of Solving Differential Equations: Study methods such as the variable separable method and other standard techniques used to solve differential equations.

2.0NCERT Class 12 Maths Chapter 9 Differential Equations : Detailed Solutions

EXERCISE - 9.1

Determine order and degree (if defined) of differential equations given in Exercises 1 to 10.

$\frac{d ^{4} y}{dx ^{4}} + sin (y^{'''}) = 0$ Sol. $\frac{d ^{4} y}{dx ^{4}} + sin (y^{'''}) = 0$ $\Rightarrow y^{''''} + sin (y^{'''}) = 0$ The highest order derivative present in the differential equation is $y^{'''}$ . Therefore, its order is four. The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.
$y^{'} + 5 y = 0$ Sol. The given differential equation is: $y^{'} + 5 y = 0$ The highest order derivative present in the differential equation is y'. Therefore, its order is one. It is a polynomial equation in y'. The highest power raised to $y^{'}$ is 1 . Hence, its degree is one.
$(\frac{ds}{dt})^{4} + 3 s \frac{d ^{2} s}{dt ^{2}} = 0$ Sol. $(\frac{ds}{dt})^{4} + 3 s \frac{d ^{2} s}{dt ^{2}} = 0$ The highest order derivative present in the given differential equation is $\frac{d ^{2} s}{dt ^{2}}$ . Therefore, its order is two. It is a polynomial equation in $\frac{d ^{2} s}{dt ^{2}}$ and $\frac{ds}{dt}$ . The power raised to $\frac{d ^{2} s}{dt ^{2}}$ is 1 . Hence, its degree is one.
$(\frac{d ^{2} y}{dx ^{2}})^{2} + cos (\frac{dy}{dx}) = 0$ Sol. $(\frac{d ^{2} y}{dx ^{2}})^{2} + cos (\frac{dy}{dx}) = 0$

The highest order derivative present in the given differential equation is $\frac{d ^{2} y}{dx ^{2}}$ . Therefore, its order is 2. The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

$\frac{d ^{2} y}{dx ^{2}} = cos 3 x + sin 3 x$ Sol. $\frac{d ^{2} y}{dx ^{2}} = cos 3 x + sin 3 x$ $\Rightarrow \frac{d ^{2} y}{dx ^{2}} - cos 3 x - sin 3 x = 0$ The highest order derivative present in the differential equation is $\frac{d ^{2} y}{dx ^{2}}$ . Therefore, its order is two. It is a polynomial equation in $\frac{d ^{2} y}{dx ^{2}}$ and the power raised to $\frac{d ^{2} y}{dx ^{2}}$ is 1. Hence, its degree is one.
$(y^{'''})^{2} + (y^{''})^{3} + (y^{'})^{4} + y^{5} = 0$ Sol. $(y^{''})^{2} + (y^{'})^{3} + (y^{'})^{4} + y^{5} = 0$ The highest order derivative present in the differential equation is $y^{'''}$ . Therefore, its order is three. The given differential equation is a polynomial equation in $y^{'''}, y^{''}$ and $y^{'}$ . The highest power raised to $y^{'''}$ is 2 . Hence, its degree is 2.
$y^{'''} + 2 y^{''} + y^{'} = 0$ Sol. $y^{'''} + 2 y^{''} + y^{'} = 0$ The highest order derivative present in the differential equation is $y^{'''}$ . Therefore, its order is three. It is a polynomial equation in $y^{'''}, y^{''}$ and $y^{'}$ . The highest power raised to $y^{'''}$ is 1 . Hence, its degree is 1.
$y^{'} + y = e^{x}$ Sol. $y^{'} + y = e^{x} \Rightarrow y^{'} + y - e^{x} = 0$ The highest order derivative present in the differential equation is y'. Therefore, its order is one. The given differential equation is a polynomial equation in y' and the highest power raised to y' is one. Hence, its degree is one.
$y^{''} + (y^{'})^{2} + 2 y = 0$ Sol. $y^{''} + (y^{'})^{2} + 2 y = 0$ The highest order derivative present in the differential equation is $y^{''}$ . Therefore, its order is two. The given differential equation is a polynomial equation in $y^{''}$ and $y^{'}$ and the highest power raised to $y^{''}$ is one. Hence, its degree is one.
$y^{''} + 2 y^{'} + sin y = 0$ Sol. $y^{''} + 2 y^{'} + sin y = 0$ The highest order derivative present in the differential equation is y". Therefore, its order is two. This is a polynomial equation in y'' and $y^{'}$ and the highest power raised to $y^{''}$ is one. Hence, its degree is one.
The degree of the differential equation $(\frac{d ^{2} y}{dx ^{2}})^{3} + (\frac{dy}{dx})^{2} + sin (\frac{dy}{dx}) + 1 = 0$ is (A) 3 (B) 2 (C) 1 (D) not defined Sol. (D) $(\frac{d ^{2} y}{dx ^{2}})^{3} + (\frac{dy}{dx})^{2} + sin (\frac{dy}{dx}) + 1 = 0$ The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined. Hence, the correct answer is D.
The order of the differential equation $2 x^{2} \frac{d ^{2} y}{dx ^{2}} - 3 \frac{dy}{dx} + y = 0$ is (A) 2 (B) 1 (C) 0 (D) not defined Sol. (A) $2 x^{2} \frac{d ^{2} y}{dx ^{2}} - 3 \frac{dy}{dx} + y = 0$ The highest order derivative present in the given differential equation is $\frac{d ^{2} y}{dx ^{2}}$ . Therefore, its order is two. Hence, the correct answer is A.

EXERCISE 9.2

In each of the Q. 1 to 10 verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation :

$y = e^{x} + 1 : y^{''} - y^{'} = 0$ Sol. $y = e^{x} + 1$ Differentiating both sides of this equation with respect to x, we get:

$\frac{d y}{d x} = \frac{d}{d x} (e^{x} + 1) \Rightarrow y^{'} = e^{x}$

Now, again differentiating equation (1) with respect to x, we get:

$\frac{d}{dx} (y^{'}) = \frac{d}{dx} (e^{x}) \Rightarrow y^{''} = e^{x}$

Substituting the values of $y^{'}$ and $y^{''}$ in the given differential equation, we get the L.H.S. as:

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

Thus, the given function is the solution of the corresponding differential equation.

$y = x^{2} + 2 x + C : y^{'} - 2 x - 2 = 0$ Sol. $y = x^{2} + 2 x + C$ Differentiating both sides of this equation with respect to x , we get:

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

Substituting the value of $y^{'}$ in the given differential equation, we get:

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

Hence, the given function is the solution of the corresponding differential equation.

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

Sol. $y = cos x + C$ Differentiating both sides of this equation with respect to x, we get:

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

$\Rightarrow y^{'} = - sin x$ Substituting the value of $y^{'}$ in the given differential equation, we get: L.H.S. $= y^{'} + sin x = - sin x + sin x = 0 =$ R.H.S. Hence, the given function is the solution of the corresponding differential equation.

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

Sol. $y = 1 + x^{2}$ Differentiating both sides of the equation with respect to x, we get:

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

$\Rightarrow y^{'} = \frac{1}{2 1 + x ^{2}} \cdot \frac{d}{dx} (1 + x^{2})$ $\Rightarrow y^{'} = \frac{2 x}{2 1 + x ^{2}}$ $\Rightarrow y^{'} = \frac{x}{1 + x ^{2}}$ $\Rightarrow y^{'} = \frac{x}{1 + x ^{2}} \times 1 + x^{2}$ $\Rightarrow y^{'} = \frac{x}{1 + x ^{2}} \cdot y$ $\Rightarrow y^{'} = \frac{x y}{1 + x ^{2}}$ $∴$ L.H.S. = R.H.S Hence, the given function is the solution of the corresponding differential equation.

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

Sol. $y = Ax$ Differentiating both sides with respect to x , we

$get: y^{'} = \frac{d}{dx} (Ax) \Rightarrow y^{'} = A$

Substituting the value of $y^{'}$ in the given differential equation, we get: L.H.S. $= x y^{'} = x \cdot A = A x = y =$ R.H.S. Hence, the given function is the solution of the corresponding differential equation.

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

$[x \neq = 0 and x > y or x < - y]$

Sol. $y = x sin x$ Differentiating both sides of this equation with respect to x, we get:

$y^{'} = \frac{d}{dx} (x sin x)$

$\Rightarrow y^{'} = sin x \cdot \frac{d}{dx} (x) + x \cdot \frac{d}{dx} (sin x)$ $\Rightarrow y^{'} = sin x + x cos x$ Substituting the value of $y^{'}$ in the given differential equation, we get: L.H.S. $= x y^{'}$

$= x (sin x + x cos x) = x sin x + x^{2} cos x = y + x^{2} \cdot 1 - sin^{2} x = y + x^{2} 1 - (\frac{y}{x})^{2} [Using Eq.(1)] = y + x x^{2} - y^{2} = R.H.S.$

Hence, the given function is the solution of the corresponding differential equation.

$xy = lo g y + C : y^{'} = \frac{y ^{2}}{1 - xy} (xy \neq = 1)$

Sol. $xy = lo g y + C$ Differentiating both sides of this equation with respect to x, we get:

$\frac{d}{d x} (x y) = \frac{d}{d x} (lo g y + C)$

$\Rightarrow y \cdot \frac{d}{dx} (x) + x \cdot \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx}$

$\Rightarrow y + x y^{'} = \frac{1}{y} y^{'} \Rightarrow (x y - 1) y^{'} = - y^{2} \Rightarrow y^{2} + x y y^{'} = y^{'} \Rightarrow y^{'} = \frac{y ^{2}}{1 - x y}$

$∴$ L.H.S. = R.H.S. Hence, the given function is the solution of the corresponding differential equation.

$y - cos y = x : (y sin y + cos y + x) y^{'} = y$

Sol. $y - cos y = x$ Differentiating both sides of the equation with respect to x, we get:

$\frac{d y}{d x} - \frac{d}{d x} (cos y) = \frac{d}{d x} (x)$

$\Rightarrow y^{'} + sin y \cdot y^{'} = 1$ $\Rightarrow y^{'} (1 + sin y) = 1$ $\Rightarrow y^{'} = \frac{1}{1 + s i n y}$ Substituting the value of $y^{'}$ in the given differential equation, we get: L.H.S.

$= (y sin y + cos y + x) y^{'} = (y sin y + cos y + y - cos y) \times \frac{1}{1 + sin y}$

[Using Eq.(1) & (2)]

$= y (1 + sin y) \cdot \frac{1}{1 + sin y} = y = R.H.S.$

Hence, the given function is the solution of the corresponding differential equation.

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

Sol. $x + y = tan^{- 1} y$ Differentiating both sides of this equation with respect to x, we get:

$\frac{d}{dx} (x + y) = \frac{d}{dx} (tan^{- 1} y)$

$\Rightarrow 1 + y^{'} = [\frac{1}{1 + y ^{2}}] y^{'} \Rightarrow y^{'} [\frac{1}{1 + y ^{2}} - 1] = 1$

$\Rightarrow y^{'} [\frac{1 - ( 1 + y ^{2} )}{1 + y ^{2}}] = 1 \Rightarrow y^{'} [\frac{- y ^{2}}{1 + y ^{2}}] = 1 \Rightarrow y^{'} = \frac{- ( 1 + y ^{2} )}{y ^{2}}$

Substituting the value of $y^{'}$ in the given differential equation, $y^{2} y^{1} + y^{2} + 1 = 0$ we get:

$L.H.S. = y^{2} y^{'} + y^{2} + 1 = y^{2} [\frac{- ( 1 + y ^{2} )}{y ^{2}}] + y^{2} + 1 = - 1 - y^{2} + y^{2} + 1 = 0$

Hence, the given function is the solution of the corresponding differential equation.

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

Sol. $y = a^{2} - x^{2}$ Differentiating both sides of this equation with respect to x, we get:

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

$\Rightarrow \frac{dy}{dx} = \frac{1}{2 a ^{2} - x ^{2}} \cdot \frac{d}{dx} (a^{2} - x^{2})$

$= \frac{1}{2 a ^{2} - x ^{2}} (- 2 x) = \frac{- x}{a ^{2} - x ^{2}}$

Substituting the value of $\frac{dy}{dx}$ in the given differential equation, we get:

$L.H.S. = x + y \frac{d y}{d x} = x + a^{2} - x^{2} \times \frac{- x}{a ^{2} - x ^{2}}$

[Using eq.(1) & (2)]

$= x - x = 0 = R.H.S.$

Hence, the given function is the solution of the corresponding differential equation.

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

Sol. (D) We know that the number of constants in the general solution of a differential equation of order n is equal to its order. Therefore, the number of constants in the general equation of fourth order differential equation is four. Hence, the correct answer is (D).

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 Sol. (D) In a particular solution of a differential equation, there are no arbitrary constants. Hence, the correct answer is (D).

EXERCISE 9.3

For each of the differential equations in Exercises 1 to 10, find the general solution :

$\frac{dy}{dx} = \frac{1 - c o s x}{1 + c o s x}$ Sol. The given differential equation is :

$\frac{d y}{d x} = \frac{1 - c o s x}{1 + c o s x}$

$\Rightarrow \frac{dy}{dx} = \frac{2 s i n ^{2} \frac{x}{2}}{2 c o s ^{2} \frac{x}{2}} = tan^{2} \frac{x}{2}$ $\Rightarrow \frac{dy}{dx} = (sec^{2} \frac{x}{2} - 1)$ Separating the variables, we get:

$dy = (sec^{2} \frac{x}{2} - 1) dx$

Now, integrating both sides of this equation, we get:

$\int dy = \int (sec^{2} \frac{x}{2} - 1) dx = \int sec^{2} \frac{x}{2} dx - \int dx$

$\Rightarrow y = 2 tan \frac{x}{2} - x + C$ This is the required general solution of the given differential equation.

$\frac{dy}{dx} = 4 - y^{2} (- 2 < y < 2)$ Sol. The given differential equation is :

$\frac{d y}{d x} = 4 - y^{2}$

Separating the variables, we get :

$\frac{dy}{4 - y ^{2}} = dx$

Now, integrating both sides of this equation, we get: $\int \frac{dy}{4 - y ^{2}} = \int dx$ $\Rightarrow sin^{- 1} \frac{y}{2} = x + C$ $\Rightarrow \frac{y}{2} = sin (x + C)$ $\Rightarrow y = 2 sin (x + C)$

This is the required general solution of the given differential equation.

$\frac{dy}{dx} + y = 1 (y \neq = 1)$ Sol. The given differential equation is:

$\frac{d y}{d x} + y = 1 \Rightarrow \frac{d y}{d x} = 1 - y$

Separating the variables, we get $\frac{dy}{1 - y} = dx$ Now, integrating both sides, we get :

$\int \frac{dy}{1 - y} = \int dx$

$\Rightarrow - lo g (1 - y) = x + lo g C$ $\Rightarrow - lo g C - lo g (1 - y) = x$ $\Rightarrow lo g C (1 - y) = - x$ $\Rightarrow C (1 - y) = e^{- x}$ $\Rightarrow 1 - y = \frac{1}{C} e^{- x}$ $\Rightarrow y = 1 - \frac{1}{C} e^{- x}$ $\Rightarrow y = 1 + Ae^{- x} ($ where $A = - \frac{1}{C})$ This is the required general solution of the given differential equation.

$sec^{2} x tan y dx + sec^{2} y tan x dy = 0$ Sol. The given differential equation is: $sec^{2} x$ tany $dx + sec^{2} y tan x dy = 0$ $\Rightarrow sec^{2} x tan y d x = - sec^{2} y tan x d y$

On separating the variables, we get:

$\Rightarrow \frac{s e c ^{2} x}{t a n x} dx = - \frac{s e c ^{2} y}{t a n y} dy$ Integrating both sides of this equation, we get:

$\int \frac{s e c ^{2} x}{t a n x} d x = - \int \frac{s e c ^{2} y}{t a n y} d y$

$\Rightarrow lo g ∣ tan x ∣ = - lo g ∣ tan y ∣ + lo g C$

$[∵ \int \frac{f ^{'} ( x )}{f ( x )} d x = lo g ∣ f (x) ∣ + C]$

$\Rightarrow lo g ∣ tan x ∣ + lo g ∣ tan y ∣ = lo g C$ $\Rightarrow tan x tan y = C$

This is the required general solution of the given differential equation.

$(e^{x} + e^{- x}) dy - (e^{x} - e^{- x}) dx = 0$ Sol. The given differential equation is:

$(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0$

$\Rightarrow (e^{x} + e^{- x}) dy = (e^{x} - e^{- x}) dx$

Separating the variables, we get:

$d y = [\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}] d x$

Integrating both sides of this equation, we get:

$\int dy = \int [\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}] dx + C$

$\frac{dy}{dx} = (1 + x^{2}) (1 + y^{2})$ Sol. The given differential equation is :

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

Integrating both sides of this equation, we get:

$\int \frac{dy}{1 + y ^{2}} = \int (1 + x^{2}) dx$

$\Rightarrow tan^{- 1} y = \int dx + \int x^{2} dx$ $\Rightarrow tan^{- 1} y = x + \frac{x ^{3}}{3} + C$

This is the required general solution of the given differential equation.

$y lo g y dx - x dy = 0$ Sol. The given differential equation is :

$y lo g y d x - x d y = 0$

$\Rightarrow y lo g y d x = x d y$

Separating the variables, we get: $\frac{dy}{y l o g y} = \frac{dx}{x}$ Integrating both sides, we get:

$\int \frac{dy}{y l o g y} = \int \frac{dx}{x}$

Let $lo g y = t \Rightarrow \frac{1}{y} dy = dt$ Substituting this value in equation (1), we get:

$\int \frac{dt}{t} = \int \frac{dx}{x} \Rightarrow lo g t = lo g x + lo g C$

$\Rightarrow lo g (lo g y) = lo g Cx$ $\Rightarrow lo g y = Cx$ $\Rightarrow y = \int [\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}] dx + C$ $\Rightarrow y = e^{Cx}$

This is the required general solution of the

$\Rightarrow y = lo g (e^{x} + e^{- x}) + C$

$[∵ \int \frac{f ^{'} ( x )}{f ( x )} dx = lo g ∣ f (x) ∣ + C]$

given differential equation.

$x^{5} \frac{dy}{dx} = - y^{5}$

This is the required general solution of the given differential equation.

Sol. The given differential equation is :

$x^{5} \frac{dy}{dx} = - y^{5}$

Separating the variables, we get :

$\frac{d y}{y ^{5}} = - \frac{d x}{x ^{5}} \Rightarrow \frac{d x}{x ^{5}} + \frac{d y}{y ^{5}} = 0$

Integrating both sides, we get:

$\int \frac{dx}{x ^{5}} + \int \frac{dy}{y ^{5}} = k (where k is any constant)$

$\Rightarrow \int x^{- 5} d x + \int y^{- 5} d y = k$

$\Rightarrow \frac{x ^{- 4}}{- 4} + \frac{y ^{- 4}}{- 4} = k \Rightarrow x^{- 4} + y^{- 4} = - 4 k$

$\Rightarrow x^{- 4} + y^{- 4} = C (C = - 4 k)$ This is the required general solution of the given differential equation.

$\frac{dy}{dx} = sin^{- 1} x$

Sol. The given differential equation is :

$\frac{d y}{d x} = sin^{- 1} x \Rightarrow d y = sin^{- 1} x d x$

Integrating both sides, we get :

$\int dy = \int sin^{- 1} x dx$

$\Rightarrow y = \int (sin^{- 1} x) \cdot 1 dx$

$\Rightarrow y = sin^{- 1} x \cdot \int (1) d x - \int [(\frac{d}{d x} (sin^{- 1} x) \cdot \int (1) d x)] d x$

$\Rightarrow y = sin^{- 1} x \cdot x - \int (\frac{1}{1 - x ^{2}} \cdot x) d x$

$\Rightarrow y = x sin^{- 1} x + \int \frac{- x}{1 - x ^{2}} d x$

Let $1 - x^{2} = t \Rightarrow - 2 x dx = dt$ Substituting this value in equation (1), we get:

$y = x sin^{- 1} x + \int \frac{1}{2 t} d t$

$\Rightarrow y = x sin^{- 1} x + \frac{1}{2} \cdot \int (t)^{- \frac{1}{2}} d t$

$\Rightarrow y = x sin^{- 1} x + \frac{1}{2} \cdot \frac{t ^{\frac{1}{2}}}{\frac{1}{2}} + C$

$\Rightarrow y = x sin^{- 1} x + t + C$

$\Rightarrow y = x sin^{- 1} x + 1 - x^{2} + C$

This is the required general solution of the given differential equation.

$e^{x} tan y dx + (1 - e^{x}) sec^{2} y dy = 0$

Sol. The given differential equation is :

$e^{x} tan y d x + (1 - e^{x}) sec^{2} y d y = 0$

$\Rightarrow (1 - e^{x}) sec^{2} y dy = - e^{x} tan ydx$ Separating the variables, we get :

$\frac{s e c ^{2} y}{t a n y} d y = \frac{- e ^{x}}{1 - e ^{x}} d x$

Integrating both sides, we get:

$\int \frac{s e c ^{2} y}{t a n y} d y = \int \frac{- e ^{x}}{1 - e ^{x}} d x$

$\Rightarrow lo g (tan y) = lo g (1 - e^{x}) + lo g C$

$[∵ \int \frac{f ^{'} ( x )}{f ( x )} dx = lo g ∣ f (x) ∣ + C]$

$\Rightarrow lo g (tan y) = lo g [C (1 - e^{x})]$ $\Rightarrow tan y = C (1 - e^{x})$ This is the required general solution of the given differential equation.

For each of the differential equations in Q. 11 to 14, find a particular solution satisfying the given condition :

$(x^{3} + x^{2} + x + 1) \frac{dy}{dx} = 2 x^{2} + x; y = 1$ when $x = 0$

Sol. The given differential equation is :

$(x^{3} + x^{2} + x + 1) \frac{d y}{d x} = 2 x^{2} + x$

$\Rightarrow \frac{dy}{dx} = \frac{2 x ^{2} + x}{( x ^{3} + x ^{2} + x + 1 )}$ $\Rightarrow dy = \frac{2 x ^{2} + x}{( x + 1 ) ( x ^{2} + 1 )} dx$ Integrating both sides, we get:

$\int dy = \int \frac{2 x ^{2} + x}{( x + 1 ) ( x ^{2} + 1 )} dx$

Let $\frac{2 x ^{2} + x}{( x + 1 ) ( x ^{2} + 1 )} = \frac{A}{x + 1} + \frac{Bx + C}{x ^{2} + 1}$ $\Rightarrow \frac{2 x ^{2} + x}{( x + 1 ) ( x ^{2} + 1 )} = \frac{Ax ^{2} + A + ( Bx + C ) ( x + 1 )}{( x + 1 ) ( x ^{2} + 1 )}$ $\Rightarrow 2 x^{2} + x = Ax^{2} + A + Bx^{2} + Bx + Cx + C$ $\Rightarrow 2 x^{2} + x = (A + B) x^{2} + (B + C) x + (A + C)$

Comparing the coefficients of $x^{2}$ and x , we get:

$A + B = 2, B + C = 1, A + C = 0$

Solving these equations, we get :

$A = \frac{1}{2}, B = \frac{3}{2} and C = \frac{- 1}{2}$

Substituting the values of A, B, and C in equation (2), we get:

$\frac{2 x ^{2} + x}{( x + 1 ) ( x ^{2} + 1 )} = \frac{1}{2} \cdot \frac{1}{( x + 1 )} + \frac{1}{2} \frac{( 3 x - 1 )}{( x ^{2} + 1 )}$

Therefore, equation (1) becomes:

$\int dy = \frac{1}{2} \int \frac{1}{x + 1} dx + \frac{1}{2} \int \frac{3 x - 1}{x ^{2} + 1} dx$

$\Rightarrow y = \frac{1}{2} lo g (x + 1) + \frac{3}{2} \int \frac{x}{x ^{2} + 1} dx - \frac{1}{2} \int \frac{1}{x ^{2} + 1} dx$ $\Rightarrow y = \frac{1}{2} lo g (x + 1) + \frac{3}{4} \cdot \int \frac{2 x}{x ^{2} + 1} dx - \frac{1}{2} tan^{- 1} x + C$ $\Rightarrow y = \frac{1}{2} lo g (x + 1) + \frac{3}{4} lo g (x^{2} + 1) - \frac{1}{2} tan^{- 1} x + C$ $\Rightarrow y = \frac{1}{4} [2 lo g (x + 1) + 3 lo g (x^{2} + 1)] - \frac{1}{2} tan^{- 1} x + C$ $\Rightarrow y = \frac{1}{4} [lo g (x + 1)^{2} (x^{2} + 1)^{3}] - \frac{1}{2} tan^{- 1} x + C$ Now, $y = 1$ when $x = 0$ $\Rightarrow 1 = \frac{1}{4} lo g (1) - \frac{1}{2} tan^{- 1} 0 + C$ $\Rightarrow 1 = \frac{1}{4} \times 0 - \frac{1}{2} \times 0 + C \Rightarrow C = 1$ Substituting $C = 1$ in equation (3), we get:

$y = \frac{1}{4} [lo g (x + 1)^{2} (x^{2} + 1)^{3}] - \frac{1}{2} tan^{- 1} x + 1$

which is the required particular solution.

$x (x^{2} - 1) \frac{dy}{dx} = 1; y = 0$ when $x = 2$

Sol. $x (x^{2} - 1) \frac{dy}{dx} = 1$ On separating the variables, we get :

$dy = \frac{dx}{x ( x ^{2} - 1 )} \Rightarrow \int dy = \int \frac{dx}{x ^{3} ( 1 - \frac{1}{x ^{2}} )}$

$\Rightarrow Put 1 - \frac{1}{x ^{2}} = t \Rightarrow \frac{2}{x ^{3}} dx = dt \int dy = \frac{1}{2} \int \frac{dt}{t} \Rightarrow y = \frac{1}{2} lo g (t) + lo g C y = \frac{1}{2} lo g \frac{x ^{2} - 1}{x ^{2}} + lo g C$

Now, $y = 0$ when $x = 2$

$0 = \frac{1}{2} lo g (\frac{3}{4}) + lo g C [From eq.(1)] [∵ lo g C = - \frac{1}{2} lo g (\frac{3}{4})] lo g C = \frac{1}{2} lo g (\frac{4}{3})$

$\Rightarrow y = \frac{1}{2} lo g (\frac{x ^{2} - 1}{x ^{2}}) + \frac{1}{2} lo g (\frac{4}{3})$ $\Rightarrow y = \frac{1}{2} lo g [\frac{4 ( x ^{2} - 1 )}{3 x ^{2}}]$ which is the required particular solution.

$cos (\frac{dy}{dx}) = a (a \in R); y = 1$ when $x = 0$

Sol. $cos (\frac{dy}{dx}) = a \Rightarrow \frac{dy}{dx} = cos^{- 1} a$ On separating the variables, we get,

$d y = cos^{- 1} a d x$

Integrating both sides, we get:

$\int dy = cos^{- 1} a \int dx$

$\Rightarrow y = cos^{- 1} a \cdot x + C$

$\Rightarrow y = x cos^{- 1} a + C$

Now, $y = 1$ when $x = 0$

$\Rightarrow 1 = 0 \cdot cos^{- 1} a + C \Rightarrow C = 1$

Substituting $C = 1$ in equation (1), we get:

$y = x cos^{- 1} a + 1$

$\Rightarrow \frac{y - 1}{x} = cos^{- 1} a \Rightarrow cos (\frac{y - 1}{x}) = a$ which is the required particular solution.

$\frac{dy}{dx} = y tan x; y = 1$ when $x = 0$ Sol. $\frac{dy}{dx} = y tan x$ Separating the variables, we get : $\frac{dy}{y} = tan xdx$ Integrating both sides, we get : $\int \frac{dy}{y} = \int tan xdx$ $\Rightarrow lo g y = lo g ∣ sec x ∣ + lo g C$ $\Rightarrow lo g y = lo g (C sec x)$ $\Rightarrow y = C sec x$ Now, $y = 1$ when $x = 0$ $\Rightarrow 1 = C \times sec 0 \Rightarrow 1 = C \times 1 \Rightarrow C = 1$ Substituting $C = 1$ in equation (1), we get : $y = sec x$ which is the required particular solution.
Find the equation of a curve passing through the point $(0, 0)$ and whose differential equation is $y^{'} = e^{x} sin x$ Sol. The differential equation of the curve is : $y^{'} = e^{x} sin x \Rightarrow \frac{dy}{dx} = e^{x} sin x$ Separating the variables, we get : $dy = e^{x} sin x dx$ Integrating both sides, we get :

$\int dy = \int e^{x} sin x dx$

Let $I = \int_{II}^{e_{II}^{x} s i n_{I} x dx}$ $\Rightarrow I = sin x \int e^{x} dx - \int (\frac{d}{dx} (sin x) \cdot \int e^{x} dx) dx$ $\Rightarrow I = sin x \cdot e^{x} - \int cos x \cdot e^{x} dx$ $\Rightarrow I = sin x \cdot e^{x}$

$- [cos x \cdot \int e^{x} d x - \int (\frac{d}{d x} (cos x) \cdot \int e^{x} d x) d x]$

$\Rightarrow I = sin x \cdot e^{x} - [cos x \cdot e^{x} - \int (- sin x) \cdot e^{x} dx]$ $\Rightarrow I = e^{x} sin x - e^{x} cos x - I$ $\Rightarrow 2 I = e^{x} (sin x - cos x)$ $\Rightarrow I = \frac{e ^{x} ( s i n x - c o s x )}{2}$ Substituting this value in equation (1), we get:

$y = \frac{e ^{x} ( s i n x - c o s x )}{2} + C$

Now, the curve passes through point $(0, 0)$ . $∴ 0 = \frac{e ^{0} ( s i n 0 - c o s 0 )}{2} + C$ $\Rightarrow 0 = \frac{1 ( 0 - 1 )}{2} + C \Rightarrow C = \frac{1}{2}$ Substituting $C = \frac{1}{2}$ in equation (2), we get: $y = \frac{e ^{x} ( s i n x - c o s x )}{2} + \frac{1}{2}$ $\Rightarrow 2 y = e^{x} (sin x - cos x) + 1$ $\Rightarrow 2 y - 1 = e^{x} (sin x - cos x)$ Hence, the required equation of the curve is $2 y - 1 = e^{x} (sin x - cos x)$

For the differential equation $xy \frac{dy}{dx} = (x + 2) (y + 2)$ , find the solution curve passing through the point $(1, - 1)$ . Sol. The differential equation of the given curve is : $xy \frac{dy}{dx} = (x + 2) (y + 2)$ Separating the variables, we get : $(\frac{y}{y + 2}) d y = (\frac{x + 2}{x}) d x$ $\Rightarrow (1 - \frac{2}{y + 2}) dy = (1 + \frac{2}{x}) dx$ Integrating both sides, we get: $\int (1 - \frac{2}{y + 2}) dy = \int (1 + \frac{2}{x}) dx$ $\Rightarrow \int dy - 2 \int \frac{1}{y + 2} dy = \int dx + 2 \int \frac{1}{x} dx$ $\Rightarrow y - 2 lo g (y + 2) = x + 2 lo g x + C$ $\Rightarrow y - x - C = lo g x^{2} + lo g (y + 2)^{2}$ $\Rightarrow y - x - C = lo g [x^{2} (y + 2)^{2}]$ Now, the curve passes through point (1, -1). $\Rightarrow - 1 - 1 - C = lo g [(1)^{2} (- 1 + 2)^{2}]$ $\Rightarrow - 2 - C = lo g 1 = 0 \Rightarrow C = - 2$ Substituting $C = - 2$ in equation (1), we get : $y - x + 2 = lo g [x^{2} (y + 2)^{2}]$ This is the required solution of the given curve.
Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point. Sol. Let x and y be the x -coordinate and y-coordinate of the curve respectively. We know that the slope of a tangent to the curve in the coordinate axis is given by the relation, $\frac{dy}{dx}$ According to the given information, we get: The product of the slope of tangent with y-coordinate = x-coordinate

$y \cdot \frac{d y}{d x} = x$

Separating the variables, we get: $y dy = x dx$ Integrating both sides, we get :

$\int y d y = \int x d x \Rightarrow \frac{y ^{2}}{2} = \frac{x ^{2}}{2} + C$

$\Rightarrow y^{2} - x^{2} = 2 C$

Now, the curve passes through the point (0, -2).

$∴ (- 2)^{2} - 0^{2} = 2 C \Rightarrow 2 C = 4$

Substituting $2 C = 4$ in equation (1), we get : $y^{2} - x^{2} = 4$ This is the required equation of the curve.

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). Find the equation of the curve given that it passes through $(- 2, 1)$ . Sol. It is given that (x, y) is the point of contact of the curve and its tangent. The slope $(m_{1})$ of the line segment joining $(x, y)$ and $(- 4, - 3)$ is $\frac{y - ( - 3 )}{x - ( - 4 )} = \frac{y + 3}{x + 4}$ .We know that the slope of the tangent to the curve is given by the relation, $\frac{dy}{dx}$ ∴ Slope $(m_{2})$ of the tangent $= \frac{dy}{dx}$ According to the given information :

$m_{2} = 2 m_{1} \frac{dy}{dx} = \frac{2 ( y + 3 )}{x + 4}$

Separating the variables, we get :

$\frac{d y}{y + 3} = \frac{2 d x}{x + 4}$

Integrating both sides, we get :

$\int \frac{dy}{y + 3} = 2 \int \frac{dx}{x + 4}$

$\Rightarrow lo g (y + 3) = 2 lo g (x + 4) + lo g C$ $\Rightarrow lo g (y + 3) = lo g C (x + 4)^{2}$ $\Rightarrow y + 3 = C (x + 4)^{2}$ .....(1)

This is the general equation of the curve. It is given that it passes through the point (-2, 1).

$\Rightarrow 1 + 3 = C (- 2 + 4)^{2} \Rightarrow 4 = 4 C \Rightarrow C = 1$ Substituting $C = 1$ in equation (1), we get : $y + 3 = (x + 4)^{2}$ This is the required equation of the curve.

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds. Sol. Let the rate of change of the volume of the balloon be k (where k is a constant). $\Rightarrow \frac{dv}{dt} = k$ $\Rightarrow \frac{d}{dt} (\frac{4}{3} π r^{3}) = k [$ Volume of sphere $= \frac{4}{3} π r^{3}]$ $\Rightarrow \frac{4}{3} π \cdot 3 r^{2} \cdot \frac{dr}{dt} = k$ $\Rightarrow 4 π r^{2} dr = kdt$

Integrating both sides, we get:

$4 π \int r^{2} dr = k \int dt \Rightarrow 4 π \cdot \frac{r ^{3}}{3} = kt + C$

$\Rightarrow 4 π r^{3} = 3 (kt + C)$ Now, at $t = 0, r = 3$

$\Rightarrow \Rightarrow \Rightarrow 4 π \times 3^{3} = 3 (k \times 0 + C) 108 π = 3 C C = 36 π$

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow 4 π \times 6^{3} = 3 (k \times 3 + C) 864 π = 3 (3 k + 36 π) 3 k = 288 π - 36 π = 252 π k = 84 π$

Substituting the values of k and C in equation (1), we get:

$4 π r^{3} = 3 [84 π t + 36 π]$

Thus, the radius of the balloon after t seconds is $(63 t + 27)^{\frac{1}{3}}$ .

In a bank, principal increases continuously at the rate of r% per year. Find the value of r if ₹100 doubles itself in 10 years $(lo g_{e} 2 = 0.6931)$ . Sol. Let p, t, and r represent the principal, time, and rate of interest respectively. It is given that the principal increases continuously at the rate of r% per year.

$\Rightarrow \frac{dp}{dt} = (\frac{r}{100}) p \Rightarrow \frac{dp}{p} = (\frac{r}{100}) dt$

Integrating both sides, we get:

$\int \frac{dp}{p} = \frac{r}{100} \int dt$

$\Rightarrow lo g p = \frac{rt}{100} + k \Rightarrow p = e^{\frac{rt}{100} + k}$

It is given that when $t = 0, p = 100$

$\Rightarrow 100 = e^{k}$

Now, if $t = 10$ , then $p = 2 \times 100 = 200$ . Therefore, equation (1) becomes :

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow 200 = e^{\frac{r}{10} + k} 200 = e^{\frac{r}{10}} \cdot 100 e^{\frac{r}{10}} = 2 \frac{r}{10} = 0.6931 r = 6.931 \Rightarrow \Rightarrow$

Hence, the value of r is 6.93%.

In a bank, principal increases continuously at the rate of 5% per year. An amount of ₹1000 is deposited with this bank, how much will it worth after 10 years $(e^{0.5} = 1.648)$ . Sol. Let p and t be the principal and time respectively. It is given that the principal increases continuously at the rate of $5%$ per year.

$\frac{dp}{dt} = (\frac{5}{100}) p \Rightarrow \frac{dp}{dt} = \frac{p}{20}$

Separating the variables, we get : $\frac{dp}{p} = \frac{dt}{20}$ Integrating both sides, we get: $\int \frac{dp}{p} = \frac{1}{20} \int dt$

$\Rightarrow lo g p = \frac{t}{20} + C \Rightarrow p = e^{\frac{t}{20} + C}$

Now, when $t = 0, p = 1000$ .

$\Rightarrow 1000 = e^{c}$

At $t = 10$ , equation (1) becomes:

$\Rightarrow \Rightarrow \Rightarrow p p p p = e^{\frac{1}{2} + C} = e^{0.5} \times e^{c} = 1.648 \times 1000 = 1648$

[From eq.(2)] Hence, after 10 years the amount will be worth ₹1648.

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

Sol. Let y be the number of bacteria at any instant t. It is given that the rate of growth of the bacteria is proportional to the number present.

$∴ \frac{dy}{dt} \propto y \Rightarrow \frac{dy}{dt} = ky$ (where k is a constant) Separating the variables, we get : $\frac{dy}{y} = kdt$ Integrating both sides, we get : $\int \frac{dy}{y} = k \int dt$ $\Rightarrow lo g y = kt + C$ Let $y_{0}$ be the number of bacteria at $t = 0$ $\Rightarrow lo g y_{0} = C$ Substituting the value of C in equation (1), we get : $lo g y = kt + lo g y_{0} \Rightarrow lo g y - lo g y_{0} = kt$ $\Rightarrow lo g (\frac{y}{y _{0}}) = kt$ Also, it is given that the number of bacteria increases by 10% in 2 hours. $\Rightarrow y = \frac{110}{100} y_{0} \Rightarrow \frac{y}{y _{0}} = \frac{11}{10}$

Substituting this value from equation (3) in equation (2), we get:

$k \cdot 2 = lo g (\frac{11}{10}) \Rightarrow k = \frac{1}{2} lo g (\frac{11}{10})$

Therefore, equation (2) becomes:

$\frac{1}{2} lo g (\frac{11}{10}) \cdot t = lo g (\frac{y}{y _{0}})$

$\Rightarrow t = \frac{2 l o g ( \frac{y}{y _{0}} )}{l o g ( \frac{11}{10} )}$ Now, let the time when the number of bacteria increases from 100000 to 200000 be $t_{1}$ .

$\Rightarrow y = 2 y_{0} at t = t_{1}$

From equation (4), we get :

$t_{1} = \frac{2 l o g ( \frac{y}{y _{0}} )}{l o g ( \frac{11}{10} )} = \frac{2 l o g 2}{l o g ( \frac{11}{10} )}$

Hence, in $\frac{2 l o g 2}{l o g ( \frac{11}{10} )}$ hours the number of bacteria increases from 100000 to 200000. 23. The general solution of the differential equation $\frac{dy}{dx} = e^{x + y}$ is (A) $e^{x} + e^{- y} = C$ (B) $e^{x} + e^{y} = C$ (C) $e^{- x} + e^{y} = C$ (D) $e^{- x} + e^{- y} = C$

Sol. (A) Given, $\frac{dy}{dx} = e^{x + y} = e^{x} \cdot e^{y}$ Separating the variables, we get :

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

$\Rightarrow e^{- y} dy = e^{x} dx$ Integrating both sides, we get:

$\int e^{- y} d y = \int e^{x} d x$

$\Rightarrow - e^{- y} = e^{x} + k$ $\Rightarrow e^{x} + e^{- y} = - k$ $\Rightarrow e^{x} + e^{- y} = C (C = - k)$ Hence, the correct answer is A.

EXERCISE 9.4

In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.

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

Sol. The given differential equation i.e.,

$(x^{2} + x y) d y = (x^{2} + y^{2}) d x can be written as: \frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x ^{2} + x y}$

Let $F (x, y) = \frac{x ^{2} + y ^{2}}{x ^{2} + x y}$ Now, $F (λ x, λ y) = \frac{( λ x ) ^{2} + ( λ y ^{2} )}{( λ x ) ^{2} + ( λ x ) ( λ y )}$

$= \frac{λ ^{2} ( x ^{2} + y ^{2} )}{λ ^{2} ( x ^{2} + xy )} = λ^{0} \cdot F (x, y)$

This shows that equation (1) is a homogeneous equation. To solve it, we make the substitution as : $y = vx$ Differentiating both sides with respect to x, we get : $\frac{dy}{dx} = v + x \frac{dv}{dx}$ Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get :

$v + x \frac{d v}{d x} = \frac{x ^{2} + ( vx ) ^{2}}{x ^{2} + x ( vx )} \Rightarrow v + x \frac{d v}{d x} = \frac{1 + v ^{2}}{1 + v}$

$\Rightarrow x \frac{dv}{dx} = \frac{1 + v ^{2}}{1 + v} - v = \frac{( 1 + v ^{2} ) - v ( 1 + v )}{1 + v}$

$\Rightarrow \Rightarrow x \frac{d v}{d x} = \frac{1 - v}{1 + v} (\frac{2 - 1 + v}{1 - v}) d v = \frac{d x}{x} \Rightarrow (\frac{1 + v}{1 - v}) d v = \frac{d x}{x} - 1) d v = \frac{d x}{x}$

Integrating both sides, we get:

$- 2 lo g (1 - v) - v = lo g x - lo g k$

This is the required solution of the given differential equation.

$y^{'} = \frac{x + y}{x}$

Sol. The given differential equation is:

$y^{'} = \frac{x + y}{x} \Rightarrow \frac{d y}{d x} = \frac{x + y}{x}$

Let $F (x, y) = \frac{x + y}{x}$ Now,

$F (λ x, λ y) = \frac{λ x + λ y}{λ x} = \frac{λ ( x + y )}{λ x} = λ^{0} F (x, y)$

Thus, the given equation is a homogeneous equation. To solve it, we make the substitution as :

$y = vx$

Differentiating both sides with respect to x, we get : $\frac{dy}{dx} = v + x \frac{dv}{dx}$ Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get :

$v + x \frac{dv}{dx} = \frac{x + vx}{x} \Rightarrow x \frac{dv}{dx} = 1 \Rightarrow \Rightarrow v + x \frac{dv}{dx} = 1 + v \int dv = \int \frac{dx}{x}$

Integrating both sides, we get:

$v = lo g x + C \Rightarrow \frac{y}{x} = lo g x + C$

$\Rightarrow y = x lo g x + Cx$ This is the required solution of the given differential equation.

$(x - y) d y - (x + y) d x = 0$

Sol. The given differential equation is :

$(x - y) d y - (x + y) d x = 0$

$\Rightarrow \frac{d y}{d x} = \frac{x + y}{x - y}$

Let $F (x, y) = \frac{x + y}{x - y}$

$∴ F (λ x, λ y) = \frac{λ x + λ y}{λ x - λ y} = \frac{λ ( x + y )}{λ ( x - y )} = λ^{0} \cdot F (x, y)$

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as :

$y = vx$

$\Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (vx) \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get:

$v + x \frac{d v}{d x} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v ( 1 - v )}{1 - v}$

$\Rightarrow x \frac{d v}{d x} = \frac{1 + v ^{2}}{1 - v}$

$\Rightarrow \int \frac{1 - v}{( 1 + v ^{2} )} d v = \int \frac{d x}{x}$

$\Rightarrow \int (\frac{1}{1 + v ^{2}} - \frac{v}{1 + v ^{2}}) d v = \int \frac{d x}{x}$

$\Rightarrow tan^{- 1} v - \frac{1}{2} lo g (1 + v^{2}) = lo g x + C$

$\Rightarrow tan^{- 1} (\frac{y}{x}) - \frac{1}{2} lo g [1 + (\frac{y}{x})^{2}] = lo g x + C$

$\Rightarrow tan^{- 1} (\frac{y}{x}) - \frac{1}{2} lo g (\frac{x ^{2} + y ^{2}}{x ^{2}}) = lo g x + C$

$\Rightarrow tan^{- 1} (\frac{y}{x}) - \frac{1}{2} [lo g (x^{2} + y^{2}) - lo g x^{2}] = lo g x + C$

$\Rightarrow tan^{- 1} (\frac{y}{x}) = \frac{1}{2} lo g (x^{2} + y^{2}) + C$

This is the required solution of the given differential equation.

$(x^{2} - y^{2}) d x + 2 x y d y = 0$

Sol. The given differential equation is :

$(x^{2} - y^{2}) d x + 2 x y d y = 0$

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

$Let F (x, y) = \frac{- ( x ^{2} - y ^{2} )}{2 x y}$

$∴ F (λ x, λ y) = - [\frac{( λ x ) ^{2} - ( λ y ) ^{2}}{2 ( λ x ) ( λ y )}] = \frac{- λ ^{2} ( x ^{2} - y ^{2} )}{λ ^{2} ( 2 xy )} = λ^{0} \cdot F (x, y)$

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

$y = vx$

$\Rightarrow \Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (vx) \frac{d y}{d x} = v + x \frac{d v}{d x}$

Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get:

$v + x \frac{dv}{dx} \Rightarrow v + x \frac{dv}{dx} = - [\frac{x ^{2} - ( vx ) ^{2}}{2 x \cdot ( vx )}] = \frac{v ^{2} - 1}{2 v}$

$\Rightarrow x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} - v = \frac{v ^{2} - 1 - 2 v ^{2}}{2 v}$

$\Rightarrow x \frac{d v}{d x} = - \frac{( 1 + v ^{2} )}{2 v}$

$\Rightarrow \frac{2 v}{1 + v ^{2}} d v = - \frac{d x}{x}$

$\Rightarrow \int \frac{2 v}{1 + v ^{2}} d v lo g (1 + v^{2}) = - \int \frac{d x}{x} = - lo g x + lo g C = lo g \frac{C}{x}$

$\Rightarrow 1 + v^{2} = \frac{C}{x}$

$\Rightarrow [1 + \frac{y ^{2}}{x ^{2}}] = \frac{C}{x}$

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

This is the required solution of the given differential equation.

$x^{2} \frac{dy}{dx} = x^{2} - 2 y^{2} + xy$ Sol. The given differential equation is :

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

Let $F (x, y) = \frac{x ^{2} - 2 y ^{2} + xy}{x ^{2}}$ $∴ F (λ x, λ y) = \frac{( λ x ) ^{2} - 2 ( λ y ) ^{2} + ( λ x ) ( λ y )}{( λ x ) ^{2}}$

$= \frac{λ ^{2} ( x ^{2} - 2 y ^{2} + xy )}{λ ^{2} ( x ^{2} )} = λ^{0} \cdot F (x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

$y = vx \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get:

$v + x \frac{d v}{d x} = \frac{x ^{2} - 2 ( vx ) ^{2} + x \cdot ( vx )}{x ^{2}}$

$\Rightarrow v + x \frac{dv}{dx} = 1 - 2 v^{2} + v \Rightarrow x \frac{dv}{dx} = 1 - 2 v^{2}$

$\Rightarrow \frac{d v}{1 - 2 v ^{2}} = \frac{d x}{x}$

$\Rightarrow \frac{1}{2} \cdot \int \frac{dv}{\frac{1}{2} - v ^{2}} = \int \frac{dx}{x}$

$\Rightarrow \frac{1}{2} \cdot [\frac{dv}{( \frac{1}{2} ) ^{2} - v ^{2}}] = \int \frac{dx}{x}$

$\Rightarrow \frac{1}{2} \cdot \frac{1}{2 \times \frac{1}{2}} lo g \frac{\frac{1}{2} + v}{\frac{1}{2} - v} = lo g ∣ x ∣ + C$

$(∵ \int \frac{dx}{a ^{2} - x ^{2}} = \frac{1}{2 a} lo g \frac{a + x}{a - x} + C)$

$\Rightarrow \frac{1}{2 2} lo g \frac{\frac{1}{2} + \frac{y}{x}}{\frac{1}{2} - \frac{y}{x}} = lo g ∣ x ∣ + C \Rightarrow \frac{1}{2 2} lo g \frac{x + 2 y}{x - 2 y} = lo g ∣ x ∣ + C$

This is the required solution for the given differential equation.

$x dy - ydx = x^{2} + y^{2} dx$ Sol. $xdy - ydx = x^{2} + y^{2} dx$

$\Rightarrow x d y = [y + x^{2} + y^{2}] d x \frac{d y}{d x} = \frac{y + x ^{2} + y ^{2}}{x}$

Let $F (x, y) = \frac{y + x ^{2} + y ^{2}}{x}$ $∴ F (λ x, λ y) = \frac{λ x + ( λ x ) ^{2} + ( λ y ) ^{2}}{λ x}$

$= \frac{λ ( y + x ^{2} + y ^{2} )}{λ x} = λ^{0} \cdot F (x, y)$

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as : $y = vx$

$\Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (vx) \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get:

$v + x \frac{d v}{d x} = \frac{vx + x ^{2} + ( vx ) ^{2}}{x}$

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

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

$\Rightarrow \int \frac{dv}{1 + v ^{2}} = \int \frac{dx}{x}$

$\Rightarrow lo g v + 1 + v^{2} = lo g ∣ x ∣ + lo g C$

$\Rightarrow lo g \frac{y}{x} + 1 + \frac{y ^{2}}{x ^{2}} = lo g ∣ Cx ∣$ $\Rightarrow lo g \frac{y + x ^{2} + y ^{2}}{x} = lo g ∣ Cx ∣$ $\Rightarrow y + x^{2} + y^{2} = Cx^{2}$ This is the required solution of the given differential equation.

${x cos (\frac{y}{x}) + y sin (\frac{y}{x})} y dx$

$= {y sin (\frac{y}{x}) - x cos (\frac{y}{x})} xdy$

Sol. The given differential equation is:

${x cos (\frac{y}{x}) + y sin (\frac{y}{x})} y d x = {y sin (\frac{y}{x}) - x cos (\frac{y}{x})} x d y \frac{d y}{d x} = \frac{{ x cos ( \frac{y}{x} ) + y sin ( \frac{y}{x} ) } y}{{ y sin ( \frac{y}{x} ) - x cos ( \frac{y}{x} ) } x} \dots\dots (1) Let F (x, y) = \frac{{ x cos ( \frac{y}{x} ) + y sin ( \frac{y}{x} ) } y}{{ y sin ( \frac{y}{x} ) - x cos ( \frac{y}{x} ) } x} F (λ x, λ y) = \frac{{ λ x cos ( \frac{λ y}{λ x} ) + λ y sin ( \frac{λ y}{λ x} ) } λ y}{{ λ y sin ( \frac{λ y}{λ x} ) - λ x cos ( \frac{λ y}{λ x} ) } λ x} = \frac{λ ^{2} { x cos ( \frac{y}{x} ) + y sin ( \frac{y}{x} ) } y}{λ ^{2} { y sin ( \frac{y}{x} ) - x cos ( \frac{y}{x} ) } x} = λ^{0} \cdot F (x, y)$

Therefore, the given differential equation is a homogeneous equation.To solve it, we make the substitution as:

$y = vx \Rightarrow \frac{d y}{d x} = v + x \times \frac{d v}{d x}$

Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get:

$v + x \frac{d v}{d x} = \frac{( x c o s v + vx s i n v ) \cdot vx}{( vx s i n v - x c o s v ) \cdot x}$

$\Rightarrow v + x \frac{dv}{dx} = \frac{v c o s v + v ^{2} s i n v}{v s i n v - c o s v}$ $\Rightarrow x \frac{dv}{dx} = \frac{v c o s v + v ^{2} s i n v}{v s i n v - c o s v} - v$ $\Rightarrow x \frac{dv}{dx} = \frac{v c o s v + v ^{2} s i n v - v ^{2} s i n v + v c o s v}{v s i n v - c o s v}$ $\Rightarrow x \frac{dv}{dx} = \frac{2 v c o s v}{v s i n v - c o s v}$ $\Rightarrow [\frac{v s i n v - c o s v}{v c o s v}] dv = \frac{2 dx}{x}$ $\Rightarrow (tan v - \frac{1}{v}) dv = \frac{2 dx}{x}$

Integrating both sides, we get :

$\int (tan v - \frac{1}{v}) dv = 2 \int \frac{dx}{x}$ $lo g (sec v) - lo g v = 2 lo g x + lo g C$ $\Rightarrow lo g (\frac{s e c v}{v}) = lo g (Cx^{2})$ $\Rightarrow (\frac{s e c v}{v}) = Cx^{2} \Rightarrow sec v = Cx^{2} v$ $\Rightarrow sec (\frac{y}{x}) = C \cdot x^{2} \cdot \frac{y}{x} \Rightarrow sec (\frac{y}{x}) = Cxy$ $\Rightarrow cos (\frac{y}{x}) = \frac{1}{Cxy} = \frac{1}{C} \cdot \frac{1}{xy}$ $\Rightarrow xy cos (\frac{y}{x}) = k (k = \frac{1}{C})$ This is the required solution of the given differential equation.

$x \frac{dy}{dx} - y + x sin (\frac{y}{x}) = 0$ Sol. $\frac{dy}{dx} - y + x sin (\frac{y}{x}) = 0$ $\Rightarrow x \frac{dy}{dx} = y - x sin (\frac{y}{x})$ $\Rightarrow \frac{dy}{dx} = \frac{y - x s i n ( \frac{y}{x} )}{x}$ Let $F (x, y) = \frac{y - x s i n ( \frac{y}{x} )}{x}$ $∴ F (λ x, λ y) = \frac{λ y - λ x s i n ( \frac{λ y}{λ x} )}{λ x}$ $= \frac{λ [ y - x s i n ( \frac{y}{x} ) ]}{λ x} = λ^{0} \cdot F (x, y)$ Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as: $y = vx \Rightarrow \frac{d}{dx} (y) = \frac{d}{dx} (vx) \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$ Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get: $v + x \frac{dv}{dx} = \frac{vx - x s i n v}{x} \Rightarrow v + x \frac{dv}{dx} = v - sin v$ $\Rightarrow - \frac{dv}{s i n v} = \frac{dx}{x}$ $\Rightarrow cosec vdv = - \frac{dx}{x} \Rightarrow \int cosec v dv = - \int \frac{dx}{x}$ $lo g ∣ cosec v - cot v ∣ = - lo g x + lo g C = lo g \frac{C}{x}$ $\Rightarrow cosec (\frac{y}{x}) - cot (\frac{y}{x}) = \frac{C}{x}$ $\Rightarrow \frac{1}{s i n ( \frac{y}{x} )} - \frac{c o s ( \frac{y}{x} )}{s i n ( \frac{y}{x} )} = \frac{C}{x}$ $\Rightarrow x [1 - cos (\frac{y}{x})] = C sin (\frac{y}{x})$ This is the required solution of the given differential equation.
$ydx + xlog (\frac{y}{x}) dy - 2 xdy = 0$ Sol. $ydx + x lo g (\frac{y}{x}) dy - 2 xdy = 0$ $\Rightarrow ydx = [2 x - x lo g (\frac{y}{x})] dy$ $\Rightarrow \frac{dy}{dx} = \frac{y}{2 x - x l o g ( \frac{y}{x} )}$ Let $F (x, y) = \frac{y}{2 x - x l o g ( \frac{y}{x} )}$ $∴ F (λ x, λ y) = \frac{λ y}{2 ( λ x ) - ( λ x ) l o g ( \frac{λ y}{λ x} )}$ $= \frac{λ y}{λ [ 2 x - x l o g ( \frac{y}{x} ) ]} = λ^{0} \cdot F (x, y)$ Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as : $y = vx \Rightarrow \frac{dy}{dx} = \frac{d}{dx} (vx) \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$ Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get: $v + x \frac{dv}{dx} = \frac{vx}{2 x - x l o g v}$ $\Rightarrow v + x \frac{dv}{dx} = \frac{v}{2 - l o g v}$ $\Rightarrow x \frac{dv}{dx} = \frac{v}{2 - l o g v} - v$ $\Rightarrow x \frac{dv}{dx} = \frac{v - 2 v + v l o g v}{2 - l o g v}$ $\Rightarrow x \frac{dv}{dx} = \frac{v l o g v - v}{2 - l o g v}$ $\Rightarrow \frac{2 - l o g v}{v ( l o g v - 1 )} dv = \frac{dx}{x}$ $\Rightarrow [\frac{1 + ( 1 - l o g v )}{v ( l o g v - 1 )}] dv = \frac{dx}{x}$

$\Rightarrow [\frac{1}{v ( l o g v - 1 )} - \frac{1}{v}] d v = \frac{d x}{x}$

Integrating both sides, we get:

$\int \frac{1}{v ( l o g v - 1 )} d v - \int \frac{1}{v} d v = \int \frac{1}{x} d x$

$\Rightarrow \int \frac{d v}{v ( l o g v - 1 )} - lo g v = lo g x + lo g C$

Let $lo g v - 1 = t$

$\Rightarrow \frac{d}{d v} (lo g v - 1) = \frac{d t}{d v} \Rightarrow \frac{1}{v} = \frac{d t}{d v} \Rightarrow \frac{d v}{v} = d t$

Therefore, equation (2) becomes:

$\Rightarrow \int \frac{dt}{t} - lo g v = lo g x + lo g C$

$\Rightarrow lo g [lo g (\frac{y}{x}) - 1] - lo g (\frac{y}{x}) = lo g (Cx) \Rightarrow lo g [\frac{lo g ( \frac{y}{x} ) - 1}{\frac{y}{x}}] = lo g (Cx) \Rightarrow \frac{x}{y} [lo g (\frac{y}{x}) - 1] = Cx \Rightarrow lo g (\frac{y}{x}) - 1 = Cy$

This is the required solution of the given differential equation.

$(1 + e^{\frac{x}{y}}) dx + e^{\frac{x}{y}} (1 - \frac{x}{y}) dy = 0$

Sol. $(1 + e^{\frac{x}{y}}) dx + e^{\frac{x}{y}} (1 - \frac{x}{y}) dy = 0$

$\Rightarrow (1 + e^{\frac{x}{y}}) d x = - e^{\frac{x}{y}} \Rightarrow \frac{d x}{d y} = \frac{- e ^{\frac{x}{y}} ( 1 - \frac{x}{y} )}{1 + e ^{\frac{x}{y}}}$

$Let F (x, y) = \frac{- e ^{\frac{x}{y}} ( 1 - \frac{x}{y} )}{1 + e ^{\frac{x}{y}}} ∴ F (λ x, λ y) = \frac{- e ^{\frac{λ x}{λ y}} ( 1 - \frac{λ x}{λ y} )}{1 + e ^{\frac{λ x}{λ y}}} = \frac{- e ^{\frac{x}{y}} ( 1 - \frac{x}{y} )}{1 + e ^{\frac{x}{y}}} = λ^{0} \cdot F (x, y)$

Therefore, the given differential equation is a homogeneous equation. To solve it, we make the substitution as:

$x = v y \Rightarrow \frac{d}{d y} (x) = \frac{d}{d y} (v y) \Rightarrow \frac{d x}{d y} = v + y \frac{d v}{d y}$

Substituting the values of x and $\frac{dx}{dy}$ in equation (1), we get:

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow v + y \frac{d v}{d y} = \frac{- e ^{v} ( 1 - v )}{1 + e ^{v}} y \frac{d v}{d y} = \frac{- e ^{v} + v e ^{v}}{1 + e ^{v}} - v y \frac{d v}{d y} = \frac{- e ^{v} + v e ^{v} - v - v e ^{v}}{1 + e ^{v}} y \frac{d v}{d y} = - [\frac{v + e ^{v}}{1 + e ^{v}}] [\frac{1 + e ^{v}}{v + e ^{v}}] d v = - \frac{d y}{y}$

Integrating both sides, we get:

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow lo g (v + e^{v}) = - lo g y + lo g C = lo g (\frac{C}{y}) lo g (v + e^{v}) = lo g (\frac{C}{y}) \frac{x}{y} + e^{\frac{x}{y}} = \frac{C}{y} x + ye^{\frac{x}{y}} = C$

This is the required solution of the given differential equation.

For each of the differential equations in Exercises from 11 to 15, find the particular solution satisfying the given condition:

$(x + y) d y + (x - y) d x = 0; y = 1$ when $x = 1$

Sol. $(x + y) d y + (x - y) d x = 0$

$\Rightarrow \Rightarrow (x + y) d y = - (x - y) d x \frac{d y}{d x} = \frac{- ( x - y )}{x + y}$

The given differential equation is a homogeneous equation. To solve it, we make the substitution as:

$y = vx$

$\Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (vx) \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get:

$v + x \frac{d v}{d x} = \frac{- ( x - vx )}{x + vx}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{v - 1}{v + 1}$

$\Rightarrow x \frac{d v}{d x} = \frac{v - 1}{v + 1} - v = \frac{v - 1 - v ( v + 1 )}{v + 1}$

$\Rightarrow x \frac{d v}{d x} = \frac{v - 1 - v ^{2} - v}{v + 1} = \frac{- ( 1 + v ^{2} )}{v + 1}$

$\Rightarrow \frac{( v + 1 )}{1 + v ^{2}} dv = - \frac{dx}{x}$

$\Rightarrow \int \frac{v + 1}{1 + v ^{2}} dv = - \int \frac{dx}{x}$

$\Rightarrow \int [\frac{v}{1 + v ^{2}} + \frac{1}{1 + v ^{2}}] dv = - \int \frac{dx}{x}$

$\Rightarrow \frac{1}{2} lo g (1 + v^{2}) + tan^{- 1} v = - lo g x + k$

$\Rightarrow lo g (1 + v^{2}) + 2 tan^{- 1} v = - 2 lo g x + 2 k$

$\Rightarrow lo g [(1 + v^{2}) \cdot x^{2}] + 2 tan^{- 1} v = 2 k$

$\Rightarrow lo g [(1 + \frac{y ^{2}}{x ^{2}}) \cdot x^{2}] + 2 tan^{- 1} \frac{y}{x} = 2 k$

$\Rightarrow lo g (x^{2} + y^{2}) + 2 tan^{- 1} \frac{y}{x} = 2 k$

Now, $y = 1$ at $x = 1$

$\Rightarrow lo g 2 + 2 tan^{- 1} 1 = 2 k \Rightarrow lo g 2 + 2 \times \frac{π}{4} = 2 k \Rightarrow \frac{π}{2} + lo g 2 = 2 k$

Substituting the value of 2 k in equation (2), we get:

$lo g (x^{2} + y^{2}) + 2 tan^{- 1} (\frac{y}{x}) = \frac{π}{2} + lo g 2$

This is the required solution of the given differential equation.

$x^{2} dy + (xy + y^{2}) dx = 0; y = 1$ when $x = 1$

Sol. $x^{2} d y + (x y + y^{2}) d x = 0$

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

The given differential equation is a homogeneous equation. To solve it, we make the substitution as:

$y = vx$

$\Rightarrow \Rightarrow \frac{d}{dx} (y) = \frac{d}{dx} (vx) \frac{dy}{dx} = v + x \frac{dv}{dx}$

Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get:

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow v + x \frac{d v}{d x} = \frac{- [ x \cdot vx + ( vx ) ^{2} ]}{x ^{2}} = - v - v^{2} x \frac{d v}{d x} = - v^{2} - 2 v = - v (v + 2) \frac{d v}{v ( v + 2 )} = - \frac{d x}{x} \int \frac{d v}{v ( v + 2 )} = - \int \frac{d x}{x} \int \frac{1}{2} [\frac{( v + 2 ) - v}{v ( v + 2 )}] d v = - \int \frac{d x}{x} \int \frac{1}{2} [\frac{1}{v} - \frac{1}{v + 2}] d v = - \int \frac{d x}{x}$

$\Rightarrow \frac{1}{2} [lo g v - lo g (v + 2)] = - lo g x + lo g C \Rightarrow \frac{1}{2} lo g (\frac{v}{v + 2}) = lo g \frac{C}{x} \Rightarrow \frac{v}{v + 2} = (\frac{C}{x})^{2} \Rightarrow \frac{\frac{y}{x}}{\frac{y}{x} + 2} = (\frac{C}{x})^{2} \Rightarrow \frac{y}{y + 2 x} = \frac{C ^{2}}{x ^{2}} \Rightarrow \frac{x ^{2} y}{y + 2 x} = C^{2} \dots$

Put, $y = 1$ and $x = 1$ in eq.(2) , we get

$\frac{1}{1 + 2} = C^{2} \Rightarrow C^{2} = \frac{1}{3}$

Substituting $C^{2} = \frac{1}{3}$ in equation (2), we get:

$\frac{x ^{2} y}{y + 2 x} = \frac{1}{3} \Rightarrow y + 2 x = 3 x^{2} y$

This is the required solution of the given differential equation.

$[x sin^{2} (\frac{y}{x}) - y] dx + x dy = 0$ ;

$y = \frac{π}{4} when x = 1$

Sol. $[x sin^{2} (\frac{y}{x}) - y] d x + x d y = 0$

$\Rightarrow \frac{dy}{dx} = \frac{- [ x s i n ^{2} ( \frac{y}{x} ) - y ]}{x}$

The given differential equation is a homogeneous equation. To solve this differential equation, we make the substitution as:

$y = vx \Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (vx) \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$

Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get:

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow v + x \frac{d v}{d x} = \frac{- [ x s i n ^{2} v - vx ]}{x} v + x \frac{d v}{d x} = - [sin^{2} v - v] = v - sin^{2} v x \frac{d v}{d x} = - sin^{2} v \frac{d v}{s i n ^{2} v} = - \frac{d x}{x} cosec^{2} v d v = - \frac{d x}{x} \int cosec^{2} v d v = - \int \frac{d x}{x} - cot v = - lo g ∣ x ∣ + lo g C cot (\frac{y}{x}) = lo g ∣ x ∣ - lo g C \frac{x}{C}$

Now, $y = \frac{π}{4}$ at $x = 1$

$\Rightarrow cot (\frac{π}{4}) = lo g \frac{1}{C} \Rightarrow 1 = - lo g C \Rightarrow C = e^{- 1}$

Substituting $C = e^{- 1}$ in equation (2), we get :

$cot (\frac{y}{x}) = lo g ∣ ex ∣$

This is the required solution of the given differential equation.

$\frac{dy}{dx} - \frac{y}{x} + cosec (\frac{y}{x}) = 0; y = 0$ when $x = 1$ Sol. $\frac{dy}{dx} = \frac{y}{x} - cosec (\frac{y}{x})$ The given differential equation is a homogeneous equation. To solve it, we make the substitution as:

$y = vx \Rightarrow \frac{d}{dx} (y) = \frac{d}{dx} (vx)$

$\Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}$

Substituting the values of y and $\frac{dy}{dx}$ in equation (1), we get:

$v + x \frac{d v}{d x} = v - cosec v$

$\Rightarrow - \frac{d v}{cosec v} = \frac{d x}{x}$

$\Rightarrow - sin v d v = \frac{d x}{x}$

$\Rightarrow - \int sin v d v = \int \frac{d x}{x}$

$cos v = lo g x + lo g C = lo g ∣ Cx ∣$

$\Rightarrow cos (\frac{y}{x}) = lo g ∣ Cx ∣$

This is the required solution of the given differential equation. Now, $y = 0$ at $x = 1$

$\Rightarrow cos (0) = lo g C$

$\Rightarrow 1 = lo g C \Rightarrow C = e^{1} = e$

Substituting $C = e$ in equation (2), we get:

$cos (\frac{y}{x}) = lo g ∣ (ex) ∣$

This is the required solution of the given differential equation.

$2 xy + y^{2} - 2 x^{2} \frac{dy}{dx} = 0; y = 2$ when $x = 1$

Sol. $2 xy + y^{2} - 2 x^{2} \frac{dy}{dx} = 0$

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

The given differential equation is a homogeneous equation. To solve it, we make the substitution as:

$y = vx$

$\Rightarrow \Rightarrow \frac{d}{dx} (y) = \frac{d}{dx} (vx) \frac{dy}{dx} = v + x \frac{dv}{dx}$

Substituting the value of y and $\frac{dy}{dx}$ in equation (1), we get:

$v + x \frac{d v}{d x} = \frac{2 x ( vx ) + ( vx ) ^{2}}{2 x ^{2}}$

$\Rightarrow v + x \frac{d v}{d x} = \frac{2 v + v ^{2}}{2}$

$\Rightarrow v + x \frac{d v}{d x} = v + \frac{v ^{2}}{2}$

$\Rightarrow - \frac{2}{v} = lo g ∣ x ∣ + C \Rightarrow - \frac{2}{\frac{y}{x}} = lo g ∣ x ∣ + C$

Now, $y = 2$ at $x = 1$ .

$\Rightarrow \Rightarrow - 1 = lo g (1) + C C = - 1$

Substituting $C = - 1$ in equation (2), we get:

$- \frac{2 x}{y} = lo g ∣ x ∣ - 1$

$\Rightarrow \frac{2 x}{y} = 1 - lo g ∣ x ∣$

$\Rightarrow y = \frac{2 x}{1 - l o g ∣ x ∣}, (x \neq = 0, x \neq = e)$

This is the required solution of the given differential equation.

A homogeneous differential equation of the form $\frac{dx}{dy} = h (\frac{x}{y})$ can be solved by making the substitution (A) $y = vx$ (B) $v = yx$ (C) $x = v y$ (D) $x = v$

Sol. For solving the homogeneous equation of the form $\frac{dx}{dy} = h (\frac{x}{y})$ , we need to make the substitution as $x = vy$ . Hence, the correct answer is C.

Which of the following is a homogeneous differential equation？（A） $(4 x + 6 y + 5) dy - (3 y + 2 x + 4) dx = 0$ （B） $(xy) dx - (x^{3} + y^{3}) dy = 0$ （C） $(x^{3} + 2 y^{2}) dx + 2 xydy = 0$ （D） $y^{2} dx + (x^{2} - xy - y^{2}) dy = 0$ Sol．Out of the given four options；option（D）is the only option in which all coefficieints of dx and dy are of same degree，therefore function F（x， y）is said to be the homogenous function of degree n ，if $F (λ x, λ y) = λ^{n} F (x, y)$ for any non－zero constant（ $λ$ ）． Consider the equation given in option D：

$y^{2} d x + (x^{2} - x y - y^{2}) d y = 0$

$\Rightarrow \frac{d y}{d x} = \frac{- y ^{2}}{x ^{2} - x y - y ^{2}} = \frac{y ^{2}}{y ^{2} + x y - x ^{2}}$ Let $F (x, y) = \frac{y ^{2}}{y ^{2} + xy - x ^{2}}$

$\Rightarrow F (λ x, λ y) = \frac{( λ y ) ^{2}}{( λ y ) ^{2} + ( λ x ) ( λ y ) - ( λ x ) ^{2}} = \frac{λ ^{2} y ^{2}}{λ ^{2} ( y ^{2} + xy - x ^{2} )} = λ^{0} (\frac{y ^{2}}{y ^{2} + xy - x ^{2}}) = λ^{0} \cdot F (x, y)$

Hence，the differential equation given in option D is a homogenous equation．

EXERCISE 9.5

For each of the differential equations given in Q. 1 to 12，find the general solution ：

$\frac{dy}{dx} + 2 y = sin x$ Sol．The given differential equation is $\frac{d y}{d x} + 2 y = sin x$ This is linear differential equation of the form $\frac{dy}{dx} + Py = Q ($ where $P = 2$ and $Q = sin x)$ Now，I．F． $= e^{\int Pdx} = e^{\int 2 dx} = e^{2 x}$ The solution of the given differential equation is given by the relation， $y (I . F) = . \int (Q \times I . F) dx +$ . $\Rightarrow ye^{2 x} = \int sin x \cdot e^{2 x} dx + C$ ．．．．．（1） Let $I = \int I \int sin x \cdot e^{2 x} dx$ I II $\Rightarrow I = sin x \cdot \int e^{2 x} dx - \int (\frac{d}{dx} (sin x) \cdot \int e^{2 x} dx) dx$ $\Rightarrow I = sin x \cdot \frac{e ^{2 x}}{2} - \int (cos x \cdot \frac{e ^{2 x}}{2}) dx$ $\Rightarrow I = \frac{e ^{2 x} s i n x}{2} - \frac{1}{2} [cos x \cdot \int e^{2 x} - \int (\frac{d}{dx} (cos x) \cdot \int e^{2 x} dx) dx]$ $\Rightarrow I = \frac{e ^{2 x} s i n x}{2} - \frac{1}{2} [cos x \cdot \frac{e ^{2 x}}{2} - \int [(- sin x) \cdot \frac{e ^{2 x}}{2}] dx]$ $\Rightarrow I = \frac{e ^{2 x} s i n x}{2} - \frac{e ^{2 x} c o s x}{4} - \frac{1}{4} \int (sin x \cdot e^{2 x}) dx$ $\Rightarrow I = \frac{e ^{2 x}}{4} (2 sin x - cos x) - \frac{1}{4} I$ $\Rightarrow \frac{5}{4} I = \frac{e ^{2 x}}{4} (2 sin x - cos x)$ $\Rightarrow I = \frac{e ^{2 x}}{5} (2 sin x - cos x)$ Therefore，equation（1）becomes： $ye^{2 x} = \frac{e ^{2 x}}{5} (2 sin x - cos x) + C$ $\Rightarrow y = \frac{1}{5} (2 sin x - cos x) + Ce^{- 2 x}$ This is the required general solution of the given differential equation．
$\frac{dy}{dx} + 3 y = e^{- 2 x}$ Sol. The given differential equation is $\frac{d y}{d x} + 3 y = e^{- 2 x}$ . This is linear differential equation of the form $\frac{dy}{dx} + Py = Q ($ where $P = 3$ and $Q = e^{- 2 x})$ Now, I.F. $= e^{\int Pdx} = e^{\int 3 dx} = e^{3 x}$ The solution of the given differential equation is given by the relation, $y (I . F) = . \int (Q \times I . F) dx +$ . $\Rightarrow ye^{3 x} = \int (e^{- 2 x} \times e^{3 x}) + C$ $\Rightarrow ye^{3 x} = \int e^{x} dx + C$ $\Rightarrow ye^{3 x} = e^{x} + C \Rightarrow y = e^{- 2 x} + Ce^{- 3 x}$

This is the required general solution of the given differential equation.

$\frac{dy}{dx} + \frac{y}{x} = x^{2}$ Sol. The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^{2}$ . This is linear differential equation of the form $\frac{dy}{dx} + Py = Q$ (where $P = \frac{1}{x}$ and $Q = x^{2}$ ) Now, I.F. $= e^{\int Pdx} = e^{\int \frac{1}{x} dx} = e^{l o g x} = x$ The solution of the given differential equation is given by,

$y (I.F.) = \int (Q \times I.F.) d x + C$

$\Rightarrow y (x) = \int (x^{2} \cdot x) dx + C$ $\Rightarrow xy = \int x^{3} dx + C \Rightarrow xy = \frac{x ^{4}}{4} + C$

This is the required general solution of the given differential equation.

$\frac{dy}{dx} + (sec x) y = tan x (0 \leq x < \frac{π}{2})$ Sol. The given differential equation is $\frac{dy}{dx} + (sec x) y = tan x$ . This is linear differential equation of the form $\frac{dy}{dx} + Py = Q$ (where $P = sec x$ and $Q = tan x$ )

Now,

$I.F. = e^{\int Pdx} = e^{\int s e c xdx} = e^{l o g (s e c x + t a n x)} = sec x + tan x$

The general solution of the given differential equation is given by the relation,

$y (I.F.) = \int (Q \times I.F.) d x + C$

$\Rightarrow y (sec x + tan x) = \int tan x (sec x + tan x) dx + C$ $\Rightarrow y (sec x + tan x) = \int sec x tan x d x + \int tan^{2} x d x + C$ $\Rightarrow y (sec x + tan x) = sec x + \int (sec^{2} x - 1) d x + C$ $\Rightarrow y (sec x + tan x) = sec x + tan x - x + C$

$cos^{2} x \frac{d y}{d x} + y = tan x (0 \leq x < \frac{π}{2})$ Sol. It is given that $cos^{2} x \frac{dy}{dx} + y = tan x$ $\Rightarrow \frac{dy}{dx} + sec^{2} x \cdot y = sec^{2} x tan x$

This is differential equation in the form of $\frac{d y}{d x} + P y = Q$ (where, $P = sec^{2} x$ and $Q = sec^{2} x tan x$ ) Now, I.F. $= e^{\int Pdx} = e^{\int sec^{2} xdx} = e^{t a n x}$ Thus, the solution of the given differential equation is given by the relation :

$y (I . F .) = \int (Q \times I . F) d x + C$

$\Rightarrow y \cdot e^{t a n x} = \int e^{t a n x} sec^{2} x tan x d x + C$

Now, Let $tan x = t \Rightarrow sec^{2} x dx = dt$ Thus, the equation (1) becomes

$\Rightarrow y \cdot e^{t a n x} = \int (e^{t} \cdot t) d t + C$

$\Rightarrow y \cdot e^{t an x} = t \cdot \int e^{t} d t - \int (\frac{d}{d t} (t) \cdot \int e^{t} d t) d t + C$ $\Rightarrow y \cdot e^{tan x} = t \cdot e^{t} - \int e^{t} dt + C$ $\Rightarrow y e^{t an x} = (t - 1) e^{t} + C$ $\Rightarrow ye^{t a n x} = (tan x - 1) e^{t a n x} + C$ $\Rightarrow y = (tan x - 1) + Ce^{- t a n x}$ Therefore, the required general solution of the given differential equation

$y = (tan x - 1) + Ce^{- t a n x}$

$x \frac{d y}{d x} + 2 y = x^{2} lo g x$ Sol. The given differential equation is :

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

$\Rightarrow \frac{dy}{dx} + \frac{2}{x} y = x lo g x$

This equation is in the form of a linear differential equation as:

$\frac{d y}{d x} + P y = Q (where P = \frac{2}{x} and Q = x lo g x)$

Now, I.F. $= e^{\int Pdx} = e^{\int \frac{2}{x} dx} = e^{2 l o g x} = e^{l o g x^{2}} = x^{2}$ The general solution of the given differential equation is given by the relation,

$y (I . F .) = \int (Q \times I . F .) d x + C$

$\Rightarrow y \cdot x^{2} = \int (x lo g x \cdot x^{2}) dx + C$ $\Rightarrow x^{2} y = \int (x^{3} lo g x) d x + C$ $\Rightarrow x^{2} y = lo g x \cdot \int x^{3} d x - \int [\frac{d}{d x} (lo g x) \cdot \int x^{3} d x] d x + C$ $\Rightarrow x^{2} y = lo g x \cdot \frac{x ^{4}}{4} - \int (\frac{1}{x} \cdot \frac{x ^{4}}{4}) d x + C$ $\Rightarrow x^{2} y = \frac{x ^{4} l o g x}{4} - \frac{1}{4} \int x^{3} dx + C$ $\Rightarrow x^{2} y = \frac{x ^{4} l o g x}{4} - \frac{1}{4} \cdot \frac{x ^{4}}{4} + C$ $\Rightarrow x^{2} y = \frac{1}{16} x^{4} (4 lo g x - 1) + C$ $\Rightarrow y = \frac{1}{16} x^{2} (4 lo g x - 1) + Cx^{- 2}$

$x lo g x \frac{d y}{d x} + y = \frac{2}{x} lo g x$

Sol. The given differential equation is:

$x lo g x \frac{d y}{d x} + y = \frac{2}{x} lo g x$

$\Rightarrow \frac{dy}{dx} + \frac{y}{x l o g x} = \frac{2}{x ^{2}}$

This equation is the form of a linear differential equation as:

$\frac{d y}{d x} + P y = Q (where P = \frac{1}{x l o g x} and Q = \frac{2}{x ^{2}})$

Now, I.F. $= e^{\int Pdx} = e^{\int \frac{1}{x l o g x} dx} = e^{l o g (l o g x)} = lo g x$ The general solution of the given differential equation is given by the relation,

$y (I . F .) = \int (Q \times I . F .) d x + C$

$\Rightarrow y lo g x = \int (\frac{2}{x ^{2}} lo g x) d x + C$

$Now, \int (\frac{2}{x ^{2}} lo g x) dx$

$= 2 \int (lo g x \cdot \frac{1}{x ^{2}}) dx$

$= 2 [lo g x \cdot \int \frac{1}{x ^{2}} d x - \int {\frac{d}{d x} (lo g x) \cdot \int \frac{1}{x ^{2}} d x} d x]$

$= 2 [lo g x (- \frac{1}{x}) - \int (\frac{1}{x} \cdot (- \frac{1}{x})) d x]$

$= 2 [- \frac{l o g x}{x} + \int \frac{1}{x ^{2}} dx]$

$= 2 [- \frac{l o g x}{x} - \frac{1}{x}]$

$= - \frac{2}{x} (1 + lo g x)$

Substituting the value of $\int (\frac{2}{x ^{2}} lo g x) d x$ in equation (1), we get :

$y lo g x = - \frac{2}{x} (1 + lo g x) + C$

This is the required general solution of the given differential equation.

$(1 + x^{2}) dy + 2 xy dx = cot xdx (x \neq = 0)$ Sol. $(1 + x^{2}) dy + 2 xy dx = cot xdx$ $\Rightarrow \frac{dy}{dx} + \frac{2 xy}{1 + x ^{2}} = \frac{c o t x}{1 + x ^{2}}$

This equation is a linear differential equation of the form:

$\frac{d y}{d x} + P y = Q (where P = \frac{2 x}{1 + x ^{2}} and Q = \frac{c o t x}{1 + x ^{2}})$

Now, I.F. $= e^{\int Pdx} = e^{\int \frac{2 x}{1 + x ^{2}} dx} = e^{l o g (1 + x^{2})} = 1 + x^{2}$ The general solution of the given differential equation is given by the relation,

$y (I.F.) = \int (Q \times I.F.) d x + C$

$\Rightarrow y (1 + x^{2}) = \int [\frac{c o t x}{1 + x ^{2}} \times (1 + x^{2})] dx + C$ $\Rightarrow y (1 + x^{2}) = \int cot x dx + C$ $\Rightarrow y (1 + x^{2}) = lo g ∣ sin x ∣ + C$

$x \frac{dy}{dx} + y - x + xy cot x = 0 (x \neq = 0)$

Sol. $x \frac{dy}{dx} + y - x + xy cot x = 0$

$\Rightarrow x \frac{d y}{d x} + y (1 + x cot x) = x$

$\Rightarrow \frac{dy}{dx} + (\frac{1}{x} + cot x) y = 1$ This equation is a linear differential equation of the form:

$\frac{d y}{d x} + P y = Q (where P = \frac{1}{x} + cot x and Q = 1)$

Now,

$I.F. = e^{\int Pdx} = e^{\int (\frac{1}{x} + c o t x) dx} = e^{l o g x + l o g (s i n x)} = e^{l o g (x s i n x)} = x sin x$

The general solution of the given differential equation is given by the relation,

$y (I.F.) = \int (Q \times I.F.) d x + C$

$\Rightarrow y (x sin x) = \int (1 \times x sin x) d x + C$

$\Rightarrow y (x sin x) = \int (x sin x) d x + C$

$\Rightarrow y (x sin x) = x \int sin xdx - \int [\frac{d}{dx} (x) \cdot \int sin xdx] + C$

$\Rightarrow y (x sin x) = x (- cos x) - \int 1 \cdot (- cos x) d x + C$

$\Rightarrow y (x sin x) = - x cos x + sin x + C$ $\Rightarrow y = \frac{- x c o s x}{x s i n x} + \frac{s i n x}{x s i n x} + \frac{C}{x s i n x}$ $\Rightarrow y = - cot \cdot x + \frac{1}{x} + \frac{C}{x s i n x}$ (which is the required solution)

$(x + y) \frac{dy}{dx} = 1$

Sol. $(x + y) \frac{dy}{dx} = 1$

$\Rightarrow \frac{d y}{d x} = \frac{1}{x + y}$

$\Rightarrow \frac{dx}{dy} = x + y$ $\Rightarrow \frac{dx}{dy} - x = y$ This is a linear differential equation of the form:

$\frac{d x}{d y} + P_{1} x = Q_{1} (where P_{1} = - 1 and Q_{1} = y)$

Now, I.F. $= e^{\int P_{1} dy} = e^{\int - dy} = e^{- y}$ The general solution of the given differential equation is given by the relation,

$x (I.F.) = \int (Q_{1} \times I.F.) d y + C$

$\Rightarrow xe^{- y} = \int (y \cdot e^{- y}) dy + C$ $\Rightarrow x e^{- y} = y \cdot \int e^{- y} d y - \int [\frac{d}{d y} (y) \int e^{- y} d y] d y + C$ $\Rightarrow xe^{- y} = y (- e^{- y}) - \int (- e^{- y}) dy + C$ $\Rightarrow xe^{- y} = - ye^{- y} + \int e^{- y} dy + C$ $\Rightarrow xe^{- y} = - ye^{- y} - e^{- y} + C$ $\Rightarrow x = - y - 1 + Ce^{y}$ $\Rightarrow x + y + 1 = Ce^{y}$ (which is the required solution)

$y dx + (x - y^{2}) dy = 0$

Sol. $y dx + (x - y^{2}) dy = 0$

$\Rightarrow \Rightarrow \Rightarrow ydx = (y^{2} - x) dy \frac{dx}{dy} = \frac{y ^{2} - x}{y} = y - \frac{x}{y} \frac{dx}{dy} + \frac{x}{y} = y$

This is a linear differential equation of the form:

$\frac{dx}{dy} + P_{1} x = Q_{1} (where P_{1} = \frac{1}{y} and Q_{1} = y)$

Now, I.F. $= e^{\int P_{1} dy} = e^{\int \frac{1}{y} dy} = e^{logy} = y$ The general solution of the given differential equation is given by the relation,

$x (I.F.) = \int (Q_{1} \times I.F.) d y + C$

$\Rightarrow xy = \int (y \cdot y) dy + C$ $\Rightarrow xy = \int y^{2} dy + C$ $\Rightarrow xy = \frac{y ^{3}}{3} + C \Rightarrow x = \frac{y ^{2}}{3} + \frac{C}{y}$

$(x + 3 y^{2}) \frac{d y}{d x} = y (y > 0)$

Sol. $(x + 3 y^{2}) \frac{dy}{dx} = y \Rightarrow \frac{dy}{dx} = \frac{y}{x + 3 y ^{2}}$

$\Rightarrow \frac{d x}{d y} = \frac{x + 3 y ^{2}}{y} = \frac{x}{y} + 3 y \Rightarrow \frac{d x}{d y} - \frac{x}{y} = 3 y$

This is a linear differential equation of the form:

$\frac{dx}{dy} + P_{1} x = Q_{1} (where P_{1} = - \frac{1}{y} and Q_{1} = 3 y)$

Now, I.F. $= e^{\int P_{i} dy} = e^{- \int \frac{dy}{y}} = e^{- l o g y} = e^{l o g (\frac{1}{y})} = \frac{1}{y}$ The general solution of the given differential equation is given by the relation,

$x (I.F.) = \int (Q_{1} \times I.F.) d y + C$

$\Rightarrow x \times \frac{1}{y} = \int (3 y \times \frac{1}{y}) dy + C$

$\Rightarrow \frac{x}{y} = 3 y + C$

$\Rightarrow x = 3 y^{2} + Cy$

For each of the differential equations given in Exercises 13 to 15, find the particular solution satisfying the given conditions :

$\frac{dy}{dx} + 2 y tan x = sin x; y = 0$ when $x = \frac{π}{3}$

Sol. The given differential equation is

$\frac{d y}{d x} + 2 y tan x = sin x$

This is a linear equation of the form:

$\frac{d y}{d x} + P y = Q (where P = 2 tan x and Q = sin x)$

Now,

$I.F. = e^{\int Pdx} = e^{\int 2 t a n xdx} = e^{2 l o g ∣ s e c x ∣} = e^{l o g (s e c^{2} x)} = sec^{2} x$

The general solution of the given differential equation is given by the relation,

$y (I . F .) = \int (Q \times I . F .) d x + C$

$\Rightarrow y (sec^{2} x) = \int (sin x \cdot sec^{2} x) dx + C$ $\Rightarrow y sec^{2} x = \int (sec x \cdot tan x) d x + C$

$\Rightarrow y sec^{2} x = sec x + C$

Put, $y = 0$ and $x = \frac{π}{3}$ , in eq.(1) Therefore, $0 \times sec^{2} \frac{π}{3} = sec \frac{π}{3} + C$ $\Rightarrow 0 = 2 + C \Rightarrow C = - 2$ Substituting $C = - 2$ in equation (1), we get:

$y sec^{2} x = sec x - 2$

$\Rightarrow y = cos x - 2 cos^{2} x$ Hence, the required solution of the given differential equation is $y = cos x - 2 cos^{2} x$ .

$(1 + x^{2}) \frac{dy}{dx} + 2 xy = \frac{1}{1 + x ^{2}}; y = 0$ when $x = 1$ Sol. $(1 + x^{2}) \frac{dy}{dx} + 2 xy = \frac{1}{1 + x ^{2}}$ $\Rightarrow \frac{dy}{dx} + \frac{2 xy}{1 + x ^{2}} = \frac{1}{( 1 + x ^{2} ) ^{2}}$

This is a linear differential equation of the form:

$\frac{d y}{d x} + P y = Q (where P = \frac{2 x}{1 + x ^{2}} and Q = \frac{1}{( 1 + x ^{2} ) ^{2}})$

Now, I.F. $= e^{\int Pdx} = e^{\int \frac{2 xdx}{1 + x ^{2}}} = e^{l o g (1 + x^{2})} = 1 + x^{2}$ The general solution of the given differential equation is given by the relation,

$y (I.F.) = \int (Q \times I.F.) d x + C$

$\Rightarrow y (1 + x^{2}) = \int [\frac{1}{( 1 + x ^{2} ) ^{2}} \cdot (1 + x^{2})] dx + C$ $\Rightarrow y (1 + x^{2}) = \int \frac{1}{1 + x ^{2}} d x + C$

$\Rightarrow y (1 + x^{2}) = tan^{- 1} x + C$

Put $y = 0$ and $x = 1$ in eq.(1), we get

$0 = tan^{- 1} 1 + C \Rightarrow C = - \frac{π}{4}$

Substituting $C = - \frac{π}{4}$ in equation (1), we get :

$y (1 + x^{2}) = tan^{- 1} x - \frac{π}{4}$

This is the required general solution of the given differential equation.

$\frac{dy}{dx} - 3 y cot x = sin 2 x; y = 2$ when $x = \frac{π}{2}$

Sol. The given differential equation is $\frac{dy}{dx} - 3 y cot x = sin 2 x$ This is a linear differential equation of the form:

$\frac{d y}{d x} + P y = Q$

(where $P = - 3 cot x$ and $Q = sin 2 x$ ) Now, I.F. $= e^{\int Pdx} = e^{- 3 \int c o t xdx}$

$= e^{- 3 l o g ∣ s i n x ∣} = e^{l o g ∣ \frac{1}{s i n ^{3} x} ∣} = \frac{1}{s i n ^{3} x}$

The general solution of the given differential equation is given by the relation,

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow y(I.F.) = \int (Q \times I.F.) d x + C y \cdot \frac{1}{sin ^{3} x} = \int [sin 2 x \cdot \frac{1}{sin ^{3} x}] d x + C y cosec^{3} x = 2 \int (cot x cosec x) d x + C y cosec^{3} x = - 2 cosec x + C y = - \frac{2}{cosec ^{2} x} + \frac{C}{cosec ^{3} x} y = - 2 sin^{2} x + csin^{3} x .$

Put $y = 2$ and $x = \frac{π}{2}$ in eq.(1) we get

$2 = - 2 sin^{2} \frac{π}{2} + C sin^{3} \frac{π}{2}$

$\Rightarrow 2 = - 2 + C \Rightarrow C = 4$ Substituting $C = 4$ in equation (1), we get:

$y = - 2 sin^{2} x + 4 sin^{3} x \Rightarrow y = 4 sin^{3} x - 2 sin^{2} x$

This is the required particular solution of the given differential equation.

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point. Sol. Let F (x, y) be the curve passing through the origin. At point (x, y), the slope of the curve will be $\frac{dy}{dx}$ According to the given information:

$\frac{d y}{d x} = x + y \Rightarrow \frac{d y}{d x} - y = x$

This is a linear differential equation of the form:

$\frac{d y}{d x} + P y = Q (where P = - 1 and Q = x)$

Now, I.F. $= e^{\int Pdx} = e^{\int (- 1) dx} = e^{- x}$ The general solution of the given differential equation is given by the relation,

$y (I.F.) = \int (Q \times I.F.) d x + C$

$\Rightarrow y e^{- x} = \int x e^{- x} d x + C$

Now,

$\int x e^{- x} d x = x \int e^{- x} d x - \int [\frac{d}{d x} (x) \cdot \int e^{- x} d x] d x = - x e^{- x} - \int - e^{- x} d x = - x e^{- x} + (- e^{- x}) = - e^{- x} (x + 1)$

Substituting in equation (1), we get:

$\Rightarrow \Rightarrow ye^{- x} = - e^{- x} (x) + y = - (x + 1) + Ce^{x} x + y + 1 = Ce^{x}$

The curve passes through the origin. So, put $x = 0, y = 0$ and equation (2) becomes:

$0 + 0 + 1 = Ce^{0} \Rightarrow C = 1$

Substituting $C = 1$ in equation (2), we get:

$x + y + 1 = e^{x}$

Hence, the required equation of curve passing through the origin is $x + y + 1 = e^{x}$ .

Find the equation of a curve passing through the point $(0, 2)$ given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5 . Sol. Let F (x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent to the curve at $(x, y)$ is $\frac{dy}{dx}$ . According to the given information: $\frac{dy}{dx} + 5 = x + y$

$\Rightarrow \frac{d y}{d x} - y = x - 5$

This is a linear differential equation of the form:

$\frac{d y}{d x} + P y = Q (where P = - 1 and Q = x - 5)$

Now, I.F. $= e^{\int Pdx} = e^{\int (- 1) dx} = e^{- x}$

The general equation of the curve is given by the relation,

$\Rightarrow y(I.F.) = \int (Q \times I.F.) d x + C y. e^{- x} = \int (x - 5) e^{- x} d x + C$

Now, $\int (x - 5) e^{- x} dx$

$= (x - 5) \int e^{- x} dx - \int [\frac{d}{dx} (x - 5) \cdot \int e^{- x} dx] dx = (x - 5) (- e^{- x}) - \int (- e^{- x}) dx = (5 - x) e^{- x} + (- e^{- x}) = (4 - x) e^{- x}$

Therefore, equation (1) becomes:

$\Rightarrow \Rightarrow y e^{- x} = (4 - x) e^{- x} y = 4 - x + Ce^{x} x + y - 4 = Ce^{x}$

The curve passes through point $(0, 2)$ Therefore, equation (2) becomes :

$0 + 2 - 4 = Ce^{0}$

$\Rightarrow - 2 = C \Rightarrow C = - 2$

Substituting $C = - 2$ in equation (2), we get:

$x + y - 4 = - 2 e^{x} \Rightarrow y = 4 - x - 2 e^{x}$

This is the required equation of the curve.

Choose the correct answer in the following Q. 18 and 19

The integrating factor of the differential equation $x \frac{d y}{d x} - y = 2 x^{2}$ is (A) $e^{- x}$ (B) $e^{- y}$ (C) $\frac{1}{x}$ (D) x

Sol. (C) The given differential equation is:

$x \frac{d y}{d x} - y = 2 x^{2} \Rightarrow \frac{d y}{d x} - \frac{y}{x} = 2 x$

This is a linear differential equation of the form:

$\frac{d y}{d x} + P y = Q (where P = - \frac{1}{x} and Q = 2 x)$

The integrating factor (I.F.) is given by the relation, $e^{\int Pdx}$

$∴$ I.F. $= e^{\int - \frac{1}{x} dx} = e^{- l o g x} = e^{l o g (x^{- 1})} = x^{- 1} = \frac{1}{x}$

Hence, the correct answer is C.

The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y (- 1 < y < 1)$ is (A) $\frac{1}{y ^{2} - 1}$ (B) $\frac{1}{y ^{2} - 1}$ (C) $\frac{1}{1 - y ^{2}}$ (D) $\frac{1}{1 - y ^{2}}$ Sol. (D) The given differential equation is:

$\Rightarrow (1 - y^{2}) \frac{d x}{d y} + y x = a y \frac{d x}{d y} + \frac{y x}{1 - y ^{2}} = \frac{a y}{1 - y ^{2}}$

This is a linear differential equation of the form:

$\frac{d x}{d y} + P x = Q (where P = \frac{y}{1 - y ^{2}} and Q = \frac{a y}{1 - y ^{2}})$

The integrating factor (I.F.) is given by the relation, $∴$ I.F. $= e^{\int Pdy} = e^{\int \frac{y}{1 - y ^{y}} dy}$

$= e^{- \frac{1}{2} l o g (1 - y^{2})} = e^{l o g [\frac{1}{1 - y ^{2}}]} = \frac{1}{1 - y ^{2}}$

Hence, the correct answer is D.

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Access NCERT Solutions for Class 12 Maths for all chapters, with exercise-wise answers, useful formulas, and easy-to-follow solutions for understanding concepts and practising questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 4	Determinants
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

4.0Class 12 Maths Chapter 9 Differential Equations Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 9.1	12 Questions and Solutions	General and particular solutions of differential equations
Exercise 9.2	12 Questions and Solutions	Order and degree, and formation of differential equations
Exercise 9.3	23 Questions and Solutions	Solving first-order, first-degree differential equations
Exercise 9.4	17 Questions and Solutions	Solving first-order differential equations of higher degree
Exercise 9.5	19 Questionsand Solutions	Solving second-order differential equations of higher degree

5.0Key Features and Benefits of Class 12 Maths Chapter 9 Differential Equations

Clear Explanation of Methods: Each solution explains the method clearly, helping students understand when and how to apply it while solving differential equations.
Step-by-Step Solutions: Detailed steps make it easier to identify the order and degree of a differential equation and solve questions correctly in exams.
Better Problem-Solving Skills: Regular practice with the solutions helps students improve accuracy and confidence while solving calculus-based equation problems.
Useful for Competitive Exams: A strong understanding of differential equations can support preparation for Mathematics Olympiads and other competitive examinations.
Application in Further Studies: Clear concepts help students understand how differential equations are used in physics and applied mathematics.

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Are these NCERT Solutions based on the latest CBSE syllabus?

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NCERT Solutions Class 12 Maths Chapter 9 Differential Equations

1.0Key Concepts of Class 12 Maths Chapter 9 Differential Equations

2.0NCERT Class 12 Maths Chapter 9 Differential Equations : Detailed Solutions

EXERCISE - 9.1

EXERCISE 9.2

EXERCISE 9.3

EXERCISE 9.4

EXERCISE 9.5

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 9 Differential Equations Exercise-wise Solutions

5.0Key Features and Benefits of Class 12 Maths Chapter 9 Differential Equations

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