RD Sharma Class 12 Exercise 21.9 Differential Equation Solutions Maths-Download PDF Online

Updated on Jan 24, 2022 04:20 PM IST

It is common for class 12 students to encounter doubts while they are in the middle of doing their homework. Most of these students do not have time to attend tuition after their school hours. Hence, a proper solution guide will be of great help for them. Especially for the chapters like Differential Equation, reference material is a must. Therefore, the importance of RD Sharma Class 12th Exercise 21.9 book is inevitable.

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RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
Differential Equations Excercise: 21.9
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise: 21.9

Differential Equations Exercise 21.9 Question 1

Answer:

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

Given: Here given that

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

To solve: We have to solve the given differential equation.
Hint: In homogeneous differential equation put

y = v x

and
Solution: We have,

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

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

Clearly since each of the functions

x y + y^{2}

is and

x^{2}

homogeneous function of degree

c_{i}

the given equation is homogeneous.
Putting

y = v x

and

\frac{d y}{d x} = v + \frac{x d v}{d x}

the given
Equation becomes

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

Integrating on both side, we get

$\begin{aligned} \int \frac{d v}{v^{2} + 2 v} = - \int \frac{d x}{x} \\ \Rightarrow \int \frac{d v}{v^{2} - 2 v + 1 - 1} = - \int \frac{d x}{x} \\ \Rightarrow \int \frac{d v}{(v - 1)^{2} - (1)^{2}} = \int \frac{d x}{x} \\ \Rightarrow \frac{1}{2} \log | \frac{v + 1 - 1}{v + 1 + 1} | = - \log x + \log c \\ \Rightarrow \log {| \frac{v}{v + 2} |}^{1 / 2} = - \log | \frac{c}{x} | [∴ \log a - \log b = {\log |}^{a} / b ∣] \\ \Rightarrow {(\frac{v}{v + 2})}^{1 / 2} = \frac{c^{2}}{x^{2}} \end{aligned}$

$\begin{aligned} ∴ y = v x \Rightarrow v = \frac{y}{x} \\ \Rightarrow \frac{y / x}{\frac{y}{x} + 2} = \frac{c^{2}}{x^{2}} \\ \Rightarrow \frac{y}{y + 2 x} = \frac{c^{2}}{x^{2}} \\ \Rightarrow x^{2} y = c^{2} (y + 2 x) \\ \Rightarrow x^{2} y = c (y + 2 x) \end{aligned}$ where C=c²
This is required solution.

Differential Equations Exercise 21.9 Question 2

Answer:

l o g (x^{2} + y^{2}) + 2 \tan^{- 1} \frac{y}{x} = k

Given:

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

To solve: we have to solve the given differential equation

Hint: In homogeneous differential equation put

$y = v x$ and $\frac{d y}{d x} = v + \frac{x d v}{d x}$

Solution: we have

$\frac{d y}{d x} = \frac{y - x}{y + x}$
Cleary since each of the functions $y - x$ and $y + x$ is homogeneous function of degree $\partial$ the given equation is homogenous equation.
Putting $y = v x$ and $\frac{d y}{d x} = v + \frac{x d v}{d x}$
The given equation becomes
$\begin{aligned} v + x \frac{d v}{d x} = \frac{v x - x}{v x + x} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{v - 1}{v + 1} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{v - 1}{v + 1} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{v - 1 - v^{2} - v}{v + 1} \\ \Rightarrow x \frac{d v}{d x} = \frac{(1 + v^{2})}{v + 1} \end{aligned}$

Separating variables, we have

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

Integrating on both side, we get.

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

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

$\Rightarrow \frac{1}{2} \log | v^{2} + 1 | + \tan^{- 1} v - \log x + \log c [∴ \int \frac{1 d x}{1 + x^{2}}] = \tan^{- 1} x$

$\Rightarrow \frac{1}{2} \log | v^{2} + 1 | + 2 \tan^{- 1} v = 2 \log (\frac{c}{x})$

putting $v = y / x$

$\Rightarrow \log | \frac{y^{2}}{x^{2}} + 1 | + 2 \tan^{- 1} (\frac{y}{x}) = 2 \log (\frac{c}{x})$

$\Rightarrow \log | \frac{y^{2} + x^{2}}{x^{2}} + 1 | + 2 \tan^{- 1} (\frac{y}{x}) = 2 \log (\frac{c}{x})$

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

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

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

$\Rightarrow \log (y^{2} + x^{2}) + 2 \tan^{- 1} (\frac{y}{x}) = K,$ $w h e r e K = 2 l o g c$

It is required solution.

Differential Equations Exercise 21.9 Question 3

Answer:

x^{2} + y^{2} = c x

Given:

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

To find: we have to solve the given differential equation.
Hint: in homogeneous differential equation put

y = v x

and

\frac{d y}{d x} = v + 2 \frac{d v}{d x}

Solution: Here,

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

Clearly since each of the function

y^{2} - x^{2}

and

2 x y

is a homogeneous function of degree 2, the given equation is homogeneous
Putting

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

\begin{aligned} v + x \frac{d v}{d x} = \frac{v^{2} x^{2} - x^{2}}{2 v x^{2}} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{v^{2} - 1}{2 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{v^{2} - 1}{2 v} - 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} \end{aligned}

Integrating on both side we get

$\begin{aligned} \Rightarrow \int \frac{2 v}{1 + v^{2}} d v = - \int \frac{d x}{x} \\ \Rightarrow \log | 1 + v^{2} | = - \log x + \log c \\ \Rightarrow \log | 1 + v^{2} | + \log x = \log c \\ \Rightarrow \log | x (1 + v^{2}) | = \log c \\ \Rightarrow x (1 + v^{2}) = c \\ \Rightarrow x (1 + \frac{y^{2}}{x^{2}}) = c \end{aligned}$ $[∴ v = y / x]$

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

This is required solution.

Differential Equations Exercise 21.9 Question 4

Answer:

y = x l o g | x | + c x

Given: here ,

x \frac{d y}{d x} = x + y

To find: we have to find the solution of given differential equation.
Hint: in homogeneous differential equation
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

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

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

Clearly It is homogeneous equation

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

So,

$\begin{aligned} v + x \frac{d v}{d x} = \frac{x + v x}{x} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{x (1 + v)}{x} \\ \Rightarrow x \frac{d v}{d x} = 1 + v - v \\ \Rightarrow x \frac{d v}{d x} = 1 \\ \Rightarrow \int d v = \int \frac{1}{x} d x \\ \Rightarrow v = \log x + c \\ \Rightarrow \frac{y}{x} = \log x + c \\ \Rightarrow y = x \log x + c x \end{aligned}$

This is required solution.

Differential Equations Exercise 21.9 Question 5

Answer:

x (x^{2} - 3 y^{2}) = c

Given: here,

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

To find: we have to find the solution of given differential equation.
Hint: in homogeneous differential equation
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

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

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

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

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

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

$\Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{2 v}$

$\Rightarrow \int \frac{2 v}{1 - 3 v^{2}} d v = \int \frac{d x}{x}$ $[I n t e r g r a t i n g b o t h s i d e]$

$\begin{aligned} \Rightarrow \frac{1}{- 3} \int \frac{- 6 v}{1 - 3 v^{2}} d v = \int \frac{d x}{x} \\ \Rightarrow \int \frac{- 6 v}{1 - 3 v^{2}} = - 3 \int \frac{d x}{x} \\ \Rightarrow \log | 1 - 3 v^{2} | = - 3 \log | x | + \log | c | \\ \Rightarrow \log | 1 - 3 v^{2} | = - \log | x^{3} | + \log | c | \\ \Rightarrow 1 - 3 v^{2} = \frac{c}{x^{3}} \\ \Rightarrow x^{3} (1 - \frac{3 y^{2}}{x^{2}}) = C \\ \Rightarrow x^{3} \frac{(x^{2} - 3 y^{2})}{x^{2}} = C \\ \Rightarrow x (x^{2} - 3 y^{2}) = C \end{aligned}$
This is required solution.

Differential Equations Exercise 21.9 Question 6

Answer:

\tan^{- 1} (\frac{y}{x}) = \frac{1}{2} l o g (x^{2} + y^{2}) + C

Given:
Hint: in homogeneous differential equation put

y = v x

To solve: we have to solve the given differential Eqn.
Solution: we have

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

Here it is a homogeneous equation.
Put

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

So,

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

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

\begin{aligned} \Rightarrow \frac{1 - v}{1 + v^{2}} d v = \frac{d v}{d x} \\ \Rightarrow \int \frac{1 - v}{1 + v^{2}} d v = \int \frac{d x}{x} \end{aligned}

[I n t e g r a t i n g o n b o t h s i d e]

\begin{aligned} \Rightarrow \int \frac{1}{1 + v^{2}} d v - \frac{1}{2} \int \frac{2 v}{1 + v^{2}} d v = \int \frac{d x}{x} \\ \Rightarrow \tan^{- 1} v - \frac{1}{2} \log (1 + v^{2}) = \log x + c . [∴ \int \frac{d x}{1 + x^{2}} = \tan^{- 1} x] \\ \Rightarrow \tan^{- 1} \frac{y}{x} - \frac{1}{2} \log (1 +^{y^{2}} /_{x^{2}}) = \log x + c . [∴ v = y / x] \\ \Rightarrow \tan^{- 1} \frac{y}{x} - \frac{1}{2} \log (\frac{x^{2} + y^{2}}{x^{2}}) = \log x + c \\ \Rightarrow \tan^{- 1} \frac{y}{x} = \frac{1}{2} \log (x^{2} + y^{2}) + c . \end{aligned}

This is required solution.

Differential Equations Exercise 21.9 Question 7

Answer:

x = c (x^{2} + y^{2})

Given:

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

To find: we have to find the solution of given differential equation.

Hint: IN homogeneous differential equation

Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

Solution: We have

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

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

It is homogeneous equation

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

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

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

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

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

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

$\Rightarrow \int \frac{- 2 v}{1 - v^{2}} d v = - \int \frac{d x}{x}$ $[I n t e g r a t i n g o n b o t h s i d e]$

$\Rightarrow \log | 1 - v^{2} | = - \log x + \log c ∣$

$\Rightarrow 1 - v^{2} = \frac{c}{x} [∴ \log c - \log x = \log \frac{c}{x}]$

$\Rightarrow x (1 - \frac{y^{2}}{x^{2}}) = C [∴ put v = \frac{y}{x}]$

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

$\Rightarrow x^{2} - y^{2} = c x$ which is required solution.

Differential Equations Exercise 21.9 Question 8

Answer:

\frac{x + \sqrt{2 y}}{x - \sqrt{2 y}} = {(c x^{2})}^{\sqrt{2}}

Given:

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

To find: we have to find the solution of given differential equation
Hint: In homogeneous differential equation put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

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}}

This is a homogeneous differential equation.
Substitute

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

We have,

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

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

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

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

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

\Rightarrow \frac{\sqrt{2}}{2} \log | \frac{\frac{1}{\sqrt{2}} + v}{\frac{1}{\sqrt{2}} - v} | = 2 \log x + \log c

\Rightarrow \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} \log | \frac{a + x}{a - x} |]

\Rightarrow \frac{1}{\sqrt{2}} \log (\frac{\frac{1}{\sqrt{2}} +^{y} / x}{\frac{1}{\sqrt{2}} -^{y} / x}) = 2 \log x + \log c

\begin{aligned} \Rightarrow \frac{1}{\sqrt{2}} \log (\frac{x + y \sqrt{2}}{x - y \sqrt{2}}) = \log x^{2} + \log c \\ \Rightarrow \log {(\frac{x + y \sqrt{2}}{x - y \sqrt{2}})}^{\frac{1}{\sqrt{2}}} = \log c x^{2} \\ \Rightarrow \frac{x + y \sqrt{2}}{x - y \sqrt{2}} = {(c x^{2})}^{\sqrt{2}} \end{aligned}

Hence this is required solution.

Differential Equations Exercise 21.9 Question 9

Answer:

x^{2} (x^{2} - 2 y^{2}) = C .

Given:

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

To solve: we have to solve the given differential equation
Hint: In homogeneous differential equation put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

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

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

It is a homogeneous equation.

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

So,

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

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

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

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

\Rightarrow \int \frac{- 4 v}{1 - 2 v^{2}} d v = - 4 \int \frac{d x}{x}

\Rightarrow \log | 1 - 2 v^{2} | = - 4 \log x + \log c

\Rightarrow 1 - 2 \frac{y^{2}}{x^{2}} = \frac{C}{x^{4}} [put v = y / x]

\Rightarrow x^{2} (x^{2} - 2 y^{2}) = C

This is required solution

Differential Equations Exercise 21.9 Question 10

Answer:

e^{x / y} = l o g y + c .

Given:

y e^{x / y} d x = (x e^{x / y} + y) d y

y e^{x / y} d x = (x e^{x / y} + y) d y

To solve: we have to solve the given differential equation
Hint: In homogeneous differential equation put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

y e^{x / y} d x = (x e^{x / y} + y) d y

\Rightarrow \frac{d y}{d x} = \frac{x e^{x / y} + y}{y e^{x / y}}

It is a homogeneous equation.
Put

x = v y

and

\frac{d y}{d x} = v + y \frac{d v}{d y}

So,

\begin{aligned} v + y \frac{d v}{d y} = \frac{v y e^{v y} / y + y}{y e^{v y} / y} \\ \Rightarrow v + y \frac{d v}{d y} = \frac{v e^{v} + 1}{e^{v}} \\ \Rightarrow y \frac{d v}{d y} = \frac{v e^{v} + 1}{e^{v}} - v \\ \Rightarrow y \frac{d v}{d y} = \frac{v e^{v} + 1 - v e^{v}}{e^{v}} \end{aligned}

\Rightarrow y \frac{d v}{d y} = \frac{1}{e^{v}}

\Rightarrow \int e^{v} d v = \int \frac{d y}{e y}

(∴ I n t e g r a t i n g o n b o t h s i d e)

\Rightarrow e^{v} = \log y + c

\Rightarrow e^{x} / y = \log y + c

[∴ v = y / x]

Hence this is required solution

Differential Equations Exercise 21.9 Question 11

Answer:

\tan^{- 1} (\frac{y}{x}) = c + l o g | x |

Given:

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

To find: we have to find the solution of differential equation.
Hint: In homogeneous differential equation put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

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

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

It is a homogeneous equation of degree 2.
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

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

\begin{aligned} \Rightarrow x \frac{d v}{d x} = 1 + v + v^{2} - v \\ \Rightarrow x \frac{d v}{d x} = 1 + v + v^{2} - v \\ \Rightarrow x \frac{d v}{d x} = 1 + v^{2} \\ \Rightarrow \int \frac{d v}{1 + v^{2}} = \int \frac{d x}{x} \end{aligned}

\begin{aligned} \tan^{- 1} v = \log x + c [∴ \int \frac{d v}{1 + v^{2}} = \tan^{- 1} v] \\ \Rightarrow \tan^{- 1} \frac{y}{x} = \log x + c \end{aligned}

[∴ v = y / x]

This is required solution.

Differential Equations Exercise 21.9 Question 12

Answer:

x^{2} y - x y^{2} = C

Given:

(y^{2} - 2 x y) d x = (x^{2} - 2 x y) d y

To find: we have to solve the given differential equation
Hint: In homogeneous differential equation put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have

(y^{2} - 2 x y) d x = (x^{2} - 2 x y) d y

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

It is homogeneous differential equation.
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

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

\begin{aligned} \Rightarrow v + x \frac{d v}{d x} = \frac{v^{2} - 2 v}{1 - 2 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{v^{2} - 2 v}{1 - 2 v} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{v^{2} - 2 v - v + 2 v^{2}}{1 - 2 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{3 v^{2} - 3 v}{1 - 2 v} \end{aligned}

\Rightarrow \frac{1 - 2 v}{3 (v^{2} - v)} d v = \frac{d v}{x}

\Rightarrow \frac{- (2 v - 1)}{3 (v^{2} - v)} d v = \frac{d x}{d x}

\Rightarrow \int \frac{2 v - 1)}{v^{2} - v)} d v = - 3 \int \frac{d x}{x}

[I n t e g r a t i n g o n b o t h s i d e]

\Rightarrow \log | v^{2} - v | = - 3 \log | x | + \log c .

\Rightarrow v^{2} - v = \frac{c}{x^{3}}

\Rightarrow \frac{y^{2}}{x^{2}} - \frac{y}{x} = \frac{c}{x^{3}}

\Rightarrow \frac{y^{2} - x y}{x^{2}} = \frac{c}{x^{3}}

\Rightarrow y^{2} - x y = \frac{c}{x}

\Rightarrow x y^{2} - x^{2} y = c

Differential Equations Exercise 21.9 Question 13

Answer:

3 x^{2} y + 2 y^{3} = c, x y \neq 0

Given:

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

To solve: we have to solve given differential equation
Hint: In homogeneous differential equation put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

or put

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

Solution: we have,

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

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

This is homogeneous equation

Put x = y v \Rightarrow \frac{d x}{d y} = v + x \frac{d v}{d y}

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

= \frac{y^{2} (v^{2} + 2)}{- 2 v y^{2}}

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

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

Separating the variables we get

\int \frac{2 v}{- 3 v^{2} + 2} d v = - \int \frac{d y}{y} =

(I n t e g r a t i n g b o t h s i d e)

Let

I = \int \frac{2 v}{- 3 v^{2} + 2}

Put t = 3 v^{2} + 2

\Rightarrow d t = 6 v d v

\Rightarrow \frac{d t}{3} = 2 v d v

\Rightarrow I = \frac{1}{3} \int \frac{d t}{t} = \frac{1}{3} \log t = \log t^{1 / 3}

\Rightarrow t^{1 / 3} = \frac{c}{y}

\Rightarrow \log t^{1 / 3} = \log y + \log c

\Rightarrow t = \frac{c^{3}}{y^{3}}

\Rightarrow (3 v^{2} + 2) = \frac{c^{3}}{y^{3}} [∴ t = 3 v^{2} + 2]

\Rightarrow y^{3} (\frac{3 x^{2}}{y^{2}} + 2) = c^{3} [∴ v = \frac{x}{y}]

\begin{aligned} \Rightarrow y^{3} (\frac{3 x^{2} + 2 y^{2}}{y^{2}}) = c^{3} \\ \Rightarrow (3 x^{2} y + 2 y^{3}) = c where C = c^{3} \end{aligned}

This is required solution

Differential Equations Exercise 21.9 Question 14

Answer:

\frac{- 3 x}{y} = l o g | x | + c

Given:

3 x^{2} d y = (3 x y + y^{2}) d x

To solve: we have to solve given differential equation
Hint: put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

in homogeneous differential equation
Solution: we have,

3 x^{2} d y = (3 x y + y^{2}) d x

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

Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

v + x \frac{d v}{d x} = \frac{3 v x^{2} + v^{2} x^{2}}{3 x^{2}}

\begin{aligned} \Rightarrow x \frac{d v}{d y} = \frac{3 v + v^{2}}{3} - v \\ \Rightarrow x \frac{d v}{d y} = \frac{v^{2}}{3} \\ \Rightarrow 3 (\frac{- 1}{v}) = \log | x | + c [∴ \int \frac{x^{2} + 1}{3 + 1} + c] \\ \Rightarrow \frac{- 3 x}{y} = \log | x | + c [∴ v = y / x] \end{aligned}

This is required solution

Differential Equations Exercise 21.9 Question 15

Answer:

(x + y) {(2 y - x)}^{2} = C .

Given: Here,

\frac{d v}{d x} = \frac{x}{2 y + x}

To solve: we have to solve given differential equation
Hint: put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

in homogeneous differential equation
Solution: we have,

\frac{d v}{d x} = \frac{x}{2 y + x}

It is homogeneous equation.
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

v + x \frac{d v}{d x} = \frac{x}{2 v x + x}

\begin{aligned} \Rightarrow x \frac{d v}{d x} = \frac{1}{2 v + 1} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 2 v^{2} - v}{2 v + 1} \\ \Rightarrow \int \frac{2 v + 1}{2 v^{2} + v - 1} d v = - \int \frac{d x}{x} \\ \int \frac{2 v + 1}{(2 v - 1) + (v + 1)} d v = - \log x + \log c (1) \end{aligned}

Using partial fraction

$\frac{2 v + 1}{(2 v - 1) + (v + 1)} = \frac{A}{(2 v - 1)} + \frac{B}{v + 1}$

$2 v + 1 = A (v + 1) + B (2 v - 1)$

$Put v = \frac{1}{2}$

$\Rightarrow 2 \times \frac{1}{2} + 1 = A | \frac{1}{2} + 1 | + 0$

$\Rightarrow A = 4 / 3$
Again Put $v = - 1$

$2 (- 1) + 1 = 0 + 3 (2 x - 1 - 1)$

$- 2 + 1 = B (- 3) \Rightarrow B = \frac{1}{3}$
Putting the value of A and B in equation

$\Rightarrow \frac{4}{3} \int \frac{d v}{2 v - 1} + \frac{1}{3} \int \frac{d v}{v + 1} = - \log x + \log c$

$\Rightarrow \frac{4 \log | 2 v - 1 |}{2} + \log | v + 1 | = \log c^{3} - \log x^{3}$

$\Rightarrow \log | (2 v - 1)^{2} | + \log | v + 1 | = \log C - \log x^{3}$

$\Rightarrow \log | (2 v - 1)^{2} (v + 1) | = \log | \frac{C}{x^{3}} ∣$

$\Rightarrow (2 v - 1)^{2} (v + 1) = \frac{C}{x^{3}}$

$\Rightarrow {(\frac{2 y}{x} - 1)}^{2} (\frac{2 y}{x} + 1) = \frac{C}{x^{3}} [∴ v =^{y} / x]$

$\Rightarrow {(\frac{2 y - x}{x^{2}})}^{2} (\frac{y + x}{x}) = \frac{C}{x^{3}}$

$\Rightarrow (2 y - x)^{2} (x + y) = C$
Hence this is required solution.

Differential Equations Exercise 21.9 Question 16

Answer:

\sqrt{x^{2} + y^{2}} = c e^{2 \tan - 1} \frac{y}{x}, x \neq 0

Given:

(x + 2 y) d x - (2 x - y) d y = 0

To find: We have to find the solution of given differential equation.
Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

in homogeneous equation.
Solution: We have,

(x + 2 y) d x - (2 x - y) d y = 0

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

It is a homogeneous equation.
Put

y = v x ∴ \frac{d y}{d x} = v + x \frac{d v}{d x}

So,

v + x \frac{d v}{d x} = \frac{x + 2 v x}{2 x - v x}

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

Separating the variables

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

\Rightarrow \int \frac{d v}{1 + v^{2}} - \int \frac{v}{1 + v^{2}} = \int \frac{d x}{x} [∴ Integrating]

\Rightarrow 2 \tan^{- 1} d v - \frac{1}{2} \log | 1 + v^{2} | = \log | x | + \log c

\Rightarrow 2 \tan^{- 1} v = \log c x + \log {| 1 + v^{2} |}_{2}^{\frac{1}{2}}

\Rightarrow 2 \tan^{- 1} v = (1 + v)^{\frac{1}{2}} x c

\Rightarrow e^{2 \tan^{- 1}} \frac{y}{x} = {1 + \frac{y^{2}}{x^{2}}}^{2} x c

[∴ v = \frac{y}{x}]

\Rightarrow e^{2 \tan^{- 1}} \frac{y}{x} = {\frac{x^{2} + y^{2}}{x}}^{\frac{1}{2}} x c

\Rightarrow e^{2 \tan^{- 1}} \frac{y}{x} = {(x^{2} + y^{2})}^{\frac{1}{2}} c

\Rightarrow \sqrt{x^{2} + y^{2}} = \frac{1}{c} e^{2 \tan^{- 1}} \frac{y}{x}

\Rightarrow \sqrt{x^{2} + y^{2}} = C e^{2 \tan^{- 1} \frac{y}{x}}; Where = \frac{1}{c}

This is required solution.

Differential Equations Exercise 21.9 Question 17

Answer:

y + \sqrt{y^{2} - x^{2}} = c

Given:

\frac{d y}{d x} = \frac{y}{x} - \sqrt{\frac{y^{2}}{x^{2}} - 1}

To solve: We have to solve the given differential equation.
Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

in homogeneous equation.
Solution:

\frac{d y}{d x} = \frac{y}{x} - \sqrt{\frac{y^{2}}{x^{2}} - 1}

It is a homogeneous equation
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

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

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

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

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

\Rightarrow \log | v + \sqrt{v^{2} - 1} | = - \log x + \log c

[∴ \int \frac{d v}{\sqrt{v^{2} - 1}} = \log | v + \sqrt{v^{2}} 1 |]

\Rightarrow (v + \sqrt{v^{2} - 1}) = \frac{c}{x}

\Rightarrow \frac{y}{x} + \sqrt{\frac{y^{2}}{x^{2}} - 1} = \frac{c}{x} [∴ v = \frac{y}{x}]

\Rightarrow \frac{y}{x} + \sqrt{\frac{y^{2} - x^{2}}{x}} = \frac{c}{x}

\Rightarrow y + \sqrt{y^{2} - x^{2}} = c

Differential Equations Exercise 21.9 Question 18

Answer:

l o g (\frac{y}{x}) = c x

Given:

\frac{d y}{d x} = \frac{y}{x} {l o g y - l o g x + 1}

Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution:

\frac{d y}{d x} = \frac{y}{x} {l o g y - l o g x + 1}

\Rightarrow \frac{d y}{d x} = \frac{y}{x} {\log \frac{y}{x} - 1} [∴ \log y - \log x = \log \frac{y}{x}]

It is homogeneous equation.

\begin{aligned} Put y = v x and \frac{d y}{d x} = v + x \frac{d v}{d x} \\ So, v + x \frac{d v}{d x} = \frac{v x}{x} {\log \frac{v x}{x} + 1} \\ \Rightarrow x \frac{d v}{d x} = v \log v \end{aligned}

Separating the variables we have,

\Rightarrow \int \frac{1}{v l o g v} d v = \int \frac{d x}{d x}

∴ \int \frac{1}{v \log v} d v

Put \log v = t

\Rightarrow \frac{1}{v} d v = d t

∴ \int \frac{1}{t} d t = \log t + c

\Rightarrow \log | \log v | + c

∴ \log | \log | = \log | x | + \log c

\Rightarrow logv = x c

\Rightarrow \log \frac{y}{x} = C x

[∴ v = \frac{y}{x}]

This is required solution.

Differential Equations Exercise 21.9 Question 19

Answer:

\tan (\frac{y}{2 x}) = c x

Given:

\frac{d y}{d x} = \frac{y}{x} + \sin (\frac{y}{x})

To find: We have to solve the given differential equation.
Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: We have,

\frac{d y}{d x} = \frac{y}{x} + \sin (\frac{y}{x})

It is a homogeneous equation
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

v + x \frac{d v}{d x} = v + \sin v

\begin{aligned} \Rightarrow x \frac{d v}{d x} = \sin v \\ \Rightarrow \int cosecv d v = \int \frac{d x}{x} \\ \Rightarrow \log | \tan (\frac{v}{2}) | = \log x + \log c \\ \Rightarrow \tan \frac{v}{2} = c x \\ \Rightarrow \tan (\frac{y}{2 x}) = c x \end{aligned}

[∴ v = \frac{y}{x}]

This is required solution.

Differential Equations Exercise 21 .9 Question 20 .

Answer:

y = c e^{\tan^{- 1} (\frac{y}{x})}

Given:

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

Hint:Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have

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}}

It is homogeneous equation.
Putting

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

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

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

\begin{aligned} \Rightarrow x \frac{d v}{d x} = \frac{- v^{2} - v + v^{2} - v^{3}}{1 - v + v^{2}} \\ \Rightarrow x \frac{d v}{d x} = \frac{- v - v^{3}}{1 - v + v^{2}} \\ \Rightarrow x \frac{d v}{d x} = \frac{- v (v^{2} + 1)}{v^{2} - v + 1} \end{aligned}

Separating the variables we have,

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

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

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

\Rightarrow - logv + \tan^{- 1} v = \log x = \log c

\Rightarrow - \log (\frac{y}{x}) + \tan^{- 1} (\frac{y}{x}) = \log x k

[∴ v = \frac{y}{x}]

\Rightarrow \log (\frac{x}{y}) + \tan^{- 1} (\frac{y}{x}) = \log x c

\Rightarrow \tan^{- 1} (\frac{y}{x}) = \log x k - \log (\frac{x}{y})

\Rightarrow \tan^{- 1} (\frac{y}{x}) = \log (\frac{x k y}{x})

\Rightarrow e^{\tan^{- 1} (\frac{y}{x})} = k y

\Rightarrow \frac{1}{k} e^{\tan^{- 1} (\frac{y}{x})} = k

\Rightarrow y = c e^{\tan^{- 1} (\frac{y}{x})} Where c = \frac{1}{k}

This is required solution.

Differential Equations Exercise 21.9 Question 21

Answer:

\sqrt{x^{2} + y^{2}} = x l o g | \frac{c}{x} |

Given:

[x \sqrt{x^{2} + y^{2}} - y^{2}] d x + x y d y = 0

To solve: we have to solve the given differential equation

Hint: Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$
Solution: $[x \sqrt{x^{2} + y^{2}} - y^{2}] d x + x y d y = 0$
$\Rightarrow \frac{d y}{d x} = \frac{[y^{2} - x \sqrt{x^{2} + y^{2}}]}{x y}$

It is a homogeneous equation.
Putting $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$
So, $v + x \frac{d v}{d x} = \frac{v^{2} x^{2} - x \sqrt{x^{2} + v^{2}} x^{2}}{v x^{2}}$
$\Rightarrow x \frac{d v}{d x} = \frac{v^{2} x^{2} - x \sqrt{x^{2} (1 + v^{2})}}{v x^{2}} - v$
$\Rightarrow x \frac{d v}{d x} = \frac{v^{2} - \sqrt{(1 + v^{2})}}{v} - v$
$\Rightarrow x \frac{d v}{d x} = \frac{- \sqrt{(1 + v^{2})}}{v}$
$\Rightarrow \int \frac{v}{\sqrt{(1 + v^{2})}} d v = - \frac{d x}{x}$ $(s e p e r a t i n g t h e v a r i a b l e s a n d i n t e r s e c t i n g b o t h s i d e s)$
$\Rightarrow \frac{1}{2} \int \frac{2 v}{\sqrt{(1 + v^{2})}} d v = - \int \frac{d x}{x}$
Let $1 + v^{2} = t$
$2 v d v = d t$
$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{t}} d t = - \log x + \log c$

$\Rightarrow \sqrt{1 + v^{2}} = \log (\frac{c}{x}) [∴ t = 1 + v^{2}]$
$\Rightarrow \frac{\sqrt{x^{2} + y^{2}}}{x} = \log (\frac{c}{x})$ $[∴ v = \frac{y}{x}]$
$\Rightarrow \sqrt{x^{2} + y^{2}} = x \log (\frac{c}{x})$
This is required solution.

Differential Equations Exercise 21.9 Question 22

Answer:

\tan (\frac{y}{x}) = l o g (\frac{c}{x})

Given:

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

To Solve: We have to solve the given differential equation.
Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: Here, given that

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

\Rightarrow \frac{d y}{d x} = \frac{y - x \cos^{2} (\frac{y}{x})}{x} \to (1)

It is homogeneous equation.
Putting $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

So, equation $(1)$ becomes

$\begin{aligned} \Rightarrow v + x \frac{d v}{d x} = \frac{v x - x \cos^{2} (\frac{v x}{x})}{x} \\ \Rightarrow v + x \frac{d v}{d x} = v - \cos^{2} v \\ \Rightarrow x \frac{d v}{d x} = - \cos^{2} v \end{aligned}$

Separating the variables we get

$\begin{aligned} \Rightarrow \int \frac{d v}{\cos^{2} v} = - \int \frac{d x}{x} \\ \Rightarrow \int \sec^{2} v d v = - \int \frac{d x}{x} \\ \Rightarrow \tan v = - \log x + \log c \end{aligned}$

$\Rightarrow \tan (\frac{y}{x}) = \log (\frac{c}{x}) [∴ \log c - \log x = \log (\frac{c}{x})]$

This is required solution.

Differential Equations Exercise 21.9 Question 23

Answer:

| y \sin (\frac{y}{x}) | = c

Given:

\frac{y}{x} \cos (\frac{y}{x}) d x - {\frac{x}{y} \sin (\frac{y}{x}) + \cos (\frac{y}{x})} d y = 0

To find: We have to find the solution of given differential equation.
Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: We have,

\begin{aligned} \frac{y}{x} \cos (\frac{y}{x}) d x - {\frac{x}{y} \sin (\frac{y}{x}) + \cos (\frac{y}{x})} d y = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{\frac{y}{x} \cos (\frac{y}{x})}{\frac{x}{y} \sin (\frac{y}{x}) + \cos (\frac{y}{x})} \end{aligned}

It is homogeneous equation.
Putting

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

v + x \frac{d v}{d x} = \frac{\frac{v x}{x} - \cos (\frac{v x}{x})}{\frac{x}{v x} \sin (\frac{v x}{x}) + \cos (\frac{v x}{x})}

\Rightarrow v + x \frac{d v}{d x} = \frac{v \cos x}{\frac{1}{v} \sin v + \cos v}

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

\Rightarrow x \frac{d v}{d x} = \frac{v^{2} \cos x}{\sin v + v \cos v} - v

\Rightarrow x \frac{d v}{d x} = \frac{v^{2} \cos v - v \sin v - v^{2} \cos v}{\sin v + v \cos v} - v

\Rightarrow x \frac{d v}{d x} = \frac{- v \sin v}{\sin v + v \cos v}

Separating the variables, we get

$\frac{\sin v + v \cos v}{v \sin v} d v = - \frac{d x}{x}$

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

$\Rightarrow \log v + \log | \sin v | = - \log x + \log c$

$\Rightarrow \log | v \sin v | = \log \frac{c}{x} [∴ \log x + \log y = \log x y and \log x - \log y = \log \frac{x}{y}]$

$\Rightarrow | v \sin v | = \frac{c}{x}$

$\Rightarrow | x (\frac{y}{x}) \sin (\frac{y}{x}) | = c [∴ v = \frac{y}{x}]$

$\Rightarrow | y \sin (\frac{y}{x}) | = c$
This is required solution.

Differential Equations Exercise 21.9 Question 24

Answer:

\frac{x^{2}}{y^{2}} {(\frac{x}{y}) - \frac{1}{2}} + l o g y^{2} = c

Given:

x y l o g (\frac{x}{y}) d x + {y^{2} - x^{2} l o g (\frac{x}{y})} d y = 0

To solve:We have to solve the given differential equation
Hint:

(1)

Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x} o r

(2)

Put

y = v x

and

\frac{d y}{d x} = v + y \frac{d v}{d x}

Solution: We have,

\begin{aligned} x y \log (\frac{x}{y}) d x + {y^{2} - x^{2} \log (\frac{x}{y})} d y = 0 \\ \Rightarrow \frac{d x}{d y} = \frac{x^{2} \log (\frac{x}{y}) - y^{2}}{x y \log (\frac{x}{y})} \end{aligned}

It is a homogeneous equation.

We put $x = v y$

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

So, $v + y \frac{d v}{d y} = \frac{v^{2} y^{2} \log (v) - y^{2}}{v y^{2} \log v}$

$\Rightarrow y \frac{d v}{d y} = \frac{v^{2} \log (v) - 1}{v \log v} - v$

$\Rightarrow y \frac{d v}{d y} = \frac{v^{2} \log (v) - 1 - v^{2} \log v}{v \log v}$

$\Rightarrow y \frac{d v}{d y} = \frac{- 1}{v \log v}$

Separating the variables, we get

$\Rightarrow v l o g v = \frac{1}{y} d y$

On integrating both sides, we get

$\Rightarrow \int \Rightarrow v l o g d v = - \int \frac{1}{y} d y$

$\Rightarrow \log v \int v \log v - \int (\frac{d}{d v} (\log v) \int v d v) = - \log y + \log c$ $[I n t e g r a t i n g u s i n g b y p a r t s]$

$\Rightarrow \frac{v^{2}}{2} \log v - \int \frac{v}{2} d v = - \log y + c$

$\Rightarrow \frac{v^{2}}{2} \log v - \frac{v^{2}}{4} = - \log y + c$

$\Rightarrow \frac{v^{2}}{2} [\log v - \frac{1}{2}] = - \log y + c$

$\Rightarrow v^{2} [\log v - \frac{1}{2}] = - 2 \log y + c$
Now, putting back the value of $v$ as $\frac{x}{y}$ , we get

$\Rightarrow \frac{x^{2}}{y^{2}} [\log (\frac{x}{y}) - \frac{1}{2}] + \log y^{2} = c [∴ 2 \log = \log y^{2}]$
Hence this is required solution.

Differential Equations Exercise 21.9 Question 25

Answer:

x + y e^{\frac{x}{y}} = c

Given:

1 + e^{\frac{x}{y}} d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0

To solve: We have to solve the given differential equation.
Hint: We have to put

x = v y

and

\frac{d x}{d y} = v + x \frac{d v}{d x}

Solution: We have,

1 + e^{\frac{x}{y}} d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0

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

It is homogeneous equation.
Put

x = v y

and

\frac{d x}{d y} = v + x \frac{d v}{d x}

So,

v + x \frac{d v}{d x} = \frac{- e^{v} (1 - v)}{1 + e^{v}} - v

\begin{array}{r} \Rightarrow y \frac{d v}{d y} = \frac{- e^{v} (1 - v)}{1 + e^{v}} - v \end{array}

\begin{array}{r} \Rightarrow y \frac{d v}{d y} = \frac{- e^{v} + v e^{v} - v - v e^{v}}{1 + e^{v}} \end{array}

\Rightarrow y \frac{d v}{d y} = \frac{- e^{v} - v}{1 + e^{v}}

\Rightarrow \frac{1 + e^{v}}{v + e^{v}} = - \frac{d y}{y} (on seperating the variables)

Integrating both sides

\int \frac{1 + e^{v}}{v + e^{v}} d v = - \frac{d y}{y}

Put

v + e^{v} = t

\Rightarrow (1 + e^{v}) d v = d t

So equation becomes

\Rightarrow \int \frac{d t}{t} = - \log y + \log c

\Rightarrow \log t = - \log y + \log c

\Rightarrow \log | v + e^{v} | + \log y = \log c [∴ t = v + e^{v}]

\Rightarrow \log | y (v + e^{v}) | = \log c

\Rightarrow y (\frac{x}{y} + e^{\frac{x}{y}}) = c [∴ v = \frac{x}{y}]

\Rightarrow x y e^{\frac{x}{y}} = c

Hence this is required solution.

Differential Equations Exercise 21.9 Question 26

Answer:

c | 2 x - y |^{\frac{5}{8}} {(4 x^{2} + y^{2})}^{\frac{3}{16}} = e^{- \frac{3}{8} \tan^{- 1 (\frac{y}{2 x})}}

Given:

(x^{2} + y^{2}) \frac{d y}{d x} = 8 x^{2} - 3 x y + 2 y^{2}

To solve: We have to solve the given differential equation.
Hint: In homogeneous equation we put

x = v y

and

\frac{d x}{d y} = v + y \frac{d v}{d y}

Solution: Here,

(x^{2} + y^{2}) \frac{d y}{d x} = 8 x^{2} - 3 x y + 2 y^{2}

\Rightarrow \frac{d x}{d y} = \frac{8 x^{2} - 3 x y + 2 y^{2}}{(x^{2} + y^{2})}

It is homogeneous equation.
Put

y = v y

and

\frac{d x}{d y} = v + x \frac{d v}{d x}

So,

v + x \frac{d v}{d x} = \frac{8 x^{2} - 3 x v x + 2 x^{2} v^{2}}{x^{2} + x^{2} v^{2}}

\Rightarrow x \frac{d v}{d x} = \frac{8 - 3 v + 2 v^{2}}{1 + v^{2}} - v

\Rightarrow x \frac{d v}{d x} = \frac{8 - 3 v + 2 v^{2} - v^{3}}{1 + v^{2}} - v

Separating the variables and integrating we get

\Rightarrow \int \frac{1 + v^{2}}{8 - 4 v + 2 v^{2} - v^{3}} d v = \int \frac{1}{x} d x

\Rightarrow \int \frac{1 + v^{2}}{(2 - v) (v^{2} + 4)} d v = \int \frac{1}{x} d x \to (A)

\Rightarrow \frac{1 + v^{2}}{(2 - v) (v^{2} + 4)} = \frac{A x + B}{4 + v^{2}} + \frac{c}{2 - v} [Using partial fraction]

\Rightarrow 1 + v^{2} = v^{2} (- A + C) + v (2 A + B) + 2 B + 4 C

Comparing the coefficient of like power V.

\begin{aligned} - A + C = 1 \to (1) \\ 2 A - B = 0 \\ \Rightarrow B = 2 A \to (2) \\ And 2 B + 4 C = 1 \to (3) \end{aligned}

Solving equation $(1), (2)$ & $(3)$ we get

A = \frac{3}{8}, B = \frac{3}{4}, C = \frac{5}{8}

Using equation $(A)$ we get
$\begin{aligned} \Rightarrow \int \frac{- \frac{3}{8} x - \frac{3}{4}}{4 + v^{2}} d v + \frac{5}{8} \int \frac{c}{2 - v} d v = \int \frac{d x}{x} \\ \Rightarrow - \frac{3}{8} \int \frac{v + 2}{4 + v^{2}} d v + \frac{5}{8} \int \frac{1}{2 - v} d v = \int \frac{d x}{x} \end{aligned}$
$\Rightarrow - \frac{3}{8} \int \frac{v}{4 + v^{2}} d v + \frac{2 x 3}{8} \int \frac{1}{4 + v^{2}} d v + \frac{5}{8} \int \frac{1}{2 - v} d v = \int \frac{d x}{x}$
$\Rightarrow - \frac{3}{16} \log | 4 + v^{2} | - \frac{3}{8} \tan^{- 1} \frac{v}{2} + \frac{5}{8} \log | 2 - v | = \log x + \log c$
$\Rightarrow [\log {[4 + v^{2}]}_{16}^{\frac{8}{16}} + \log e^{\frac{3}{8} \tan^{- 1} (\frac{v}{2})} + \log | 2 - v |^{\frac{5}{8}}] = \log c x$
$\Rightarrow {(4 + v^{2})}^{\frac{3}{16}} \times e^{\frac{8}{s} \tan^{- 1} (\frac{v}{2})} \times (2 - v)^{\frac{5}{8}} = \frac{c}{x}$
Put $y = v x$
$\Rightarrow {(4 + \frac{y^{2}}{x^{2}})}^{\frac{3}{16}} \times e^{\frac{3}{8} \tan^{- 1} (\frac{y}{2 x})} \times (2 - v)^{\frac{5}{8}} = \frac{c}{x}$
$\Rightarrow {(\frac{4 x^{2} + y^{2}}{{(x^{2})}^{\frac{8}{16}}})}^{\frac{8}{16}} \times \frac{(2 - y)^{\frac{5}{8}}}{x^{\frac{5}{8}}} = e^{\frac{a}{8} \tan^{- 1} (\frac{y}{2 x})} \times \frac{c}{x}$
$\Rightarrow \frac{{(4 x^{2} + y^{2})}^{\frac{3}{16}} (2 - y)^{\frac{5}{8}}}{x} = e^{\frac{3}{8} \tan^{- 1} (\frac{y}{2 x})} \times \frac{c}{x} [∴ x^{\frac{6}{16}, x^{\frac{5}{8}} = x^{\frac{6}{16}} + \frac{5}{8} = x^{\frac{8}{8}} = x]}$
$\Rightarrow {(4 x^{2} + y^{2})}^{\frac{3}{16}} (2 - y)^{\frac{5}{8}} = c e^{\frac{3}{8} \tan^{- 1} (\frac{y}{2 x})}$
$\Rightarrow c | 2 x - y |^{\frac{5}{8}} {(4 x^{2} + y^{2})}^{\frac{3}{16}} = e^{\frac{3}{8} \tan^{- 1} (\frac{y}{2 x})} where c = \frac{1}{c}$
This is required solution.

Differential Equations Exercise 21.9 Question 27

Answer:

\tan (\frac{y}{x}) = l o g | \frac{c}{x} |

Given:

x \frac{d y}{d x} = y - x c o s^{2} (\frac{y}{x})

To Find: We have to find the solution of the given differential equation.
Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: We have,

\Rightarrow x \frac{d y}{d x} = y - x c o s^{2} (\frac{y}{x})

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

It is homogeneous equation.

Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

So,

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

$\begin{aligned} \Rightarrow x \frac{d v}{d x} = v - \cos^{2} v - v \\ \Rightarrow x \frac{d v}{d x} = - \cos^{2} v \end{aligned}$

Separating the variables and integrating both sides we get

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

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

$\Rightarrow \tan v = - \log x + \log c$ $[∴ \int \sec^{2} v d v = \tan v]$

$\Rightarrow \tan \frac{y}{x} = \log \frac{c}{x}$ $[∴ v = \frac{y}{x}]$
This is required solution.

Differential Equations Exercise 21.9 Question 29

Answer:

y + \sqrt{y^{2} - x^{2}} = c x^{3}

Given:

x \frac{d y}{d x} - y = 2 \sqrt{y^{2} - x^{2}}

To Find: We have to find the solution of the given differential equation.
Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: Here,

x \frac{d y}{d x} - y = 2 \sqrt{y^{2} - x^{2}}

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

It is homogeneous equation.
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} \Rightarrow v + x \frac{d v}{d x} = \frac{2 \sqrt{v^{2} x^{2} - x^{2}} + v x}{x} \\ \Rightarrow x \frac{d v}{d x} = 2 \sqrt{v^{2} - 1} + v - v \\ \Rightarrow x \frac{d v}{d x} = 2 \sqrt{v^{2} - 1} \end{aligned}

Separating the variables and integrating both sides we get

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

$\Rightarrow \log | v + \sqrt{v^{2} - 1} |$

$= 2 \log x + \log c [∴ \int \frac{d v}{\sqrt{v^{2} - 1}} = \log | v + \sqrt{v^{2} - 1} |]$

$\Rightarrow \log | v + \sqrt{v^{2} - 1} | = \log c x^{2}$

$\Rightarrow v + \sqrt{v^{2} - 1} = c x^{2}$

$\Rightarrow \frac{y}{x} + \sqrt{{(\frac{y}{x})}^{2} - 1} = c x^{2} [∴ v = \frac{y}{x}]$

$\Rightarrow y + \sqrt{y^{2} - x^{2}} = c x^{3}$
This is required solution.

Differential Equations Exercise 21.9 Question 30

Answer:

\sec (\frac{y}{x}) = c x y

Given:

x \cos (\frac{y}{x}) \cdot (y d x + x d y) = y \sin (\frac{y}{x}) \cdot (x d y - y d x)

To solve: We have to solve the given differential equation.

Hint: Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

Solution: Here, we have,

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

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

It is homogeneous equation.

Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

So,

$\begin{aligned} \Rightarrow v + x \frac{d v}{d x} = \frac{- x v x \cos (\frac{v x}{x}) - v^{2} x^{2} \sin (\frac{v x}{x})}{- v x^{2} \sin (\frac{v x}{x}) + x^{2} \cos (\frac{v x}{x})} \\ \Rightarrow x \frac{d v}{d x} = \frac{(v \cos v + v^{2} \sin v)}{(v \sin v - \cos v)} - v \end{aligned}$

$\begin{aligned} \Rightarrow x \frac{d v}{d x} = \frac{v \cos v + v^{2} \sin v - v^{2} \sin v + v \cos v}{v \sin v - \cos v} \\ \Rightarrow x \frac{d v}{d x} = \frac{2 v \cos v}{- (\cos v - v \sin v)} \end{aligned}$

Separating the variables and integrating both sides we get

$\Rightarrow \int \frac{\cos v - v \sin v}{v \cos v} d v = - 2 \int \frac{1}{x} d x$

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

$\Rightarrow \log v - \log | \sec v | = - 2 \log x + \log c$

$\Rightarrow \log | \frac{v}{| \sec v |} | = - \log x^{2} + \log c$

$\Rightarrow \frac{v}{| \sec v |} = \frac{c}{x^{2}}$

$\Rightarrow \frac{\frac{y}{2}}{\sec (\frac{y}{x})} = \frac{c}{x^{2}} [∴ v = \frac{y}{x}]$

$\Rightarrow \frac{y}{x \sec (\frac{y}{x})} = \frac{c}{x^{2}}$

$\Rightarrow \sec (\frac{y}{x}) = \frac{x y}{c}$

$\Rightarrow \sec (\frac{y}{x}) = c x y where c = \frac{1}{c}$

This is required solution.

Differential Equations Exercise 21.9 Question 31

Answer:

\frac{x}{x + y} + l o g x = c

Given:

(x^{2} + 3 x y + y^{2}) d x - x^{2} d y = 0

To find: we have to find the solution of given differential equation.

Hint: Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

Solution: we have,

$\begin{aligned} (x^{2} + 3 x y + y^{2}) d x - x^{2} d y = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{x^{2} + 3 x y + y^{2}}{x^{2}} \end{aligned}$

It is a homogeneous equation Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

So,

$\begin{aligned} x \frac{d v}{d x} + v = \frac{x^{2} + 3 x^{2} + v^{2} x^{2}}{x^{2}} \\ \Rightarrow x \frac{d v}{d x} + v = \frac{1 + 3 v + v^{2}}{1} - v \\ \Rightarrow x \frac{d v}{d x} + v = 1 + 2 v + v^{2} \\ \Rightarrow x \frac{d v}{d x} + v = (1 + v)^{2} [∴ (v + 1)^{2} = 1 + 2 v + v^{2}] \end{aligned}$

Separating the variable and Integrating bot side we get

$\begin{aligned} \Rightarrow \int \frac{1}{(v + 1)^{2}} d v = \int \frac{d x}{x} \\ - \frac{1}{v + 1} = \log x + k [∴ \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c] \\ \Rightarrow \frac{- 1}{y / x + 1} = \log x + k \\ \Rightarrow \frac{- x}{x + y} = \log x + k \\ \Rightarrow \frac{x}{x + y} = \log x = - k \\ \Rightarrow \frac{x}{x + y} = \log x = c where c = - k \end{aligned}$
This is a required solution.

Differential Equations Exercise 21.9 Question 32

Answer:

\log | x^{2} + x y + y^{2} | = 2 \sqrt{3} \tan^{- 1} (\frac{x + 2 y}{\sqrt{3 x}}) + c

Given:

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

To solve: have to solve the given differential equation.
Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: Here,

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

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

It is homogeneous equation
Put

y = v x

\Rightarrow

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

v + x \frac{d y}{d x} = \frac{x + 2 v x}{x - v x}

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

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

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

Separating the variable and Integrating both side we get

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

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

$\Rightarrow \int \frac{(2 v + 1) - 3}{1 + v + v^{2}} d v = - 2 \int \frac{1}{x} d x$

$\Rightarrow \int \frac{(2 v + 1) - 3}{1 + v + v^{2}} d v = - \int \frac{3}{v^{2} + 2 v (\frac{1}{2}) + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} d x = - 2 \int \frac{1}{x} d x$

$\Rightarrow \log | 1 + v + v^{2} | - 3 x \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{v + \frac{1}{2}}{\sqrt{3 / 2}}) = - 2 \log x + c$

$\Rightarrow \log | 1 + v + v^{2} | + \log x^{2} - 2 \sqrt{3} \tan^{- 1} (\frac{v + \frac{1}{2}}{\sqrt{3 / 2}}) = c .$

$\Rightarrow \log [x^{2} (1 + \frac{y}{x} +^{y^{2}} /_{x^{2}}] - 2 \sqrt{3} \tan^{- 1} (\frac{\frac{2 y}{x} + 1}{\sqrt{3}}) = c . [∴ v = \frac{y}{x}]$

$\Rightarrow \log (y^{2} + x y + x^{2}) = 2 \sqrt{3} \tan^{- 1} (\frac{2 y + x}{\sqrt{3 x}}) + c$

This is required solution.

Differential Equations Exercise 21.9 Question 33

Answer:

x^{2} y^{12} = c^{4} {(2 y^{2} x^{2})}^{5}

Given:

(2 x^{2} y + y^{3}) d x + (x y^{2} - 3 x^{3}) d y = 0

To solve: we have to solve differential equation
Hint: Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

(2 x^{2} y + y^{3}) d x + (x y^{2} - 3 x^{3}) d y = 0

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

Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} \Rightarrow x \frac{d v}{d x} = \frac{2 v + v^{3}}{- v^{2} + 3} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{2 v^{3} - v}{- v^{2} + 3} \end{aligned}

Separating and Integrating both side we get

$\begin{aligned} \Rightarrow \int \frac{- v^{2} + 3}{2 v^{3} - v} d v = \int \frac{1}{x} d x \\ ∴ \frac{- v^{2} + 3}{v (2 v^{2} - 1)} = \frac{A}{v} + \frac{B V + c}{2 v^{2} - 1} (using partial fraction) \\ \Rightarrow 3 - v^{2} = A (2 v^{2} - 1) + (B v + c) v \\ \Rightarrow 3 - v^{2} = (2 A + B) v^{2} + c v - A \end{aligned}$

Comparing the coefficient of like power of v, we get

$A = - 3, B, C = 0$

And $2 A + B = - 1$

$\Rightarrow B = 5$

So,

$\Rightarrow \int \frac{- 3}{v} d v + \int \frac{5 v}{2 v^{2} - 1} d v = \int \frac{1}{x} d x$

$\Rightarrow - 3 \int \frac{1}{v} d v + \frac{5}{4} \int \frac{5 v}{2 v^{2} - 1} d v = \int \frac{1}{x} d x$

$\Rightarrow - 3 \log v + \frac{5}{4} \log (2 v^{2} - 1) = \log x + \log c$

$\Rightarrow - 12 \log v + 5 \log (2 v^{2} - 1) = 4 \log x + 4 \log c$

$\Rightarrow \frac{{(2 v^{2} - 1)}^{2}}{v^{2}} = x^{4} c^{4}$

$\Rightarrow \frac{{[2 y^{2} - x^{2} / x^{2}]}^{5}}{y^{12} /_{x^{12}}} = x^{4} c^{4} [∴ v = \frac{y}{x}]$

$\Rightarrow \frac{{(2 y^{2} - x^{2})}^{5} x x^{2}}{y^{2}} = x^{4} c^{4}$

$\Rightarrow x^{2} y^{12} = \frac{{(2 y^{2} x^{2})}^{5}}{c^{4}}$

$\Rightarrow x^{4} y^{12} = c^{4} {(2 y^{2} x^{2})}^{5} [where \frac{1}{c^{4}} = c^{4}]$

This is required solution.

Differential Equations Exercise 21.9 Question 34

Answer:

x \sin (\frac{y}{x}) = c (1 + \cos \frac{y}{x})

Given:

x \frac{d y}{d x} - y + x \sin (\frac{y}{x}) = 0

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

x \frac{d y}{d x} - y + x \sin (\frac{y}{x}) = 0

\Rightarrow \frac{d y}{d x} = \frac{y + \sin \frac{y}{x}}{x}

It is homogeneous equation.
Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$
So,
$x \frac{d y}{d x} = \frac{v x - xsin (\frac{y}{x})}{x}$
$\Rightarrow x \frac{d v}{d x} = v - \sin v - v$

Separating the variables and integrating both side we get
$\int \frac{d v}{\sin v} = - \int \frac{1}{x} d x$
$\Rightarrow \int cosec v d v = - \int \frac{1}{x} d x$
$\Rightarrow \log (cosec v + \cot v) = - \log x + \log c$
$\Rightarrow cosec v + \cot v = \frac{c}{x}$
$\Rightarrow \frac{c}{x} = \frac{1}{cosec v + \cot v}$
$\Rightarrow \frac{x}{c} = \frac{1}{\frac{1}{\sin v} + \frac{\cos v}{\sin v}}$
$\Rightarrow \frac{x}{c} = \frac{\sin v}{1 + \cos v}$
$(1 + \cos v) x = c \sin v$
$\Rightarrow x (1 + \cos (\frac{y}{x}) = c \sin (\frac{y}{x}))$
$[∴ v = \frac{y}{x}]$

Differential Equations Exercise 21.9 Question 35

Answer:

C y = l o g | \frac{y}{x} | - 1

Given:

y d x + {x l o g \frac{y}{x}} d y - 2 x d y = 0

To solve: we have to solve differential equation

Hint: Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

Solution: we have,

$y d x + {x l o g \frac{y}{x}} d y - 2 x d y = 0$

Divided by dx both side we get

$\begin{aligned} \Rightarrow y + x \log (\frac{y}{x}) \frac{d y}{d x} - 2 x \frac{d y}{d x} = 0 \\ \Rightarrow \frac{d y}{d x} = \frac{y}{2 x - x \log (\frac{y}{x})} \end{aligned}$

It is homogeneous equation

Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

So, $x \frac{d y}{d x} + v = \frac{v x}{2 x - x \log (\frac{y}{x})}$

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

$\Rightarrow x \frac{d y}{d x} = \frac{v - 2 v + 2 v x \log v}{2 - \log v}$

Separating the variables and integrating both side we get

$\begin{aligned} \Rightarrow \int \frac{2 - \log v}{v \log v - v} d v = \int \frac{d x}{x} \\ \Rightarrow \int \frac{1 + 1 - \log v}{- v (1 - \log v)} d v = \log x + \log c \\ \Rightarrow \int \frac{1}{v (\log v - 1)} - \int \frac{1}{x} d v = \log x + \log c \\ \Rightarrow \int \frac{1}{v (\log v - 1)} d v - \log v = \log x + \log c \end{aligned}$

Let $t = l o g v - 1$

$\Rightarrow d t = \frac{1}{v} - 1$

Ow equation becomes

$\Rightarrow \frac{d t}{t} - \log v = \log x + \log c$

$\Rightarrow \log t - \log v = \log x + \log c$

$\Rightarrow \log | \log v - 1 | - \log v = \log x + \log c [∴ t = \log v - 1]$

$\Rightarrow \log (\log v - 1) = \log x + \log v + \log c$

$\Rightarrow \log (\log v - 1) = \log c v x (using logrithm property)$

$\Rightarrow \log (\log \frac{y}{x} - 1) = \log c x x y^{y} / x [∴ v = \frac{y}{x}]$

$\Rightarrow \log \frac{y}{x} - 1 = c y$

$\Rightarrow c y = \log | \frac{y}{x} | - 1$

This is required solution.

Differential Equations Exercise 21.9 Question 36 (i)

Answer:

{(x^{2} - y^{2})}^{2} = x^{2}

Given:

x^{2} + y^{2} = 2 x y d y, y (1) = 0

To find: we have to find the solution of given differential equation.

Hint: we will put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

Solution: we have,

$x^{2} + y^{2} = 2 x y d y, y (1) = 0$

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

It is homogeneous equation.

Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

So,

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

Integrating both side we get

$\begin{aligned} \int \frac{2 v}{1 - v^{2}} d v = \int \frac{d x}{x} \\ \Rightarrow \log | 1 - v^{2} | = - \log | x | + \log | c | \\ \Rightarrow \log | 1 - v^{2} | = \log | \frac{c}{x} | \\ \Rightarrow | 1 - v^{2} | = | \frac{c}{x} | [Taking antilog on bothsides] \end{aligned}$

Putting $v = \frac{y}{x}$

$\begin{aligned} \Rightarrow | \frac{x^{2} - y^{2}}{x^{2}} | = | \frac{c}{x} | \\ \Rightarrow | x^{2} - y^{2} | = | c x | \end{aligned}$ .......(ii)
It is given that y(1)=0 i.e when x=1 , y=0
Putting x=1 , y=0 in (ii) we get

$1 - c = 0$

$c = 1$
Putting value of c in equation (ii) we get

$\begin{aligned} | x^{2} - y^{2} | = | x | \\ {(x^{2} - y^{2})}^{2} = x^{2} \end{aligned}$

This is required solution.

Differential Equations Exercise 21.9 Question 36 (ii)

Answer:

y = x l o g (l o g | x |)

Given:

x e^{\frac{y}{x}} - y + x \frac{d y}{d x} = 0, y (e) = 0

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

x e^{\frac{y}{x}} - y + x \frac{d y}{d x} = 0, y (e) = 0

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

......(i)
It is homogeneous equation.
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} v + x \frac{d v}{d x} = \frac{v x - x e^{\frac{y}{x}}}{x} \\ \Rightarrow x \frac{d v}{d x} = v - e^{v} - v \\ \Rightarrow x \frac{d v}{d x} = - e^{v} \\ \Rightarrow - e^{- v} = \frac{d x}{x} \end{aligned}

Separating the variables and integrating both side we get

$\begin{aligned} \int - e^{- v} d v = \int \frac{d x}{x} \\ \Rightarrow e^{r} = \log | x | + \log | c | \\ \Rightarrow v = \log (\log | x |) + \log (\log | c |) \end{aligned}$

Putting $v = \frac{y}{x}$

$\Rightarrow \frac{y}{x} = l o g (l o g | x |) + k$ ....(ii)

it is given that $y = 0$ when $x = e$

Putting $x = e$ , $y = 0$ in $(i i)$ we get

$\begin{aligned} \Rightarrow y = x \log (\log | x |) + k \\ \Rightarrow 0 = e \log (\log | x |) + k \\ \Rightarrow k = 0 \end{aligned}$
Putting value of c in equation (ii) we get

$\begin{aligned} \Rightarrow \frac{y}{x} = \log (\log | x |) + 0 \\ \Rightarrow y = x \log (\log | x |) \end{aligned}$
This is required solution.

Differential Equations Exercise 21.9 Question 36 (iii)

Answer:

l o g | x | = \cos \frac{y}{x} - 1

Given:

\frac{d y}{d x} = \frac{y}{x} + \cos e c \frac{y}{x}, y (1) = 0

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

\frac{d y}{d x} = \frac{y}{x} + \cos e c \frac{y}{x}, y (1) = 0

....(i)
It is homogeneous equation.
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} v + x \frac{d v}{d x} = \frac{v x}{x} + cosec \frac{v x}{x} \\ \Rightarrow x \frac{d v}{d x} = v - \cos e c v - v \\ \Rightarrow x \frac{d v}{d x} = - cosec v \\ \Rightarrow \sin v d v = \frac{d x}{x} \end{aligned}

Separating the variables and integrating both side we get

$\begin{aligned} \int \sin v d v = - \int \frac{d x}{x} \\ \Rightarrow - \cos v = - \log | x | + c \\ Putting v = \frac{y}{x} \\ \Rightarrow - \cos \frac{y}{x} = - \log | x | + c \end{aligned}$ ...(ii)

It is given that y=0 when x=0

Putting y=0, x=1 in equation (ii) we get

$\Rightarrow - \cos (\frac{0}{1}) = 0 + c$

$\Rightarrow c = - 1$

Putting value of c in equation (ii) we get

$\Rightarrow - \cos \frac{y}{x} = - l o g | x | + 1$

$\Rightarrow l o g | x | = \cos \frac{y}{x} - 1$

This is required solution.

Differential Equations Exercise 21.9 Question 36 (iv)

Answer:

y = \frac{x}{1 + l o g | x |}

Given:

(x y - y^{2}) d x = x^{2} d y, y = (1) =!

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

(x y - y^{2}) d x = x^{2} d y, y = (1) =!

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

....(i)
It is homogeneous equation.
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} v + x \frac{d v}{d x} = \frac{v x}{x} + \cos e c \frac{v x}{x} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{x^{2} (v - v^{2})}{x^{2}} \\ \Rightarrow x \frac{d v}{d x} = v - v^{2} - v \\ \Rightarrow x \frac{d v}{d x} = - v^{2} \end{aligned}

Separating the variables and integrating both side we get

$\begin{aligned} - \int \frac{d v}{v^{2}} = \int \frac{d x}{x} \\ \Rightarrow - (- \frac{1}{v}) = \log | x | + c \\ \Rightarrow \frac{1}{v} = \log | x | + c \end{aligned}$

Putting $v = \frac{y}{x}$

$\Rightarrow \frac{x}{y} = l o g | x | + c$ ...(ii)

It is given that y=1 when x=!

Putting y=1, x=! in equation (ii) we get

$\Rightarrow 1 = l o g | x | + c$

$\Rightarrow c +!$

Putting value of c in equation (ii) we get

$\begin{aligned} \Rightarrow \frac{x}{y} = \log | x | + c \\ y = \frac{x}{1 + \log | x |} \end{aligned}$

This is required solution.

Differential Equations Exercise 21.9 Question 36 (v)

Answer:

x y = 2 | y - x |^{\frac{3}{2}}

Given:

\frac{d y}{d x} = \frac{y (x + 2 y)}{x (2 x + y)}, y (1) = 2

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

\frac{d y}{d x} = \frac{y (x + 2 y)}{x (2 x + y)}, y (1) = 2

...(i)
It is homogeneous equation.
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

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

Separating the variables and integrating both side we get

$\begin{aligned} \int \frac{2 + v}{v - v^{2}} = \int \frac{d x}{x} \\ \int \frac{2 + v}{v (v - 1)} = \int \frac{d x}{x} \\ take, \\ \frac{2 + v}{v (v - 1)} = \frac{A}{v} + \frac{B}{v - 1} \\ \Rightarrow \frac{2 + v}{v (v - 1)} = \frac{A (v - 1) + B v}{v (v - 1)} \\ \Rightarrow 2 + v = A (v - 1) + B v \end{aligned}$

comparing the coefficient of like powers of v,

$\begin{aligned} A = - 2 \\ A + B = 1 \Rightarrow B = 3 \\ Using (ii) :- \\ \int (\frac{- 2}{v} + \frac{3}{v - 1}) d v = \int \frac{d x}{x} \end{aligned}$

$\begin{aligned} \Rightarrow \int \frac{- 2}{v} d x + \int \frac{3}{v - 1} d v = \int \frac{d x}{x} \\ \Rightarrow - 2 \log | v | + 3 \log | v - 1 | = \log x + \log c \\ \Rightarrow - \log | ϑ |^{2} + \log | v - 1 |^{3} = \log x c \\ \Rightarrow | v - 1 |^{3} = v^{2} x c \\ Putting v = \frac{y}{x} \\ \Rightarrow {| \frac{y - x}{x} |}^{3} = \frac{y^{2}}{x} c \end{aligned}$

It is given that y=0 when x=0

Putting y=0, x=1 in equation (iii) we get

$\Rightarrow | \frac{2 - 1}{2} |^{3} = \frac{4}{1} c$

$\Rightarrow c = \frac{1}{4}$

Putting value of c in equation (iii) we get

$\Rightarrow∣ \frac{y - x}{x} ∣^{\frac{3}{2}} = \frac{y^{2}}{x} (\frac{1}{4})$

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

This is required solution.

Differential Equations Exercise 21.9 Question 36 (vi)

Answer:

{(x^{3} + y^{3})}^{2} = 4 x^{2} y^{2}

Given:

(y^{4} - 2 x^{3} y) d x + (x^{4} - 2 x y^{3}) d x = 0, y (1) = 1

To find: we have to find the solution of given differential equation.

Hint: we will put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

Solution: we have,

$(y^{4} - 2 x^{3} y) d x + (x^{4} - 2 x y^{3}) d x = 0, y (1) = 1$

$\frac{d y}{d x} = \frac{2 x^{3} y - y^{4}}{x^{4} - 2 x y^{3}}$ .....(i)

It is homogeneous equation.

Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$

So,

$\begin{aligned} v + x \frac{d v}{d x} = \frac{2 x^{3} y - y^{4}}{x^{4} - 2 x y^{3}} \\ \Rightarrow x \frac{d v}{d x} = \frac{2 v - v^{4}}{1 - 2 v^{3}} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{v^{4} + v}{1 - 2 v^{3}} \end{aligned}$

Separating the variables and integrating both side we get

$\int \frac{1 - 2 v^{3}}{v (v^{3} + 1)} = \int \frac{d x}{x}$ ....(ii)

$take,$

$\frac{1 - 2 v^{3}}{v (v^{3} + 1)} = \frac{1 - 2 v^{3}}{v (v + 1) (v^{2} - v + 1)} = \frac{A}{v} + \frac{B}{v - 1} + \frac{v + 1}{v^{2} - v + 1}$

$1 - 2 v^{3} = v^{3} (A + B + C) + v^{2} (- B + C + 1) + v (B + D) + A$

$C o m p a r i n g t h e c o e f f i c i e n t o f l i k e p o w e r s o f v,$

$A = 1$

$B + D = 0$

$- B + C + 1 = 0$

$A + B + C = - 2$

$O n s o l v i n g$

$A = 1, B = - 1, C = - 2, x =!$

Using (ii)

$\begin{aligned} \int (\frac{1}{v} - \frac{1}{v - 1} - \frac{2 v + 1}{v^{2} - v + 1}) d v = \int \frac{d x}{x} \\ \Rightarrow \log | ↓ | - \log | v + 1 | - \log | v^{2} - v + 1 | = \log x + \log c \\ \Rightarrow \log | \frac{v}{v^{3} + 1} | = \log | x c | \\ Putting v = \frac{y}{x} \\ \Rightarrow \frac{\frac{x}{y}}{y^{3} + x^{3}} \times x^{3} = x c \\ \Rightarrow \frac{x y}{y^{3} + x^{3}} = c \end{aligned}$ ....(iii)

It is given that y=! when x=1

Putting y=1, x=1 in equation (iii) we get

Differential Equations Exercise 21.9 Question 36 (vii)

Answer:

x^{4} + 6 x^{2} y^{2} + y^{4} = 8

Given:

x (x^{2} + 3 y^{2}) d x + y (y^{2} + 3 x^{2}) d y = 0, y (1) =!

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

x (x^{2} + 3 y^{2}) d x + y (y^{2} + 3 x^{2}) d y = 0, y (1) =!

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

It is homogeneous equation.
Put

y = v x

and

\frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} v + x \frac{d v}{d x} = \frac{x (x^{2} + 3 y^{2})}{y (y^{2} + 3 x^{2})} \\ \Rightarrow x \frac{d v}{d x} = - \frac{x (x^{2} + 3 v^{2} x^{2})}{v x (v^{2} x^{2} + 3 x^{2})} - v \\ \Rightarrow x \frac{d v}{d x} = - \frac{1 + 3 v^{2}}{v (v^{2} + 3)} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{- v^{4} - 6 v^{2} - 1}{v (v^{2} + 3)} \end{aligned}

Separating the variables and integrating both side we get

$\begin{aligned} \int \frac{v (v^{2} + 3)}{v^{4} + 6 v^{2} + 1} = - \int \frac{d x}{x} \\ \Rightarrow \int \frac{4 v (v^{2} + 3)}{v^{4} + 6 v^{2} + 1} = - 4 \int \frac{d x}{x} [Multiply4] \\ \Rightarrow \log | v^{4} + 6 v^{2} + 1 | = - 4 \log | x | + \log | c | \\ \Rightarrow \log | v^{4} + 6 v^{2} + 1 | = \log | \frac{c}{x^{4}} | \\ \Rightarrow | v^{4} + 6 v^{2} + 1 | = | \frac{c}{x^{4}} | \end{aligned}$

Putting $v = \frac{y}{x}$

$\begin{aligned} \Rightarrow | \frac{y^{4}}{x^{4}} + 6 \frac{y^{2}}{x^{2}} + 1 | = | \frac{c}{x^{4}} | \\ \Rightarrow | y^{4} + 6 x^{2} y^{2} + 1 | = | c | \end{aligned}$ ....(iii)

It is given that y(1)=!

Putting y=1, x=1 in equation (ii) we get

$\Rightarrow 1 + 6 + 1 = c$

$\Rightarrow c = 8$
Putting value of c in equation (ii) we get

$\begin{aligned} \Rightarrow | y^{4} + 6 x^{2} y^{2} + 1 | = 8 \\ \Rightarrow y^{4} + 6 x^{2} y^{2} + 1 = 8 \end{aligned}$
This is required solution.

Differential Equations Exercise 21.9 Question 36 (viii)

Answer:

\cot (\frac{y}{x}) = l o g ∣ e x ∣

Given:

{x \sin^{2} \frac{y}{x} - y} d x + x d y = 0, y (1) = \frac{π}{4}

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

{x \sin^{2} \frac{y}{x} - y} d x + x d y = 0, y (1) = \frac{π}{4}

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

.....(i)
It is homogeneous equation.
put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} v + x \frac{d v}{d x} = - \sin^{2} v + v \\ \Rightarrow x \frac{d v}{d x} = - \sin^{2} v \end{aligned}

Separating the variables and integrating both side we get

\begin{aligned} \int \frac{- d v}{\sin^{2} v} = \int \frac{d x}{x} \\ \Rightarrow \cot v = \log | x | + \log | c | \\ Putting v = \frac{y}{x} \\ \Rightarrow \cot (\frac{y}{x}) = \log | x c | \end{aligned}

....(ii)

It is given that $y = \frac{π}{4}$ when $x = 1$

Putting $y = \frac{π}{4}$ , $x = 1$ in equation (ii) we get

$\Rightarrow \cot (\frac{π}{4}) = l o g ∣ c ∣$

$\Rightarrow c = e$

Putting value of c in equation (ii) we get

$\Rightarrow \cot (\frac{π}{4}) = l o g ∣ e x ∣$

This is required solution.

Differential Equations Exercise 21.9 Question 36 (ix)

Answer:

\tan \frac{y}{2 x} = \frac{2}{x}

Given:

x \frac{d y}{d x} - y + x \sin (\frac{y}{x}) = o, y (2) = x

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

x \frac{d y}{d x} - y + x \sin (\frac{y}{x}) = o, y (2) = x

\frac{d y}{d x} = \frac{y}{x} - \sin (\frac{y}{x})

...(i)
It is homogeneous equation.
put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} v + x \frac{d v}{d x} = v - \sin v \\ \Rightarrow x \frac{d v}{d x} = - \sin v \end{aligned}

Separating the variables and integrating both side we get

$\begin{aligned} \int \frac{d v}{- \sin v} = \int \frac{d x}{x} \\ \Rightarrow \int cosecv d v = - \int \frac{d x}{x} \\ \Rightarrow \log | \tan \frac{v}{2} | = - \log | x | + \log | c | \\ Putting v = \frac{y}{x} \\ \Rightarrow | \tan \frac{y}{2 x} | = \frac{c}{x} \end{aligned}$ ...(ii)
It is given that $y (2) = π$
Putting $y (2) = π, x = 2$ in equation (ii) we get
$\begin{aligned} \Rightarrow \tan (\frac{π}{4}) = \frac{c}{2} \\ \Rightarrow 1 = \frac{c}{2} \\ \Rightarrow c = 2 \end{aligned}$

Putting value of c in equation (ii) we get

$\tan \frac{y}{2 x} = \frac{2}{x}$

This is required solution.

Differential Equations Exercise 21.9 Question 37

Answer:

\begin{array}{r} \sin (\frac{y}{x}) = \log | x | + \frac{1}{\sqrt{2}} \end{array}

Given:

\begin{aligned} x \cos (\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x \end{aligned}

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

\begin{aligned} x \cos (\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x \end{aligned}

....(i)
It is homogeneous equation.
put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} x \cos (\frac{v x}{x}) (v + x \frac{d v}{d x}) = v x \cos (\frac{v x}{x}) + x \\ \Rightarrow x \cos (v) (v + x \frac{d v}{d x}) = x (v \cos v + 1) \\ \Rightarrow \cos v (v + x \frac{d v}{d x}) = v \cos v + 1 \\ \Rightarrow v \cos v + x \cos v \frac{d v}{d x} = v \cos v + 1 \\ \Rightarrow x \cos v \frac{d v}{d x} = 1 \end{aligned}

Separating the variables and integrating both side we get

$\begin{aligned} \int \cos v d v = \int \frac{d x}{x} \\ \Rightarrow \sin v = \log | x | + c \\ Putting v = \frac{y}{x} \\ \Rightarrow \sin (\frac{y}{x}) = \log | x | + c \end{aligned}$ ...(ii)
It is given that $y = \frac{π}{4}$ when $x = 1$
Putting $y = \frac{π}{4}$ , $x = 1$ in equation (ii) we get
$\Rightarrow \sin (\frac{π}{4}) = c$
$\Rightarrow c = \frac{1}{\sqrt{2}}$

Putting value of c in equation (ii) we get

$\Rightarrow \sin (\frac{y}{x}) = l o g ∣ x ∣ + \frac{1}{\sqrt{2}}$

This is required solution.

Differential Equations Exercise 21.9 Question 38

Answer:

\frac{2}{3} \tan^{- 1} (\frac{2 y + x}{x \sqrt{3}}) - \frac{π}{\sqrt{3}} = \log (x^{2} + x y + y^{2})

Given:

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

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

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

..(i)
It is homogeneous equation.
put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

So,

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

Separating the variables and integrating both side we get

$\begin{aligned} \int \frac{v - 1}{1 + 2 v + v^{2}} d v = - \int \frac{d x}{x} \\ \Rightarrow \int \frac{\frac{1}{2} (2 v + 1) - \frac{3}{2}}{1 + v + v^{2}} d v = - \log | x | \\ \Rightarrow \frac{1}{2} \int \frac{(2 v + 1)}{1 + v + v^{2}} - \frac{3}{2} \int \frac{1}{1 + v + v^{2}} d v = - \log | x | \\ \Rightarrow \log | 1 + v + v^{2} | - 3 \int \frac{1}{v^{2} + v + 1 + {(\frac{1}{2})}^{4} - {(\frac{1}{2})}^{2}} = - 2 \log | x | \end{aligned}$

$\begin{aligned} \Rightarrow \log | v^{2} + v + 1 | - 3 \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} ∣ \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}}) = - 2 \log | x | \\ Putting v = \frac{y}{x} \\ \Rightarrow \log | \frac{y^{2}}{x^{2}} + \frac{y}{x} + 1 | - 2 \sqrt{3} \tan^{- 1} (\frac{\frac{2 y}{x} + 1}{\sqrt{3}}) = - \log | x | + c \\ \Rightarrow \log | y^{2} + x y + x^{2} | - \log x^{2} - 2 \sqrt{3} \tan^{- 1} (\frac{2 y + x}{x \sqrt{3}}) = - \log x^{2} + c \\ \Rightarrow \log | y^{2} + x y + x^{2} | = c + 2 \sqrt{3} \tan^{- 1} (\frac{2 y + x}{x \sqrt{3}}) \end{aligned}$
It is given that y=0 when x=1
Putting y=0, x=1 in equation (ii) we get
$\begin{aligned} \Rightarrow \log | 1 | = c + 2 \sqrt{3} \tan^{- 1} (\frac{1}{\sqrt{3}}) \\ \Rightarrow 0 = c + 2 \sqrt{3} \times \frac{π}{6} \\ \Rightarrow c = - \frac{π}{\sqrt{3}} \end{aligned}$
Putting value of c in equation (ii) we get
$\begin{aligned} \Rightarrow \log | y^{2} + x y + x^{2} | = 2 \sqrt{3} \tan^{- 1} (\frac{2 y + x}{x \sqrt{3}}) - \frac{π}{\sqrt{3}} \end{aligned}$
This is required solution.

Differential Equations Exercise 21.9 Question 39

Answer:

\frac{x^{2}}{2 y^{2}} = l o g ∣ y ∣

Given:

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

To find: we have to find the solution of given differential equation.
Hint: we will put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

Solution: we have,

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

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

...(i)
It is homogeneous equation.
put

y = v x a n d \frac{d y}{d x} = v + x \frac{d v}{d x}

So,

\begin{aligned} v + x \frac{d v}{d x} = (\frac{1}{\frac{1}{v} + v}) \\ \Rightarrow x \frac{d v}{d x} = \frac{x}{1 + v^{2}} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{v - v - v^{3}}{1 + v^{2}} \\ \Rightarrow x \frac{d v}{d x} = \frac{- v^{3}}{1 + v^{2}} \end{aligned}

Separating the variables and integrating both side we get

$\begin{aligned} - (\frac{1 + v^{2}}{v^{3}}) d v = \frac{d x}{x} \\ \Rightarrow \int (- \frac{1}{v^{3}} - \frac{1}{v}) d v = \int \frac{d x}{x} \\ \Rightarrow \frac{1}{2 v^{2}} - \log | v | = \log | x | + c \end{aligned}$

Putting $v = \frac{y}{x}$

$\begin{aligned} \Rightarrow \frac{x^{2}}{2 y^{2}} - \log | \frac{y}{x} | = \log | x | + c \\ \Rightarrow \frac{x^{2}}{2 y^{2}} = \log | x | + \log | \frac{y}{x} | + c \\ \Rightarrow \frac{x^{2}}{2 y^{2}} = \log | y | + c \end{aligned}$ ...(ii)

It is given that y=1 when x=0

Putting y=1, x=0 in equation (ii) we get

$\begin{aligned} \Rightarrow 0 = \log (1) + c \\ \Rightarrow c = 0 \end{aligned}$

Putting value of c in equation (ii) we get

$\begin{aligned} \Rightarrow \frac{x^{2}}{2 y^{2}} = \log | y | \end{aligned}$

This is required solution.

The 21st chapter of mathematics, Differential Equations for the class 12 students, is one of the essential portions where students need to score good marks. Exercise 21.9 in his chapter has 48 questions, including its subparts. The concepts that this exercise deals with are to solve the differential equations and to find the particular solution of the differential equations. The RD Sharma Class 12 Chapter 21 Exercise 21.9 is readily available to clarify their doubts.

There are various methods by which these sums can be solved. The RD Sharma Class 12th Exercise 21.9 book consists of all these methods to adapt the ones that they feel are easy. Additionally, there are many practice questions given apart from the ones given in the textbook. With its help, the students can practice as much as they want to be clear in the concept. As this book follows the NCERT pattern, the CBSE board students have a tremendous advantage over it.

Many experts have spent scads of time providing answers for the Class 12 RD Sharma Chapter 21 Exercise 21.9 Solution book and the other RD Sharma books. However, students of every mark category like toppers, average learners, and slow learners can find their respective way of solving the problems with the help of this book. This is possible due to each sum's shortcut and elaborated methods present in RD Sharma Class 12 Solutions Differential Equation Ex 21.9 material.

The other reason why many students use RD Sharma books is free PDF copies at the Career360 website. Everyone can access the authorized RD Sharma Class 12th Exercise 21.9 for free of cost from this website. As no payment is required to own this best set of solution books, it is used by most of the students.

Another great advantage is that the staff members use the RD Sharma Solutions Chapter 21 Ex 21.9 to pick questions for the tests and exams. Therefore, when a student practices every solution given in this book, there are higher chances for the occurrence of more marks effortlessly. Therefore, visit the career 360 website to own a copy of RD Sharma solution materials.

RD Sharma Chapter wise Solutions

Frequently Asked Questions (FAQs)

1. What is the optimal solution book to learn the differential equation concept for class 12?

The RD Sharma Class 12th Exercise 21.9 is the optimal solution book for the students to clear their doubts and learn the differential integration concepts.

2. How much should I pay to download the RD Sharma book?

The career 360 website lets visitors download any RD Sharma solution books without charging even a minimal fee. Hence, it is free if you download it from this site.

3. How many questions are given in the maths textbook for exercise 21.9?

There are 48 questions, including the subparts present in the textbook; you can use the RD Sharma Class 12th Exercise 21.9 book to refer to the solutions.

4. Is the RD Sharma solution material available in PDF format?

The RD Sharma solution books are present on the Career 360 website in PDF format to download and use as a soft copy.

5. Do the RD Sharma books follow the NCERT pattern?

Yes, the RD Sharma books follow the NCERT pattern making it strongly recommendable for the CBSE board students.

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RD Sharma Class 12 Exercise 21.9 Differential Equation Solutions Maths-Download PDF Online

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise: 21.9

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