RD Sharma Class 12 Ex 22.6 Solutions Chapter 22 Differential Equations

Here we provide RD Sharma Class 12 Ex 22.6 Solutions Chapter 22 Differential Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 22.6 Solutions Chapter 22 Differential Equations book pdf download. Now you will get step-by-step solutions to each question.

Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	22
Exercise	22.6
Category	RD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 22.6 Solutions Chapter 22 Differential Equations

Question 1. Solve the following differential equation

$\frac{dv}{dx}+\frac{1+y^2}{y}=0$

Solution:

From the question it is given that,

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

Transposing we get,

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

By cross multiplication,

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

Integrating on both side, we will get,

$∫\frac{y}{(1 + y^2)} dy = ∫-dx$

$∫\frac{2y}{(1 + y^2)} dy = -2 ∫dx$

log (1 + y²) = – 2x + c₁

Therefore, $\frac{1}{2}$ log [1 + y²] + x = c

Question 2. Solve the following differential equation

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

Solution:

From the question it is given that,

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

By cross multiplication,

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

Integrating on both side, we will get,

$∫(y - \frac{y}{(1 + y^2)} dy = ∫dx$

$∫ydy - ∫\frac{y}{(1 + y^2)} dy = ∫ dx\\ ∫ydy - \frac{1}{2} ∫\frac{2y}{(1 + y2)} dy = ∫ dx\\ \frac{y^2}{2} - \frac{1}{2} log [y^2 + 1] = x + c$

Question 3. Solve the following differential equation:

$\frac{dy}{dx} = sin^2 y$

Solution:

From the question it is given that,

$\frac{dy}{dx} = sin^2 y$

By cross multiplication,

$\frac{dy}{sin^2 y} = dx$

As we know that, $\frac{1}{sin x}$ = cosec x

cosec²y dy = dx

Integrating on both side, we will get,

∫cosec² y dy = ∫dx + c

– cot y = x + c

Question 4. Solve the following differential equation:

$\frac{dy}{dx} = \frac{(1 - cos 2y)}{(1 + cos 2y)}$

Solution:

From the question it is given that,

$\frac{dy}{dx} = \frac{(1 - cos 2y)}{(1 + cos 2y)}$

We know that, 1 – cos 2y = 2sin²y and 1 + cos 2y = 2 cos²y

So, $\frac{dy}{dx} = \frac{(2 sin^2 y)}{(2 cos^2 y)}$

Also we know that, $\frac{sin θ}{cos θ}$ = tan θ

By cross multiplication,

$\frac{dy}{tan^2 y} = dx$

Integrating on both side, we get,

∫cot²y dy = ∫dx

∫ (cosec²y – 1) dy = ∫dx

– cot y- y + c = x

c = x + y + cot y

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RD Sharma Class 12 Ex 22.6 Solutions Chapter 22 Differential Equations

Question 1. Solve the following differential equation

Question 2. Solve the following differential equation

Question 3. Solve the following differential equation:

Question 4. Solve the following differential equation:

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