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.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter22
Exercise22.6
CategoryRD Sharma Solutions

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

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

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