# Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.6

• Last Updated : 19 Jan, 2021

### Question 1. Solve the following differential equation

Solution:

From the question it is given that,

Transposing we get,

By cross multiplication,

Integrating on both side, we will get,

log (1 + y2) = – 2x + c1

Therefore, log [1 + y2] + x = c

### Question 2. Solve the following differential equation

Solution:

From the question it is given that,

By cross multiplication,

Integrating on both side, we will get,

### Question 3. Solve the following differential equation:

Solution:

From the question it is given that,

By cross multiplication,

As we know that,  = 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:

Solution:

From the question it is given that,

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

So,

Also we know that,  = tan θ

By cross multiplication,

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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