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

**Question 1. Solve the following differential equation**

**Solution:**

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From the question it is given that,

Transposing we get,

By cross multiplication,

Integrating on both side, we will get,

log (1 + y

^{2}) = – 2x + c_{1}Therefore, log [1 + y

^{2}] + 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

cosec

^{2}y dy = dxIntegrating on both side, we will get,

∫cosec

^{2}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 = 2sin

^{2}y and 1 + cos 2y = 2 cos^{2}ySo,

Also we know that, = tan θ

By cross multiplication,

Integrating on both side, we get,

∫cot

^{2}y dy = ∫dx∫ (cosec

^{2}y – 1) dy = ∫dx– cot y- y + c = x

c = x + y + cot y