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

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