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

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

Summary

Exercise 22.6 focuses on solving homogeneous differential equations of the first order. These equations are characterized by terms that maintain the same degree when x and y are replaced by kx and ky respectively. The problems range from simple rational expressions to more complex forms involving trigonometric functions. Students are asked to find both general and particular solutions. The primary method used to solve these equations involves the substitution y = vx, where v is a new function of x, which transforms the homogeneous equation into a separable one. This exercise helps students recognize homogeneous forms, apply the appropriate substitution, and solve the resulting separable equations.