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Class 12 RD Sharma Solutions – Chapter 22 Differential Equations – Exercise 22.6
  • Last Updated : 19 Jan, 2021

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

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