# (X + Tan Y) Dy = Sin 2y Dx - Mathematics

Sum

(x + tan y) dy = sin 2y dx

#### Solution

We have,
$\left( x + \tan y \right)dy = \sin 2y \text{ dx }$
$\Rightarrow \frac{dx}{dy} = x\text{ cosec }2y + \frac{1}{2} \sec^2 y$
$\Rightarrow \frac{dx}{dy} - x\text{ cosec }2y = \frac{1}{2} \sec^2 y . . . . . \left( 1 \right)$
Clearly, it is a linear differential equation of the form
$\frac{dx}{dy} + Px = Q$
where
$P = -\text{ cosec }2y$
$Q = \frac{1}{2} \sec^2 y$
$\therefore I . F . = e^{\int P dy }$
$= e^{- \int \text{ cosec }\text{ 2y } dy}$
$= e^{- \frac{1}{2}\log\left| \tan y \right|} = \frac{1}{\sqrt{\tan y}}$
$\text{ Multiplying both sides of }\left( 1 \right)\text{ by }\frac{1}{\sqrt{\tan y}},\text{ we get }$
$\frac{1}{\sqrt{\tan y}}\left( \frac{dx}{dy} - x\text{ cosec }2y \right) = \frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^2 y$
$\Rightarrow \frac{1}{\sqrt{\tan y}}\frac{dx}{dy} - x\text{ cosec }2y\frac{1}{\sqrt{\tan y}} = \frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^2 y$
Integrating both sides with respect to y, we get
$\frac{1}{\sqrt{\tan y}}x = \int \frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^2 \text{ y }dy + C$
$\Rightarrow \frac{x}{\sqrt{\tan y}} = I + C . . . . . \left( 2 \right)$
$\text{ where }I = \int\frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^2 y dy$
$\text{ Putting }t = \tan y,\text{ we get }$
$dt = \sec^2 y dy$
$\therefore I = \frac{1}{2}\int\frac{1}{\sqrt{t}} \times dt$
$= \sqrt{t}$
$= \sqrt{\tan y}$
$\text{ Putting the value of I in }\left( 2 \right),\text{ we get }$
$\frac{x}{\sqrt{\tan y}} = \sqrt{\tan y} + C$
$\Rightarrow x = \tan y + C\sqrt{\tan y}$
$\text{ Hence,} x = \tan y + C\sqrt{\tan y}\text{ is the required solution .}$

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Chapter 22: Differential Equations - Exercise 22.10 [Page 106]

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Exercise 22.10 | Q 25 | Page 106

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