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(x + tan y) dy = sin 2y dx

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#### 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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