Question

(x + tan y) dy = sin 2y dx

Answer 1

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
\[\frac{1}{\sqrt{\tan y}}x = \int \frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^2 \text{ y }dy + C\]
\[ \Rightarrow \frac{x}{\sqrt{\tan y}} = I + C . . . . . \left( 2 \right)\]
\[\text{ where }I = \int\frac{1}{2} \frac{1}{\sqrt{\tan y}} \times \sec^2 y dy\]
\[\text{ Putting }t = \tan y,\text{ we get }\]
\[dt = \sec^2 y dy\]
\[ \therefore I = \frac{1}{2}\int\frac{1}{\sqrt{t}} \times dt\]
\[ = \sqrt{t}\]
\[ = \sqrt{\tan y}\]
\[\text{ Putting the value of I in }\left( 2 \right),\text{ we get }\]
\[\frac{x}{\sqrt{\tan y}} = \sqrt{\tan y} + C \]
\[ \Rightarrow x = \tan y + C\sqrt{\tan y}\]
\[\text{ Hence,} x = \tan y + C\sqrt{\tan y}\text{ is the required solution .} \]