Question

$(x+\tan y)dy=\sin 2ydx$

Answer 1

Given,

$(x+\tan y)dy=\sin 2ydx$

$\frac{dx}{dy}=\frac{x+\tan y}{\sin 2y}$

$\frac{dx}{dy}=\frac{x}{\sin 2y}+\frac{\tan y}{\sin 2y}$

$\frac{dx}{dy}+(-\csc 2y)x=\frac{1}{2}{\sec }^{2}y$

$\frac{dx}{dy}+Rx=S$

$R=-\csc 2y\to I.F=e^{\int -\csc 2ydy}=e^{\log \cot y}=\cot y$

$x\times I.F=\int S\times I.Fdy$

$x\times \cot y=\int \frac{1}{2}{\sec }^{2}y\times \cot ydy$

$x\cot y=\int \csc ydy$

$x\cot y=\frac{1}{2}\log (\csc 2y-\cot 2y)+c$