Question

Find the general solution of given differential equation.

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

• A. $x\cot x=\log \tan y+C$
• B. $x\cot y=\log \tan y+C$
• C. $x\cot y=\log \tan x+C$
• D. none of these

Answer 1

The correct option is B $x\cot y=\log \tan y+C$

$(x+tany)dy=sin2ydx$
It can be written as
$\frac{dx}{dy}=\frac{x}{\sin 2y}+\frac{\tan y}{\sin 2y}$
$\frac{dx}{dy}-\frac{x}{\sin 2y}=\frac{1+{\tan }^{2}y}{2}$
which is a linear differential with x as dependent variable.
Here, $P=-\frac{1}{\sin 2y}=-cosec2y$ ; $Q=\frac{1+{\tan }^{2}y}{2}$
Integrating factor $I.F.=e^{\int Pdy}$
$=e^{\int -cosec2ydy}$
$=e^{-\log \tan y}$
$\Rightarrow I.F.=\frac{1}{\tan y}=\cot y$
Solution of given differential eqn is
$x\cot y=\int \frac{1+{\tan }^{2}y}{2\tan y}dy+C$
$\Rightarrow x\cot y=\int cosec2ydy+C$
$\Rightarrow x\cot y=\log \tan y+C$