Question

The general solution of the differential equation, $\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0$, is

• A. $y \sqrt{\tan x}=\cot x+C$
• B. $y \sqrt{\cot x}=\tan x+C$
• C. $ y \sqrt{\cot x}=x+C$
• D. $y \sqrt{\tan x}=x+C$

Answer 1

Answer: (C)

$\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0$
$\frac{d y}{d x}-(\operatorname{cosec} 2 x) y=\sqrt{\tan x}$
IF $=e^{\int-\operatorname{cosec} 2 x d x}=e^{-\frac{1}{2} \ln \tan x}=\sqrt{\cot x}$
so solution is
$y \sqrt{\cot x}=\int 1 d x$
$y \sqrt{\cot x} =x+C$