The general solution of the differential equation, sin 2 x ( d y d x-√tan x )-y=0, is :

Given, $\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0$sin2x(dxdy​−tanx​)−y=0 or, $\frac{d y}{d x}=\frac{y}{\sin 2 x}+\sqrt{\tan x}$dxdy​=sin2xy​+tanx​ or, $\frac{d y}{d x}-y \operatorname{cosec} 2 x=\sqrt{\tan x}$dxdy​−ycosec2x=tanx​ Now, integrating factor (I.F) $\displaystyle =e^{\int-\operatorname{cosec} 2 x}$=e∫−cosec2x or, I.F $=e^{-\frac{1}{2} \log |\tan x|}=e^{\log (\sqrt{\tan x})^{-1}}$=e−21​log∣tanx∣=elog(tanx​)−1 $=\frac{1}{\sqrt{\tan x}}=\sqrt{\cot x}$=tanx​1​=cotx​ Now, general solution of eq. (1) is written as $$\begin{aligned} &\text { y (I. F.) }=\int \mathrm{Q}(\text { I.F.) } d x+c \\ &\therefore y \sqrt{\cot x}=\int \sqrt{\tan x} \cdot \sqrt{\cot x} d x+c \\ &\therefore y \sqrt{\cot x}=\int 1 . d x+c \end{aligned}$$​ y (I. F.) =∫Q( I.F.) dx+c∴ycotx​=∫tanx​⋅cotx​dx+c∴ycotx​=∫1.dx+c​