The general solution of the differential equation, $\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0$,…
The general solution of the differential equation, $\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0$, is :
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$y \sqrt{\tan x}=x+c$
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$y \sqrt{\cot x}=\tan x+c$
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$y \sqrt{\tan x}=\cot x+c$
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$y \sqrt{\cot x}=x+c$
Solution
Given, $\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0$ or, $\frac{d y}{d x}=\frac{y}{\sin 2 x}+\sqrt{\tan x}$ or, $\frac{d y}{d x}-y \operatorname{cosec} 2 x=\sqrt{\tan x}$
Now, integrating factor (I.F) $=e^{\int-\operatorname{cosec} 2 x}$ or, I.F $=e^{-\frac{1}{2} \log |\tan x|}=e^{\log (\sqrt{\tan x})^{-1}}$ $=\frac{1}{\sqrt{\tan x}}=\sqrt{\cot x}$
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}
$
Asked in: JEE Main 2014 (12 Apr Online)
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