Question

Solve : $\sin 2x(\frac{dy}{dx}-\sqrt{\tan x})-y=0$

Answer 1

$\sin 2x(\frac{dy}{dx}-\sqrt{\tan x})-y=0\Rightarrow \frac{dy}{dx}-\frac{y}{\sin 2x}=\sqrt{\tan x}⟶\left(1\right)$

comparing equation1 with$\frac{dy}{dx}+py=q$, we get

$p=\frac{-1}{{\sin }^{2}x}andq=\sqrt{\tan x}$

$Now,I.F=e^{\int pdx}=e^{\int \frac{-1}{\sin 2x}dx}=e^{-\int \frac{1}{2\tan x}dx}=e^{-\int \frac{dt}{2t}}=e^{-\frac{1}{2}\log t}=e^{{\left(\log t\right)}^{\frac{-1}{2}}}=\frac{1}{\sqrt{t}}\therefore I.F=\frac{1}{\sqrt{\tan x}}Now,y.\frac{1}{\sqrt{\tan x}}=\int \frac{1}{\sqrt{tanx}}.\sqrt{\tan x}dx\Rightarrow \frac{y}{\sqrt{\tan x}}=x+C\Rightarrow y\sqrt{\cot x}=x+C$