A) \[y\sqrt{\cot x}=\tan x+C\]
B) \[y=\sqrt{\cot x}\,\,(x+C)\]
C) \[y\sqrt{tanx}=\cot x+C\text{ }\!\!~\!\!\text{ }\]
D) \[y=\sqrt{\tan x}(x+C)\]
Correct Answer: D
Solution :
[d] \[\sin 2x.\left( \frac{dy}{dx}-\sqrt{\tan x} \right)-y=0\] \[\Rightarrow \,\,\,\,\,\frac{dy}{dx}+\left( \frac{-1}{\sin 2x} \right)y=\sqrt{\tan x}\] Above is a Linear Differential Equation. \[I.F.={{e}^{-\int{\cos ec\,\,2xdx}}}={{e}^{-\frac{1}{2}\log \,\tan x}}=\sqrt{\cot x}\] Therefore, general solution is \[y\sqrt{\cot x}=\int{\sqrt{\cot x}\,\sqrt{\tan \,x\,}\,dx+C}\] or \[y\sqrt{\cot x}=x+C\]
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