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

Option 1)
$y\sqrt{\tan x}=x+c$

Option 2)
$y\sqrt{\cot x}=\tan x+c$

Option 3)
$y\sqrt{\tan x}=\cot x+c$

Option 4)
$y\sqrt{\cot x}= x+c$

Answers (1)

best_answer

As we learnt in

Linear Differential Equation -

$\frac{dy}{dx}+Py= Q$

- wherein

P, Q are functions of x alone.

$Sin2x(\frac{dy}{dx}-\sqrt{tanx})=y$

$=>\frac{dy}{dx}-\sqrt{tanx}=ycosec2x$

$=>\frac{dy}{dx}-ycosec2x=\sqrt{tanx}$

=>P=-cosec2x $\Delta Q=\sqrt{\tan x}$

$-\int cosec2xdx=-\frac{1}{2}log\left | cosec2x-cot2x \right |$

$=log\left | cosec2x+cot2x \right |^\frac{1}{2}$

I.F.= $e^{log}^\left | cosec2x+cot2x \right |^{1/2}$

$=|cosec2x+cot2x|^\frac{1}{2}$

$\therefore y\sqrt{cosec2x+cot2x}=\int \sqrt{tanx(cosec2x+cot2x)} dx$

$=\int \sqrt{tanxcotx\: }\ dx=\int 1.dx=x+c$

$\therefore y\sqrt{cotx}=x+c$

Option 1)

$y\sqrt{\tan x}=x+c$

This option is incorrect

Option 2)

$y\sqrt{\cot x}=\tan x+c$

This option is incorrect

Option 3)

$y\sqrt{\tan x}=\cot x+c$

This option is incorrect

Option 4)

$y\sqrt{\cot x}= x+c$

This option is correct

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