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)

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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