JEE Main & Advanced JEE Main Paper (Held On 12 April 2014)

  • question_answer
    The general solution of the differential equation, \[\sin 2x\left( \frac{dy}{dx}-\sqrt{\tan x} \right)-y=0,\]is:   [JEE Main Online Paper ( Held On 12 Apirl  2014 )

    A) \[y\sqrt{\tan x}=x+c\]  

    B) \[y\sqrt{\cot x}=\tan x+c\]

    C) \[y\sqrt{\tan x}=\cot x+c\]

    D) \[y\sqrt{\cot x}=x+c\]

    Correct Answer: D

    Solution :

                    Given, \[\sin 2x\left( \frac{dy}{dx}-\sqrt{\tan x} \right)-y=0\] or, \[\frac{dy}{dx}=\frac{y}{\sin 2x}+\sqrt{\tan x}\] \[\frac{dy}{dx}-y\cos ec2x=\sqrt{\tan x}\]                           ?(1) Now, integrating factor \[(I.F)={{e}^{\int_{{}}^{{}}{-\cos ec2x}}}\] or, \[I.F={{e}^{-\frac{1}{2}\log |\tan x|}}={{e}^{\log {{\left( \sqrt{\tan x} \right)}^{-1}}}}\] \[=\frac{1}{\sqrt{\tan x}}=\sqrt{\cot x}\] Now, general solution of eq. (1) is written as \[y(I.F.)=\int_{{}}^{{}}{Q(I.F.)dx+c}\] \[\therefore \]\[y\sqrt{\cot x}=\int_{{}}^{{}}{\sqrt{\tan x}.\sqrt{\cot x}dx+c}\] \[\therefore \]\[y\sqrt{\cot x}=\int_{{}}^{{}}{1.dx+c}\] \[\]

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