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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer
    If \[y=y\left( x \right)\]is the solution of the differential equation \[\frac{dy}{dx}=(tanx-y)se{{c}^{2}}x,x\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right),\]such that\[y(0)=0,\]then \[y\left( -\frac{\pi }{4} \right)\]is equal to : [JEE Main 10-4-2019 Morning]

    A) \[2+\frac{1}{e}\]                   

    B) \[\frac{1}{2}-e\]

    C) \[e-2\]

    D) \[\frac{1}{2}-e\]

    Correct Answer: C

    Solution :

    \[\frac{dy}{dx}=(tan\,x-y)se{{c}^{2}}x\] Now, \[put\,\tan x=t\Rightarrow \frac{dt}{dx}=se{{c}^{2}}x\] So\[\frac{dy}{dt}+y=t\] On solving, we get \[y{{e}^{t}}={{e}^{t}}(t-1)+c\] \[\Rightarrow y=(tanx-1)+c{{e}^{-\tan x}}\] \[\Rightarrow y(0)=0\Rightarrow c=1\] \[\Rightarrow y=\tan x-1+{{e}^{-\tan x}}\] So \[y\left( -\frac{\pi }{4} \right)=e-2\]                      


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