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JEE Main & Advanced Mathematics Differential Equations Question Bank Self Evaluation Test - Differential Equations

  • question_answer
    A continuously differentiable function \[\phi \,(x)\],\[x\in [0,\pi ]-\left\{ \frac{\pi }{2} \right\}\] satisfying \[y'=1+{{y}^{2}},y(0)=0=y(\pi )\] is

    A) \[\tan x\]

    B) \[x(x-\pi )\]

    C) \[(x-\pi )(1-{{e}^{x}})\]

    D) \[{{\sec }^{2}}x\]  

    Correct Answer: A

    Solution :

    [a] Given \[\frac{dy}{dx}=1+{{y}^{2}}\] \[\Rightarrow \frac{dy}{1+{{y}^{2}}}=dx\Rightarrow {{\tan }^{-1}}y=x+c\] \[\Rightarrow y=\tan (x+c)\], now \[y(0)=0\Rightarrow tanc=0\] \[y(\pi )=0\Rightarrow tan(\pi +c)=0\Rightarrow c=n\pi \] \[\therefore y=\tan x\]


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