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JEE Main & Advanced Mathematics Differential Equations Question Bank Formation of differential equations

  • question_answer
    Differential equation of \[y=\sec ({{\tan }^{-1}}x)\] is   [UPSEAT 2002]

    A)                 \[(1+{{x}^{2}})\frac{dy}{dx}=y+x\]           

    B)                 \[(1+{{x}^{2}})\frac{dy}{dx}=y-x\]

    C)                 \[(1+{{x}^{2}})\frac{dy}{dx}=xy\]               

    D)                 \[(1+{{x}^{2}})\frac{dy}{dx}=\frac{x}{y}\]

    Correct Answer: C

    Solution :

                       \[y=\sec ({{\tan }^{-1}}x)\]         \[\frac{dy}{dx}=\sec ({{\tan }^{-1}}x)\tan ({{\tan }^{-1}}x)\,.\,\frac{1}{1+{{x}^{2}}}\]\[=\frac{xy}{1+{{x}^{2}}}\]                                 Þ \[(1+{{x}^{2}})\frac{dy}{dx}=xy\].


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