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JEE Main & Advanced Mathematics Differential Equations Question Bank Critical Thinking

  • question_answer
    The differential equation of the family of curves for which the length of the normal is equal to a constant k, is given by [Pb. CET 2004]

    A) \[{{y}^{2}}\frac{dy}{dx}={{k}^{2}}-{{y}^{2}}\]        

    B) \[{{\left( y\frac{dy}{dx} \right)}^{2}}={{k}^{2}}-{{y}^{2}}\]

    C) \[y{{\left( \frac{dy}{dx} \right)}^{2}}={{k}^{2}}+{{y}^{2}}\]

    D) \[{{\left( y\frac{dy}{dx} \right)}^{2}}={{k}^{2}}+{{y}^{2}}\]

    Correct Answer: B

    Solution :

    • The length of normal is given by, \[y\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}\]                   
    • \[\therefore \] \[y\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}=k\] Þ \[{{y}^{2}}\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]={{k}^{2}}\]           
    • Þ \[{{y}^{2}}+{{y}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}={{k}^{2}}\] Þ \[{{y}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}={{k}^{2}}-{{y}^{2}}\].


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