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

  • question_answer
    The equation of the curve satisfying \[xdy-ydx=\sqrt{{{x}^{2}}-{{y}^{2}}}\] and \[y(1)=0\] is:

    A) \[y={{x}^{2}}\log (\sin \,x)\]

    B) \[y=x\sin (log\,x)\]

    C) \[{{y}^{2}}=x{{(x-1)}^{2}}\]

    D) \[y=2{{x}^{2}}(x-1)\]

    Correct Answer: B

    Solution :

    [b] The given equation can be rewritten as \[{{x}^{2}}\left[ \frac{xdy-ydx}{{{x}^{2}}} \right]=x\sqrt{1-\frac{{{y}^{2}}}{{{x}^{2}}}}\] \[\Rightarrow {{x}^{2}}\frac{d}{dx}\left( \frac{y}{x} \right)=x\sqrt{1-\frac{{{y}^{2}}}{{{x}^{2}}}}\] Put \[\frac{y}{x}=z,\] we get \[{{\sin }^{-1}}(z)=logx+c\Rightarrow si{{n}^{-1}}\left( \frac{y}{x} \right)=\log x+c\] Apply the boundary value y(1) = 0 \[\Rightarrow {{\sin }^{-1}}\left( \frac{0}{1} \right)=\log 1+c\Rightarrow c=0\] \[\therefore \,\,\,\,{{\sin }^{-1}}\left( \frac{y}{x} \right)=\log x+0\Rightarrow y=x\,\,\sin \,(log\,x)\]

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