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

  • question_answer
    If \[{{x}^{2}}+{{y}^{2}}=1\] then \[\left( {y}'=\frac{dy}{dx},{y}''=\frac{{{d}^{2}}y}{d{{x}^{2}}} \right)\] [IIT Screening 2000]

    A)                 \[y{y}''-2{{({y}')}^{2}}+1=0\]      

    B)                 \[y{y}''+{{({y}')}^{2}}+1=0\]

    C)                 \[y{y}''-{{({y}')}^{2}}-1=0\]           

    D)                 \[y{y}''+2{{({y}')}^{2}}+1=0\]

    Correct Answer: B

    Solution :

                       Differentiating w.r.t. \[x,\] \[\frac{dy}{dx}=1+x+y+xy\] or \[x+y{y}'=0\]                 Differentiating again w.r.t. \[x,\] \[1+{{{y}'}^{2}}+y{y}''=0\].


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