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

  • question_answer
    If \[y=a{{x}^{n+1}}+b{{x}^{-n}},\] then \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}\] equals to           [RPET 2001]

    A)                 \[n(n-1)y\]             

    B)                 \[n(n+1)y\]

    C)                 ny           

    D)                 n2y

    Correct Answer: B

    Solution :

                       \[y=a{{x}^{n+1}}+b{{x}^{-n}}\]         Differentiate with respect to x , \[\frac{dy}{dx}=a(n+1){{x}^{n}}-bn{{x}^{-n-1}}\]         Again differentiate, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=a\,n\,(n+1){{x}^{n-1}}+b\,n\,(n+1){{x}^{-n-2}}\]         Þ \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=a\,n\,(n+1){{x}^{n+1}}+b\,n\,(n+1){{x}^{-n}}\]                 Þ \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=n(n+1)y\].


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