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JEE Main & Advanced Mathematics Differential Equations Question Bank Critical Thinking

  • question_answer
    The solution of the equation \[\frac{{{x}^{2}}{{d}^{2}}y}{d{{x}^{2}}}=\ln x,\] when   \[x=1\], \[y=0\] and \[\frac{dy}{dx}=-1\] is [Orissa JEE 2003]

    A) \[\frac{1}{2}{{(\ln x)}^{2}}+\ln x\]                                      

    B) \[\frac{1}{2}{{(\ln x)}^{2}}-\ln x\]

    C) \[-\frac{1}{2}{{(\ln x)}^{2}}+\ln x\]                                     

    D) \[-\frac{1}{2}{{(\ln x)}^{2}}-\ln x\]

    Correct Answer: D

    Solution :

    • \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{\log x}{{{x}^{2}}}\Rightarrow \frac{dy}{dx}=\frac{-(\log x+1)}{x}+c\]        
    • At \[\frac{dy}{dx}=-1\], \[x=1,\,\,y=0,\] \ \[c=0\]                   
    • Þ \[y=-\int{\frac{\log x+1}{x}dx=-\frac{1}{2}{{(\log x)}^{2}}-\log x}\].


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