CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    An integrating factor of the differential equation \[x\frac{dy}{dx}+y\log x=x{{e}^{x}}{{x}^{-\frac{1}{2}\log x}}\]\[(x>0)\]is:

    A)  \[{{x}^{\log x}}\]                            

    B)  \[{{(\sqrt{x})}^{\log x}}\]

    C)  \[{{(\sqrt{e})}^{{{(\log x)}^{2}}}}\]         

    D)         \[{{e}^{{{x}^{2}}}}\]

    E)  \[\frac{{{x}^{2}}}{2}\]

    Correct Answer: C

    Solution :

    Given differential equation is \[x\frac{dy}{dx}+y\log x=x{{e}^{x}}{{x}^{-1/2\log x}}\] \[\frac{dy}{dx}+y\frac{1}{x}\log x={{e}^{x}}{{x}^{-1/2\log x}}\] Here, \[P=\frac{1}{x}\log x\]and\[Q={{e}^{x}}{{x}^{-1/2\log x}}\] \[\therefore \]  \[IF={{e}^{\int{\frac{1}{x}\log x\,dx}}}\]                 \[={{e}^{\frac{{{(\log x)}^{2}}}{2}}}\]                 \[={{(\sqrt{e})}^{{{(\log x)}^{2}}}}\]


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