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

  • question_answer
    The solution of \[\frac{dy}{dx}=\frac{{{e}^{x}}({{\sin }^{2}}x+\sin 2x)}{y(2\,\,\log \,\,y+1)}\] is

    A) \[{{y}^{2}}(\log \,y)-{{e}^{x}}{{\sin }^{2}}x+c=0\]

    B) \[{{y}^{2}}(\log \,y)-{{e}^{x}}{{\cos }^{2}}x+c=0\]

    C) \[{{y}^{2}}(\log \,y)+{{e}^{x}}{{\cos }^{2}}x+c=0\]

    D) None of these

    Correct Answer: A

    Solution :

    [a] \[\frac{dy}{dx}=\frac{{{e}^{x}}(si{{n}^{2}}x+sin2x)}{y(2logy+1)}\] \[\Rightarrow \int{(2ylogy+y)dy=\int{{{e}^{x}}(si{{n}^{2}}x+sin2x)dx}}\] On integrating by parts, we get \[{{y}^{2}}(logy)={{e}^{x}}{{\sin }^{2}}x+c.\]

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