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MCQMediumJEE Main 2025 · 29 January, Shift 1Linear Differential Equations

Mathematics Question from JEE Main 2025 · 29 January, Shift 1

Let y=y(x)y = y(x) be the solution of the differential equation cos⁡x(log⁡e(cos⁡x))2 dy+(sin⁡x−3ysin⁡xlog⁡e(cos⁡x))dx=0\cos x \left(\log_e (\cos x)\right)^2 \, dy + \left(\sin x - 3 y \sin x \log_e (\cos x)\right) dx = 0, x∈(0,π2)x \in \left( 0, \frac{\pi}{2} \right). If y(π4)=−1log⁡e2y\left( \frac{\pi}{4} \right) = \frac{-1}{\log_e 2}, then y(π6)y\left( \frac{\pi}{6} \right) is equal to:

  • A

    1log⁡e(4)−log⁡e(3)\frac{1}{\log_e (4) - \log_e (3)}

  • B

    2log⁡e(3)−log⁡e(4)\frac{2}{\log_e (3) - \log_e (4)}

  • C

    −1log⁡e(4)- \frac{1}{\log_e (4)}

  • D

    1log⁡e(3)−log⁡e(4)\frac{1}{\log_e (3) - \log_e (4)}

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