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Linear equation in log of cos: (log cos x)^2 cos x dy + (sin x - 3 y sin x log cos x) dx = 0

The same question, asked 2 times: 8 Apr 2023*, 29 Jan 2025. Every wording below, oldest first.

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  1. Question 1

    8 Apr 2023

    177
    Let the solution curve x=x(y),0<y<π2x=x(y),0 <y<\frac{\pi}{2}, of the differential equation (log⁡e(cos⁡y))2cos⁡ydx−(1+3x\left( \log_{e}(\cos y) \right)^{2}\cos ydx - (1 + 3x  log⁡e(cos⁡y))sin⁡ydy=0\left. \ \log_{e}(\cos y) \right)\sin ydy = 0 satisfy x(π3)=12log⁡e2x\left( \frac{\pi}{3} \right)=\frac{1}{2\log_{e}2}. If x(π6)=1log⁡em−log⁡enx\left( \frac{\pi}{6} \right)=\frac{1}{\log_{e}m-\log_{e}n}, where mm and nn are co-prime, then mnmn is equal to

  2. Question 2

    29 Jan 2025

    60
    Let y=y(x)y = y(x) be the solution of the differential equation cos⁡x(log⁡e(cos⁡x))2dy+(sin⁡x−3ysinxlog⁡e(cos⁡x))dx=0\cos x\left( \log_{e}(\cos x) \right)^{2}dy +\left( \sin x - 3ysinx\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 :

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