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Differential Equations question

2025 · 24 Jan · Shift 2 · Q48
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  5. /2025 · 24 Jan · Shift 2 · Q48

Differential Equations question

2025 · 24 Jan · Shift 2 · Q48

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation 2cos⁡x dy dx=sin⁡2x−4ysin⁡x,x∈(0,π2)2 \cos x \frac{\mathrm{~d} y}{\mathrm{~d} x}=\sin 2 x-4 y \sin x, x \in\left(0, \frac{\pi}{2}\right)2cosx dx dy​=sin2x−4ysinx,x∈(0,2π​). If y(π3)=0y\left(\frac{\pi}{3}\right)=0y(3π​)=0, then y′(π4)+y(π4)y^{\prime}\left(\frac{\pi}{4}\right)+y\left(\frac{\pi}{4}\right)y′(4π​)+y(4π​) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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