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Differential Equations question

2025 · 2 Apr · Shift 2 · Q49
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  5. /2025 · 2 Apr · Shift 2 · Q49

Differential Equations question

2025 · 2 Apr · Shift 2 · Q49

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution of the differential equation dy dx+2ysec⁡2x=2sec⁡2x+3tan⁡x⋅sec⁡2x\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y \sec ^2 x=2 \sec ^2 x+3 \tan x \cdot \sec ^2 x dxdy​+2ysec2x=2sec2x+3tanx⋅sec2x such that y(0)=54y(0)=\frac{5}{4}y(0)=45​. Then 12(y(π4)−e−2)12\left(y\left(\frac{\pi}{4}\right)-\mathrm{e}^{-2}\right)12(y(4π​)−e−2) is equal to ‾\underline{\hspace{2cm}}​
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