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MCQMediumJEE Main 2025 · 3 April, Shift 2Linear Differential Equations

Mathematics Question from JEE Main 2025 · 3 April, Shift 2

Let y=y(x)y = y(x) be the solution of the differential equation dydx+3(tan⁡2x)y+3y=sec⁡2x\frac{dy}{dx} + 3(\tan^2 x) y + 3y = \sec^2 x, with y(0)=13+e3y(0) = \frac{1}{3} + e^3. Then y(π4)y\left(\frac{\pi}{4}\right) is equal to

  • A

    23\frac{2}{3}

  • B

    43\frac{4}{3}

  • C

    43+e3\frac{4}{3} + e^3

  • D

    23+e3\frac{2}{3} + e^3

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