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MCQMediumJEE Main 2026 · 28 January, Shift 1Linear Differential Equations

Mathematics Question from JEE Main 2026 · 28 January, Shift 1

Let y=y(x)y = y(x) be the solution of the differential equation xdydx−sin⁡2y=x3(2−x3)cos⁡2y,  x≠0.x\frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y,\; x \ne 0. If y(2)=0y(2) = 0, then tan⁡(y(1))\tan(y(1)) is equal to:

  • A

    34\dfrac{3}{4}

  • B

    −34-\dfrac{3}{4}

  • C

    74\dfrac{7}{4}

  • D

    −74-\dfrac{7}{4}

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