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MathsJEE Main 28 Jan 2026, Shift 1·Q20·MCQ·Moderate

Differential Equations · Linear differential equations

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,\quad x \ne 0. If y(2)=0y(2) = 0, then tan⁡(y(1))\tan(y(1)) is equal to
  1. (1)34\dfrac{3}{4}
  2. (2)74\dfrac{7}{4}
  3. (3)−74-\dfrac{7}{4}
  4. (4)−34-\dfrac{3}{4}
We corrected a printing error in the original paper

Source printed “If y(2)=xy(2) = x”; corrected to the intended “y(2)=0y(2) = 0” (the source's own solution uses y(2)=0y(2) = 0, and the key (2) follows from it).

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