JEE Main 2026 — 28 January, Morning Shift — Question 19

$x \frac{d y}{d x}-\sin 2 y=x^{3}\left(2-x^{3}\right) \cos ^{2} y$

$\sec ^{2} y \frac{d y}{d x}-2 \tan y \cdot \frac{1}{x}=x^{2}\left(2-x^{3}\right)$

$\tan y=t \Rightarrow \sec ^{2} y \frac{d y}{d x}=\frac{d t}{d x}$

$\frac{\mathrm{dt}}{\mathrm{dx}}-\frac{2 \mathrm{t}}{\mathrm{x}}=\mathrm{x}^{2}\left(2-\mathrm{x}^{3}\right)$ (LDE)

I.F. $=\mathrm{e}^{\int-\frac{2}{\mathrm{x}} \mathrm{dx}}=\mathrm{e}^{-2 \ln \mathrm{x}}=\frac{1}{\mathrm{x}^{2}}$

$\therefore \frac{\mathrm{t}}{\mathrm{x}^{2}}=\int \frac{1}{\mathrm{x}^{2}} \mathrm{x}^{2}\left(2-\mathrm{x}^{3}\right) \mathrm{dx}+\mathrm{C}$

$\frac{\tan \mathrm{y}}{\mathrm{x}^{2}}=2 \mathrm{x}-\frac{\mathrm{x}^{4}}{4}+\mathrm{C}$

$y(2)=0 \Rightarrow 0=4-4+C \Rightarrow C=0$

$\tan \mathrm{y}=2 \mathrm{x}^{3}-\frac{1}{4} \mathrm{x}^{6}$

$x=1 \Rightarrow \tan y=2-\frac{1}{4}=\frac{7}{4}$