JEE Main 2024 — 31 January, Shift 2 — Question 27

$\sec ^{2} x \frac{d x}{d y}+e^{2 y} \tan ^{2} x+\tan x=0$ (Put $\left.\tan x=t \Rightarrow \sec ^{2} x \frac{d x}{d y}=\frac{d t}{d y}\right)$

$\frac{d t}{d y}+e^{2 y} \times t^{2}+t=0$

$\frac{d t}{d y}+t=-t^{2} \cdot e^{2 y}$

$\frac{1}{\mathrm{t}^{2}} \frac{\mathrm{dt}}{\mathrm{dy}}+\frac{1}{\mathrm{t}}=-\mathrm{e}^{2 \mathrm{y}}$ $\left(\right.$

Put $\left.\frac{1}{\mathrm{t}}=\mathrm{u} \frac{-1}{\mathrm{t}^{2}} \frac{\mathrm{dt}}{\mathrm{dy}}=\frac{\mathrm{du}}{\mathrm{dy}}\right)$

$\frac{-d u}{d y}+u=-e^{2 y}$

$\frac{d u}{d y}-u=e^{2 y}$ I.F. $=\mathrm{e}^{-\int \mathrm{dy}}=\mathrm{e}^{-\mathrm{y}}$

$u e^{-y}=\int e^{-y} \times e^{2 y} d y$

$\frac{1}{\tan \mathrm{x}} \times \mathrm{e}^{-\mathrm{y}}=\mathrm{e}^{\mathrm{y}}+\mathrm{c}$

$x=\frac{\pi}{4}, y=0, c=0$

$x=\frac{\pi}{6}, \quad y=\alpha$

$\sqrt{3} \mathrm{e}^{-\alpha}=\mathrm{e}^{\alpha}+0$

$\mathrm{e}^{2 \alpha}=\sqrt{3}$

$e^{8 \alpha}=9$