JEE Main 2024 — 8 April, Shift 2 — Question 10

$\quad \sec ^{2} y \frac{d y}{d x}+2 x \sin y \sec y=x^{3} \cos y \sec y$

$\sec ^{2} y \frac{d y}{d x}+2 x \tan y=x^{3}$

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

$\frac{d t}{d x}+2 x t=x^{3}$

If $=e^{\int 2 x d x}=e^{x^{2}}$

$t e^{x^{2}}=\int x^{3} \cdot e^{x^{2}} d x+c$

$\mathrm{x}^{2}=\mathrm{Z} \Rightarrow \mathrm{t} \cdot \mathrm{e}^{\mathrm{Z}}=\frac{1}{2} \int \mathrm{e}^{\mathrm{Z}} \cdot \mathrm{ZdZ}=\frac{1}{2}\left[\mathrm{e}^{\mathrm{Z}} \cdot \mathrm{Z}-\mathrm{e}^{\mathrm{Z}}\right]+\mathrm{c}$

$2 \tan \mathrm{y}=\left(\mathrm{x}^{2}-1\right)+2 \mathrm{ce}^{-\mathrm{x}^{2}}$

$\mathrm{y}(1)=0 \Rightarrow \mathrm{c}=0 \Rightarrow \mathrm{y}(\sqrt{3})=\frac{\pi}{4}$