QJEEM24Ma7158: JEE Main 2024 Mathematics - Class 12 JEEM24Ma7158 Solution: Let be the solution of the differential equation . If , then is …

$\begin{aligned}& \sec ^2 x \frac{d x}{d y}+e^{2 y} \tan ^2 x+\tan x=0 \\& \left(\text { Put } \tan x=t \Rightarrow \sec ^2 x \frac{d x}{d y}=\frac{d t}{d y}\right) \\& \frac{d t}{d y}+e^{2 y} \times t^2+t=0 \\& \frac{d t}{d y}+t=-t^2 \cdot e^{2 y} \\& \frac{1}{t^2} \frac{d t}{d y}+\frac{1}{t}=-e^{2 y} \\& \left(\text { Put } \frac{1}{t}=u \frac{-1}{t^2} \frac{d t}{d y}=\frac{d u}{d y}\right) \\& \frac{-d u}{d y}+u=-e^{2 y} \\& \frac{d u}{d y}-u=e^{2 y} \\& \text { I.F. }=e^{-\int d y}=e^{-y} \\& u e^{-y}=\int e^{-y} \times e^{2 y} d y \\& \frac{1}{\tan x} \times e^{-y}=e^y+c \\& x=\frac{\pi}{4}, y=0, c=0 \\& \begin{aligned}& \mathrm{x}=\frac{\pi}{6}, \quad \mathrm{y}=\alpha \\& \sqrt{3} \mathrm{e}^{-\alpha}=\mathrm{e}^\alpha+0 \\& \mathrm{e}^{2 \alpha}=\sqrt{3} \\& \mathrm{e}^{8 \alpha}=9\end{aligned}\end{aligned}$