JEE Main 2025 — Differential Equations Question with Solution

$\begin{array}{rl}& \text{ If }{\mathrm{e}}^{\int \tan \mathrm{xdx}}={\mathrm{e}}^{\ln (\sec \mathrm{x})}=\sec \mathrm{x}\\ & \therefore \mathrm{y}\cdot \sec \mathrm{x}=\int \left\{\frac{2+\sec \mathrm{x}}{(1+2\sec \mathrm{x}{)}^2}\right\}\sec \mathrm{xdx}\\ & =\int \frac{2\cos \mathrm{x}+1}{(\cos \mathrm{x}+2{)}^2}\mathrm{dx}\text{ Let }\cos \mathrm{x}=\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}\\ & =\int \frac{2\left(\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}\right)+1}{{\left(\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}+2\right)}^2}2\mathrm{dt}\end{array}$ $\begin{aligned} & =\int \frac{2-2 t^2+1+t^2}{\left(1-t^2+2+2 t^2\right)^2} \times 2 \mathrm{dt} \\ & =2 \int \frac{3-t^2}{\left(t^2+3\right)^2} \mathrm{dt} \end{aligned}$ Let $\mathrm{t}+\frac{3}{\mathrm{t}}=\mathrm{u}$ $\begin{aligned} & \left(1-\frac{3}{\mathrm{t}^2}\right) \mathrm{dt}=\mathrm{du} \\ & =-2 \int \frac{\mathrm{du}}{\mathrm{u}^2} \end{aligned}$ $\begin{aligned} & y \cdot(\sec x)=\frac{2}{u}+c \\ & y \cdot \sec x=\frac{2}{t+\frac{3}{t}}+c...(I) \end{aligned}$ $\text{ At }\mathrm{x}=\frac{\pi }{3},\mathrm{t}=\tan \frac{\mathrm{x}}{2}=\frac{1}{\sqrt{3}}$ $2\cdot \frac{\sqrt{3}}{10}=\frac{2}{\frac{1}{\sqrt{3}}+3\sqrt{3}}+\mathrm{c}$ $\begin{array}{rl}& \text{ 2. }\frac{\sqrt{3}}{10}=\frac{2\sqrt{3}}{10}+\mathrm{c}\Rightarrow \mathrm{C}=0\\ & \text{ At }\mathrm{x}=\frac{\pi }{4},\mathrm{t}=\tan \frac{\mathrm{x}}{2}=\sqrt{2}-1\\ & \therefore \mathrm{y}\cdot \sqrt{2}=\frac{2}{\sqrt{2}-1+\frac{3}{\sqrt{2}-1}}\\ & \mathrm{y}\cdot \sqrt{2}=\frac{2(\sqrt{2}-1)}{6-2\sqrt{2}}\\ & \mathrm{y}=\frac{\sqrt{2}(\sqrt{2}-1)}{2(3-\sqrt{2})}=\frac{1}{\sqrt{2}}\times \frac{2\sqrt{2}-1}{7}\\ & =\frac{4-\sqrt{2}}{14}\end{array}$