JEE Main 2025 — 2 April, Evening Shift — Question 45

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

I.F. $=e^{\int 2 \sec ^{2} x d x}$

I.F. $=e^{2 \tan x}$

$y \cdot e^{2 \tan x}=\int e^{2 \tan x}(2+3 \tan x) \sec ^{2} x d x$

Put $\tan x=u$

$$\begin{aligned} & \sec ^{2} x d x=d u \\& y \cdot e^{2 u}=\int e^{2 u}(2+3 u) d u \\& y \cdot e^{2 u} \Rightarrow \frac{2 e^{2 u}}{2}+3 \int e^{2 u} \cdot u d u \\& y \cdot e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\int \frac{e^{2 u}}{2}\right] \\& y e^{2 u}=e^{2 u}+3\left[\frac{u e^{2 u}}{2}-\frac{e^{2 u}}{4}\right]+C \\& y e^{2 \tan x}=e^{2 \tan x}+3\left[\frac{\tan x e^{2 \tan x}}{2}-\frac{e^{2 \tan x}}{4}\right]+C \end{aligned}$$ $$F(0)=\frac{5}{4}$$ $$\frac{5}{4}=1-\frac{3}{4}+C$$ $$\frac{5}{4}-\frac{1}{4}=C$$ $$1=C$$ $$y=1+3\left(\frac{\tan x}{2}-\frac{1}{4}\right)+1 \cdot e^{-2 \tan x}$$ $$y\left(\frac{\pi}{4}\right)=1+3\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{e^{2}}$$ $$y\left(\frac{\pi}{4}\right)=\frac{7}{4}+\frac{1}{e^{2}}$$ $$12\left(y\left(\frac{x}{4}\right)-\frac{1}{e^{2}}\right)=12\left(\frac{7}{4}+\frac{1}{e^{2}}-\frac{1}{e^{2}}\right)=21$$