JEE Main 2025 — 3 April, Evening Shift — Question 39

$\frac{d y}{d x}+3\left(\tan ^{2} x\right) y+3 y=\sec ^{2} x$

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

I.F $=e^{\int 3 \sec ^{2} x d x}$

$$=e^{3 \tan x}$$

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

$y \cdot e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+c$

Also $f(0)=\frac{1}{3}+e^{3}$

$\Rightarrow\left(\frac{1}{3}+e^{3}\right)=\frac{1}{3}+c$

$\Rightarrow c=e^{3}$

$\therefore y . e^{3 \tan x}=\frac{e^{3 \tan x}}{3}+e^{3}$

Put $x=\frac{\pi}{4}$

$y e^{3}=\frac{e^{3}}{3}+e^{3} \Rightarrow y=\frac{4}{3}$