JEE Main 2025 (Online) 3rd April Evening ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+3\left(\tan^{2}x\right)y+3y=\sec^{2}x$, $y(0)=\frac{1}{3}+e^{3}$. Then $y\left(\frac{\pi}{4}\right)$ is equal to
Differential Equation
Solution
Concept. $3\tan^2x+3=3\sec^2x$, so the IF is $e^{3\tan x}$. Formulas used. (1) $1+\tan^2x=\sec^2x$(2) $\int\sec^2x\,e^{3\tan x}\,dx=\frac13e^{3\tan x}$Step 1 - Linear form. By (1), $y'+3\sec^2x\cdot y=\sec^2x$, IF $=e^{3\tan x}$. Step 2 - Integrate. By (2), $$y\,e^{3\tan x}=\frac13e^{3\tan x}+C\ \Longrightarrow\ y=\frac13+Ce^{-3\tan x}.$$ Step 3 - Constant. $y(0)=\frac13+C=\frac13+e^3$, so $C=e^3$. Step 4 - At $x=\frac\pi4$. $y=\frac13+e^3e^{-3}=\frac43$. Answer: (D)