JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx}+(\tan x)y=\sin x$, $0\le x\le\frac{\pi}{3}$, with $y(0)=0$, then $y\left(\frac{\pi}{4}\right)$ equal to :
Differential Equation
Solution
Concept. Linear equation with $P=\tan x$, so IF $=\sec x$. Formulas used. (1) $e^{\int\tan x\,dx}=\sec x$ and $\int\tan x\,dx=\ln\sec x$Step 1 - Integrate. By (1), $(y\sec x)'=\sin x\sec x=\tan x$, so $$y\sec x=\ln\sec x+C.$$ Step 2 - Use $y(0)=0$. $C=0$, so $y=\cos x\,\ln\sec x$. Step 3 - At $x=\frac\pi4$. $$y=\frac{1}{\sqrt2}\ln\sqrt2=\frac{1}{2\sqrt2}\ln2.$$ Answer: (A)