QJEEM24Maf7b9: JEE Main 2024 Mathematics - Class 12 JEEM24Maf7b9 Solution: Let be the solution of the differential equation satisfying the …

$\begin{aligned}& \frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)} \\& =\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^2 x\right)} \\& \frac{d y}{d x}=\sec ^2 x+y \cdot 2(\operatorname{cosec} 2 x) \\& \frac{d y}{d x}-2 \operatorname{cosec}(2 x) \cdot y=\sec ^2 x \\& \frac{d y}{d x}+p \cdot y=Q\end{aligned}$

$\text { I.F. }=\mathrm{e}^{\int p \mathrm{dx}}=\mathrm{e}^{\int-2 \operatorname{cosec}(2 \mathrm{x}) \mathrm{dx}}$

Let $2 \mathrm{x}=\mathrm{t}$

$2 \frac{\mathrm{dx}}{\mathrm{dt}}=1$

$\mathrm{dx}=\frac{\mathrm{dt}}{2}$

$=\mathrm{e}^{-\int \operatorname{cosec}(\mathrm{t}) \mathrm{dt}}$

$=\mathrm{e}^{-\ln \left|\tan \frac{t}{2}\right|}$

$=\mathrm{e}^{-\ln |\tan x|}=\frac{1}{|\tan x|}$

$\begin{aligned}& y(\text { IF })=\int Q(I F) d x+c \\& \Rightarrow y \frac{1}{|\tan x|}=\int \sec ^2 x \cdot \frac{1}{|\tan x|}+c \\& y \cdot \frac{1}{|\tan x|}=\int \frac{d t}{|t|}+c \quad \text { for } \tan x=t \\& y \cdot \frac{1}{|\tan x|}=\ln |t|+c \\& y=|\tan x|(\ln |\tan x|+c) \\& \text { Put } x=\frac{\pi}{4}, y=2 \\& 2=\ln 1+c \Rightarrow c=2 \\& y=|\tan x|(\ln |\tan x|+2) \\& y\left(\frac{\pi}{3}\right)=\sqrt{3}(\ln \sqrt{3}+2)\end{aligned}$