Let $y=y(x)$ be the solution of the differential equation $$\frac{d y}{d x}=\frac{(\tan x)+y}{\sin x(\sec x-\sin x \tan x)}, x \in\left(0, \frac{\pi}{2}\right)$$ satisfying the condition $y\left(\frac{\pi}{4}\right)=2$. Then, $y\left(\frac{\pi}{3}\right)$ is

<p>$$\begin{aligned} & \frac{d y}{d x}=\frac{\sin x+y \cos x}{\sin x \cdot \cos x\left(\frac{1}{\cos x}-\sin x \cdot \frac{\sin x}{\cos x}\right)} \\ & =\frac{\sin x+y \cos x}{\sin x\left(1-\sin ^2 x\right)} \\ & \frac{d y}{d x}=\sec ^2 x+y \cdot 2(\operatorname{cosec} 2 x) \\ & \frac{d y}{d x}-2 \operatorname{cosec}(2 x) \cdot y=\sec ^2 x \\ & \frac{d y}{d x}+p \cdot y=Q \end{aligned}$$</p> <p>$$\text { I.F. }=\mathrm{e}^{\int p \mathrm{dx}}=\mathrm{e}^{\int-2 \operatorname{cosec}(2 \mathrm{x}) \mathrm{dx}}$$</p> <p>Let $2 \mathrm{x}=\mathrm{t}$</p> <p>$2 \frac{\mathrm{dx}}{\mathrm{dt}}=1$</p> <p>$\mathrm{dx}=\frac{\mathrm{dt}}{2}$</p> <p>$=\mathrm{e}^{-\int \operatorname{cosec}(\mathrm{t}) \mathrm{dt}}$</p> <p>$=\mathrm{e}^{-\ln \left|\tan \frac{t}{2}\right|}$</p> <p>$=\mathrm{e}^{-\ln |\tan x|}=\frac{1}{|\tan x|}$</p> <p>$$\begin{aligned} & y(\text { IF })=\int Q(I F) d x+c \\ & \Rightarrow y \frac{1}{|\tan x|}=\int \sec ^2 x \cdot \frac{1}{|\tan x|}+c \\ & y \cdot \frac{1}{|\tan x|}=\int \frac{d t}{|t|}+c \quad \text { for } \tan x=t \\ & y \cdot \frac{1}{|\tan x|}=\ln |t|+c \\ & y=|\tan x|(\ln |\tan x|+c) \\ & \text { Put } x=\frac{\pi}{4}, y=2 \\ & 2=\ln 1+c \Rightarrow c=2 \\ & y=|\tan x|(\ln |\tan x|+2) \\ & y\left(\frac{\pi}{3}\right)=\sqrt{3}(\ln \sqrt{3}+2) \end{aligned}$$</p>