If $$y = yx$$ is the solution curve of the differential equation $$\frac{dy}{dx} + y\tan x = x\sec x$$, $$0 \leq x \leq \frac{\pi}{3}$$, $$y(0) = 1$$, then $$y(\frac{\pi}{6})$$ is equal to

Given: $$\frac{dy}{dx} + y\tan x = x\sec x$$, $$y(0) = 1$$.

This is a first-order linear ODE. The integrating factor is:

$$ \text{IF} = e^{\int \tan x\, dx} = e^{-\ln|\cos x|} = \sec x $$

Multiplying both sides by $$\sec x$$:

$$ \frac{d}{dx}(y \sec x) = x \sec^2 x $$

Integrating:

$$ y \sec x = \int x \sec^2 x\, dx $$

Using integration by parts with $$u = x$$, $$dv = \sec^2 x\, dx$$:

$$ \int x \sec^2 x\, dx = x \tan x - \int \tan x\, dx = x \tan x + \ln|\cos x| + C $$

So: $$y \sec x = x \tan x + \ln(\cos x) + C$$

Using $$y(0) = 1$$: $$1 \cdot 1 = 0 + 0 + C \Rightarrow C = 1$$

$$ y = x \sin x + \cos x \ln(\cos x) + \cos x $$

At $$x = \frac{\pi}{6}$$:

$$ y\left(\frac{\pi}{6}\right) = \frac{\pi}{6} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \ln\left(\frac{\sqrt{3}}{2}\right) + \frac{\sqrt{3}}{2} $$ $$ = \frac{\pi}{12} + \frac{\sqrt{3}}{2}\left[\ln\left(\frac{\sqrt{3}}{2}\right) + 1\right] $$ $$ = \frac{\pi}{12} + \frac{\sqrt{3}}{2}\ln\left(\frac{e\sqrt{3}}{2}\right) $$ $$ = \frac{\pi}{12} - \frac{\sqrt{3}}{2}\ln\left(\frac{2}{e\sqrt{3}}\right) $$

Therefore, $$y\left(\frac{\pi}{6}\right) = \frac{\pi}{12} - \frac{\sqrt{3}}{2}\log_e\frac{2}{e\sqrt{3}}$$.