If $$y = y(x)$$ is the solution of the differential equation $$\frac{dy}{dx} + (\tan x)y = \sin x$$, $$0 \leq x \leq \frac{\pi}{3}$$, with $$y(0) = 0$$, then $$y\left(\frac{\pi}{4}\right)$$ is equal to:

The differential equation is $$\frac{dy}{dx} + (\tan x) y = \sin x$$ with $$y(0) = 0$$. This is a first-order linear ODE.

The integrating factor is $$\mu = e^{\int \tan x \, dx} = e^{\ln |\sec x|} = \sec x$$.

Multiplying both sides by $$\sec x$$: $$\frac{d}{dx}(y \sec x) = \sin x \cdot \sec x = \tan x$$.

Integrating: $$y \sec x = \int \tan x \, dx = -\ln|\cos x| + C = \ln|\sec x| + C$$.

Applying the initial condition $$y(0) = 0$$: $$0 \cdot 1 = \ln 1 + C$$, so $$C = 0$$.

Therefore, $$y \sec x = \ln(\sec x)$$, giving $$y = \cos x \cdot \ln(\sec x)$$.

At $$x = \frac{\pi}{4}$$: $$y\left(\frac{\pi}{4}\right) = \cos\frac{\pi}{4} \cdot \ln\left(\sec\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \cdot \ln(\sqrt{2}) = \frac{1}{\sqrt{2}} \cdot \frac{1}{2}\ln 2 = \frac{\ln 2}{2\sqrt{2}}$$.