Let g be a differentiable function such that $$\int_0^x g(t)\,dt = x - \int_0^x tg(t)\,dt$$, $$x \geq 0$$ and let $$y = y(x)$$ satisfy the differential equation $$\frac{dy}{dx} - y\tan x = 2(x+1)\sec x \cdot g(x)$$, $$x \in \left[0, \frac{\pi}{2}\right)$$. If $$y(0) = 0$$, then $$y\left(\frac{\pi}{3}\right)$$ is equal to

Let $$F(x)=\displaystyle\int_{0}^{x} g(t)\,dt$$. The given relation is
$$F(x)=x-\int_{0}^{x} t\,g(t)\,dt,\qquad x\ge 0.$$

Differentiate both sides with respect to $$x$$:
$$F'(x)=g(x)$$, while by Leibniz rule
$$\frac{d}{dx}\left[x-\int_{0}^{x} t\,g(t)\,dt\right]=1-x\,g(x).$$

Hence
$$g(x)=1-x\,g(x)\;\Longrightarrow\;(1+x)\,g(x)=1\;\Longrightarrow\;g(x)=\frac{1}{1+x}.$$

The differential equation for $$y=y(x)$$ is
$$\frac{dy}{dx}-y\,\tan x=2(x+1)\sec x\cdot g(x).$$

Substituting $$g(x)=\dfrac{1}{1+x}$$ gives
$$\frac{dy}{dx}-y\,\tan x=2\sec x.$$

This is linear of the form $$\dfrac{dy}{dx}+P(x)\,y=Q(x)$$ with $$P(x)=-\tan x$$ and $$Q(x)=2\sec x.$$
The integrating factor (I.F.) is
$$\text{I.F.}=e^{\int P(x)\,dx}=e^{\int -\tan x\,dx}=e^{\ln\cos x}= \cos x.$$

Multiply the equation by the integrating factor:
$$\cos x\,\frac{dy}{dx}-y\sin x=\cos x\cdot 2\sec x=2.$$
The left side is $$\dfrac{d}{dx}\bigl(y\,\cos x\bigr).$$ Therefore
$$\frac{d}{dx}\bigl(y\,\cos x\bigr)=2.$$

Integrate: $$y\,\cos x = 2x + C.$$

Using the condition $$y(0)=0$$:
$$0\cdot 1 = 2\cdot 0 + C \;\Longrightarrow\; C=0.$$

Thus $$y(x)=\frac{2x}{\cos x}=2x\,\sec x.$$

Finally, at $$x=\dfrac{\pi}{3}$$:
$$y\!\left(\frac{\pi}{3}\right)=2\left(\frac{\pi}{3}\right)\sec\!\left(\frac{\pi}{3}\right) = \frac{2\pi}{3}\times 2 = \frac{4\pi}{3}.$$

Hence the required value is $$\dfrac{4\pi}{3}$$, which corresponds to Option B.