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(B) Given $\int_0^x g(t) dt = x - \int_0^x tg(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $g(x) = 1 - xg(x)$. Rearr

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$\frac{4 \pi}{3}$

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(B) Given $\int_0^x g(t) dt = x - \int_0^x tg(t) dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $g(x) = 1 - xg(x)$. Rearranging gives $g(x)(1+x) = 1$,so $g(x) = \frac{1}{1+x}$. Substitute $g(x)$ into the differential equation: $\frac{dy}{dx} - y 	an x = 2(x+1) \sec x \cdot \frac{1}{1+x} = 2 \sec x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -	an x$ and $Q = 2 \sec x$. The integrating factor $IF = e^{\int P dx} = e^{-\int 	an x dx} = e^{\ln(\cos x)} = \cos x$. The solution is $y \cdot IF = \int Q \cdot IF dx + C$. $y \cos x = \int 2 \sec x \cdot \cos x dx + C = \int 2 dx + C = 2x + C$. Given $y(0) = 0$,we have $0 \cdot \cos(0) = 2(0) + C$,so $C = 0$. Thus,$y \cos x = 2x$,which means $y = 2x \sec x$. At $x = \frac{\pi}{3}$,$y(\frac{\pi}{3}) = 2 \cdot \frac{\pi}{3} \cdot \sec(\frac{\pi}{3}) = 2 \cdot \frac{\pi}{3} \cdot 2 = \frac{4 \pi}{3}$.

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

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