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

<p>To solve the given problem, let's start by considering the equation: </p> <p>$ \int_0^x g(t) \, dt = x - \int_0^x \tan g(t) \, dt $</p> <p>Differentiate both sides with respect to $ x $:</p> <p>$ g(x) = 1 - xg(x) $</p> <p>Rearranging gives:</p> <p>$ g(x)(1 + x) = 1 \implies g(x) = \frac{1}{1 + x} $</p> <p>Now, consider the differential equation:</p> <p>$ \frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x) $</p> <p>Substitute $ g(x) = \frac{1}{1 + x} $:</p> <p>$ \frac{dy}{dx} - y \tan x = 2(x+1) \sec x \cdot \frac{1}{1 + x} $</p> <p>This simplifies to:</p> <p>$ \frac{dy}{dx} - y \tan x = 2 \sec x $</p> <p>To solve this, we use an Integrating Factor (I.F):</p> <p>$ \text{I.F} = e^{\int \tan x \, dx} = e^{-\ln \cos x} = \cos x $</p> <p>Multiply the entire differential equation by the Integrating Factor:</p> <p>$ \cos x \cdot \frac{dy}{dx} - y \cdot \sin x = 2 $</p> <p>This implies:</p> <p>$ \frac{d}{dx}(y \cos x) = 2 $</p> <p>Integrate both sides with respect to $ x $:</p> <p>$ y \cos x = \int 2 \, dx = 2x + c $</p> <p>Given the initial condition $ y(0) = 0 $:</p> <p>$ 0 \cdot 1 = 2 \cdot 0 + c \implies c = 0 $</p> <p>Thus:</p> <p>$ y \cos x = 2x $</p> <p>Evaluate at $ x = \frac{\pi}{3} $:</p> <p>$ y \cdot \frac{1}{2} = 2 \cdot \frac{\pi}{3} $</p> <p>Solving for $ y $:</p> <p>$ y = \frac{4 \pi}{3} $ </p> <p>Therefore, $ y\left(\frac{\pi}{3}\right) = \frac{4 \pi}{3} $.</p>