Let g be a differentiable function such that _0^x g(t) d t=x-_0^x tg(t) d t, x 0 and let y…

Question

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 $\frac{d y}{d x}-y 	an x=2(x+1) \sec x g(x), x \in\left[0, \frac{\pi}{2}ight)$. If $y(0)=0$, then $y\left(\frac{\pi}{3}ight)$ is equal to

Accepted Answer

To solve the given problem, let's start by considering the equation:

$ \int_0^x g(t) \, dt = x - \int_0^x 	an g(t) \, dt $

Differentiate both sides with respect to $ x $:

$ g(x) = 1 - xg(x) $

Rearranging gives:

$ g(x)(1 + x) = 1 \implies g(x) = \frac{1}{1 + x} $

Now, consider the differential equation:

$ \frac{dy}{dx} - y 	an x = 2(x+1) \sec x g(x) $

Substitute $ g(x) = \frac{1}{1 + x} $:

$ \frac{dy}{dx} - y 	an x = 2(x+1) \sec x \cdot \frac{1}{1 + x} $

This simplifies to:

$ \frac{dy}{dx} - y 	an x = 2 \sec x $

To solve this, we use an Integrating Factor (I.F):

$ 	ext{I.F} = e^{\int 	an x \, dx} = e^{-\ln \cos x} = \cos x $

Multiply the entire differential equation by the Integrating Factor:

$ \cos x \cdot \frac{dy}{dx} - y \cdot \sin x = 2 $

This implies:

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

Integrate both sides with respect to $ x $:

$ y \cos x = \int 2 \, dx = 2x + c $

Given the initial condition $ y(0) = 0 $:

$ 0 \cdot 1 = 2 \cdot 0 + c \implies c = 0 $

Thus:

$ y \cos x = 2x $

Evaluate at $ x = \frac{\pi}{3} $:

$ y \cdot \frac{1}{2} = 2 \cdot \frac{\pi}{3} $

Solving for $ y $:

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

Therefore, $ y\left(\frac{\pi}{3}ight) = \frac{4 \pi}{3} $.

Solutions · 1