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

Question

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

Accepted Answer

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

${\int }_{x}^{0}g(t)\, dt=x-{\int }_{x}^{0}	an g(t)\, dt$

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:

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

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):

$={e}^{\int 	an x\, dx}={e}^{-\ln \cos x}=\cos x$

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

Solutions · 1