Let y = y ( x ) satisfies the equation (dy)/(dx) - |A| = 0, for all...

We have,

A = \left[\begin{cases}y & \sin x & 1 \\ 0 & - 1 & 1 \\ 2 & 0 & \dfrac{1}{x}\end{cases}\right]

$⇒ \left|A\right| = - \dfrac{y}{x} + 2\sin x + 2$

Now,

$\dfrac{dy}{dx} = \left|A\right|$

$⇒ \dfrac{dy}{dx} = - \dfrac{y}{x} + 2\sin x + 2$

$⇒ \dfrac{dy}{dx} + \dfrac{y}{x} = 2\sin x + 2$

$\text{I.F.} = {e}^{∫ \dfrac{1}{x}dx} = x$

Solution is

$y \times \left(I . F .\right) = ∫ \left(I . F .\right) \times \left(2\sin x + 2\right)dx$

$⇒ yx = ∫ x ( 2\sin x + 2 ) dx$

$⇒ yx = 2 ∫ x\left(\sin x\right)dx + 2 ∫ xdx$

$⇒ xy = {x}^{2} - 2xcosx + 2\sin x + c \ldots \left(i\right)$

Now $x = \pi, y = \pi + 2$, hence

$⇒ \pi\left(\pi + 2\right) = {\pi}^{2} - 2\pi\left(- 1\right) + 0 + c$

$⇒ c = 0$

Now \left(i\right)$becomes</p><p>$xy = {x}^{2} - 2xcosx + 2\sin x$</p><p>Put$x = \dfrac{\pi}{2}$, then</p><p>\left(\dfrac{\pi}{2}\right)y = {\left(\dfrac{\pi}{2}\right)}^{2} - 2 \cdot \dfrac{\pi}{2}\cos \left(\dfrac{\pi}{2}\right) + 2\sin \left(\dfrac{\pi}{2}\right)$

$⇒ \left(\dfrac{\pi}{2}\right)y = \dfrac{{\pi}^{2}}{4} + 2$

$⇒ y = \dfrac{\pi}{2} + \dfrac{4}{\pi}$