Let y=y(x) satisfy the equation frac{dy}{dx}- | vedclass

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(B) First,calculate the determinant $|A|$:$|A| = y(-1 \cdot \frac{1}{x} - 0) - \sin x(0 \cdot \frac{1}{x} - 2) + 1(0 \cdot 0 - 2(-1))$$|A| = -\frac{y}

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$\frac{\pi}{2}+\frac{4}{\pi}$

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(B) First,calculate the determinant $|A|$:$|A| = y(-1 \cdot \frac{1}{x} - 0) - \sin x(0 \cdot \frac{1}{x} - 2) + 1(0 \cdot 0 - 2(-1))$$|A| = -\frac{y}{x} + 2 \sin x + 2$Given $\frac{dy}{dx} - |A| = 0$,we have $\frac{dy}{dx} = -\frac{y}{x} + 2 \sin x + 2$,which rearranges to $\frac{dy}{dx} + \frac{y}{x} = 2 \sin x + 2$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{1}{x}$ and $Q = 2 \sin x + 2$.The integrating factor $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.The general solution is $y \cdot I.F. = \int Q \cdot I.F. dx + C$.$yx = \int x(2 \sin x + 2) dx = 2 \int x \sin x dx + \int 2x dx$.Using integration by parts for $\int x \sin x dx = -x \cos x + \sin x$,we get:$yx = 2(-x \cos x + \sin x) + x^2 + C = x^2 - 2x \cos x + 2 \sin x + C$.Given $y(\pi) = \pi + 2$,substitute $x = \pi$:$(\pi + 2)\pi = \pi^2 - 2\pi \cos(\pi) + 2 \sin(\pi) + C$$\pi^2 + 2\pi = \pi^2 - 2\pi(-1) + 0 + C \Rightarrow \pi^2 + 2\pi = \pi^2 + 2\pi + C \Rightarrow C = 0$.Thus,$yx = x^2 - 2x \cos x + 2 \sin x$.For $x = \frac{\pi}{2}$:$y(\frac{\pi}{2}) \cdot \frac{\pi}{2} = (\frac{\pi}{2})^2 - 2(\frac{\pi}{2}) \cos(\frac{\pi}{2}) + 2 \sin(\frac{\pi}{2})$$y(\frac{\pi}{2}) \cdot \frac{\pi}{2} = \frac{\pi^2}{4} - 0 + 2(1) = \frac{\pi^2}{4} + 2$.$y(\frac{\pi}{2}) = \frac{\pi}{2} + \frac{4}{\pi}$.

Let $y=y(x)$ satisfy the equation $\frac{dy}{dx}-|A|=0$,for all $x>0$,where $A=\begin{bmatrix} y & \sin x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x} \end{bmatrix}$. If $y(\pi)=\pi+2$,then the value of $y\left(\frac{\pi}{2}\right)$ is:

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