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(B) Given the differential equation: $x dy = (y + x^3 \cos x) dx$Divide by $x^2$ (assuming $x 
eq 0$): $\frac{x dy - y dx}{x^2} = x \cos x dx$This is

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

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(B) Given the differential equation: $x dy = (y + x^3 \cos x) dx$Divide by $x^2$ (assuming $x 
eq 0$): $\frac{x dy - y dx}{x^2} = x \cos x dx$This is the derivative of the quotient: $d(\frac{y}{x}) = x \cos x dx$Integrating both sides: $\int d(\frac{y}{x}) = \int x \cos x dx$Using integration by parts for the right side: $\frac{y}{x} = x \sin x - \int \sin x dx$$\frac{y}{x} = x \sin x + \cos x + C$Given $y(\pi) = 0$,substitute $x = \pi$ and $y = 0$: $0 = \pi \sin(\pi) + \cos(\pi) + C$$0 = 0 - 1 + C \implies C = 1$So,$\frac{y}{x} = x \sin x + \cos x + 1$$y = x^2 \sin x + x \cos x + x$Now,calculate $y(\frac{\pi}{2})$: $y(\frac{\pi}{2}) = (\frac{\pi}{2})^2 \sin(\frac{\pi}{2}) + \frac{\pi}{2} \cos(\frac{\pi}{2}) + \frac{\pi}{2}$$y(\frac{\pi}{2}) = \frac{\pi^2}{4}(1) + \frac{\pi}{2}(0) + \frac{\pi}{2} = \frac{\pi^2}{4} + \frac{\pi}{2}$

Let $y=y(x)$ be the solution of the differential equation $x dy = (y + x^3 \cos x) dx$ with $y(\pi) = 0$. Then $y(\frac{\pi}{2})$ is equal to:

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