Let y=y(x) be the solution of the differential equation

$x d y=\left(y+x^{3} \cos x\right) d x$

$x d y=y d x+x^{3} \cos x d x$

$\frac{x d y-y d x}{x^{2}}=\frac{x^{3} \cos x d x}{x^{2}}$

$\frac{d}{d x}\left(\frac{y}{x}\right)=\int x \cos x d x$

$\Rightarrow \frac{y}{x}=x \sin x-\int 1 \cdot \sin x d x$

$\frac{y}{x}=x \sin x+\cos x+C$

$\Rightarrow 0=-1+C \Rightarrow C=1, x=\pi, y=0$

so $\frac{y}{x}=x \sin x+\cos x+1$

$y=x^{2} \sin x+x \cos x+x \quad x=\frac{\pi}{2}$

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