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

$$x\frac{dy}{dx}=y+x^3\cos x$$

$$\frac{dy}{dx}-\frac{y}{x}=x^2\cos x$$

This is a linear differential equation.

Integrating factor:

$$I.F.=e^{\int-\frac1x\,dx}$$

$$=\frac1x$$

Multiplying throughout by

$$\frac1x,$$

$$\frac1x\frac{dy}{dx}-\frac{y}{x^2}=x\cos x$$

Hence,

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

Integrating,

$$\frac{y}{x}=\int x\cos x\,dx+C$$

Using integration by parts,

$$\int x\cos x\,dx=x\sin x+\cos x$$

Therefore,

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

$$y=x^2\sin x+x\cos x+Cx$$

Using

$$y(\pi)=0,$$

$$0=\pi^2\sin\pi+\pi\cos\pi+C\pi$$

$$0=0-\pi+C\pi$$

$$C=1$$

Hence,

$$y=x^2\sin x+x\cos x+x$$

Now substitute

$$x=\frac{\pi}{2}$$

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

$$=\frac{\pi^2}{4}+0+\frac{\pi}{2}$$

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