Let y=y(x) be the solution of the differential equation xdy=(y+x^3 cosx)dx with y(π )=0, then y( π 2) is equal to:

We have, $xdy=(y+{x}^{3}\mathrm{cos}x)dx$xdy=(y+x3cosx)dx

$$\Rightarrow xdy=ydx+{x}^{3}\mathrm{cos}xdx$$⇒xdy=ydx+x3cosxdx

$$\Rightarrow \frac{xdy-ydx}{{x}^{2}}=\frac{{x}^{3}\mathrm{cos}xdx}{{x}^{2}}$$⇒x2xdy−ydx​=x2x3cosxdx​

$$\Rightarrow \frac{d}{dx}(\frac{y}{x})=\int x\mathrm{cos}xdx$$⇒dxd​(xy​)=∫xcosxdx

$$\Rightarrow \frac{y}{x}=x\mathrm{sin}x-\int 1.\mathrm{sin}xdx$$⇒xy​=xsinx−∫1.sinxdx

Therefore, $\frac{y}{x}=x\mathrm{sin}x+\mathrm{cos}x+C$xy​=xsinx+cosx+C

At $x=\pi ,y=0,$x=π,y=0,

$$0=-1+C$$0=−1+C

$$\Rightarrow C=1,x=\pi ,y=0$$⇒C=1,x=π,y=0

So, $\frac{y}{x}=x\mathrm{sin}x+\mathrm{cos}x+1$xy​=xsinx+cosx+1

$$\Rightarrow y={x}^{2}\mathrm{sin}x+x\mathrm{cos}x+x$$⇒y=x2sinx+xcosx+x

Hence, $y(\frac{\pi }{2})=\frac{{\pi }^{2}}{4}+\frac{\pi }{2}.$y(2π​)=4π2​+2π​.