Mathematics - Differential Equation Question with Solution | TestHub

We have, $xdy=\left(y+x^3\cos x\right)dx$

$\Rightarrow xdy=ydx+x^3\cos xdx$

$\Rightarrow \frac{xdy-ydx}{x^2}=\frac{x^3\cos xdx}{x^2}$

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

$\Rightarrow \frac{y}{x}=x\sin x-\int 1.\sin xdx$

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

At $x=\mathrm{\pi }, y=0,$

$0=-1+C$

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

So, $\frac{y}{x}=x\sin x+\cos x+1$

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

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