Question

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

• A. $\frac{{\pi }^2}{2}+\frac{\pi }{4}$
• B. $\frac{{\pi }^2}{2}-\frac{\pi }{4}$
• C. $\frac{{\pi }^2}{4}-\frac{\pi }{2}$
• D. $\frac{{\pi }^2}{4}+\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{{\pi }^2}{4}+\frac{\pi }{2}$

Given: $xdy=(y+x^3\cos x)dx$
$\Rightarrow \frac{xdy-ydx}{x^2}=x\cos xdx$
$\Rightarrow \int d\left(\frac{y}{x}\right)=\int x\cos xdx$
$\Rightarrow \frac{y}{x}=x\sin x+\cos x+C$
if $y\left(\pi \right)=0\Rightarrow C=1$
$\therefore y=x^2\sin x+x\cos x+x$
At $x=\frac{\pi }{2},$ we have
$y\left(\frac{\pi }{2}\right)=\frac{{\pi }^2}{4}+\frac{\pi }{2}$