Question

Let $y=y(x)$ be the solution of the differential equation, $x y'-y=x^{2}(x \cos x+\sin x), x>0$ If $y (\pi)=\pi,$ then $y''\left(\frac{\pi}{2}\right)+ y \left(\frac{\pi}{2}\right)$ is equal to :

• A. $2+\frac{\pi}{2}$
• B. $1+\frac{\pi}{2}$
• C. $1+\frac{\pi}{2}+\frac{\pi^{2}}{4}$
• D. $2+\frac{\pi}{2}+\frac{\pi^{2}}{4}$

Answer 1

Answer: (A)

$x \frac{d y}{d x}-y=x^{2}(x \cos x+\sin x), x > 0$
$\frac{d y}{d x}-\frac{y}{x}=x(x \cos x+\sin x)$
$ \Rightarrow \frac{d y}{d x}+P y=Q$
so, $I . F .=e^{\int-\frac{1}{x} d x}=\frac{1}{|x|}=\frac{1}{x}(x>0)$
Thus, $\frac{y}{x}=\int \frac{1}{x}(x(x \cos x+\sin x)) d x$
$\Rightarrow \frac{y}{x}=x \sin x+C$
$\because y(\pi)=\pi $
$\Rightarrow C=1$
so, $y=x^{2} \sin x+x$
$ \Rightarrow (y)_{\pi / 2}=\frac{\pi^{2}}{4}+\frac{\pi}{2}$
Also, $\frac{d y}{d x}=x^{2} \cos x+2 x \sin x+1$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=-x^{2} \sin x+4 x \cos x+2 \sin x$
$\left.\Rightarrow \frac{d^{2} y}{d x^{2}}\right|_{\frac{\pi}{2}}=-\frac{\pi^{2}}{4}+2$
Thus, $y\left(\frac{\pi}{2}\right)+\frac{d^{2} y}{d x^{2}\left(\frac{\pi}{2}\right)}=\frac{\pi}{2}+2$