Question

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

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