# Let y=y(x) be the solution of the differential equation,

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:

1. (1) $2+\frac{\pi}{2}$

2. (2) $1+\frac{\pi}{2}+\frac{\pi^{2}}{4}$

3. (3) $2+\frac{\pi}{2}+\frac{\pi^{2}}{4}$

4. (4) $1+\frac{\pi}{2}$

Correct Option: 1

Solution:

$\frac{d y}{d x}-\frac{y}{x}=x(x \cos x+\sin x)$

I.F. $=e^{-\int \frac{1}{x} d x}=\frac{1}{x}$

$\therefore \int d\left(\frac{y}{x}\right)=\int(x \cos x+\sin x) d x$

$\Rightarrow \frac{y}{x}=x \sin x+C \quad \because y(\pi)=\pi \Rightarrow C=1$

$y=x^{2} \sin x+x \Rightarrow y\left(\frac{\pi}{2}\right)=\frac{\pi^{2}}{4}+\frac{\pi}{2}$

$y^{\prime}=2 x \sin x+x^{2} \cos x+1$

$y^{\prime \prime}=2 \sin x-x^{2} \sin x \Rightarrow y^{\prime \prime}\left(\frac{\pi}{2}\right)=2-\frac{\pi^{2}}{4}$

$\therefore y^{\prime \prime}\left(\frac{\pi}{2}\right)+y\left(\frac{\pi}{2}\right)=2-\frac{\pi^{2}}{4}+\frac{\pi^{2}}{4}+\frac{\pi}{2}=2+\frac{\pi}{2}$