# Solve this

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. $2+\frac{\pi}{2}$

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

3. $1+\frac{\pi}{2}+\frac{\pi^{2}}{4}$

4. $2+\frac{\pi}{2}+\frac{\pi^{2}}{4}$

Correct Option: 1

Solution:

$x \frac{d y}{d x}-y=x^{2}(x \cos x+\sin x), x>0$

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

so, I.F. $=e^{\int-\frac{1}{x} d x}=\frac{1}{|x|}=\frac{1}{x}(x>0)$

Thus, $\frac{y}{x}=\int \frac{1}{x}(x(x \cos x+\sin x)) d x$

$\Rightarrow \frac{y}{x}=x \sin x+C$

$\because \mathrm{y}(\pi)=\pi \Rightarrow \mathrm{C}=1$

so, $\mathrm{y}=\mathrm{x}^{2} \sin \mathrm{x}+\mathrm{x} \Rightarrow(\mathrm{y})_{\pi / 2}=\frac{\pi^{2}}{4}+\frac{\pi}{2}$

Also, $\frac{d y}{d x}=x^{2} \cos x+2 x \sin x+1$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=-x^{2} \sin x+4 x \cos x+2 \sin x$

$\left.\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right|_{\frac{\pi}{2}}=-\frac{\pi^{2}}{4}+2$

Thus, $\mathrm{y}_{\left(\frac{\pi}{2}\right)}+\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}\left(\frac{\pi}{2}\right)}=\frac{\pi}{2}+2$