Question

Let $y=y\left(x\right)$ satisfies the equation $\frac{dy}{dx}-|A|=0,$ for all $x>0,$ where $A=\left[\begin{array}{ccc}y& \sin x& 1\\ 0& -1& 1\\ 2& 0& \frac{1}{x}\end{array}\right]$.
If $y\left(\pi \right)=\pi +2,$ then the value of $y\left(\frac{\pi }{2}\right)$ is:

• A. $\frac{\pi }{2}-\frac{4}{\pi }$
• B. $\frac{\pi }{2}+\frac{4}{\pi }$
• C. $\frac{3\pi }{2}-\frac{1}{\pi }$
• D. $\frac{\pi }{2}-\frac{1}{\pi }$

Answer 1

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

From the given matrix, we have: $|A|=\frac{-y}{x}-\sin x(-2)+1\left(2\right)$
$\Rightarrow |A|=2+2\sin x-\frac{y}{x}$
Hence we have: $\frac{dy}{dx}+\frac{y}{x}=2+2\sin x$
$\text{I.f}=e^{\ln x}=x$
$\int d\left(xy\right)=\int 2x(1+\sin x)dx$
$\Rightarrow xy=x^2-2x\cos x+\int 2\cos xdx$
$\Rightarrow xy=x^2-2x\cos x+2\sin x+c$
$\therefore y\left(\pi \right)=\pi +2$
$\Rightarrow \pi (\pi +2)={\pi }^2-2\pi (-1)+0+c$
$\Rightarrow c=0$
For $y\left(\frac{\pi }{2}\right)$
$\frac{\pi }{2}y=\frac{{\pi }^2}{4}-\frac{2\pi }{2}\left(0\right)+2$
$\Rightarrow y\left(\frac{\pi }{2}\right)=\frac{\pi }{2}+\frac{4}{\pi }$