Let y = y(x)


Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ satisfies the equation $\frac{\mathrm{dy}}{\mathrm{dx}}-|\mathrm{A}|=0$

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 $y(\pi)=\pi+2$, then the value of $y\left(\frac{\pi}{2}\right)$ is :

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

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

  3. $\frac{3 \pi}{2}-\frac{1}{\pi}$

  4. $\frac{\pi}{2}-\frac{4}{\pi}$

Correct Option: 1


$|A|=-\frac{y}{x}+2 \sin x+2$

$\frac{d y}{d x}=|A|$

$\frac{d y}{d x}=-\frac{y}{x}+2 \sin x+2$

$\frac{d y}{d x}+\frac{y}{x}=2 \sin x+2$

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

$\Rightarrow y x=\int x(2 \sin x+2) d x$

$x y=x^{2}-2 x \cos x+2 \sin x+c \ldots \ldots$ (i)

Now $\mathrm{x}=\pi, \mathrm{y}=\pi+2$

Use in (i)


Now (i) be comes

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

put $x=\pi / 2$

$\frac{\pi}{2} \mathrm{y}=\left(\frac{\pi}{2}\right)^{2}-2 \cdot \frac{\pi}{2} \cos \frac{\pi}{2}+2 \sin \frac{\pi}{2}$

$\frac{\pi}{2} y=\frac{\pi^{2}}{4}+2$

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