Answer:
Option A
Explanation:
We have
$(2+sin x)\frac{\text{d}y}{\text{d}x}+(y+1)\cos x=0$
$\Rightarrow \frac{\text{d}y}{\text{d}x}+\frac{cosx}{2+sinx}y=\frac{-cosx}{2+sinx}$
which is a linear differential equation
$\therefore$ IF= $e^{\int_{}^{} \frac{cosx}{2+sin x}dx}=e^{\log(2+sin x)}$
= 2+ sin x
$\therefore$ Required solution is given by
y+(2+sinx) = $\int_{}^{} \frac{-cos x}{2+sin x}.(2+sin x)dx +C$
$\Rightarrow y(2+sin x)=-\sin x+C$
Also y(0) =1
$\therefore$ $1(2+\sin 0)=-\sin 0 +C$
$\Rightarrow$ C=2
$\therefore$ $y= \frac{2-\sin x}{2+\sin x}$
$\Rightarrow$ $y(\frac{\pi}{2})= \frac{2-\sin \frac{\pi}{2}}{2+\sin \frac{\pi}{2}}=\frac{1}{3}$
