Let $y=y(x)$ be the solution of the differential equation,
$\frac{2+\sin x}{y+1} \cdot \frac{d y}{d x}=-\cos x, y>0, y(0)=1 .$ If $y(\pi)=a$
and $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=\pi$ is $\mathrm{b}$, then the ordered pair
(a, b) is equal to :
Correct Option: , 4
$\frac{2+\sin x}{y+1} \frac{d y}{d x}=-\cos x, y>0$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}+1}=\frac{-\cos \mathrm{x}}{2+\sin \mathrm{x}} \mathrm{dx}$
By integrating both sides :
$\ell \mathrm{n}|\mathrm{y}+1|=-\ell \mathrm{n}|2+\sin \mathrm{x}|+\ell \mathrm{nK}$
$\Rightarrow y+1=\frac{K}{2+\sin x} \quad(y+1>0)$
$\Rightarrow \mathrm{y}(\mathrm{x})=\frac{\mathrm{K}}{2+\sin \mathrm{x}}-1$
Given $\mathrm{y}(0)=1 \Rightarrow \mathrm{K}=4$
So, $y(x)=\frac{4}{2+\sin x}-1$
$a=y(\pi)=1$
$\left.\left.\mathrm{b}=\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=\pi}=\frac{-\cos \mathrm{x}}{2+\sin \mathrm{x}}(\mathrm{y}(\mathrm{x})+1)\right]_{\mathrm{x}=\pi}=1$
So, $(a, b)=(1,1)$
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