# Let y=y(x) be the solution of the differential equation,

Question:

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 :

1. $(2,1)$

2. $\left(2, \frac{3}{2}\right)$

3. $(1,-1)$

4. $(1,1)$

Correct Option: , 4

Solution:

$\frac{2+\sin x}{y+1} \frac{d y}{d x}=-\cos x, y>0$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}+1}=\frac{-\cos \mathrm{x}}{2+\sin \mathrm{x}} \mathrm{dx}$

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, $(a, b)=(1,1)$