Let $y = y(x)$ be the solution of the differential equation $\frac{2+sin\, x}{y+1}\frac{dy}{dx} = - cos \, x, y > 0, y(0) =1.$ If $y(\pi ) = a $ and $\frac{dy}{dx} $ at $ x = \pi $ is b, then the ordered pair (a, b) is equal to

(1, 1)

The correct answer is option (4) : (1, 1)

We have,

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

$⇒\frac{1}{y+1}dy =-\frac{cosx}{2+sinx}dx$

Integrating both sides, we obtain

$log ( y + 1) = - log (2 + sin x) + log C$

$⇒y+1=\frac{C}{2+sinx}$

It is given that $y(0) = 1$ i.e. $y = 1$ when x= 0.

$2=\frac{C}{2} ⇒C=4$

Putting $C+4$ in (i), we obtain

$y + 1=\frac{4}{2+sinx}$

$⇒y = \frac{2-sinx}{2+sinx} $

$⇒y (\pi ) =\frac{2-sin \pi }{2+sin \pi } = 1 ⇒ a= 1.$

Now, $y + 1 =\frac{4}{2+ sin x}$

$⇒\frac{dy}{dx} = -\frac{4cosx}{(2+ sin x)^2}$

$⇒\left(\frac{dy}{dx}\right)_{x=\pi }= \frac{-4×-1}{(2+0)^2}= 1 ⇒ b = 1.$

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