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

${{2 + \sin x} \over {y + 1}}.{{dy} \over {dx}} = - \cos x$ <br><br>$\Rightarrow$ $$\int {{{dy} \over {y + 1}}} = \int {{{\left( { - \cos x} \right)dx} \over {2 + \sin x}}} $$ <br><br>$\Rightarrow$ $\ln \left| {y + 1} \right| = - \ln \left| {2 + \sin x} \right| + \ln c$ <br><br>$\Rightarrow$ $\ln \left| {\left( {y + 1} \right)\left( {2 + \sin x} \right)} \right| = \ln c$ <br><br>As y(0) = 1 <br><br>$\therefore$ $\ln \left| {\left( {1 + 1} \right)\left( {2 + 0} \right)} \right| = \ln c$ <br><br>$\Rightarrow$ $\therefore$ c = 4 <br><br>$\therefore$ ${\left( {y + 1} \right)\left( {2 + \sin x} \right)}$ = 4 <br><br>$\Rightarrow$ y = ${{{2 - \sin x} \over {2 + \sin x}}}$ <br><br>Given y($\pi$) = a <br><br>$\therefore$ y($\pi$) = ${{{2 - \sin \pi } \over {2 + \sin \pi }}}$ = 1 = a <br><br>$${{dy} \over {dx}} = {{\left( {2 + \sin x} \right)\left( { - \cos x} \right) - \left( {2 - \sin x} \right).\left( {\cos x} \right)} \over {{{\left( {2 + \sin x} \right)}^2}}}$$ <br><br>$\Rightarrow$ ${\left. {{{dy} \over {dx}}} \right|_{x = \pi }} =$ 1 = b <br><br>So, (a, b) = (1, 1)