Question
If $$y = y\left( x \right)$$ and $$\frac{{2 + \sin \,x}}{{1 + y}}\left( {\frac{{dy}}{{dx}}} \right) = - \cos \,x,\,y\left( 0 \right) = 1$$ then $$y\left( {\frac{\pi }{2}} \right)$$ equals :
A.
$$\frac{1}{3}$$
B.
$$\frac{2}{3}$$
C.
$$ - \frac{1}{3}$$
D.
$$1$$
Answer :
$$\frac{1}{3}$$
Solution :
$$\eqalign{
& \frac{{dy}}{{dx}}\left( {\frac{{2 + \sin \,x}}{{1 + y}}} \right) = - \cos \,x,\,y\left( 0 \right) = 1 \cr
& \Rightarrow \frac{{dy}}{{\left( {1 + y} \right)}} = \frac{{ - \cos \,x}}{{2 + \sin \,x}}dx \cr
& {\text{Integrating both sides}} \cr
& \Rightarrow \ln \left( {1 + y} \right) = - \ln \left( {2 + \sin \,x} \right) + C \cr
& {\text{Put}}\,x = 0\,{\text{and}}\,y = 1 \cr
& \Rightarrow \ln \left( 2 \right) = - \ln \,2 + C \cr
& \Rightarrow C = \ln \,4 \cr
& {\text{Put }}x = \frac{\pi }{2} \cr
& \ln \left( {1 + y} \right) = - \ln \,3 + \ln \,4 \cr
& \Rightarrow \ln \left( {1 + y} \right) = \ln \frac{4}{3} \cr
& \Rightarrow y = \frac{1}{3} \cr} $$