question_answer

               If    y(x) is a solution of \[\left( \frac{2+\sin x}{1+y} \right)\,\frac{dy}{dx}=-\cos x\] and y(0) = 1, find the value of \[y\left( \frac{\pi }{2} \right).\]

Answer:

Given differential equation is \[\frac{(2+\sin x)}{(1+y)}\frac{dy}{dx}=-\cos x\] which can be rewritten as \[\frac{dy}{1+y}=-\frac{\cos x}{(2+\sin \,x)}dx\] On integrating both sides, we get \[\int{\frac{dy}{1+y}=-\int{\frac{\cos x}{2+\sin \,x}}}\,dx\] \[\Rightarrow \]   \[\log |1+y|\,=-\int{\frac{\cos x}{2+\sin x}dx}\] Put \[2+\sin =t\] \[\Rightarrow \] \[\cos x\,dx=dt\] Now, \[\log |1+y|\,\,=-\int{\frac{dt}{t}\,\,\,\Rightarrow \,\,\,\log |1+y|}\,\,=-\log |t|+{{C}_{1}}\] \[\Rightarrow \]   \[\log |1+y|+\log |2+\sin x|\,\,={{C}_{1}}\] \[[\because \,t=2+\sin \,x]\] \[\Rightarrow \]   \[\log \,|(1+y)(2+\sin x)|\,\,={{C}_{1}}\]     \[[\because \log m+\log n=\log (m\cdot n)]\] \[\Rightarrow \]   \[(1+y)(2+\sin \,x)=C,\] where \[C={{e}^{{{c}_{1}}}}\] Since, it is given that y(0) = 1, therefore             \[(1+1)(2+\sin \,0)=C\,\,\,\Rightarrow \,\,\,4=C\] Hence, \[1+y=\frac{4}{2+\sin x}\] \[\Rightarrow \] \[y=\frac{4}{2+\sin x}-1\] Now, \[y\left( \frac{\pi }{2} \right)=\frac{4}{2+\sin \frac{\pi }{2}}-1=\frac{4}{3}-1=\frac{1}{3}\]