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If $\big(\large\frac{2+\sin x}{1+y}\big)\frac{dy}{dx}$$=-\cos x,y(0)=1$ then $y\big(\large\frac{\pi}{2}\big)=$

$(a)\;1\qquad(b)\;\large\frac{1}{2}$$\qquad(c)\;\large\frac{1}{3}$$\qquad(d)\;\large\frac{1}{4}$

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The given equation can be written as $\large\frac{\cos x}{2+\sin x}$$dx+\large\frac{dy}{y+1}$$=0$
$\Rightarrow \log(2+\sin x)+\log(y+1)=\log c$
$\Rightarrow \log(y+1)(2+\sin x)=\log c$
$\Rightarrow (y+1)(2+\sin x)=0$
$\Rightarrow 2\times 2=c$
$\Rightarrow c=4$
$\therefore y+1=\large\frac{4}{2+\sin x}$
$\Rightarrow y=\large\frac{2-\sin x}{2+\sin x}$
$\Rightarrow y\big(\large\frac{\pi}{2}\big)=\frac{1}{3}$
Hence (c) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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