If y=y(x) & 2+sin⁡xy+1dydx=−cos⁡x, y(0)=1, then yπ2=

2+sin⁡xy+1dydx=−cos⁡x,y(0)=1 ⇒∫dyy+1=∫−cos⁡x2+sin⁡xdx ⇒logy+1=-log2+sin x+logc ⇒y+1=c2+sinx x=0 ,y=1 then c=4⇒(y+1)=4sin⁡x+2atx=π2⇒y+1=4sin⁡π2+2=43⇒y=43−1=13⇒yπ2=13

If y=y(x) & 2+sin⁡xy+1dydx=−cos⁡x, y(0)=1, then yπ2=

2+sin⁡xy+1dydx=−cos⁡x,y(0)=1 ⇒∫dyy+1=∫−cos⁡x2+sin⁡xdx ⇒logy+1=-log2+sin x+logc ⇒y+1=c2+sinx x=0 ,y=1 then c=4⇒(y+1)=4sin⁡x+2atx=π2⇒y+1=4sin⁡π2+2=43⇒y=43−1=13⇒yπ2=13

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

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$\frac{1}{3}$

$\frac{2}{3}$

$- \frac{1}{3}$

answer is A.

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Detailed Solution

$(\frac{2 + \sin x}{y + 1}) \frac{dy}{dx} = - \cos x, y (0) = 1 \Rightarrow \int \frac{dy}{y + 1} = \int \frac{- \cos x}{2 + \sin x} dx \Rightarrow \log (y + 1) = - \log (2 + \sin x) + \log c \Rightarrow y + 1 = \frac{c}{2 + \sin x} x = 0, y = 1 t h e n c = 4$

$\Rightarrow (y + 1) = \frac{4}{\sin x + 2} atx = \frac{π}{2} \Rightarrow y + 1 = \frac{4}{\sin \frac{π}{2} + 2} = \frac{4}{3} \Rightarrow y = \frac{4}{3} - 1 = \frac{1}{3} \Rightarrow y (\frac{π}{2}) = \frac{1}{3}$