If y=y(x) and 2+sinxy+1dydx=−cosx,y(0)=1 then y(π/2) is
1/3
2/3
-1/3
1
dyy+1=−cosx2+sinxdx⇒ln(y+1)=−log(2+sinx)+logc⇒y+1=c2+sinx, put x=0,y=1→c=4y(π2)+1=42+sinπ2=43⇒y(π2)=13