If `(2+sinx)(dy)/(dx)+(y+1)cosx=0` and `y(0)=1`, then `y((pi)/(2))` is equal to

Options

A. `1/3`
B. `1/2`
C. 1
D. 0

Answer

a

Explanation

Given that , ` ((2+sinx)/(1+y) ) (dy)/(dx) = - cos x `
` rArr (dy)/(dx) = - cos x ((1+y)/(2+sinx))`
` rArr (dy)/(dx) + (cosx )/(2+sin x) * y = - (cosx)/(2+sin x)`
which is a linear differential equation ,
` :. IF = e^(int (cosx)/(2+sinx)dx)=e^(log (2sinx ) = 2 + sin x)`
So, its general solution ,
` y * (IF) = int (-cos x)/(2+sinx )* (IF) dx +C`
` y (2 + sin x) = int (-cosx)/(2+sinx) * (2 + sin x)dx+C`
` y ( 2 + sin x) = - int cos x dx + C`
` rArr y (2 + sin x ) = - sin x + C " "` ...(i)
Now, y = 1 , when x = 0
` rArr 1* (2+sin 0) = - sin 0 + C`
` rArr 2 = C rArr C = 2`
` :. y (2 + sin x) = - sin x +2`
` rArr y = (2-sinx)/(2+sinx) " " ` ...(ii)
` :. y (pi/2) = (2 -sin. pi/2)/(2 + sin.pi/2) = (2-1)/(2+1) =1/3`