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. `(4)/(3)`
C. `(1)/(2)`
D. `-(2)/(3)`

Answer

A

Explanation

The given differential equation is
`(2+sinx)(dy)/(dx)+(y+1)cosx=0`
`rArr" "(2+sinx)dy(y+1)cosx dx =0`
`rArr" "(1)/(y+1)dy+(cosx)/(2+sinx)dx=0`
Integrating both sides, we get
`log(y+1)+log(2+sinx)=logC`
`rArr" "(y+1)(2+sinx)=C" ...(i)"`
It is given that `y(0)=1` i.e. y = 1 when x = 0. Putting x = 0, y = 1 in (i), we get
`2(2+0)=CrArr C=4`
Putting C = 4 in (i), we obtain
`(y+1)(2+sinx)=4" ...(ii)"`
Putting `x=(pi)/(2)` in (ii), we get `y=(1)/(3)`.