If `y(x)` is a solution of the differential equation `((2+sinx)/(1+y))(dy)/(dx)=-cosx` and `y(0)=1`, then find the value of `y(pi/2)`.

Explanation

Given that, `((2+sin x)/(1+y))(dy)/(dx)=-cos x`
`Rightarrow (dy)/(1+y)=-(cosx)/(2+sinx)dx`
On integrating both sides, we get
`int(1)/(1+y)dy=-int (cos x)/(2+sinx)dx`
`Rightarrowlog(1+y)=-log(2+sinx)+logC`
`Rightarrow log(1+y)+log(2+sinx)=logC`
`Rightarrow log(1+y)(2+sinx)=logC`
`Rightarrow (1+y)(2+sinx)=C`
`Rightarrow 1+y=(C)/(2+sinx)`
When, x=0 and y=1, then `1=(C)/(2+sin x)-1`
`Rightarrow C=4`
On putting C=4 in Eq (i) we get `y=(4)/(2+sinx)-1`
`y((pi)/(2))=(4)/(2+"sin"(pi)/(2))-1=(4)/(2+1)-1`
`(4)/(3)-1=(1)/(3)`