Solve the differential equation `"dy=cos x(2-y cosec x)dx"` given that `y=2, "when x" d=(pi)/(2)`

Explanation

Given differential equation,
`" "dy=cosx(2-y" cosec"x)dx`
`rArr" "(dy)/(dx)=cosx(2-y" cosec"x)`
`rArr" "(dy)/(dx)=2cosx-y" cosec"x*cosx`
`rArr" "(dy)/(dx)=2cosx-ycotx`
`rArr" "(dy)/(dx)+ycotx=2cosx`
which is linear differential equation.
On comparing it with `(dy)/(dx)+Py=Q`, we get
`" "P=cotx, Q=2cosx`
`" "IF=e^(intPdx)=e^(intcotxdx)=e^(logsinx)=sinx`
The general solution is
`" "y*sinx=int2cosx*sinxdx +C`
`rArr" "y*sinx=intsin2xdx+C" "[because sin2x=2sinxcosx]`
`rArr" "y*sinx=-(cos2x)/(2)+C" "...(i)`
When `x=(pi)/(2) and y=2`, then
`" "2*sin""(pi)/(2)=-(cos(2xx(pi)/(2)))/(2)+C`
`rArr" "2*1=+(1)/(2)+C`
`rArr" "2-(1)/(2)=CrArr(4-1)/(2)=C`
`rArr therefore" "C=(3)/(2)`
On subsituting the value of C in Eq. (i), we get
`" "ysinx=-(1)/(2)cos2x+(3)/(2)`