If the solution of the differential equation `(dy)/(dx)=1/(xcosy+sin2y)` is `x=c e^(siny)-k(1+siny),` then the value of `k` is_______

Answer

2

Explanation

`(dy)/(dx) = 1/(xcosy+2sinycosy)`
`therefore (dx)/(dy) = xcosy+2sinycosy`
`therefore (dx)/(dy) +(-cosy)x=2sinycosy`
`therefore I.F. =e^(-intcosydx)=e^(-siny)`
Thus, the solution is
`x.e^(-siny)=2inte^(-siny).sinycosydy = -2sinye^(-siny)+2inte^(-siny)cosydy`
`=-2sinye^(-siny)+2inte^(-siny)cosydy`
`=-2sinye^(-siny)-2e^(-siny)+c`
i.e., `x=-2siny-2+ce^(siny)=ce^(siny)-2(1+siny)`
`therefore k=2`