Find the general solution of the differential equation `(1+tany)(dx-dy)+2x dy=0`

Explanation

Given differential equation is `(1+tany)(dx-dy)+2xdy=0`
On dividing throughout by `dy`, we get
`" "(1+tany)((dx)/(dy)-1)+2x=0`
`rArr" "(1+tany)(dx)/(dy)-(1+tany)+2x=0`
`rArr" "(1+tany)(dx)/(dy)+2x=(1+tany)`
`rArr" "(dx)/(dy)+(2x)/(1+tany)=1`
which is a linear differential equation.
On comparing it with `(dx)/(dy)+Px=Q`, we get
`" "P=(2)/(1+tany),Q=1`
`" "IF=e^(int(2)/(1+tany)dy)=e^(int(2cosy)/(cosy+siny)dy)`
`" "=e^(int(cosy+siny+cosy-siny)/(cosy+siny)dy)`
`" "=e^(int(1+(cosy-siny)/(cosy+sinx))dy)=e^(y+log(cosy+siny)`
`" "=e^(y)*(cosy+siny)" "[becausee^(logx)=x]`
The general solution is
`" "x*e^(y)(cosy+siny)=int1*e^(y)(cosy+siny)dy+C`
`rArr" "x*e^(y)(cosy+siny)=inte^(y)(siny+cosy)dy+C`
`rArr" "x*e^(y)(cosy+siny)=e^(y)siny+C" "[becauseinte^(x){f(x)=f'(x)}dx=e^(x)f(x)]`
`rArr" "x(siny+cosy)=siny+Ce^(-y)`