The integrating factor of linear differential equation` (dy)/(dx) + y sec x = tan x ` is

Explanation

` (dy)/(dx) + y sec x = tanx …(1)`
This is the linear differential equation of the form :
`(dy)/(dx) + Py = Q `
` therefore ` Here , P sec x and Q = tan x
` therefore I.F. = e^(intPdx)=e^(int"sec x dx")`
` = e^(log ("sec x + tan x)`
` therefore ` I.F. = sec x + tan x
Now , multiplying both sides of (i) by I.F .
` (sec x + tan x) [(dy)/(dx) + y (secx)] = tan x (sec x + tan x )`
`rArr sec x tan x (dy)/(dx) + y (sec^(2) x + sec x tan x)`
` = tan x (sec x + tan x)`
` rArr (d)/(dx) [(sec x + tan x'y] = tan x (sec x + tan x)`
On intergrating both sides, we get
` therefore y (sec x + tan x) = int tan x (sec x + tan x )dx + C `
` = int (sec x . tan x + tan^(2) x )dx+ C `
` = int (sec x * tan x + sec^(2) x - 1) dx + c`
` therefore y (sec x + tan x) = sec x + tan x - x + c`
` therefore y =( (sec x + tanx)+ (c - x))/(sec x + tan x)`
` therefore y = 1 + (c-x)/(secx + tan x)`
This is a general solution.