Solve: `dy / dx + y secx = tanx`

Explanation

The given equation is of the form
` (dy)/(dx) + Py = Q`, where ` P = sec x and Q = tan x `
Thus, the given equation is linear.
`IF =e ^( int Pdx) = e ^( int sec x dx ) = e ^(log (sec x + tan x ) ) = (sec x + tanx )`
So, the required solution is
` y xx IF = int { Q xx( IF ) }dx + C`,
i.e, `y (secx + tan x ) = int tan x (sec x + tan x ) dx + C `
` " " = int sec x tan x dx = int tan ^(2) x dx + C `
` " " = sec x + int (sec ^(2) x - 1 ) dx + C `
` " " =sec x + tanx - x + C`.
Hence, ` y (sec x +tan x ) =sec x + tanx - x+ C ` is the required solution.