The general solution of the differential equation `(dy)/(dx) = y tan x - y^(2) sec x` is

Options

A. `tan x = (c + sec x)y`
B. `sec y = (c + tan y)x`
C. `sec x = (c + tan x) y`
D. None of these

Answer

C

Explanation

We have `(dy)/(dx) = y tan x - y^(2) sec x`
`rArr" "(1)/(y^(2))(dy)/(dx)-(1)/(y) tan x = -sec x`
Putting `(1)/(y) = v rArr (-1)/(y^(2))(dy)/(dx)=(dv)/(dx)`, we get
`(dv)/(dx) + tan x * v = sec x` which is linear diff. equation
`I.F. = e^(int tan xdx) = e^(log sec x) = sec x`
`therefore" "`The solution is
`v sec x = int sec^(2) xdx + c`
`rArr" "(1)/(y) sec x = tan x + c`
`rArr" "sec x = y (c + tan x)`