Question

Solution of the differential equation `cosx dy= y(sinx -y)dx , 0 lt x lt pi/2` (A) `secx=(tanx+c)y` (B) `ysecx=tanx+c` (C) `ytanx=secx+c` (D) `tanx=(secx+c)y`

Answer 1

`dy/dx = (y(sinx - y))/cosx = y tan x - y^2 secx`
`1/y^2 dy/dx = 1/y tan x - secx`
let `1/y = t`
`-1/y^2 dy/dt = dt/dx`
now, `-dt/dx = t tan x - secx`
`- dt/dx - t tan x = -sec x`
`dt/dx + t tan x = sec x `
`IF = e^(int tan x dx)= e^(ln |secx|)= sec x`
`t(IF) = int (IF) sec x `
`t * sec x = int(sec^2 x)`
`t sec x = int sec^2 x = tan x+ c`
`1/y sec x = tan x + c`
`secx = y tanx + cy`
option 1 is correct