Trig Differential Equation

[Thomas154321]

I've been having problems with this question:

(dy/dx)sinx + y*secx = (cosx)^2

Any help please?

[Glaysher]

(dy/dx) + y*sec x*cosec x = cosec x (cos x)^2

sec x * cosec x = 1/[sin x * cos x] = 1/[(1/2) sin 2x] = 2 cosec 2x

Integral of 2 cosec 2x = ln |tan x|

So integrating factor = tan x

tan x (dy/dx) + y*sec² x = cos x

Integrating

y tan x = sin x + c

[Thomas154321]

It all make sense now! I thought the integral of 2 cosec 2x was 2ln |tan x|, which gave me a tan sqaured. oops. Cheers.

[Glaysher]

I had to think about that one too but I differentiated ln |tan x| to check