Solution of the differential equation t any sec2x dx + tanx sec2 ydy = 0 is:
The given differential equation is
⇒ tanysec2xdx + tanxsec2ydy = 0
Divide throughout by tanxtany
Integrate
Put tanx = t hence that is sec2xdx = dt
Put tany = z hence that is sec2ydy = dt
⇒ log t + log z + c = 0
Resubstitute t and z
⇒ log(tan x) + log(tan y) + c = 0
Using log a + log b = log ab
⇒ log(tan x tan y) = -c
⇒ tan x tan y = e-c
e is a constant -c is a constant hence e-c is a constant, let it be denoted as k hence k = e-c
⇒ tan x tan y = k