Question

Solution of the differential equation t any sec²x dx + tanx sec² ydy = 0 is:

A. tanx + tany = k
B. tanx – tan y = k
C. tanx/tany = k
D. tanx . tany = k

Answer 1

The given differential equation is

⇒ tanysec²xdx + tanxsec²ydy = 0

Divide throughout by tanxtany

⇒ 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