Question

For the differential equation, find the general solution:

sec² x tan y dx + sec² y tan x dy = 0

Answer 1

The given differential equation is:

`sec^2x tany dx + sec^2y tanx dy = 0`

⇒ `(sec^2x tany  dx + sec^2y tanx  dy)/(tanx tany) = 0`

⇒ `sec^2x/tanx dx + sec^2y/tanydy = 0`

⇒ `sec^2x/tanx dx = -sec^2y/tany dy`

Integrating both sides of this equation , we get : 

`∫sec^2x/tanx dx = -∫sec^2y/tany dy`   ...(1)

Let tanx = t .

∴ `d/dx(tan x) = dt/dx`

⇒ `sec^2x = dt/dx`

⇒ `sec^2x dx = dt`

Now , `∫sec^2x/tanx dx = ∫1/t dt.`

= log t

= log (tan x)

Similarly , `∫sec^2y/tany dy = log(tan y).`

Substituting these values in equation (1) , we get :

log(tan x) = - log(tan y) + log C

⇒ `log(tan x) + log(tan y) = log C`

⇒ `log(tan x tany) = log C`   [∵ loga + logb = log(ab)]

⇒ tanx tany = C

Answer 2

sec² x tan y dx = - sec² y tan x dy

`(sec^2 x)/tan x  dx = (- sec^2 y)/tan y  dy`

On integrating

`int  (sec^2 x)/tan x  dx =  - int (sec^2 y)/tan y  dy`

`log abs (tan x) = log abs (tan y) + log C`

`log abs (tan x) = log abs (C/tan y)`

tan x . tan y = C