Question

The solution of the D.E. sec²x . tan ydx + sec²y tan xdy = 0 is ______.

विकल्प

• tan x . cot y = c

• cot x – cot y = c

• tan x . tan y = c

• cot x – tan y = c

Answer 1

The solution of the D.E. sec²x. tan ydx + sec²y tan xdy = 0 is tan x . tan y = c.

Explanation:

Given D.E.:

sec²x tan y dx + sec²y tan x dy = 0

Step 1: Rearrange

sec²x tan y dx = – sec²y tan x dy

Divide both sides by tan x tan y:

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

On integrating both sides:

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

Put v = tan x and u = tan y

`(dv)/dx = sec^2x` and `(du)/dy = sec^2 y`

dv = sec²x dx and du = sec²y dy

⇒ `int (dv)/v = - int (du)/u`

⇒ log v = − log u + log c

⇒ log tan x = − log tan y + log c

⇒ log tan x + log tan y = log c

⇒ log (tan x . tan y) = log c

⇒ tan x . tan y = c