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`

Step 2: Simplify

We know:

`(sec^2x)/(tan x) = (1//cos^2x)/(sin x//cos x)`

= `1/(sin x cos x)`

So the equation becomes:

`(dx)/(sin x cos x) = - (dy)/(sin y cos y)`

Step 3: Integrate

`int (dx)/(sin x cos x) = - int (dy)/(sin y cos y)`

Now, `1/(sin x cos x) = (sec^2x)/(tan x)`

And `int (sec^2x)/(tan x) dx = log |tan x|`

Similarly for y:

`int (sec^2y)/(tan y) dy = log | tan y|`

So, log |tan x| = – log |tan y|  + C

Step 4: Simplify

log |tan x| + log |tan y| = C

log |tan x tan y| = C

tan x tan y = C