Question

4. sec2 x tan y dx + sec2 y tan x dy 0

Answer 1

The given differential equation is,

sec 2 xtanydx+ sec 2 ytanxdy=0

Divide the above differential equation with tanxtany on both the sides.

sec 2 xtanydx+ sec 2 ytanxdy tanxtany =0 sec 2 xtany tanxtany dx+ sec 2 ytanx tanxtany dy=0 sec 2 x tanx dx=− sec 2 y tany dy

Integrating both sides, we get,

∫ sec 2 x tanx dx =− ∫ sec 2 y tany dy (1)

Let,

I 1 = ∫ sec 2 x tanx dx I 2 =− ∫ sec 2 y tany dy

Let, tanx=t

Differentiating both sides, we get,

sec 2 xdx=dt sec 2 x= dt dx sec 2 xdx=dt

Similarly, for the variable y.

tany=r sec 2 ydy=dr sec 2 y= dr dx sec 2 ydy=dr

Substitute the value of sec 2 xdx and tany in I 1 .

∫ sec 2 x tanx dx = ∫ dt t =log| t | =log| tanx | (2)

Similarly, substitute the value of sec 2 ydy and tanx in I 2 .

− ∫ sec 2 y tany dy =−log| tany |+log C 2 (3)

Substitute the values of equations (2) and (3) in equation (1).

log| tanx |=−log| tany |+logC log| tanx |=log| C tany | tanx= C tany tanxtany=C