Question

Find the general solution of the differential equation:
sec²x.tanydx-sec²ydy=0​

Answer 1

$\large\underline{\bf{Solution-}}$

$\bf :\longmapsto\: {sec}^{2}x \: tany \: dx - {sec}^{2}y \: dy = 0$

$\rm :\longmapsto\: {sec}^{2}x \: tany \: dx = {sec}^{2}y \: dy$

$\rm :\longmapsto\: {sec}^{2}x \: dx = \dfrac{ {sec}^{2}y }{tany}dy$

☆ On integrating both sides, we get

$\displaystyle\rm :\longmapsto\: \int {sec}^{2}x \: dx = \int\dfrac{ {sec}^{2}y }{tany}dy$

$\displaystyle\rm :\longmapsto\: tanx= \int\dfrac{ \dfrac{d}{dx}tany }{tany}dy$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \displaystyle\red{ \bigg \{ \because \: \int \: {sec}^{2} x = tanx + c \bigg \}}$

$\rm :\longmapsto\:tanx = log(tany) + c$

$\: \: \: \: \: \: \: \: \: \: \: \displaystyle\red{ \bigg \{ \because \: \int \: \dfrac{f'(x)}{f(x)}dx = logf(x) + c \bigg \}}$

$\bf :\longmapsto\:tanx - log(tany) = c$

Additional Information :-

Solution for Linear Differential Equation :-

The equation of the form

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: is \: linear \: differential \: equation$

☆ Step 1 :- Find Integrating Factor, I.F.

$\rm :\longmapsto\:I.F. = {e} \: \: ^{\displaystyle \int \: p \: dx}$

☆ Step 2 :- Solution is given by

$\rm :\longmapsto\:y \: \times \: I.F. \: = \: \displaystyle \int \: (q \: \times \: I.F.) \: dx$