Question

3e^xtany dx + (1-e^x) sec^2ydy=0​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\rm :\longmapsto\:Solve : \: {3e}^{x}tany \: dx + (1 - {e}^{x}) {sec}^{2}y \: dy = 0$

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

Given that

$\rm :\longmapsto\:\: {3e}^{x}tany \: dx + (1 - {e}^{x}) {sec}^{2}y \: dy = 0$

can be rewritten as

$\rm :\longmapsto\:\: {3e}^{x}tany \: dx = - \: (1 - {e}^{x}) {sec}^{2}y \: dy$

$\rm :\longmapsto\:\: {3e}^{x}tany \: dx = \: ({e}^{x} - 1) {sec}^{2}y \: dy$

$\rm :\longmapsto\:\dfrac{ {3e}^{x} }{ {e}^{x} - 1}dx = \dfrac{ {sec}^{2} y}{tany} dy$

On integrating both sides, we get

$\rm :\longmapsto\:\displaystyle\int\sf \dfrac{ {3e}^{x} }{ {e}^{x} - 1}dx =\displaystyle\int\sf \dfrac{ {sec}^{2} y}{tany} dy$

$\rm :\longmapsto\:3\displaystyle\int\sf \dfrac{ \dfrac{d}{dx}({e}^{x} - 1) }{ {e}^{x} - 1}dx =\displaystyle\int\sf \dfrac{ \dfrac{d}{dx}{tan} y}{tany} dy$

$\rm :\longmapsto\:3 log( {e}^{x} - 1 ) = log(tany) + log(c)$

$\: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf \because\: \displaystyle\int\sf \dfrac{\dfrac{d}{dx}f(x)}{f(x)} = log(f(x)) + c }}$

$\rm :\longmapsto\:log( {e}^{x} - 1 )^{3} = log(tany \times c)$

$\green{\boxed{ \sf \because \: logx + logy = logxy \: \: \: and \: \: \: log {x}^{y} = y \: logx\: }}$

$\bf\implies \: {( {e}^{x} - 1) }^{3} = c \: tany$

Additional Information :-

Linear Differential equation :-

The linear differential equation is given by

$\rm :\longmapsto\:\dfrac{dy}{dx} + py = q \: wherep, \: q \: \in \: f(x)$

Solution is given by

$\rm :\longmapsto\:y \times IF = \displaystyle\int\sf (q \times IF) \: dx$

where,

$\rm :\longmapsto\:IF = {e} \: \: ^{\displaystyle\int\sf p \: dx}$