Question

solve differential equation 3*e^x*tany dx+(1-e^x)( sec)^2 y dy=0​

Answer 1

Answer:

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Answer 2

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

Given Differential equation is

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

can be rewritten as

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

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

On separate the variables, we get

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

On integrating both sides, we get

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

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

We know,

$\: \: \: \: \: \: \: \: \: \: \purple{\boxed{\displaystyle\int\tt \: \dfrac{f'(x)}{f(x)}dx \: = \: log(f(x)) + c}}$

So, using this identity, we get

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

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

$\: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \:log(xy) = logx + logy \bigg \}} \\ \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: log( {x}^{y} ) = y log(x) \bigg \}}$

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

Hence,

Solution of differential equation

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

is

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

Additional Information :-

The solution of Homogeneous differential equation

$\rm :\longmapsto\:\dfrac{dy}{dx} = f\bigg(\dfrac{y}{x} \bigg) \: is \: given \: as$

Step :- 1

$\rm :\longmapsto\:Let \: y \: = \: vx$

Step :- 2

$\rm :\longmapsto\:\: \dfrac{d}{dx}y \: = \dfrac{d}{dx}\: vx$

$\rm :\longmapsto\:\: \dfrac{dy}{dx} \: = v\dfrac{d}{dx}\:x \: + \: x\dfrac{d}{dx}v$

$\rm :\longmapsto\:\: \dfrac{dy}{dx} \: = v \: + \: x\dfrac{dv}{dx}$

Thus,

Given differential equation reduced to

$\rm :\longmapsto\:\: \ v \: + \: x\dfrac{dv}{dx} = f(v)$

and solved by variable separation method.