Question

The integrating factor of dy/dx+y tan x=sec x

Answer 1

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

Given Differential equation is

$\rm :\longmapsto\:\dfrac{dy}{dx} + y \: tanx \: = \: secx$

↝ Its a linear differential equation,

So, on comparing with

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

We get now,

$\red{\rm :\longmapsto\:p = tanx}$

and

$\red{\rm :\longmapsto\:q = secx}$

We know,

↝ Integrating Factor is evaluated as

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

So, on substituting the value of p, we get

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

$\rm :\longmapsto\:I.F. \: = \: {e} \: ^{log \: secx \: }$

$\rm \implies\:I.F. = secx$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: {e}^{logx} \: = \: x \bigg \}}$

Hence,

$\purple{\rm \implies\:\boxed{ \tt{ \: I.F. = secx}}}$

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Additional Information :-

1. Linear Differential equation is of the form

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

having Integrating Factor is

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

and Solution is

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: y \times I.F. = \displaystyle\int\sf (q \times I.F.) \: dx \: }}}$

2. Linear Differential equation of the form

$\red{\rm :\longmapsto\:\dfrac{dx}{dy} + px = q, \: where \: p \: and \: q \: \in \: f(y)}$

having Integrating Factor

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: I.F. = {e} \: ^{\displaystyle\int\sf p \: dy} \: }}}$

and Solution is

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: x \times I.F. = \displaystyle\int\sf (q \times I.F.) \: dy \: }}}$