Question

The integrating factor of (1+y²)+ (x-etan-¹y) dy/dx=0​

Answer 1

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Answer 2

Answer:

Solution

$\: \: \: \: \: \: \: \: \: \tt \: (1 + {y}^{2} ) +( x - {e}^{ { \tan }^{ - 1} y}) \dfrac{dy}{dx} = 0$

$\implies \rm \dfrac{(1 + {y}^{2})dx +( x - {e}^{ { \tan}^{ - 1}y)} dy}{dx} = 0$

$\implies\tt \: (1 + {y}^{2} ) dx+( x - {e}^{ { \tan }^{ - 1} y}) dy= 0$

$\implies \rm \dfrac{(1 + {y}^{2})dx }{(1 + {y}^{2} )}=\bigg( -\dfrac{x}{1 + {y}^{2} } +\dfrac{{e}^{ { \tan}^{ - 1}y } }{1 + {y}^{2} } \bigg)dy$

[ 1 + y² is divided in each term ]

$\implies \rm \: \dfrac{dx}{dy} + \rm\dfrac{1}{1 + {y}^{2} } .x = \dfrac{ {e}^{ { \tan}^{ - 1} y} }{1 + {y}^{2} } ....(1)$

This is linear differential equation

Where

$\rm \: P = \dfrac{1}{1 + {y}^{2} }$

$\rm \: Q = \dfrac{ {e}^{ { \tan}^{ - 1}y } }{1 + {y}^{2} }$

So, Integrating factor ( I F ) =

$\rm= {e}^{ \displaystyle\int \rm P \ dy }$

$\rm \: = {e}^{ \displaystyle\int \rm\dfrac{dy}{1 + {y}^{2} } }$

$\rm= {e}^{ { \tan }^{ - 1}y }$

So Integrating factor is

$\boxed{\rm= {e}^{ { \tan }^{ - 1}y }}$

Let- us find general solution also,

The Solution of given differential equation is

$\implies\rm \: x. {e}^{ \tan^{ - 1} y } = \displaystyle \int \rm\frac{{e}^{ \tan^{ - 1} y }.{e}^{ - \tan^{ - 1} y }}{1 + {y}^{2} }$

In R.H.S side put

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm{e}^{ { \tan}^{ - 1}y } = t$

$\implies \rm \dfrac{{e}^{ { \tan}^{ - 1}y } }{1 + {y}^{2} } = \dfrac{dt}{dy}$

$\implies \rm \dfrac{{e}^{ { \tan}^{ - 1}y } } {1 + {y}^{2} }.dy = dt$

Put it in R.H.S we get,

$\implies\rm \: x. {e}^{ \tan^{ - 1} y } = \displaystyle \int \rm \: t \: dt \: + c$

$\implies\rm \: x {e}^{ \tan^{ - 1} y } = \dfrac{ {t}^{2} }{2} + c$

$\underline{\boxed{\implies\rm \: x {e}^{ \tan^{ - 1} y } = \dfrac{ {{e}^{ {2 \tan}^{ - 1}y } }}{2} + c }}$

This us the required differential equation

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