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Question

The solution of the equation $fraction numerator d y over denominator d x end fraction equals fraction numerator x plus y over denominator x minus y end fraction$ is

$c (x^{2} + y^{2})^{1 / 2} + e^{\tan^{- 1} (y / x)} = 0$
$c (x^{2} + y^{2})^{1 / 2} = e^{\tan^{- 1} (y / x)}$
$c (x^{2} - y^{2}) = e^{\tan^{- 1} (y / x)}$
None of these

The correct answer is: $c left parenthesis x to the power of 2 end exponent plus y to the power of 2 end exponent right parenthesis to the power of 1 divided by 2 end exponent equals e to the power of tan to the power of negative 1 end exponent invisible function application left parenthesis y divided by x right parenthesis end exponent$

Given equation, $fraction numerator d y over denominator d x end fraction equals fraction numerator x plus y over denominator x minus y end fraction$
It is a homogeneous equation so putting $y equals v x$
and $fraction numerator d y over denominator d x end fraction equals v plus x fraction numerator d v over denominator d x end fraction comma$ we get
$v plus x fraction numerator d v over denominator d x end fraction equals fraction numerator x plus v x over denominator x minus v x end fraction equals fraction numerator 1 plus v over denominator 1 minus v end fraction$
Þ $x fraction numerator d v over denominator d x end fraction equals fraction numerator 1 plus v to the power of 2 end exponent over denominator 1 minus v end fraction$
Þ $fraction numerator 1 over denominator x end fraction d x equals open parentheses fraction numerator 1 over denominator 1 plus v to the power of 2 end exponent end fraction minus fraction numerator v over denominator 1 plus v to the power of 2 end exponent end fraction close parentheses d v$
Þ $log subscript e end subscript invisible function application x equals tan to the power of negative 1 end exponent invisible function application v minus fraction numerator 1 over denominator 2 end fraction log invisible function application left parenthesis 1 plus v to the power of 2 end exponent right parenthesis plus log subscript e end subscript invisible function application c$
Substituting $v equals fraction numerator y over denominator x end fraction comma$ we get
$log subscript e end subscript invisible function application x equals tan to the power of negative 1 end exponent invisible function application fraction numerator y over denominator x end fraction minus fraction numerator 1 over denominator 2 end fraction log invisible function application open square brackets 1 plus open parentheses fraction numerator y over denominator x end fraction close parentheses to the power of 2 end exponent close square brackets plus log subscript e end subscript invisible function application c$
Þ $c left parenthesis x to the power of 2 end exponent plus y to the power of 2 end exponent right parenthesis to the power of 1 divided by 2 end exponent equals e to the power of tan to the power of negative 1 end exponent invisible function application left parenthesis y divided by x right parenthesis end exponent$ .

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parallel

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