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Question

Solution of differential equation $fraction numerator d y over denominator d x end fraction equals fraction numerator y minus x over denominator y plus x end fraction$ is

$\log_{e} (x^{2} + y^{2}) + 2 \tan^{- 1} \frac{y}{x} + c = 0$
$\frac{y^{2}}{2} + x y = x y - \frac{x^{2}}{2} + c$
$(1 + \frac{x}{y}) y = (1 - \frac{x}{y}) x + c$
$y = x - 2 \log_{e} y + c$

The correct answer is: $log subscript e end subscript invisible function application left parenthesis x to the power of 2 end exponent plus y to the power of 2 end exponent right parenthesis plus 2 tan to the power of negative 1 end exponent invisible function application fraction numerator y over denominator x end fraction plus c equals 0$

Put $y equals v x$ Þ $fraction numerator d y over denominator d x end fraction equals v plus x fraction numerator d v over denominator d x end fraction$
$therefore v plus x fraction numerator d v over denominator d x end fraction equals fraction numerator v minus 1 over denominator v plus 1 end fraction$ Þ $x fraction numerator d v over denominator d x end fraction equals fraction numerator v minus 1 over denominator v plus 1 end fraction minus v$
Þ $x fraction numerator d v over denominator d x end fraction equals negative fraction numerator v to the power of 2 end exponent plus 1 over denominator v plus 1 end fraction$ Þ $not stretchy integral fraction numerator d x over denominator x end fraction equals negative not stretchy integral fraction numerator v plus 1 over denominator v to the power of 2 end exponent plus 1 end fraction blank d v$
Þ $negative log subscript e end subscript invisible function application x equals fraction numerator 1 over denominator 2 end fraction not stretchy integral fraction numerator 2 v over denominator v to the power of 2 end exponent plus 1 end fraction d v plus not stretchy integral fraction numerator 1 over denominator v to the power of 2 end exponent plus 1 end fraction d v$
Þ $negative log subscript e end subscript invisible function application x equals fraction numerator 1 over denominator 2 end fraction log invisible function application left parenthesis v to the power of 2 end exponent plus 1 right parenthesis plus tan to the power of negative 1 end exponent invisible function application v plus c$
Þ $negative 2 log subscript e end subscript invisible function application x equals log invisible function application open parentheses fraction numerator x to the power of 2 end exponent plus y to the power of 2 end exponent over denominator x to the power of 2 end exponent end fraction close parentheses plus 2 tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator y over denominator x end fraction close parentheses plus c$
Þ $log subscript e end subscript invisible function application left parenthesis x to the power of 2 end exponent plus y to the power of 2 end exponent right parenthesis plus 2 tan to the power of negative 1 end exponent invisible function application fraction numerator y over denominator x end fraction plus c equals 0$ .

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