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Question

The solution of the equation $fraction numerator d y over denominator d x end fraction equals fraction numerator x over denominator 2 y minus x end fraction$ is

$(x - y) (x + 2 y)^{2} = c$
$y = x + c$
$y = (2 y - x) + c$
$y = \frac{x}{2 y - x} + c$

The correct answer is: $left parenthesis x minus y right parenthesis left parenthesis x plus 2 y right parenthesis to the power of 2 end exponent equals c$

$fraction numerator d y over denominator d x end fraction equals fraction numerator x over denominator 2 y minus x end fraction$ . Put $y equals v x$ Þ $v plus x fraction numerator d v over denominator d x end fraction equals fraction numerator d y over denominator d x end fraction$
$v plus x fraction numerator d v over denominator d x end fraction equals fraction numerator x over denominator 2 v minus x end fraction equals fraction numerator 1 over denominator 2 v minus 1 end fraction$
$x fraction numerator d v over denominator d x end fraction equals fraction numerator 1 over denominator 2 v minus 1 end fraction minus v equals fraction numerator 1 minus 2 v to the power of 2 end exponent plus v over denominator 2 v minus 1 end fraction equals negative fraction numerator left parenthesis v minus 1 right parenthesis left parenthesis 2 v plus 1 right parenthesis over denominator 2 v minus 1 end fraction$
$fraction numerator left parenthesis 2 v minus 1 right parenthesis over denominator left parenthesis 2 v plus 1 right parenthesis left parenthesis v minus 1 right parenthesis end fraction equals fraction numerator negative d x over denominator x end fraction$ ; $fraction numerator 1 over denominator 3 left parenthesis v minus 1 right parenthesis end fraction plus fraction numerator 4 over denominator 3 left parenthesis 2 v plus 1 right parenthesis end fraction equals fraction numerator negative d x over denominator x end fraction$
$fraction numerator 1 over denominator 3 end fraction log invisible function application left parenthesis v minus 1 right parenthesis plus fraction numerator 4 over denominator 3 end fraction. fraction numerator 1 over denominator 2 end fraction log invisible function application left parenthesis 2 v plus 1 right parenthesis equals log invisible function application fraction numerator 1 over denominator x end fraction plus log invisible function application c$
$log invisible function application left parenthesis v minus 1 right parenthesis to the power of 1 divided by 3 end exponent plus log invisible function application left parenthesis 2 v plus 1 right parenthesis to the power of 2 divided by 3 end exponent equals log invisible function application fraction numerator c over denominator x end fraction blank equals left parenthesis v minus 1 right parenthesis to the power of 1 divided by 3 end exponent left parenthesis 2 v plus 1 right parenthesis to the power of 2 divided by 3 end exponent equals fraction numerator c over denominator x end fraction$
$open parentheses fraction numerator y minus x over denominator x end fraction close parentheses blank open parentheses fraction numerator 2 y plus x over denominator x end fraction close parentheses to the power of 2 end exponent equals fraction numerator c to the power of 3 end exponent over denominator x to the power of 3 end exponent end fraction$ Þ $left parenthesis x minus y right parenthesis left parenthesis x plus 2 y right parenthesis to the power of 2 end exponent equals c$ .

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Statement II: $stack P Q with rightwards arrow on top cross times stack R S with rightwards arrow on top equals 0$ and $stack P Q with rightwards arrow on top cross times stack S T with rightwards arrow on top not equal to stack 0 with rightwards arrow on top$

parallel

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