The differential equation of the curve xA−1+yA+1=1, is given by (A is arbitrary constant, y'=dydx )

1A−1+yA+1=0 y'=1+A1−A⇒2A2=y'−1y1+1Now put in curve to get (y'+1)(y−xy')=2y'

The differential equation of the curve xA−1+yA+1=1, is given by (A is arbitrary constant, y'=dydx )

1A−1+yA+1=0 y'=1+A1−A⇒2A2=y'−1y1+1Now put in curve to get (y'+1)(y−xy')=2y'

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The differential equation of the curve $\frac{x}{A - 1} + \frac{y}{A + 1} = 1,$ is given by (A is arbitrary constant, $y' = \frac{d y}{d x}$ )

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$(y' - 1) (y + x y') = 2 y'$

$(y' + 1) (y + x y') = y'$

$(y' + 1) (y - x y') = 2 y'$

$(y' + 1) (y - x y') = y'$

answer is C.

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$\frac{1}{A - 1} + \frac{y}{A + 1} = 0$
$y' = \frac{1 + A}{1 - A} \Rightarrow \frac{2 A}{2} = \frac{y' - 1}{y^{1} + 1}$
Now put in curve to get $(y' + 1) (y - x y') = 2 y'$

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The differential equation of the curve xA−1+yA+1=1, is given by (A is arbitrary constant, y'=dydx )

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The differential equation of the curve $\frac{x}{A - 1} + \frac{y}{A + 1} = 1,$ is given by (A is arbitrary constant, $y' = \frac{d y}{d x}$ )