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The differential equation corresponding to the family of curves given by $a x^{2} + b y^{2} = 1,$ where a and b are arbitrary constants is

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$x \frac{d^{2} y}{{dx}^{2}} = \frac{dy}{dx}$

$x y \frac{d^{2} y}{{dx}^{2}} + x {(\frac{dy}{dx})}^{2} -y \frac{dy}{dx} =0$

$x y \frac{d^{2} y}{{dx}^{2}} + y \frac{dy}{dx} -x \frac{dy}{dx} = 0$

$x y \frac{d^{2} y}{{dx}^{2}} - x {(\frac{dy}{dx})}^{2} + y \frac{dy}{dx} = 0$

answer is B.

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Detailed Solution

$a x^{2} + b y^{2} = 1$

$2 a x + 2 b y y_{1} = 0$

$\Rightarrow \frac{{yy}_{1}}{x} = \frac{-a}{b}$

$\Rightarrow \frac{x [{yy}_{2} {+y}_{1}^{2}] - {yy}_{1}}{x^{2}} = 0$

$\Rightarrow {xyy}_{2} {+ xy}_{1}^{2} {- yy}_{1} = 0$

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