If $x^{2} + y^{2} = 1$ , then

{y y}^{''} - {(2 y^{'})}^{2} + 1 = 0

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{y y}^{''} + {(y^{'})}^{2} + 1 = 0

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{y y}^{''} - {(y^{'})}^{2} - 1 = 0

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{y y}^{''} + 2 {(y^{'})}^{2} + 1 = 0

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Solution

The correct option is B ${y y}^{''} + {(y^{'})}^{2} + 1 = 0$
Given, $x^{2} + y^{2} = 1$
On differentiating with respect to $x$ , we get
$2 x + 2 {y y}^{'} = 0$
$\Rightarrow x + {y y}^{'} = 0$
Again, differentiating, we get
$1 + {y y}^{''} + {(y^{'})}^{2} = 0$

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