If $x^{2} - 2 x^{2} y^{2} + 5 x + y - 5 = 0$ and $y (1) = 1$ , then

y^{'} (1) = 1

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y^{''} (1) = - \frac{4}{3}

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y^{''} (1) = - \frac{22}{3}

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y^{'} (1) = \frac{2}{3}

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Solution

The correct options are
A $y^{'} (1) = 1$
C $y^{''} (1) = - \frac{22}{3}$
$2 x - 4 x y^{2} - 4 x^{2} y y^{'} + 5 + y^{'} = 0$
Substitute $x = 1, y = 1$
$2 - 4 - 4 y^{'} + 5 + y^{'} = 0$
$\Rightarrow 3 y^{'} = 3 \Rightarrow y^{'} = 1$
$\Rightarrow 2 - 4 y^{2} - 8 x y y^{'} - 4 x^{2} y^{' 2} - 8 x y y^{'} - 4 x^{2} y y " + y "= 0$
Substitute $x = 1, y = 1, y^{'} = 1$
$3 y^{''} = - 22$
$\Rightarrow y^{''} = - \frac{22}{3}$

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If x2−2x2y2+5x+y−5=0 and y(1)=1, then

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