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Implicit Differentiation

lf y=e-x2 t...

Question

lf $y = e^{- x^{2}}$ then $y_{n + 2} + 2 x . y_{n + 1} + 2 (n + 1) y_{n} =$

A

0

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B

1

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C

y

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D

- y

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Solution

The correct option is A $0$
Given, $y = e^{- x^{2}}$
$y^{'} = - 2 x e^{- x^{2}}$
$y^{''} = - 2 x e^{- x^{2}}$ $+ 4 x^{2} e^{- x^{2}}$
$y^{''} = - 2 y - 2 x y^{'}$
$y^{''} + 2 y + 2 x y^{'} = 0$
$y^{'''} + 2 y^{'} + 2 y^{'} + 2 x y^{''} = 0$
$∴ y_{n + 2} + 2 x y_{n + 1} + 2 (n + 1) y_{n} = 0$

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