If y=x+1+x2n, then 1+x2d2ydx2+xdydx=

y=x+1+x2n dydx=nx+1+x2n-11+1(2x)21+x2 =nx+1+x21+x2n=ny1+x2 1+x2y12=n2 y2 1+x22y1 y2+y122x=n2 2y·y1 1+x2y2+xy1=n2y

If y=x+1+x2n, then 1+x2d2ydx2+xdydx=

y=x+1+x2n dydx=nx+1+x2n-11+1(2x)21+x2 =nx+1+x21+x2n=ny1+x2 1+x2y12=n2 y2 1+x22y1 y2+y122x=n2 2y·y1 1+x2y2+xy1=n2y

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If $y = {(x + \sqrt{1 + x^{2}})}^{n},$ then $(1 + x^{2}) \frac{d^{2} y}{{dx}^{2}} + x \frac{dy}{dx} =$

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$2 n^{2} y$

$n^{2} y$

$- n^{2} y$

$- y$

answer is A.

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Detailed Solution

$y = {(x + \sqrt{1 + x^{2}})}^{n} \frac{dy}{dx} = n {(x + \sqrt{1 + x^{2}})}^{n - 1} (1 + \frac{1 (2 x)}{2 \sqrt{1 + x^{2}}}) = n {(\frac{x + \sqrt{1 + x^{2}}}{\sqrt{1 + x^{2}}})}^{n} = \frac{ny}{\sqrt{1 + x^{2}}} (1 + x^{2}) {(y_{1})}^{2} = n^{2} y^{2} (1 + x^{2}) 2 y_{1} y_{2} + {y_{1}}^{2} (2 x) = n^{2} 2 y \cdot y_{1} (1 + x^{2}) y_{2} + {xy}_{1} = n^{2} y$