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If $y=\left(x+\sqrt{1+x^{2}}\right)^{n},$ then $\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}$ is
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Verified Answer
The correct answer is:
$n^{2} y$
$y=\left(x+\sqrt{1+x^{2}}\right)^{n}$
$$
\begin{array}{c}
\frac{d y}{d x}=n\left(x+\sqrt{1+x^{2}}\right)^{n-1}\left(1+\frac{1}{2}\left(1+x^{2}\right)^{-1 / 2} \cdot 2 x\right) \\
\qquad \begin{array}{c}
\frac{d y}{d x}=n\left(x+\sqrt{1+x^{2}}\right)^{n-1} \frac{\left(\sqrt{1+x^{2}}+x\right)}{\sqrt{1+x^{2}}} \\
=\frac{n\left(\sqrt{1+x^{2}}+x\right)^{n}}{\sqrt{1+x^{2}}}
\end{array}
\end{array}
$$
or $\sqrt{1+x^{2}} \frac{d y}{d x}=n y$ or $\sqrt{1+x^{2}} y_{1}=n y$
$$
\left(y_{1}=\frac{d y}{d x}\right)
$$
Squaring, $\left(1+x^{2}\right) y_{1}^{2}=n^{2} y^{2}$
Differentiating,
$$
\left(1+x^{2}\right) 2 y_{1} y_{2}+y_{1}^{2} \cdot 2 x=n^{2} \cdot 2 y y_{1}
$$
or $\left(1+x^{2}\right) y_{2}+x y_{1}=n^{2} y$
$$
\begin{array}{c}
\frac{d y}{d x}=n\left(x+\sqrt{1+x^{2}}\right)^{n-1}\left(1+\frac{1}{2}\left(1+x^{2}\right)^{-1 / 2} \cdot 2 x\right) \\
\qquad \begin{array}{c}
\frac{d y}{d x}=n\left(x+\sqrt{1+x^{2}}\right)^{n-1} \frac{\left(\sqrt{1+x^{2}}+x\right)}{\sqrt{1+x^{2}}} \\
=\frac{n\left(\sqrt{1+x^{2}}+x\right)^{n}}{\sqrt{1+x^{2}}}
\end{array}
\end{array}
$$
or $\sqrt{1+x^{2}} \frac{d y}{d x}=n y$ or $\sqrt{1+x^{2}} y_{1}=n y$
$$
\left(y_{1}=\frac{d y}{d x}\right)
$$
Squaring, $\left(1+x^{2}\right) y_{1}^{2}=n^{2} y^{2}$
Differentiating,
$$
\left(1+x^{2}\right) 2 y_{1} y_{2}+y_{1}^{2} \cdot 2 x=n^{2} \cdot 2 y y_{1}
$$
or $\left(1+x^{2}\right) y_{2}+x y_{1}=n^{2} y$
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