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
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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$\mathrm{n}^2 \mathrm{y}$
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$-n^2 y$
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$-y$
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$2 x^2 y$
Solution
$
y = (x + \sqrt{1 + x^2})^n\\
\frac{dy}{dx} = n(x + \sqrt{1 + x^2})^{n-1} \left( 1 + \frac{1}{2}(1 + x^2)^{-\frac{1}{2}} \cdot 2x \right)\\
\sqrt{1 + x^2} \frac{dy}{dx} = ny \quad \text{or} \quad \sqrt{1 + x^2}y_1 = ny \quad \text{(Here, } y_1 = \frac{dy}{dx} \text{)}
\text{Squaring, } (1 + x^2)y_1^2 = n^2y^2
\\
\text{Differentiating, } (1 + x^2)2y_1y_2 + y_1^2 \cdot 2x = n^2 \cdot 2yy_1 \quad \\
\text{(Here, } y_2 = \frac{d^2y}{dx^2} \text{)}
\text{or } (1 + x^2)y_2 + xy_1 = x^2y
$
Asked in: JEE Main 2002
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