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प्रश्न
\[\text { If y } = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} , \text { prove that }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \]
Disclaimer: There is a misprint in the question,
\[\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0\] must be written instead of
\[\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \] ?
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उत्तर
\[\text { We have,} \]
\[y = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} . . . (1)\]
\[\text { Differentiating y with respect to x, we get }\]
\[\frac{d y}{d x} =\text { an} \left\{ x + \sqrt{x^2 + 1} \right\}^{n - 1} \left( 1 + \frac{1}{2\sqrt{x^2 + 1}} \times 2x \right) - bn \left\{ x - \sqrt{x^2 + 1} \right\}^{- n - 1} \left( 1 - \frac{1}{2\sqrt{x^2 + 1}} \times 2x \right)\]
\[ = \text { an }\left\{ x + \sqrt{x^2 + 1} \right\}^{n - 1} \left( 1 + \frac{x}{\sqrt{x^2 + 1}} \right) - bn \left\{ x - \sqrt{x^2 + 1} \right\}^{- n - 1} \left( 1 - \frac{x}{\sqrt{x^2 + 1}} \right)\]
\[ = \text { an }\left\{ x + \sqrt{x^2 + 1} \right\}^{n - 1} \left( \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} \right) - bn \left\{ x - \sqrt{x^2 + 1} \right\}^{- n - 1} \left( \frac{\sqrt{x^2 + 1} - x}{\sqrt{x^2 + 1}} \right)\]
\[ = \text { an } \left\{ x + \sqrt{x^2 + 1} \right\}^{n - 1} \left( \frac{x + \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right) + bn \left\{ x - \sqrt{x^2 + 1} \right\}^{- n - 1} \left( \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right)\]
\[ = \left\{ a \left\{ x + \sqrt{x^2 + 1} \right\}^n \left( \frac{n}{\sqrt{x^2 + 1}} \right) + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} \right\}\left( \frac{n}{\sqrt{x^2 + 1}} \right)\]
\[ = \left( \frac{n}{\sqrt{x^2 + 1}} \right)y \left[ \text { From }(1) \right]\]
\[ \Rightarrow \sqrt{x^2 + 1}\frac{d y}{d x} = ny\]
\[\text { Squaring both sides, we get }\]
\[\left( x^2 + 1 \right) \left( \frac{d y}{d x} \right)^2 = n^2 y^2 . . . (2)\]
\[\text{ Differentiating (2) with respect to x, we get }\]
\[\left( x^2 + 1 \right)2\frac{d y}{d x} \times \frac{d^2 y}{d x^2} + 2x \left( \frac{d y}{d x} \right)^2 = n^2 \left( 2y\frac{d y}{d x} \right)\]
\[ \Rightarrow \left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\left( \frac{d y}{d x} \right) = n^2 \left( y \right)\]
\[ \Rightarrow \left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\left( \frac{d y}{d x} \right) - n^2 y = 0\]
\[\text { Hence, }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 .\]
