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If Y = a { X + √ X 2 + 1 } N + B { X − √ X 2 + 1 } − N , Prove that ( X 2 + 1 ) D 2 Y D X 2 + X D Y D X − N 2 Y = 0 Disclaimer: There is a Misprint in the Question, - Mathematics

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Question

\[\text { If y } = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} , \text { prove that }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \]

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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Solution

\[\text { We have,} \]

\[y = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} . . . (1)\]

\[\text { Differentiating y with respect to x, we get }\]

\[\frac{d y}{d x} =\text { an} \left\{ x + \sqrt{x^2 + 1} \right\}^{n - 1} \left( 1 + \frac{1}{2\sqrt{x^2 + 1}} \times 2x \right) - bn \left\{ x - \sqrt{x^2 + 1} \right\}^{- n - 1} \left( 1 - \frac{1}{2\sqrt{x^2 + 1}} \times 2x \right)\]

\[ = \text { an }\left\{ x + \sqrt{x^2 + 1} \right\}^{n - 1} \left( 1 + \frac{x}{\sqrt{x^2 + 1}} \right) - bn \left\{ x - \sqrt{x^2 + 1} \right\}^{- n - 1} \left( 1 - \frac{x}{\sqrt{x^2 + 1}} \right)\]

\[ = \text { an }\left\{ x + \sqrt{x^2 + 1} \right\}^{n - 1} \left( \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} \right) - bn \left\{ x - \sqrt{x^2 + 1} \right\}^{- n - 1} \left( \frac{\sqrt{x^2 + 1} - x}{\sqrt{x^2 + 1}} \right)\]

\[ = \text { an } \left\{ x + \sqrt{x^2 + 1} \right\}^{n - 1} \left( \frac{x + \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right) + bn \left\{ x - \sqrt{x^2 + 1} \right\}^{- n - 1} \left( \frac{x - \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right)\]

\[ = \left\{ a \left\{ x + \sqrt{x^2 + 1} \right\}^n \left( \frac{n}{\sqrt{x^2 + 1}} \right) + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} \right\}\left( \frac{n}{\sqrt{x^2 + 1}} \right)\]

\[ = \left( \frac{n}{\sqrt{x^2 + 1}} \right)y \left[ \text { From }(1) \right]\]

\[ \Rightarrow \sqrt{x^2 + 1}\frac{d y}{d x} = ny\]

\[\text { Squaring both sides, we get }\]

\[\left( x^2 + 1 \right) \left( \frac{d y}{d x} \right)^2 = n^2 y^2 . . . (2)\]

\[\text{ Differentiating (2) with respect to x, we get }\]

\[\left( x^2 + 1 \right)2\frac{d y}{d x} \times \frac{d^2 y}{d x^2} + 2x \left( \frac{d y}{d x} \right)^2 = n^2 \left( 2y\frac{d y}{d x} \right)\]

\[ \Rightarrow \left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\left( \frac{d y}{d x} \right) = n^2 \left( y \right)\]

\[ \Rightarrow \left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\left( \frac{d y}{d x} \right) - n^2 y = 0\]

\[\text { Hence, }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 .\]

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Chapter 12: Higher Order Derivatives - Exercise 12.1 [Page 18]

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RD Sharma Mathematics [English] Class 12
Chapter 12 Higher Order Derivatives
Exercise 12.1 | Q 53 | Page 18

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