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If Y = X N { a Cos ( Log X ) + B Sin ( Log X ) } , Prove that X 2 D 2 Y D X 2 + ( 1 − 2 N ) X D Y D X + ( 1 + N 2 ) Y = 0 Disclaimer: There is a Misprint in the Question. It Must Be - Mathematics

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प्रश्न

\[\text { If y } = x^n \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\}, \text { prove that } x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0 \] Disclaimer: There is a misprint in the question. It must be 

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

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

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उत्तर

\[\text { We have,} \]

\[y = x^n \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\} . . . (1)\]

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

\[\frac{d y}{d x} = n x^{n - 1} \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\} + x^n \left\{ - a \sin\left( \log x \right) \times \frac{1}{x} + b \cos\left( \log x \right) \times \frac{1}{x} \right\}\]

\[ = \frac{n}{x} x^n \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\} + x^{n - 1} \left\{ - a \sin\left( \log x \right) + b \cos\left( \log x \right) \right\}\]

\[ = \frac{n}{x}y + x^{n - 1} \left\{ - a \sin\left( \log x \right) + b \cos\left( \log x \right) \right\} \left[\text {  From }(1) \right]\]

\[ \Rightarrow x^{n - 1} \left\{ - a \sin\left( \log x \right) + b \cos\left( \log x \right) \right\} = \frac{d y}{d x} - \frac{n}{x}y . . . (2)\]

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

\[\frac{d^2 y}{d x^2} = \frac{n}{x}\frac{d y}{d x} - \frac{ny}{x^2} + \left( n - 1 \right) x^{n - 2} \left\{ - a \sin\left( \log x \right) + b \cos\left( \log x \right) \right\} + x^{n - 1} \left\{ - a \cos\left( \log x \right) \times \frac{1}{x} - b \sin\left( \log x \right) \times \frac{1}{x} \right\}\]

\[ = \frac{n}{x}\frac{d y}{d x} - \frac{ny}{x^2} + \left( n - 1 \right)\frac{x^{n - 1}}{x}\left\{ - a \sin\left( \log x \right) + b \cos\left( \log x \right) \right\} - \frac{x^n}{x^2}\left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\}\]

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

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

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

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

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

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

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अध्याय 12: Higher Order Derivatives - Exercise 12.1 [पृष्ठ १८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 12 Higher Order Derivatives
Exercise 12.1 | Q 52 | पृष्ठ १८

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