$If y = x^{n} \{a \cos (\log x) + b \sin (\log x)\}, prove that x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) x \frac{d y}{d x} + (1 + n^{2}) y = 0 .$

Disclaimer: There is a misprint in the question. It must be $x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) x \frac{d y}{d x} + (1 + n^{2}) y = 0$ instead of $x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) \frac{d y}{d x} + (1 + n^{2}) y = 0$ .

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Solution

$We have, y = x^{n} \{a \cos (\log x) + b \sin (\log x)\} . . . (1) Differentiating y with respect to x, we get \frac{d y}{d x} = n x^{n - 1} \{a \cos (\log x) + b \sin (\log x)\} + x^{n} \{- a \sin (\log x) \times \frac{1}{x} + b \cos (\log x) \times \frac{1}{x}\} = \frac{n}{x} x^{n} \{a \cos (\log x) + b \sin (\log x)\} + x^{n - 1} \{- a \sin (\log x) + b \cos (\log x)\} = \frac{n}{x} y + x^{n - 1} \{- a \sin (\log x) + b \cos (\log x)\} [From (1)] \Rightarrow x^{n - 1} \{- a \sin (\log x) + b \cos (\log x)\} = \frac{d y}{d x} - \frac{n}{x} y . . . (2) Differentiating \frac{d y}{d x} with respect to x, we get \frac{d^{2} y}{d x^{2}} = \frac{n}{x} \frac{d y}{d x} - \frac{n y}{x^{2}} + (n - 1) x^{n - 2} \{- a \sin (\log x) + b \cos (\log x)\} + x^{n - 1} \{- a \cos (\log x) \times \frac{1}{x} - b \sin (\log x) \times \frac{1}{x}\} = \frac{n}{x} \frac{d y}{d x} - \frac{n y}{x^{2}} + (n - 1) \frac{x^{n - 1}}{x} \{- a \sin (\log x) + b \cos (\log x)\} - \frac{x^{n}}{x^{2}} \{a \cos (\log x) + b \sin (\log x)\} = \frac{n}{x} \frac{d y}{d x} - \frac{n y}{x^{2}} + (\frac{n - 1}{x}) (\frac{d y}{d x} - \frac{n}{x} y) - \frac{y}{x^{2}} [From (1) and (2)] = \frac{n}{x} \frac{d y}{d x} - \frac{n y}{x^{2}} + (\frac{n - 1}{x}) \frac{d y}{d x} - \frac{n (n - 1) y}{x^{2}} - \frac{y}{x^{2}} = \frac{d y}{d x} (\frac{n + n - 1}{x}) - \frac{(n + n^{2} - n + 1) y}{x^{2}} = (\frac{2 n - 1}{x}) \frac{d y}{d x} - \frac{(n^{2} + 1) y}{x^{2}} \Rightarrow x^{2} \frac{d^{2} y}{d x^{2}} - x (2 n - 1) \frac{d y}{d x} + (n^{2} + 1) y = 0 Hence, x^{2} \frac{d^{2} y}{d x^{2}} + (1 - 2 n) x \frac{d y}{d x} + (1 + n^{2}) y = 0 .$

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