Let $y = a cos (log x) + b sin (log x)$ is a solution of the differential equation $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0$ .Prove.

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Solution

$\begin{matrix} y = a cos (log x) + b sin (log x) \frac{d y}{d x} = - a sin (log x) \frac{1}{x} + b cos (log x) \frac{1}{x} \frac{d y}{d x} = \frac{1}{x} (b cos (log x) - a sin (log x)) \frac{d^{2} y}{d x^{2}} = - \frac{1}{x^{2}} (- b cos (log x) + a sin (log x)) + \frac{1}{x} (b sin (log x) \frac{1}{x} - a cos (log x) \frac{1}{x}) \frac{d^{2} y}{d x^{2}} = \frac{1}{x^{2}} (- b cos (log x) + a sin (log x) - b sin (log x) - a cos (log x)) x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = - b cos (log x) + a sin (log x) - b sin (log x) - a cos (log x) + b cos (log x) - a sin (log x) + a cos (log x) + b sin (log x) = 0 \end{matrix}$

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y = a c o s (l o g x) + b s i n (l o g x)

x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0.

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