Prove that $y = a \cos \left(\log x\right) + b \sin \left(\log x\right)$ is a solution of the differential equation $x^2 \; \dfrac{d^2 y}{dx^2} + x \; \dfrac{dy}{dx} + y = 0$ where a and b are arbitrary constants.
Given: $y = a \cos \left(\log x\right) + b \sin \left(\log x\right)$ $\;\;\; \cdots \; (1)$
Differentiating equation $(1)$ w.r.t x gives
$\dfrac{dy}{dx} = - a \sin \left(\log x\right) \times \dfrac{1}{x} + b \cos \left(\log x\right) \times \dfrac{1}{x}$ $\;\;\; \cdots \; (2)$
Differentiating equation $(2)$ w.r.t x gives
$\begin{aligned}
\dfrac{d^2 y}{dx^2} & = \dfrac{-a}{x} \cos \left(\log x\right) \times \dfrac{1}{x} + \dfrac{a}{x^2} \sin \left(\log x\right) - \dfrac{b}{x^2} \cos \left(\log x\right) - \dfrac{b}{x} \sin \left(\log x\right) \times \dfrac{1}{x} \\\\
& = \dfrac{-1}{x^2} \left[a \cos \left(\log x\right) + b \sin \left(\log x\right)\right] - \dfrac{1}{x} \left[\dfrac{b}{x} \cos \left(\log x\right) - \dfrac{a}{x} \sin \left(\log x\right)\right] \;\;\; \cdots \; (3)
\end{aligned}$
$\therefore$ $\;$ In view of equations $(1)$ and $(2)$, equation $(3)$ becomes
$\dfrac{d^2 y}{dx^2} = \dfrac{-1}{x^2} \times y - \dfrac{1}{x} \times \dfrac{dy}{dx}$
i.e. $x^2 \; \dfrac{d^2 y}{dx^2} + x \; \dfrac{dy}{dx} + y = 0$
is the required differential equation.