Answer
$y = {x^3}{\text{ is a solution}}$
Work Step by Step
$$\eqalign{
& y = {x^3},{\text{ }}{x^2}y'' - 6y = 0 \cr
& {\text{Differentiate the given function}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^3}} \right] \cr
& \frac{{dy}}{{dx}} = 3{x^2} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {3{x^2}} \right] = 6x \cr
& or \cr
& y' = 3{x^2}{\text{ and }}y'' = 6x \cr
& \cr
& {\text{Substitute }}y''{\text{ and }}y{\text{ into the given differential equation}} \cr
& \underbrace {{x^2}y'' - 6y = 0}_ \downarrow \cr
& {x^2}\left( {6x} \right) - 6\left( {{x^3}} \right) = 0 \cr
& 6{x^3} - 6{x^3} = 0 \cr
& {\text{ }}0 = 0 \cr
& {\text{The statement is true}}{\text{, then }}y = {x^3}{\text{ is a solution}} \cr
& {\text{of the given differential equation}}{\text{.}} \cr} $$
