Solution: Rewriting the given differential equation, by multiplying with $y^2$ we get \[ x^2 y^{\prime\prime} = 2xy^\prime \implies x^2 y^{\prime\prime} - 2xy^\prime = 0. \] The above differential equation is in the Cauchy-Euler form which has indicial equation \[ m(m-1) - 2m = 0. \] The roots of the above indicial equations are $m = 0$ and $m=3$. Therefore, the general solution of the given differential equation will be \[ y(x) = c_1 + c_2 t^3. \]