Using Euler-Cauchy equation, we know that $x^{m}$ is a solution. Substituiting and taking out the common factor $x^{m}$, \begin{align} m^{2} - 4m + 4 &= 0\newline \implies m &= 2 \end{align}

Since this is the case of double root, the general solution is \begin{align} y = \roundbr{c_{1} + c_{2}\ln x}x^{2} \end{align}