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

Since this is the case of double root, the general solution is $$\begin{array}{r}y=(c_1+c_2\ln x)x^2\end{array}$$