Answer

Using Euler-Cauchy equation, we know that \(x^{m}\) is a solution. Substituiting and taking out the common factor \(x^{m}\),

\[ \begin{aligned} m^{2} - 4m + 4 &= 0\newline \implies m &= 2 \end{aligned} \]

Since this is the case of double root, the general solution is

\[ \begin{aligned} y = \roundbr{c_{1} + c_{2}\ln x}x^{2} \end{aligned} \]