As with the constant coefficient equations, we begin by writing down the characteristic equation. Doing a simple computation,

\[ \begin{aligned} 0 &=r(r-1)+5 r+12 \\[4pt] &=r^{2}+4 r+12 \\[4pt] &=(r+2)^{2}+8, \\[4pt] -8 &=(r+2)^{2}, \end{aligned} \label{2.97} \]

one determines the roots are \(r=-2 \pm 2 \sqrt{2} i\). Therefore, the general solution is

\[y(x)=\left[c_{1} \cos (2 \sqrt{2} \ln |x|)+c_{2} \sin (2 \sqrt{2} \ln |x|)\right] x^{-2} \nonumber\]

Deriving the solution for Case 2 for the Cauchy-Euler equations works in the same way as the second for constant coefficient equations, but it is a bit messier. First note that for the real root, \(r=r_{1}\), the characteristic equation has to factor as \(\left(r-r_{1}\right)^{2}=0 .\) Expanding, we have

\(r^{2}-2 r_{1} r+r_{1}^{2}=0\)

The general characteristic equation is

\(a r(r-1)+b r+c=0\)

Dividing this equation by \(a\) and rewriting, we have

\(r^{2}+\left(\dfrac{b}{a}-1\right) r+\dfrac{c}{a}=0\)

Comparing equations, we find

\(\dfrac{b}{a}=1-2 r_{1}, \quad \dfrac{c}{a}=r_{1}^{2}\)

So, the Cauchy-Euler equation for this case can be written in the form

\(x^{2} y^{\prime \prime}+\left(1-2 r_{1}\right) x y^{\prime}+r_{1}^{2} y=0\)

Now we seek the second linearly independent solution in the form \(y_{2}(x)=\) \(v(x) x^{r_{1}}\). We first list this function and its derivatives,

\[\begin{equation} \begin{aligned} &y_{2}(x)=v x^{r_{1}} \\[4pt] &y_{2}^{\prime}(x)=\left(x v^{\prime}+r_{1} v\right) x^{r_{1}-1} \\[4pt] &y_{2}^{\prime \prime}(x)=\left(x^{2} v^{\prime \prime}+2 r_{1} x v^{\prime}+r_{1}\left(r_{1}-1\right) v\right) x^{r_{1}-2} \end{aligned}\end{equation}\label{2.98} \]

Inserting these forms into the differential equation, we have

\[ \begin{aligned} 0 &=x^{2} y^{\prime \prime}+\left(1-2 r_{1}\right) x y^{\prime}+r_{1}^{2} y \\[4pt] &=\left(x v^{\prime \prime}+v^{\prime}\right) x^{r_{1}+1} \end{aligned} \label{2.99} \]

Thus, we need to solve the equation

\(x v^{\prime \prime}+v^{\prime}=0\)

Or

\(\dfrac{v^{\prime \prime}}{v^{\prime}}=-\dfrac{1}{x}\)

Integrating, we have

\(\ln \left|v^{\prime}\right|=-\ln |x|+C\)

where \(A=\pm e^{C}\) absorbs \(C\) and the signs from the absolute values. Exponentiating, we obtain one last differential equation to solve,

\(v^{\prime}=\dfrac{A}{x}\)

Thus,

\(v(x)=A \ln |x|+k\)

So, we have found that the second linearly independent equation can be written as

\(y_{2}(x)=x^{r_{1}} \ln |x|.\)

Therefore, the general solution is found as \(y(x)=\left(c_{1}+c_{2} \ln |x|\right) x^{r}\).