1.2.4 Example 4

\begin{align*} r & =-\frac {3}{16x^{2}}-\frac {2}{9\left ( x-1\right ) ^{2}}+\frac {3}{16x\left ( x-1\right ) }\\ & =-\frac {-32x^{2}+27x-27}{144x^{2}\left ( x-1\right ) ^{2}}\end{align*}

There is pole at \(x=0\) of order 2, and pole at \(x=1\) of order 2. And \(O\left ( \infty \right ) =4-2=2.\)we see that \(O\left ( \infty \right ) \) is satisfied for case 1 and case 3. Recall that case 2 has no \(O\left ( \infty \right ) \) conditions. The pole order is satisfied for case 1 (must have even order or order 1), also the pole order is satisfied for case 2 (have at least one pole of order 2), and pole order is satisfied for case 3 (can only have poles of order 1 or 2). So all three cases are satisfied. Remember that just because the necessary conditions are met, this does not mean a Liouvillian solution exists. Hence \(L=\left [ 1,2,4,6,12\right ] \).