1.1 Examples how to determine which case the ode belongs into

1.1.1 Example 1

$$r=\frac{4x^6-8x^5+12x^4+4x^3+7x^2-20x+4}{4x^4}$$ There is one pole at $x=0$of order $4$. And $O\left(\mathrm{\infty }\right)=4-6=-2$. Conditions for case 1 are met. Since it has a pole of even order. Also $O\left(\mathrm{\infty }\right)$ is even. Case 2 are not satisfied, since there is no pole of order $2$ and no odd pole of order greater than $2$ exist. Case 3 is also not met, since the pole is order $4$ and case 3 will only work if pole is order 1 or 2. Hence $L=\left[1\right]$

1.1.2 Example 2

$$r=x$$ There is one pole of order zero (an even pole). So case 1 or 3 qualify. But $O\left(\mathrm{\infty }\right)=0-1=-1$ which is odd. But case 1 and 3 require $O\left(\mathrm{\infty }\right)$ be even. Hence case 1,2,3 all fail. This is case 4 where there is no Liouvillian solution. This is known already, because this is the known Airy ode $y^{\prime \prime }=xy$. Its solution are the Airy special functions. These are not Liouvillian solutions. Hence $L=\left[\right]$

1.1.3 Example 3

$$\begin{array}{rl}r& =\frac{1}{x}-\frac{3}{16x^2}\\ & =\frac{16x-3}{16x^2}\end{array}$$

There is pole at $x=0$ of order $2$. And $O\left(\mathrm{\infty }\right)=2-1=1$. Case 1 is not satisfied, since $O\left(\mathrm{\infty }\right)$ is not greater than 2. Also case 3 can not hold, since case 3 requires $O\left(\mathrm{\infty }\right)$ be at least order 2 and here it is 1. Only possibility left is case 2. There is one pole of order 2. Since case 2 have no conditions on $O\left(\mathrm{\infty }\right)$ to satisfy, then case 2 has been met. So this is case 2 only. $L=\left[2\right]$

1.1.4 Example 4

$$\begin{array}{rl}r& =-\frac{3}{16x^2}-\frac{2}{9{(x-1)}^2}+\frac{3}{16x(x-1)}\\ & =-\frac{-32x^2+27x-27}{144x^2{(x-1)}^2}\end{array}$$

There is pole at $x=0$ of order 2, and pole at $x=1$ of order 2. And $O\left(\mathrm{\infty }\right)=4-2=2.$we see that $O\left(\mathrm{\infty }\right)$ is satisfied for case 1 and case 3. Recall that case 2 has no $O\left(\mathrm{\infty }\right)$ 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 $L=[1,2,4,6,12]$.

1.1.5 Example 5

$$r=\frac{-5x+27}{36{(x-1)}^2}$$ $O\left(\mathrm{\infty }\right)=2-1=1$. And $r$ has pole at $x=1$ of order 2. We see that $O\left(\mathrm{\infty }\right)$ is not satisfied for case 1 and case 3 (case 1 requires even or greater than 2 for $O\left(\mathrm{\infty }\right)$ and case 3 requires $O\left(\mathrm{\infty }\right)=2$.). So our only hope is case 2. Case 2 has no $O\left(\mathrm{\infty }\right)$ conditions. But it needs to have at least one pole of order 2 or a pole which is odd order and greater than 2. This is satisfied here, since pole is order 2. Hence only case 2 is possible. Hence $L=[2]$. I do not understand why paper says all three cases are possible for this. This seems to be an error in the paper (1).

1.1.6 Example 6

$$r=(x^2+3)$$ There is zero order pole. (even order). $O\left(\mathrm{\infty }\right)=0-2=-2$. Hence only case 1 is possible. $L=\left[1\right]$

1.1.7 Example 7

$$r=\frac{1}{x^2}$$ One pole at $x=0$ of order 2. And $O\left(\mathrm{\infty }\right)=2-0=2$. Case 1 is satisfied. Also case 2 since pole is even order. Also case 3 is satisfied. Hence all three cases are satisfied. $L=[1,2,4,6,12]$

1.1.8 Example 8

$$r=\frac{4x^2-15}{4x^2}$$ We see that $O\left(\mathrm{\infty }\right)=0$. From the table this means only case 1 and 2 are possible. (since case 2 has no conditions on $O\left(\mathrm{\infty }\right)$ and only case 1 allows zero order for $O\left(\mathrm{\infty }\right)$). We see there is a pole at $x=0$ of order 2. This is allowed by both case 1 and case 2. Hence case 1,2 are possible and $L=[1,2]$