1.1.1 Example 1

1.1.2 Example 2

1.1.3 Example 3

1.1.4 Example 4

1.1.5 Example 5

1.1.6 Example 6

1.1.7 Example 7

1.1.8 Example 8

1.1.2 Example 2

1.1.3 Example 3

1.1.4 Example 4

1.1.5 Example 5

1.1.6 Example 6

1.1.7 Example 7

1.1.8 Example 8

\[ r=\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\] There is one pole at \(x=0\,\ \)of order \(4\). And \(O\left ( \infty \right ) =4-6=-2\). Conditions for case 1 are met. Since it has a pole of even order. Also \(O\left ( \infty \right ) \) 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 \(L=\left [ 1\right ] \)

\[ r=x \] There is one pole of order zero (an even pole). So case 1 or 3 qualify. But \(O\left ( \infty \right ) =0-1=-1\) which is odd. But case 1 and 3 require \(O\left ( \infty \right ) \) 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 \(L=\left [ {}\right ] \)

\begin {align*} r & =\frac {1}{x}-\frac {3}{16x^{2}}\\ & =\frac {16x-3}{16x^{2}} \end {align*}

There is pole at \(x=0\) of order \(2\). And \(O\left ( \infty \right ) =2-1=1\). Case 1 is not satisﬁed, since \(O\left ( \infty \right ) \) is not greater than 2. Also case 3 can not hold, since case 3 requires \(O\left ( \infty \right ) \) 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 \(O\left ( \infty \right ) \) to satisfy, then case 2 has been met. So this is case 2 only. \(L=\left [ 2\right ] \)

\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 \(L=\left [ 1,2,4,6,12\right ] \).

\[ r=\frac {-5x+27}{36\left ( x-1\right ) ^{2}}\] \(O\left ( \infty \right ) =2-1=1\). And \(r\) has pole at \(x=1\) of order 2. We see that \(O\left ( \infty \right ) \) is not satisﬁed for case 1 and case 3 (case 1 requires even or greater than 2 for \(O\left ( \infty \right ) \) and case 3 requires \(O\left ( \infty \right ) =2\).). So our only hope is case 2. Case 2 has no \(O\left ( \infty \right ) \) 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).

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

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

\[ r=\frac {4x^{2}-15}{4x^{2}}\] We see that \(O\left ( \infty \right ) =0\). From the table this means only case 1 and 2 are possible. (since case 2 has no conditions on \(O\left ( \infty \right ) \) and only case 1 allows zero order for \(O\left ( \infty \right ) \)). 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 \(L=\left [ 1,2\right ] \)