4.0.1 Step 0

This step is similar to Kovacic algorithm. In it we determine necessary conditions for each case but it is done is more direct way in this version. Given \(y^{\prime \prime }=ry\), we write \(d=\frac {s}{t}\) then now we do square free factorization on \(t\) which gives

\[ t=t_{1}t_{2}^{2}t_{3}^{3}\cdots t_{m}^{m}\]

For example, if \(t=x^{2}\), then \(t_{1}=1,t_{2}=x\). And if \(t=3-x^{3}\) then \(t_{1}=-1,t_{2}=x^{3}-3\). And \(O\left ( \infty \right ) =\deg \left ( t\right ) -\deg \left ( s\right ) \). Then now we determine which case we are in by finding necessary conditions, This is done slightly different from the original Kovacic. So at the end of this step we know if \(L=\left [ 1\right ] \) (case 1) or \(L=\left [ 1,2\right ] \) (case 1 and case 2), or \(L=\left [ 2\right ] \) (case 2 only) or \(L=\left [ 4,6,12\right ] \) (case 3 only) and so on.

The necessary conditions are based on the square free factorization on \(t=t_{1}t_{2}^{2}t_{3}^{3}\cdots t_{m}^{m}\) and is summarized in Carolyn J. Smith paper (3) as (these are all the same necessary conditions as from original Kovacic paper) but expressed in terms of the square free factorization of \(t=t_{1}t_{2}^{2}\cdots t_{m}^{m}\) where \(r=\frac {s}{t}\).

1. \(L=\left [ 1\right ] \) (meaning case 1) if \(t_{i}=1\) for all odd \(i\geq 3\) (i.e. no odd order poles allowed other than \(1\)) and \(O\left ( \infty \right ) \) is even only. (i.e. allowed \(O\left ( \infty \right ) \) are \(\cdots ,-6,-4,-2,0,2,4,6,\cdots \).
2. \(L=\left [ 2\right ] \) (meaning case 2) if \(t_{2}\neq 1\) or \(t_{i}\neq 1\) for any odd \(i\geq 3\). (i.e. only pole of order 2 is allowed, and then poles of order \(3,5,7,\cdots \) are allowed.
3. \(L=\left [ 4,6,12\right ] \) (meaning case 3), if \(t_{i}=1\) for all \(i>2\) and \(O\left ( \infty \right ) \geq 2\). (i.e. poles of order 1,2 is only allowed).