5.3.1.3 Step 3

In this step we first attempt to find a polynomial \(p\left ( x\right ) \) of degree \(d\), for \(\omega \) found in step 2. This is done by solving

\[ p^{\prime \prime }+2\omega p^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) p=0 \]

For example, if \(d=2\), then we let \(p\left ( x\right ) =x^{2}+ax+b\) and if \(\omega \) happened to be say \(\frac {1}{x^{2}}-\frac {3}{2x}+x-1\), then by substituting these in the above, we can solve for \(a,b\) (if a solution exist). Then the solution to \(y^{\prime \prime }=ry\) is \(y=p\left ( x\right ) e^{\int \omega dx}\). If the degree \(d=1\) then we guess \(p\left ( x\right ) =x+a\) and try to solve for \(a\). If the degree \(d=0\), then we let \(p\left ( x\right ) =1\), a constant. In the special case of \(p\left ( x\right ) =1\), there is no coefficients \(a_{i}\) to solve for. So we would just need to verify that

\[ \omega ^{\prime }+\omega ^{2}-r=0 \]

In this case.

This completes the full algorithm for case 1. We will now go over many examples for case 1, showing how to implement this algorithm for each example.

The hardest part of the kovacic algorithm is just finding all the \(\left [ \sqrt {r}\right ] _{c},\alpha _{c}^{\pm },\left [ \sqrt {r}\right ] _{\infty },\alpha _{\infty }^{\pm }\). Once these are found, the rest of the algorithm is much more direct.