4.0.4 step 3

The input to this step is the integer \(d\) and \(\Theta \) found from Eq (7,8) as described in step 2 and also \(r\) which comes from \(z^{\prime \prime }=rz\).

This step is broken into these parts. First we find the \(p_{-1}\left ( x\right ) \) polynomial. If we are to solve for its coefficients, then next we build the minimal polynomial from the \(p_{i}\left ( x\right ) \) polynomials constructed during finding \(p_{-1}\left ( x\right ) \). The minimal polynomial \(p_{\min }\left ( x\right ) \) will be a function of \(\omega \). Next we solve for \(\omega \) from \(p_{\min }\left ( x\right ) =0\). If this is successful, then we have found \(\omega \) and the first solution to the ode \(z^{\prime \prime }=rz\) is \(e^{\int \omega dx}\,\). Below shows how this is done.

We start by forming a polynomial

\[ p\left ( x\right ) =x^{d}+a_{d-1}x^{d-1}+\cdots +a_{0}\]

Notice that \(x^{d}\) has coefficient \(1\). The goal is to solve for the \(a_{i}\) coefficient. Now depending on case  number \(n\), we do the following. If case \(n=1\) then

\begin{align} p_{1} & =-p\tag {9}\\ p_{0} & =-p_{1}^{\prime }-\Theta p_{1}\nonumber \\ p_{-1} & =-p_{0}^{\prime }-\Theta p_{0}-\left ( 1\right ) \left ( 1\right ) rp_{1}\nonumber \end{align}

And it is \(p_{-1}=0\) which is solved for the coefficients \(a_{i}\). For the case \(n=2\) we find \(p\) as follows

\begin{align} p_{2} & =-p\tag {10}\\ p_{1} & =-p_{2}^{\prime }-\Theta p_{2}\nonumber \\ p_{0} & =-p_{1}^{\prime }-\Theta p_{1}-\left ( 1\right ) \left ( 2\right ) rp_{2}\nonumber \\ p_{-1} & =-p_{0}^{\prime }-\Theta p_{0}-\left ( 2\right ) \left ( 1\right ) rp_{1}\nonumber \end{align}

And it is \(p_{-1}=0\) which is solved for the coefficients \(a_{i}\). For the case \(n=4\) we find \(p\) as follows

\begin{align} p_{4} & =-p\tag {11}\\ p_{3} & =-p_{4}^{\prime }-\Theta p_{4}\nonumber \\ p_{2} & =-p_{3}^{\prime }-\Theta p_{3}-\left ( 1\right ) \left ( 4\right ) rp_{4}\nonumber \\ p_{1} & =-p_{2}^{\prime }-\Theta p_{2}-\left ( 2\right ) \left ( 3\right ) rp_{3}\nonumber \\ p_{0} & =-p_{1}^{\prime }-\Theta p_{1}-\left ( 3\right ) \left ( 2\right ) rp_{2}\nonumber \\ p_{-1} & =-p_{0}^{\prime }-\Theta p_{0}-\left ( 4\right ) \left ( 1\right ) rp_{1}\nonumber \end{align}

And it is \(p_{-1}=0\) which is solved for the coefficients \(a_{i}\). For the case \(n=6\) we find \(p\) as follows

\begin{align} p_{6} & =-p\tag {12}\\ p_{5} & =-p_{6}^{\prime }-\Theta p_{6}\nonumber \\ p_{4} & =-p_{5}^{\prime }-\Theta p_{5}-\left ( 1\right ) \left ( 6\right ) rp_{6}\nonumber \\ p_{3} & =-p_{4}^{\prime }-\Theta p_{4}-\left ( 2\right ) \left ( 5\right ) rp_{5}\nonumber \\ p_{2} & =-p_{3}^{\prime }-\Theta p_{3}-\left ( 3\right ) \left ( 4\right ) rp_{4}\nonumber \\ p_{1} & =-p_{2}^{\prime }-\Theta p_{2}-\left ( 4\right ) \left ( 3\right ) rp_{3}\nonumber \\ p_{0} & =-p_{1}^{\prime }-\Theta p_{1}-\left ( 5\right ) \left ( 2\right ) rp_{2}\nonumber \\ p_{-1} & =-p_{0}^{\prime }-\Theta p_{0}-\left ( 6\right ) \left ( 1\right ) rp_{1}\nonumber \end{align}

And it is \(p_{-1}=0\) which is solved for the coefficients \(a_{i}\). For the case \(n=12\) we find \(p\) as follows

\begin{align} p_{12} & =-p\tag {13}\\ p_{11} & =-p_{12}^{\prime }-\Theta p_{12}\nonumber \\ p_{10} & =-p_{11}^{\prime }-\Theta p_{11}-\left ( 1\right ) \left ( 12\right ) rp_{12}\nonumber \\ p_{9} & =-p_{10}^{\prime }-\Theta p_{10}-\left ( 2\right ) \left ( 11\right ) rp_{11}\nonumber \\ p_{8} & =-p_{9}^{\prime }-\Theta p_{9}-\left ( 3\right ) \left ( 10\right ) rp_{10}\nonumber \\ p_{7} & =-p_{8}^{\prime }-\Theta p_{8}-\left ( 4\right ) \left ( 9\right ) rp_{9}\nonumber \\ p_{6} & =-p_{7}^{\prime }-\Theta p_{7}-\left ( 5\right ) \left ( 8\right ) rp_{8}\nonumber \\ p_{5} & =-p_{6}^{\prime }-\Theta p_{6}-\left ( 6\right ) \left ( 7\right ) rp_{7}\nonumber \\ p_{4} & =-p_{5}^{\prime }-\Theta p_{5}-\left ( 7\right ) \left ( 6\right ) rp_{6}\nonumber \\ p_{3} & =-p_{4}^{\prime }-\Theta p_{4}-\left ( 8\right ) \left ( 5\right ) rp_{5}\nonumber \\ p_{2} & =-p_{3}^{\prime }-\Theta p_{3}-\left ( 9\right ) \left ( 4\right ) rp_{4}\nonumber \\ p_{1} & =-p_{2}^{\prime }-\Theta p_{2}-\left ( 10\right ) \left ( 3\right ) rp_{3}\nonumber \\ p_{0} & =-p_{1}^{\prime }-\Theta p_{1}-\left ( 11\right ) \left ( 2\right ) rp_{2}\nonumber \\ p_{-1} & =-p_{0}^{\prime }-\Theta p_{0}-\left ( 12\right ) \left ( 1\right ) rp_{1}\nonumber \end{align}

If we are able to solve for all the \(a_{i}\) by solving

\[ p_{-1}\left ( x\right ) =0 \]

Then we now have determined \(p\left ( x\right ) \). This is used to find \(\omega \) as follows. For the case \(n=1\)

\[ \omega =\frac {p^{\prime }}{p}+\Theta \]

For the case \(n=2\)  \(\omega \) is found by solving

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

Where \(\phi =\frac {p^{\prime }}{p}+\Theta \).

Where \(p_{0},p_{1},p_{2}\) are from Eq (10) above. For the case \(n=4\) then

\[ p_{\min }=\frac {1}{4!}p_{0}+\frac {1}{3!}p_{1}\omega +\frac {1}{2!}p_{2}\omega ^{2}+p_{3}\omega ^{3}+p_{4}\omega ^{4}\]

Where \(p_{0},p_{1},p_{2},p_{3},p_{4}\) are from Eq (11) above. For the case \(n=6\) then

\[ p_{\min }=\frac {1}{6!}p_{0}+\frac {1}{5!}p_{1}\omega +\frac {1}{4!}p_{2}\omega ^{2}+\frac {1}{3!}p_{3}\omega ^{3}+\frac {1}{2!}p_{4}\omega ^{4}+p_{5}\omega ^{5}+p_{6}\omega ^{6}\]

Where \(p_{0},p_{1},p_{2},p_{3},p_{4},p_{5},p_{6}\) are from Eq (12) above. For the case \(n=12\) then

\[ p_{\min }=\frac {1}{12!}p_{0}+\frac {1}{11!}p_{1}\omega +\frac {1}{10!}p_{2}\omega ^{2}+\frac {1}{9!}p_{3}\omega ^{3}+\frac {1}{8!}p_{4}\omega ^{4}+\frac {1}{7!}p_{5}\omega ^{5}+\frac {1}{6!}p_{6}\omega ^{6}+\frac {1}{5!}p_{7}\omega ^{7}+\frac {1}{4!}p_{8}\omega ^{8}+\frac {1}{3!}p_{9}\omega ^{9}+\frac {1}{2!}p_{10}\omega ^{10}+p_{11}\omega ^{11}+p_{12}\omega ^{12}\]

Where \(p_{i}\) for \(i=0\cdots 12\) are from Eq (13) above.  In each case, we now solve for

\[ p_{\min }\left ( x\right ) =0 \]

For \(\omega \). If this is successful, then now we have to verify the solution satisfies the Riccati ODE. \(\omega \) must satisfy in all case the following equation

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

If it does, then we have solved the \(z^{\prime \prime }=rz\) ode \(z=e^{\int \omega dx}\) and also the original \(y^{\prime \prime }+ay^{\prime }+by=0\) ode. This completes the Kovacic algorithm. Examples are given below showing how to implement the above to solve number of ode’s.