6.2.2.3 step 3

In this step the algorithm determines a polynomial \(p(x)=a_0+a_1 x+ a_2 x^2 + \dots + x^d\) of degree \(d\). This is done by solving for the coefficients \(a_i\) from

\begin{align*} p'''+3 \theta p'' + \left (3 \theta ^2 + 3 \theta ' - 4 r \right ) p' + \left (\theta ''+ 3 \theta \theta ' + \theta ^3 - 4 r \theta -2 r' \right )p=0 \tag {1} \end{align*}

Where \(\theta \) was found in step \(2\) and \(r\) is from \(z''=rz\). If \(p(x)\) can be found that satisfies (1) then

\begin{align*} \phi = \theta + \frac {p'}{p} \tag {2} \end{align*}

\(\omega \) is then solved for from

\begin{align*} \omega ^2 - \phi \omega + \left (\frac {1}{2} \phi ' +\frac {1}{2} \phi ^2 -r\right ) &=0 \tag {3} \end{align*}

If solution \(\omega \) to (3) can be found, then the solution to \(z''=rz\) is given by

\begin{align*} z = e^{\int { \omega \,dx}} \end{align*}

This completes the full algorithm for case two. The general solution to the original ode is now determined as outlined at the end of case one above.