6.2.1.3 step 3

In this step the algorithm finds 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^{\prime \prime }+2\omega p^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) p&=0 \tag {7} \end{align*}

Where \(\omega \) is from the second step above and \(r\) is from \(z''=rz\).

For an example, if \(d=2\), then \(p\left ( x\right ) =x^{2}+a_1 x + a_0\) is substituted in (3) and \(a_0,a_1\) are solved for. If solution exists, then the solution to \(z''=r z\) will be

\begin{align*} z&=p(x) e^{\int \omega dx} \end{align*}

If the degree \(d=1\) then \(p\left ( x\right )=x+a_0\) and the same process is applied. If the degree \(d=0\), then \(p\left (x\right )=1\).

The first basis solution to the original ode is now be found from

\begin{align*} y_1&=z e^{-\frac {1}{2} \int a \, dx} \end{align*}

And the second basis solution using reduction of order formula is

\begin{align*} y_2&=y_1 \int { \frac { e^{-\int a \, dx} }{y_{1}^{2}} \, dx} \end{align*}

Hence the general solution to the original ode is

\begin{align*} y(x) = c_1 y_1 + c_2 y_2 \end{align*}

This completes the full algorithm for case 1. The part that needs most care is in finding \(\left \{\left [\sqrt {r}\right ] _{c},\alpha _{c}^{\pm },\left [ \sqrt {r}\right ] _{\infty },\alpha _{\infty }^{\pm }\right \}\). Once these are calculated, the rest of the algorithm is much more direct.