6.3.3.1 Example 1
This is the same ode used in second example above for case two as the necessary
conditions for case three are also satisfied.
\begin{align*} (1-x) x^2 y''+(5 x -4) x y'+(6-9 x) y=0 \end{align*}
As shown earlier, this ode is transformed to \(z''=rz\) where
\begin{align*} r &= \frac {s}{t} \\ &= \frac {4-x }{4 x (x -1)^2} \end{align*}
There is a pole at \(x=0\) of order \(1\) and a pole at \(x=1\) of order \(2\). For the pole of order \(1\) at \(x=0\), \(E_0=\{12\}\). For the
pole of order \(2\) at \(x=1\)
\begin{alignat*}{2} E_1 &=\left \{ 6 + \frac {12 k}{n} \sqrt {1+ 4 b}\right \} \qquad \text {for} \quad k&&=-\frac {n}{2}\cdots \frac {n}{2} \tag {1} \end{alignat*}
Where \(k\) is incremented by \(1\) each time, and \(n\) is any of \(\{4,6,12\}\) and \(b\) is the coefficient of \(\frac {1}{(x-1)^2}\) in the
partial fraction decomposition of \(r\) which is
\begin{align*} r = \frac {3}{4 \left (x -1\right )^{2}}-\frac {1}{x -1}+\frac {1}{x} \end{align*}
The above shows that \(b=\frac {3}{4}\). Starting with \(n=4\) (if this \(n\) produces no solution then \(n=6,12\) will be tried
also). Equation (1) now becomes
\begin{alignat*}{2} E_1 &=\left \{ 6 + \frac {12 k}{4} \sqrt {1+ 4 \left (\frac {3}{4}\right ) } \right \} \qquad \text {for} \quad k&&=-2\cdots 2 \end{alignat*}
Which simplifies to
\begin{alignat*}{2} E_1 &=\left \{ 6 + 6 k \right \} \qquad \text {for} \quad k&&=-2\cdots 2 \\ &=\left \{ -6, 0, 6, 12, 18 \right \} \end{alignat*}
\(E_\infty \) is found using (1) but with different \(b\). In this case \(b\) is given by \(b=\frac {\operatorname {lcoef}(s)}{\operatorname {lcoeff}(t)}\) where \(r=\frac {s}{t}\). \(\operatorname {lcoef}(s)\) is the leading
coefficient of \(s\) and \(\operatorname {lcoef}(t)\) is the leading coefficient of \(t\). Since \(r=\frac {-x+4 }{4 x^3-8x^2+4 x}\) then \(b = -\frac {1}{4}\). Equation (1) becomes
\begin{alignat*}{2} E_\infty &=\left \{ 6 + \frac {12 k}{4} \sqrt {1- 4 \left (\frac {1}{4}\right ) } \right \} \qquad \text {for} \quad k&&=-2\cdots 2 \end{alignat*}
This simplifies to
\begin{alignat*}{2} E_\infty &=\left \{6 \right \} \end{alignat*}
The following table summarizes step \(1\) results using \(n=4\).
The next step is to determine a non negative integer \(d\) using
\begin{align*} d &= \frac {n}{12} \left ( e_\infty - \sum _{c \in \Gamma } e_c \right ) \end{align*}
Where in the above \(e_c\) is a distinct element from each corresponding \(E_c\). This means all
possible tuples \(\{e_{c_1},e_{c_2},\dots ,e_{c_n}\}\) are tried in the sum above, where \(e_{c_i}\) is one element of each \(E_c\) found
earlier.
This results in the following values for \(d\) using \(n=4\) and \(e_\infty =6\).
\begin{alignat*}{2} d &= \frac {1}{3} \left ( 6 - (12 - 6) \right ) &&= 0 \\ &= \frac {1}{3} \left ( 6 - (12 + 0) \right ) &&= -2 \\ &= \frac {1}{3} \left ( 6 - (12 + 6) \right ) &&= -4 \\ &= \frac {1}{3} \left ( 6 - (12 + 12) \right ) &&= -6 \\ &= \frac {1}{3} \left ( 6 - (12 + 18) \right ) &&= -8 \\ \end{alignat*}
Therefore only the first case using \(e_0=12,e_1=-6\) generated non-negative integer \(d\). The following rational
function is now formed
\begin{align*} \theta &= \frac {n}{12} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {4}{12} \left (\frac {12}{\left (x-\left (0\right )\right )}+\frac {-6}{\left (x-\left (1\right )\right )}\right ) \\ &= \frac {2 x -4}{\left (x -1\right ) x} \end{align*}
And
\begin{align*} S &= \prod _{c \in \Gamma } (x-c) \\ &= (x-0) (x-1) \\ &= x(x-1) \end{align*}
This completes the step 2 of the algorithm.
Since the degree \(d=0\), then \(p(x)=1\). Now \(P_i(x)\) polynomials are generated using
\begin{align*} P_n &= - p(x) \\ P_{i-1} &= - S p'_{i} + ((n-i)S' -S\theta ) P_i - (n-1)(i+1) S^2 r P_{i+1} \qquad i=n,n-1,\dots , 0 \end{align*}
These generate the following set
\begin{align*} P_{4} &= -1\\ P_{3} &= 2 x -4\\ P_{2} &= -3 \left (x -2\right )^{2}\\ P_{1} &= 3 \left (x -2\right )^{3}\\ P_{0} &= -\frac {3 \left (x -2\right )^{4}}{2}\\ P_{-1} &= 0 \end{align*}
There is nothing to solve for from the last equation \(P_{-1} = 0\) as \(p(x)=1\) is already known because the
degree \(d\) was zero and hence there are no coefficients \(a_i\) to solve for.
Next \(\omega \) is determined as the solution to the following equation using \(n=4\).
\begin{align*} \sum _{i=0}^{n} S^i \frac {P_i}{(n-i)!} \omega ^i &= 0 \\ \frac {P_0}{4!} + \frac {S P_1}{3!} \omega +\frac {S^2 P_2}{2!} \omega ^2 +\frac {S^3 P_3}{1!} \omega ^3 + +\frac {S^4 P_4}{0!} \omega ^4 &=0\\ -\frac {1}{16}\left (2 \omega \,x^{2}-2 x \omega -x +2\right )^{4} &=0 \end{align*}
Solving the above and using any one of the roots results in
\begin{align*} \omega =\frac {1}{2 x \left (x -1\right )}\left (x -2\right ) \end{align*}
The above \(\omega \) is used to find a solution to \(z''=r z\) from
\begin{align*} z &= e^{ \int \omega \,dx}\\ &= {\mathrm e}^{\int \frac {x -2}{2 x \left (x -1\right )}d x}\\ &= \frac {x}{\sqrt {x -1}} \end{align*}
Therefore one solution to the original ode in \(y\) is
\begin{align*} y &= z e^{ \int -\frac {1}{2} a \,dx}\\ &= z e^{ -\int \frac {1}{2} \frac {5 x^{2}-4 x}{-x^{3}+x^{2}} \,dx}\\ &= z e^{2 \ln \left (x \right )+\frac {\ln \left (x -1\right )}{2}}\\ &= \left (\frac {x}{\sqrt {x -1}}\right ) \left (x^{2} \sqrt {x -1}\right ) \end{align*}
The second solution to the original ode is found using reduction of order. This completes
the solution using case \(3\) for degree \(n=4\) of \(\omega \).