6.3.2.2 Example 2

This is an ode in which the necessary conditions for all three cases are satisfied, but solved using case two to illustrate the algorithm.

\begin{align*} (1-x) x^2 y''+(5 x -4) x y'+(6-9 x) y=0 \end{align*}

Converting it \(y''+ a y' + b y=0\) gives

\begin{align*} y''+\frac {5 x -4}{(1-x) x} y'+\frac {6-9 x}{(1-x) x^2} y=0 \end{align*}

Where \(a=\frac {5 x -4}{(1-x) x}, b=\frac {6-9 x}{(1-x) x^2}\). Applying the transformation \(z=y e^{\frac {1}{2} \int {a\,dx}}\) gives \(z''=r z\) where \(r = \frac {1}{4} a^2 + \frac {1}{2} a' - b\). Therefore

\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 \(2\). Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case one are satisfied. Since there is a pole of order \(2\) then the necessary conditions for case two are also satisfied. Since pole order is not larger than \(2\) and the order at \(\infty \) is \(2\) then the necessary conditions for case three are also satisfied. Any one of the three cases algorithm could be used to solve this, but here case two will be used for illustration.

The pole of order \(1\) at \(x=0\) gives \(E_0=\{4\}\) and the pole of order \(2\) at \(x=1\) gives \(E_1=\{2, 2+ 2\sqrt {1+ 4b}, 2-2 \sqrt {1+ 4b}\}\) where \(b\) is the coefficient of \(\frac {1}{(x-1)^2}\) in the partial fraction decomposition of \(r\). The partial fractions decomposition of \(r\) 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 \(b={\frac {3}{4}}\), therefore \(E_1 =\{-2, 2, 6\}\).

\(\mathcal {O}(\infty )=2\) therefore \(E_{\infty }=\{2, 2+ 2\sqrt {1+ 4b}, 2-2 \sqrt {1+ 4b}\}\) where \(b=\frac {\operatorname {lcoef}(s)}{\operatorname {lcoeff}(t)}\) where \(r=\frac {s}{t}\). This gives \(b=-\frac {1}{4}\), hence

\begin{align*} E_{\infty }&=\{2, 2+ 2\sqrt {1+ 4b}, 2-2 \sqrt {1+ 4b}\}\\ &=\left \{2, 2+ 2\sqrt {1- 4\frac {1}{4}}, 2-2 \sqrt {1- 4-\frac {1}{4}}\right \}\\ &=\{2, 2, 2\}\\ &=\{2\} \end{align*}

The following table summarizes step \(1\) results.

pole \(c\) location	pole order	\(E_c\)
\(0\)	\(1\)	\(\{4\}\)
\(1\)	\(2\)	\(\{-2, 2, 6\}\)

Table 6.4: First step, case two. \(E_c\) set information

Order of \(r\) at \(\infty \)	\(E_\infty \)
\(2\)	\(\{2\}\)

Table 6.5: First step, case two. \(\mathcal {O}(\infty )\) information

The above completes step 1 for case two. Step 2 searches for a non-negative integer \(d\) using

\begin{align*} d&=\frac {1}{2} \left ( e_{\infty } - \sum _{c\in \Gamma } e_c \right ) \end{align*}

Where in the above \(e_c \in E_c\), \(e_\infty \in E_\infty \) were found in the first step. The following are the possible combinations to use

\begin{alignat*}{2} d&=\frac {1}{2} \left ( 2 - ( 4 -2 ) \right ) &&= 0 \\ &=\frac {1}{2} \left (2 - ( 4 +2 )\right ) &&=-2 \\ &=\frac {1}{2} \left (2 - ( 4 +6 )\right ) &&=-4 \\ \end{alignat*}

The above shows that \(E_c=\{e_0=4,e_1=-2\}\) are the family of values to use and all other values are discarded.

The following rational function \(\theta \) is determined using

\begin{align*} \theta &= \frac {1}{2} \sum _{c \in \Gamma } \frac {e_c}{x-c} \\ &= \frac {1}{2} \left (\frac {4}{\left (x-\left (0\right )\right )}+\frac {-2}{\left (x-\left (1\right )\right )}\right ) \\ &= \frac {2}{x}-\frac {1}{x -1} \end{align*}

The algorithm now searches for a monic polynomial \(p(x)\) of degree \(d=0\) such that

\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 \end{align*}

Since \(d=0\), then \(p(x)=1\). Substituting the values found in step \(2\) in the above equation and simplifying gives

\begin{align*} 0 = 0 \end{align*}

Hence \(p(x)=1\) can be used. Let

\begin{align*} \phi &= \theta + \frac {p'}{p}\\ &= \frac {2}{x}-\frac {1}{x -1} \end{align*}

And \(\omega \) be the solution of

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

Substituting the values for \(\phi \) and \(r\) into the above equation gives

\begin{align*} w^{2}-\left (\frac {2}{x}-\frac {1}{x -1}\right ) w +\frac {\left (x -2\right )^{2}}{4 \left (x -1\right )^{2} x^{2}} &= 0 \end{align*}

Solving for \(\omega \) gives two roots, either one can be used. Using

\begin{align*} \omega &= \frac {x -2}{2 \left (x -1\right ) x} \end{align*}

Therefore to \(z'' = r z\) is

\begin{align*} z &= e^{ \int \omega \,dx} \\ &= {\mathrm e}^{\int \frac {x -2}{2 \left (x -1\right ) x}d x}\\ &= \frac {x}{\sqrt {x -1}} \end{align*}

The first solution to the original ode in \(y\) is now found from

\begin{align*} y_1 &= 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}}\\ &= z \left (x^{2} \sqrt {x -1}\right )\\ &= \frac {x}{\sqrt {x -1}} \left (x^{2} \sqrt {x -1}\right )\\ &= x^3 \end{align*}

The second solution \(y_2\) to the original ode is found using reduction of order.